case, so assume $q\not=1$. With Lemma 6.14 in mind, the eigenspaces to
consider are

$\operatorname{lin}\Big{\\{}\begin{pmatrix}1&0\\\
0&0\end{pmatrix},\begin{pmatrix}0&0\\\
0&1\end{pmatrix}\Big{\\}},\quad\mathbb{C}\begin{pmatrix}0&1\\\
0&0\end{pmatrix},\quad\mathbb{C}\begin{pmatrix}0&0\\\ 1&0\end{pmatrix}.$

As $S$ should also be self-adjoint, there are just four quantum graphs over
$\mathbb{M}_{2}$ with $Q_{d}SQ_{d}^{-1}=S$ and $Q_{d}^{-1/2}\in S$ (this from
Proposition 5.25), namely

$\displaystyle S_{1}=\mathbb{C}\begin{pmatrix}1&0\\\ 0&1/q\end{pmatrix},\qquad
S_{3}=\mathbb{M}_{2},$ $\displaystyle
S_{2A}=\operatorname{lin}\Big{\\{}\begin{pmatrix}1&0\\\
0&1/q\end{pmatrix},\begin{pmatrix}0&1\\\
0&0\end{pmatrix},\begin{pmatrix}0&0\\\ 1&0\end{pmatrix}\Big{\\}},\quad
S_{2B}=\operatorname{lin}\Big{\\{}\begin{pmatrix}1&0\\\
0&0\end{pmatrix},\begin{pmatrix}0&0\\\ 0&1\end{pmatrix}\Big{\\}}.$

Here $S_{1}$ is the empty quantum graph, $S_{3}$ the complete quantum graph,
and $S_{2A},S_{2B}$ lie strictly between these, and are incomparable. A simple
computation shows that taking the complement, $S\mapsto S_{c}$ from
Proposition 6.12, interchanges $S_{1},S_{3}$ and $S_{2A},S_{2B}$. If we apply
Theorem 5.29, then $(S_{1})_{0}$ is the span of the identity, $(S_{2A})_{0}$
is the span of the identity and the off-diagonal matrices, and $S_{2B},S_{3}$
remain unchanged. Thus, in this special case, Theorem 5.29 interacts as we
might hope with the complement.

######  Remark 6.15.

We present an example where Theorem 5.29 really does not interact with the
complement. Let $B=\mathbb{M}_{2}\oplus\mathbb{M}_{2}$ acting on
$H=\mathbb{C}^{2}\oplus\mathbb{C}^{2}$, so that
$B^{\prime}\cong\mathbb{C}\oplus\mathbb{C}$. Pick $0<t<1/2$ and set

$Q_{1}=\begin{pmatrix}t&0\\\ 0&1-t\end{pmatrix},\quad
Q_{2}=\begin{pmatrix}1-t&0\\\ 0&t\end{pmatrix},\quad Q=Q_{1}\oplus Q_{2}.$

We identify $\mathcal{B}(H)$ with $2\times 2$ matrices of operators in
$\mathbb{B}(\mathbb{C}^{2})\cong\mathbb{M}_{2}$. Set

$x_{0}=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}\in\mathbb{M}_{2},\quad
x=\begin{pmatrix}0&x_{0}\\\ 0&0\end{pmatrix}\in\mathcal{B}(H).$

Finally let $T=\operatorname{lin}\\{x,x^{*}\\}$, so $T$ is a self-adjoint,
$B^{\prime}$-bimodule. Then

$Q_{1}x_{0}=\begin{pmatrix}0&t\\\
1-t&0\end{pmatrix}\not\in\mathbb{C}x_{0}\quad\implies\quad
Qx=\begin{pmatrix}0&Q_{1}x_{0}\\\ 0&0\end{pmatrix},$

while

$Q_{1}x_{0}Q_{2}^{-1}=\begin{pmatrix}0&t\\\
1-t&0\end{pmatrix}Q_{2}^{-1}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix}=x_{0}\quad\implies\quad
QxQ^{-1}=\begin{pmatrix}0&Q_{1}x_{0}Q_{2}^{-1}\\\ 0&0\end{pmatrix}\in T,$

from which it follows that $QTQ^{-1}=T$.

By the final part of Remark 6.13, $S_{0}=T^{\perp}$ also satisfies the
required conditions, and so by Theorem 5.29, $S=Q^{-1/2}S_{0}$ satisfies all
the conditions of Theorem 5.17. Towards a contradiction, suppose that
$(S_{c})_{0}=(S_{0})^{\perp}+B^{\prime}$. By Remark 6.13, always
$(S_{c})_{0}=Q(S_{0})^{\perp}+B^{\prime}$, so we obtain that
$(S_{0})^{\perp}+B^{\prime}=Q(S_{0})^{\perp}+B^{\prime}$, that is,
$T+B^{\prime}=QT+B^{\prime}$, that is, $QT\subseteq T+B^{\prime}$. However,
given the form of $Qx$ computed above, it is clear that $Qx\not\in
T+B^{\prime}$, giving the required contradiction. ∎

##  7 Homomorphisms

In this section, we look at various notions of “homomorphism” between quantum
graph. We note that there have been notions of a “quantum homomorphism”
between classical graphs (see [33, Section 4] or [31, Theorem 4.14, Remark
5.5] for example) which can further be extended to quantum homomorphisms
between quantum graphs. These notions are linked to the quantum automorphisms
we discuss in Section 9 below, and are discussed in detail in [25, Section
4.4], [30, Section 2.3], for example. In contrast, in this section we look at
what might be thought of as “classical homomorphisms” between (possibly)
quantum graphs. Such notions seem to be less firmly established than other
definitions made in this paper, so we compare and contrast them, without
offering a definitive definition.

We start with two ideas from Weaver in [46]; again we work with quantum graphs
in the sense of Definition 4.4. For exactly what we mean by “Kraus form” see
the discussion below. It is not at all clear that the following definition is
independent of the choice of Kraus form, but this follows from [46], or our,
different, argument to follow.

######  Definition 7.1 (See [46, Theorem 7.4, Theorem 8.2]).

Let $B_{i}\subseteq\mathcal{B}(H_{i})$ for $i=1,2$ for finite-dimensional
Hilbert space $H_{i}$. Let $\theta:B_{2}\rightarrow B_{1}$ be a UCP map with
Kraus form $\theta(x)=\sum_{i=1}^{n}b_{i}^{*}xb_{i}$ for some operators
$b_{i}\in\mathcal{B}(H_{1},H_{2})$. Let $S_{1}\subseteq\mathcal{B}(H_{1})$ be
a quantum graph. The _pushfoward_ of $S_{1}$ along $\theta$, denoted
$\overrightarrow{S_{1}}$ is the $B_{2}^{\prime}$-operator bimodule generated
by

$\\{b_{i}xb_{j}^{*}:x\in S_{1}\\}\subseteq\mathcal{B}(H_{2}).$

Let $S_{2}\subseteq\mathcal{B}(H_{2})$ be a quantum graph. The _pullback_ of
$S_{2}$ along $\theta$ is, denoted by $\overleftarrow{S_{2}}$ is the
$B_{1}^{\prime}$-operator bimodule generated by

$\\{b_{i}^{*}xb_{j}:x\in S_{2}\\}\subseteq\mathcal{B}(H_{1}).$

Again here we have translated from trace-preserving CP maps to UCP maps,
equivalently, have “reversed the arrows”, compare with Remark 6.3 above. We
notice that the pullback is always unital if $S_{2}$ is unital, but the
pushfoward may not be, in accordance with what happens in the classical
situation. While we stated this for quantum graphs, obviously the definition
works for any quantum relation. The reader who would like some motivation can
see Remark 7.11 below.

The paper [46] develops an “intrinsic” way to characterise quantum graphs
(making use of results from [45]), and then shows that the pushfoward and
pullback only depend on $\theta$ (and not the Kraus form chosen) when recast
using this intrinsic characterisation of $S_{1}$ and $S_{2}$. We shall show
how to use Stinespring dilations (and a duality argument) to give a direct
argument. A biproduct of our construction is that in the above definition, it
suffices to take the linear span of the sets given: they are automatically
operator bimodules over the relevant commutants.

Let us recall how to construct the Kraus form of a normal CP map between
arbitrary von Neumann algebras. This is surely known, but we are not aware of
a canonical reference. Let $M,N$ be von Neumann algebras acting on Hilbert
spaces $H_{M},H_{N}$ respectively. We start by observing that the usual
Stinespring dilation of a normal CP map $\theta:M\rightarrow N$ between von
Neumann algebras, say $V,\rho,\pi,K$ automatically has that $\pi$ is normal.
Consider now an arbitrary dilation (so we do not assume minimality). By the
structure theorem for normal $*$-homomorphisms between von Neumann algebras
(for example, [39, Theorem 5.5, Chapter IV]) there is an index set $I$ and an
isometry $W:K\rightarrow H_{M}\otimes\ell^{2}(I)$ with $W^{*}(x\otimes
1)W=\pi(x)$ for $x\in N$. Furthermore, the projection $WW^{*}$ is in
$M^{\prime}\bar{\otimes}\mathcal{B}(\ell^{2}(I))$. Then

$\theta(x)=V^{*}\pi(x)V=V^{*}W^{*}(x\otimes 1)WV\qquad(x\in M),$

and there is a family of operators $(b_{i})\subseteq\mathcal{B}(H_{N},H_{M})$
with $WV\xi=\sum_{i}b_{i}\xi\otimes\delta_{i}$ and hence
$\theta(x)=\sum_{i}b_{i}^{*}xb_{i}$ gives us a Kraus representation.

So far, it is not clear how the property that $\theta$ maps into $N$ (and not
just $\mathcal{B}(H_{N})$) is reflected in the Kraus representation.

######  Lemma 7.2.

Let $\theta:M\rightarrow\mathcal{B}(H_{N})$ be a normal CP map with some
Stinespring dilation $V,\pi,K$, meaning that $\pi:M\rightarrow\mathcal{B}(K)$
is a normal unital $*$-homomorphism, $V:H_{N}\rightarrow K$ an operator, and
$\theta(x)=V^{*}\pi(x)V$ for $x\in M$; no minimality assumption is made. The
following are equivalent:

  1. 1.

$\theta$ maps into $N$;

  2. 2.

there is a normal unital $*$-homomorphism
$\rho:N^{\prime}\rightarrow\pi(M)^{\prime}$ with
$Vx^{\prime}=\rho(x^{\prime})V$ for $x^{\prime}\in N^{\prime}$.

###### Proof.

Suppose first that $\theta$ maps into $N$. Let $V_{0},\pi_{0},K_{0},\rho_{0}$
be the minimal Stinespring dilation for $\theta$. We may define
$u:K_{0}\rightarrow K$ by $u\pi_{0}(x)V_{0}\xi=\pi(x)V\xi$ for $x\in M,\xi\in
H_{N}$. As

$(\pi(x)V\xi|\pi(y)V\eta)=(\xi|\theta(x^{*}y)\eta)=(\pi_{0}(x)V_{0}\xi|\pi_{0}(y)V_{0}\eta)\qquad(x,y\in
M,\xi,\eta\in H_{N}),$

it follows that $u$ is well-defined, an isometry, and by minimality, extends
by linearity and continuity to all of $K_{0}$. By construction,
$u\pi_{0}(x)=\pi(x)u$ for $x\in M$, and $uV_{0}=V$. Define
$\rho(x^{\prime})=u\rho_{0}(x^{\prime})u^{*}$ which is a normal
$*$-homomorphism, as $u^{*}u=1$. It follows that

$\rho(x^{\prime})V=u\rho_{0}(x^{\prime})u^{*}(uV_{0})=u\rho_{0}(x^{\prime})V_{0}=uV_{0}x^{\prime}=Vx^{\prime}\qquad(x^{\prime}\in
N^{\prime}).$

Furthermore, for $x^{\prime}\in N^{\prime},x\in M$, we calculate that
$\rho(x^{\prime})\pi(x)=u\rho_{0}(x^{\prime})u^{*}\pi(x)=u\rho_{0}(x^{\prime})\pi_{0}(x)u^{*}=u\pi_{0}(x)\rho_{0}(x^{\prime})u^{*}=\pi(x)u\rho_{0}(x^{\prime})u^{*}=\pi(x)\rho(x^{\prime})$.
Thus $\rho$ maps into $\pi(M)^{\prime}$.

Conversely, if we have such a $\rho$ then for $x^{\prime}\in N^{\prime},x\in
M$, we see that
$\theta(x)x^{\prime}=V^{*}\pi(x)Vx^{\prime}=V^{*}\pi(x)\rho(x^{\prime})V=V^{*}\rho(x^{\prime})\pi(x)V=x^{\prime}V^{*}\pi(x)V=x^{\prime}\theta(x)$.
Thus $\theta(x)\in N^{\prime\prime}=N$, as claimed. ∎

As there is a bijection between dilations $V,\pi,K$ with
$K=H_{M}\otimes\ell^{2}(I)$ and $\pi(x)=x\otimes 1$, and Kraus forms
$\theta(x)=\sum_{i\in I}b_{i}^{*}xb_{i}$, this lemma can be translated into a
(slightly cumbersome) condition on a Kraus form.

Weaver shows in [45, Theorem 2.7] that the definition of a Quantum Relation,
Definition 4.1, is independent of the ambient Hilbert space $H$ upon which our
von Neumann algebra $M$ acts. We will study the ideas of this proof, and
separate them out into a number of useful constructions. Let
$M\subseteq\mathcal{B}(H_{M})$ be a von Neumann algebra, let
$S\subseteq\mathcal{B}(H_{M})$ be a quantum relation on $M$, and let $H$ be
some other Hilbert space. Let $\pi:M\rightarrow\mathcal{B}(H_{M}\otimes
H);x\mapsto x\otimes 1$ and define

$S_{\pi}=S\bar{\otimes}\mathcal{B}(H),$

the weak∗-closed linear span of $\\{x\otimes a:x\in S,a\in\mathcal{B}(H)\\}$
in $\mathcal{B}(H_{M}\otimes H)$. It is routine that
$\pi(M)^{\prime}=M^{\prime}\bar{\otimes}\mathcal{B}(H)$, and so $S_{\pi}$ is a
quantum relation on $\pi(M)$.

######  Lemma 7.3.

Every quantum relation on $\pi(M)$ is of the form $S_{\pi}$ for some quantum
relation $S$ on $M$, and the map $S\mapsto S_{\pi}$ is an order-preserving
bijection, and preserves the properties of $S$ being self-adjoint,
respectively, $S$ being unital.

###### Proof.

The only non-obvious thing to check is that every quantum relation on $\pi(M)$
is of the form $S_{\pi}$. Let $T$ be a weak∗-closed operator bimodule for
$\pi(M)^{\prime}=M^{\prime}\bar{\otimes}\mathcal{B}(H)$. For $x\in T$,
consider

$(1\otimes\theta_{\xi_{1},\eta_{1}})x(1\otimes\theta_{\xi_{2},\eta_{2}})=(\operatorname{id}\otimes\omega_{\xi_{1},\eta_{2}})(x)\otimes\theta_{\xi_{2},\eta_{1}}\in
T\qquad(\xi_{1},\xi_{2},\eta_{1},\eta_{2}\in H).$

Let $S=\\{(\operatorname{id}\otimes\omega)(x):x\in T\\}$ a weak∗-closed
subspace of $\mathcal{B}(H_{M})$ which is seen to be an $M^{\prime}$-bimodule.
By linearity, and that $T$ is weak∗-closed, it follows that $y\otimes a\in T$
for each $y\in S,a\in\mathcal{B}(H)$. Furthermore, we can approximate $x\in T$
in the weak∗-topology by multiplying on the left and right by $1\otimes\theta$
for suitable finite-rank operators $\theta$, and so $x$ is in
$S\bar{\otimes}\mathcal{B}(H)$. Thus $T=S\bar{\otimes}\mathcal{B}(H)=S_{\pi}$,
as required. ∎

Now let $\theta:M\rightarrow\mathcal{B}(K)$ be a normal unital
$*$-homomorphism, not assumed injective. Again by the structure theorem for
such homomorphism, [39, Theorem 5.5, Chapter IV], there is some $H$ and an
isometry $u:K\rightarrow H_{M}\otimes H$ such that $u^{*}(x\otimes
1)u=\theta(x)$ for $x\in M$, and with $uu^{*}\in
M^{\prime}\bar{\otimes}\mathcal{B}(H)$; equivalently, $(x\otimes
1)u=u\theta(x)$ for $x\in M$. Define

$S_{\theta}=u^{*}S_{\pi}u=\\{u^{*}\alpha u:\alpha\in
S\bar{\otimes}\mathcal{B}(H)\\}\subseteq\mathcal{B}(K),$

where $\pi(x)=x\otimes 1$ as before. Notice that the definition of
$S_{\theta}$ generalises the definition of $S_{\pi}$.

######  Lemma 7.4.

The definition of $S_{\theta}$ is independent of the choice of $u,H$, and
$S_{\theta}$ is a quantum relation on $\theta(M)$. The map $S\mapsto
S_{\theta}$ is order preserving, and preserves the properties of $S$ being
self-adjoint, respectively, $S$ being unital.

###### Proof.

We show independence of $u,H$. Let $u_{1}:K\rightarrow H_{M}\otimes H_{1}$ be
an isometry with $u_{1}\theta(x)=(x\otimes 1)u_{1}$ for each $x\in M$. Let
$S_{1}=u_{1}^{*}(S\bar{\otimes}\mathcal{B}(H_{1}))u_{1}$.

Set $H_{0}=H\oplus H_{1}$ and let $\iota:H\rightarrow H_{0}$ be the inclusion.
For $x\otimes a\in S\otimes\mathcal{B}(H_{0})$ the algebraic tensor product,
clearly $x\otimes\iota^{*}a\iota\in S\otimes\mathcal{B}(H)$. Thus by
weak∗-continuity,
$((1\otimes\iota)u)^{*}(S\bar{\otimes}\mathcal{B}(H_{0}))(1\otimes\iota)u\subseteq
S_{\theta}$. However, the same reasoning shows that
$(1\otimes\iota)(S\bar{\otimes}\mathcal{B}(H))(1\otimes\iota^{*})\subseteq
S\bar{\otimes}\mathcal{B}(H_{0})$ and hence
$((1\otimes\iota)u)^{*}(S\bar{\otimes}\mathcal{B}(H_{0}))(1\otimes\iota)u=S_{\theta}$.
We conclude that we can identify $u$ with $(1\otimes\iota)u$ without changing
$S_{\theta}$, and similarly for $u_{1}$. So consider $u,u_{1}$ as maps
$K\rightarrow H_{M}\otimes H_{0}$.

With these identifications, still $u\theta(x)=(x\otimes 1)u$ and similarly for
$u_{1}$, and hence $uu_{1}^{*}\in(M\otimes
1)^{\prime}=M^{\prime}\bar{\otimes}\mathcal{B}(H_{0})$. As
$S\bar{\otimes}\mathcal{B}(H_{0})$ is an
$M^{\prime}\bar{\otimes}\mathcal{B}(H_{0})$-operator bimodule, we see that for
$\alpha\in S\bar{\otimes}\mathcal{B}(H_{0})$ also $\beta=u_{1}u^{*}\alpha
uu_{1}^{*}\in S\bar{\otimes}\mathcal{B}(H_{0})$. Thus $S_{\theta}\ni
u^{*}\alpha u=u_{1}^{*}\beta u_{1}\in S_{1}$, and so $S_{\theta}\subseteq
S_{1}$. By symmetry, also $S_{1}\subseteq S_{\theta}$, and so we have
equality, as claimed.

Define $\rho:\theta(M)^{\prime}\rightarrow\mathcal{B}(H_{M}\otimes H)$ by
$\rho(x^{\prime})=ux^{\prime}u^{*}$. As in the proof of Lemma 7.2, $\rho$ is a
$*$-homomorphism mapping into $(M\otimes
1)^{\prime}=M^{\prime}\bar{\otimes}\mathcal{B}(H)$. So, for
$x_{1}^{\prime},x_{2}^{\prime}\in\theta(M)^{\prime}$ and $\alpha\in
S\bar{\otimes}\mathcal{B}(H)$,

$x_{1}^{\prime}u^{*}\alpha
ux_{2}^{\prime}=u^{*}\rho(x_{1}^{\prime})\alpha\rho(x_{2}^{\prime})u\in
u^{*}(S\bar{\otimes}\mathcal{B}(H))u=S_{\theta},$

using the bimodule property of $S\bar{\otimes}\mathcal{B}(H)$. Hence
$S_{\theta}$ is a $\theta(M)^{\prime}$-operator bimodule.

To show that $S_{\theta}$ is weak∗-closed, we used the Krein-Smulian Theorem,
[13, Theorem 12.1] for example, which tells us that it suffices to show that
the (norm closed) unit ball of $S_{\theta}$ is weak∗-closed. Towards this, let
$(u^{*}\alpha_{i}u)$ be a bounded net in $S_{\theta}$ converging weak∗ to
$x\in\mathcal{B}(K)$. There seems to be no reason why $(\alpha_{i})$ is a
bounded net, but for each $i$, set
$\beta_{i}=uu^{*}\alpha_{i}uu^{*}\in\mathcal{B}(H_{M}\otimes H)$. As
$uu^{*}\in M^{\prime}\bar{\otimes}\mathcal{B}(H)$, it follows that each
$\beta_{i}\in S\bar{\otimes}\mathcal{B}(H)$. As $u$ is an isometry,
$u^{*}\beta_{i}u=u^{*}\alpha_{i}u$. Of course,
$\|\beta_{i}\|\leq\|u^{*}\alpha_{i}u\|$, and so $(\beta_{i})$ is a bounded
net. By moving to a subnet if necessary, we may suppose that
$\beta_{i}\rightarrow\beta\in S\bar{\otimes}\mathcal{B}(H)$, as
$S\bar{\otimes}\mathcal{B}(H)$ is weak∗-closed. Then
$u^{*}\beta_{i}u\rightarrow u^{*}\beta u$ and so $x=u^{*}\beta u\in
S_{\theta}$, as required.

The remaining claims are clear. ∎

The following shows how to compute the pullback from a Stinespring dilation
directly (not using the Kraus form) and shows that even the Kraus form
definition is already an $N^{\prime}$-operator bimodule. We prove this for
possibly infinite-dimensional operator bimodules, as this is not much more
work.

######  Theorem 7.5.

Let $\theta:M\rightarrow N$ be a normal CP map with Kraus form
$\theta(x)=\sum_{i\in I}b_{i}^{*}xb_{i}$. Let $S\subseteq\mathcal{B}(H_{M})$
be a quantum relation over $M$, and as before, define the pullback of $S$
along $\theta$ to be $\overleftarrow{S}$, the weak∗-closed
$N^{\prime}$-operator bimodule generated by elements $b_{i}^{*}xb_{j}$ for
$x\in S,i,j\in I$. Then:

  1. 1.

already the weak∗-closed subspace generated by elements $b_{i}^{*}xb_{j}$ for
$x\in S,i,j\in I$ is an $N^{\prime}$-operator bimodule;

  2. 2.

if $V,\pi,K$ is any Stinespring dilation of $\theta$, then $\overleftarrow{S}$
is the weak∗-closure of $V^{*}S_{\pi}V$, where $S_{\pi}$ is as defined above.

###### Proof.

We first show how to express $\overleftarrow{S}$ using a special Stinespring
dilation. For the moment, suppose that $V,\pi,K=H_{M}\otimes\ell^{2}(I)$ comes
from the Kraus representation, so $V\xi=\sum_{i}b_{i}\xi\otimes\delta_{i}$ and
$\pi(x)=x\otimes 1$. Then for each $i,j$,

$V^{*}(x\otimes
e_{ij})V(\xi)=V^{*}(xb_{j}(\xi)\otimes\delta_{i})=b_{i}^{*}xb_{j}(\xi)\qquad(x\in
S,\xi\in H_{M}),$

so that $V^{*}(x\otimes e_{ij})V=b_{i}^{*}xb_{j}$ for $x\in S$. Denote by $X$
the weak∗-closure of $\operatorname{lin}\\{V^{*}(x\otimes a)V:x\in
S,a\in\mathcal{B}(\ell^{2}(I))\\}$. It follows that
$\overleftarrow{S}\subseteq X$, but as $\operatorname{lin}\\{e_{ij}\\}$ is
weak∗ dense in $\mathcal{B}(\ell^{2}(I))$, also $X\subseteq\overleftarrow{S}$,
and hence we have equality. It follows that $\overleftarrow{S}$ equals the
weak∗-closure of $V^{*}S_{\pi}V$.

Now let
$\rho:N^{\prime}\rightarrow\pi(M)^{\prime}=M^{\prime}\bar{\otimes}\mathcal{B}(\ell^{2}(I))$
be as in Lemma 7.2. Then, for $x^{\prime}_{1},x_{2}^{\prime}\in
N^{\prime},y\in S_{\pi}$,

$x_{2}^{\prime}V^{*}yVx_{1}^{\prime}=V^{*}\rho(x_{2}^{\prime})y\rho(x_{1}^{\prime})V\in
V^{*}S_{\pi}V,$

by the operator bimodule property of $S_{\pi}$. Thus $V^{*}S_{\pi}V$ is an
$N^{\prime}$-operator bimodule, and thus so is its weak∗-closure. We have
hence shown claim (1).

Now let $V_{0},\pi_{0},K_{0}$ be the minimal Stinespring dilation for
$\theta$. There is an isometry $u:K_{0}\rightarrow K$ with $uV_{0}=V$ and
$\pi(x)u=u\pi_{0}(x)$ for $x\in M$, compare the proof of Lemma 7.2. Notice
then that $\pi_{0}(x)=u^{*}(x\otimes 1)u$ for each $x\in M$, so by definition,
$S_{\pi_{0}}=u^{*}S_{\pi}u$, and hence
$V^{*}S_{\pi}V=V_{0}^{*}u^{*}S_{\pi}uV=V_{0}^{*}S_{\pi_{0}}V_{0}$.

Finally, suppose that $V,\pi,K$ is any Stinespring dilation for $\theta$.
Again, there is an isometry $u:K_{0}\rightarrow K$ with $uV_{0}=V$ and
$\pi(x)u=u\pi_{0}(x)$ for $x\in M$. For a suitable Hilbert space $K^{\prime}$,
we can find an isometry $v:K\rightarrow H_{M}\otimes K^{\prime}$ with
$v\pi(x)=(x\otimes 1)v$. Thus
$S_{\pi}=v^{*}(S\bar{\otimes}\mathcal{B}(K^{\prime}))v$, and as
$\pi_{0}(x)=u^{*}\pi(x)u=u^{*}v^{*}(x\otimes 1)vu$, also
$S_{\pi_{0}}=u^{*}v^{*}(S\bar{\otimes}\mathcal{B}(K^{\prime}))vu=u^{*}S_{\pi}u$.
Hence $V_{0}^{*}S_{\pi_{0}}V_{0}=V_{0}^{*}u^{*}S_{\pi}uV_{0}=V^{*}S_{\pi}V$.
Thus (2) is shown. ∎

######  Remark 7.6.

Let us compare this with the quantum graph we constructed in Theorem 6.4.
There we started with $\theta:C\rightarrow B$ a UCP map and constructed a
quantum graph over $B$ which is $S=V^{*}\pi(C)^{\prime}V$. Here $V,\pi,K$ is a
Stinespring dilation of $\theta$. As this is independent of the choice, we can
choose $K=H_{C}\otimes K_{0}$ for a suitable $K_{0}$, and $\pi(x)=x\otimes 1$,
so that $\pi(C)^{\prime}=C^{\prime}\bar{\otimes}\mathcal{B}(K_{0})=T_{\pi}$,
where $T=C^{\prime}$ the trivial quantum graph over $C$. Thus $S$ is actually
just the pullback of the trivial quantum graph $C^{\prime}$ for the UCP map
$\theta$. ∎

######  Corollary 7.7.

Let $B_{1},B_{2}$ be finite-dimensional $C^{*}$-algebras. Let
$\theta:B_{2}\rightarrow B_{1}$ be a UCP map, and let $S$ be a quantum graph
over $B_{2}$. Then $\overleftarrow{S}$ is a quantum graph over $B_{1}$. If
$V,\pi,K$ is any Stinespring dilation of $\theta$, then
$\overleftarrow{S}=V^{*}S_{\pi}V$.

###### Proof.

In the finite-dimensional setting, there is no need to take weak∗-closures,
and so $\overleftarrow{S}=V^{*}S_{\pi}V$, for any quantum relation $S$. If $S$
is self-adjoint, clearly also $V^{*}S_{\pi}V$ is. As $\theta$ is assumed to be
unital, $V^{*}V=1$, and so if $S$ is unital, so is $S_{\pi}$, and hence so is
$V^{*}S_{\pi}V$. ∎

While this result gives a rather pleasing form for the pullback at the level
of operator bimodules, we have not been able to formulate a simple result for
the associated projections and quantum adjacency matrices.

The pushforward seemingly does not make sense in the infinite-dimensional
case, but in our usual finite-dimensional case, we can use a duality argument.
The following should be compared to a more general construction which holds in
the infinite-dimensional case under some conditions, see [1, Proposition 3.1]
for example.

######  Proposition 7.8.

Let $B_{1},B_{2}$ be finite-dimensional $C^{*}$-algebras equipped with
faithful traces $\psi_{1},\psi_{2}$. Let $\theta:B_{2}\rightarrow B_{1}$ be a
completely positive map. Define $\hat{\theta}:B_{1}\rightarrow B_{2}$ by

$\psi_{1}(a\theta(b))=\psi_{2}(\hat{\theta}(a)b)\qquad(a\in B_{1},b\in
B_{2}).$

Then $\hat{\theta}$ exists, and is completely positive. $\hat{\theta}$ is UCP
if and only if $\theta$ is trace preserving: $\psi_{1}\circ\theta=\psi_{2}$.

###### Proof.

As $\psi_{2}$ is faithful, any functional $f$ on $B_{2}$ arises from a unique
$b_{0}\in B_{2}$ as $f(b)=\psi_{2}(b_{0}b)$ for $b\in B_{2}$, compare with the
discussion at the start of Section 5.4. Applying this to the functional
$b\mapsto\psi_{1}(a\theta(b))$ gives the existence and uniqueness of
$\hat{\theta}(a)$. Assuming that $\theta$ is positive, for $a\geq 0$,

$\psi_{2}(b^{*}\hat{\theta}(a)b)=\psi_{2}(\hat{\theta}(a)bb^{*})=\psi_{1}(a\theta(bb^{*}))=\psi_{1}(a^{1/2}\theta(bb^{*})a^{1/2})\geq
0\qquad(b\in B_{2}).$

This shows that $\hat{\theta}(a)\geq 0$, so $\hat{\theta}$ is positive.

Consider now
$\hat{\theta}\otimes\operatorname{id}:B_{1}\otimes\mathbb{M}_{n}\rightarrow
B_{2}\otimes\mathbb{M}_{n}$. Giving $\mathbb{M}_{n}$ the trace
${\operatorname{Tr}}$ and hence $B_{i}\otimes\mathbb{M}_{n}$ the faithful
trace $\psi_{i}\otimes{\operatorname{Tr}}$, it follows readily that
$\hat{\theta}\otimes\operatorname{id}$ and $\theta\otimes\operatorname{id}$
are related in the same way, and hence the previous argument shows that
$\hat{\theta}\otimes\operatorname{id}$ is positive when
$\theta\otimes\operatorname{id}$ is. It follows that $\theta$ CP implies that
$\hat{\theta}$ is CP.

Finally, $\psi_{2}(\hat{\theta}(1)b)=\psi_{1}(\theta(b))$ and so
$\hat{\theta}(1)=1$ if and only if $\theta$ is trace preserving. ∎

######  Lemma 7.9.

For $i=1,2$ let $B_{i}\subseteq\mathcal{B}(H_{i})$, equip $\mathcal{B}(H_{i})$
with ${\operatorname{Tr}}$, and let $\psi_{i}$ be the restriction to $B_{i}$.
Let $\theta:B_{2}\rightarrow B_{1}$ be a completely positive map with Kraus
form $\theta(x)=\sum_{i=1}^{n}b_{i}^{*}xb_{i}$ for some
$b_{i}\in\mathcal{B}(H_{1},H_{2})$. Then $\hat{\theta}$ has Kraus form
$\hat{\theta}(y)=\sum_{i=1}^{n}b_{i}yb_{i}^{*}$.

###### Proof.

We simply calculate that for $x\in B_{2},y\in B_{1}$ we have

$\psi_{2}(\hat{\theta}(y)x)=\sum_{i}\psi_{1}(yb_{i}^{*}xb_{i})=\sum_{i}{\operatorname{Tr}}(yb_{i}^{*}xb_{i})=\sum_{i}{\operatorname{Tr}}(b_{i}yb_{i}^{*}x),$

from which the claim follows. ∎

######  Corollary 7.10.

Let $\theta:B_{2}\rightarrow B_{1}$ be a CP map, and let $S$ be a quantum
graph over $B_{1}$. Then $\overrightarrow{S}$ is a self-adjoint quantum
relation over $B_{1}$, and $\overrightarrow{S}$ is equal to
$\operatorname{lin}\\{b_{i}xb_{j}^{*}:x\in S\\}$ for any Kraus form
$\theta(x)=\sum_{i}b_{i}^{*}xb_{i}$.

###### Proof.

Form $\hat{\theta}:B_{1}\rightarrow B_{2}$ which is CP (but maybe not unital)
and has Kraus form $\hat{\theta}(x)=\sum_{i}b_{i}xb_{i}^{*}$. Hence, the
pullback of $S$ along $\hat{\theta}$ is exactly the pushforward of $S$ along
$\theta$, and so the claims for $\overrightarrow{S}$ follow from Theorem 7.5.
∎

We have not discussed quantum graphs which arise from quantum adjacency
matrices with respect to a non-tracial state. By Theorem 5.29, this is
equivalent to studying quantum graphs $S$ with the additional property that
$QSQ^{-1}=S$, but it is not very clear what condition we might place on a CP
map $\theta$ to ensure this. Further, we have not discussed links with quantum
adjacency matrices $A$ nor projections $e$, because it seems somewhat hard to
give concrete formula for, say, $\overleftarrow{e}$ associated to
$\overleftarrow{S}$.

###  7.1 Further notions

We give a quick summary of some of the other suggestions for what a
“homomorphism” for a quantum graph should be, and related ideas.

In [46, Section 6] given a quantum graph $S\subseteq\mathcal{B}(H)$ over
$B\subseteq\mathcal{B}(H)$ and a projection $p\in B$, the _restriction_
$S_{p}$ is $pSp$, the quantum graph over $pBp\subseteq\mathcal{B}(p(H))$. The
map $\theta:B\rightarrow pBp,a\mapsto pap$ is UCP with Stinespring dilation
$\pi=\operatorname{id},K=H,V=p$, and so $pSp$ is simply the pullback of $S$
along this UCP map.

In [38, Definition 7], given quantum graphs $S_{i}\subseteq\mathcal{B}(H_{i})$
over $\mathcal{B}(H_{i})$ for $i=1,2$, a _homomorphism_ $S_{1}\rightarrow
S_{2}$ is defined to be a UCP map
$\theta:\mathcal{B}(H_{2})\rightarrow\mathcal{B}(H_{1})$ with Kraus form
$\theta(x)=\sum_{i}b_{i}^{*}xb_{i}$, with the property that
$b_{i}S_{1}b_{j}^{*}\subseteq S_{2}$ for each $i,j$. This is equivalent to
$\overrightarrow{S_{1}}\subseteq S_{2}$, which agrees with Weaver’s notion of
a _CP morphism_ , [46, Definition 7.5], once we recall that we always work
with UCP maps, not TPCP maps. In the case $B=\mathbb{M}_{n}$, links between
these ideas and TRO equivalence are explored in [22, Section 7].

We should note that here Stahkle works with trace-free quantum graphs, see the
remarks after Proposition 6.12. If we look instead at the quantum graphs
$S_{i}^{\perp}$, then the condition becomes
$b_{i}^{*}S_{2}^{\perp}b_{j}\subseteq S_{1}^{\perp}$ for each $i,j$. That is,
$\overleftarrow{S_{2}^{\perp}}\subseteq S_{1}^{\perp}$. However, it seems to
the author that these notions make sense in general, regardless of whether we
work with unital, trace-free, or more general quantum relations.

######  Remark 7.11.

We follow [38] and motivate this definition from the classical situation.
Consider a classical graph $G$ with associated
$S_{G}=\operatorname{lin}\\{e_{u,v}:(u,v)\in E(G)\\}$ over $C(V(G))$.
Similarly for a graph $H$. Let $f:G\rightarrow H$ be a homomorphism, so $f$ is
a map $V(G)\rightarrow V(H)$ with $(u,v)\in E(G)\implies(f(u),f(v))\in E(H)$.

For the moment, let $f:V(G)\rightarrow V(H)$ be any map. We obtain a unital
$*$-homomorphism $\theta:C(V(H))\rightarrow C(V(G))$ given by
$\theta(a)(u)=a(f(u))$ for $a\in C(V(H)),u\in V(G)$. Then $\theta$ is in
particular a UCP map, and we can construct a Stinespring dilation as follows.

Let $n=|V(G)|$ and define
$V:\ell^{2}(V(G))\rightarrow\ell^{2}(V(H))\otimes\mathbb{C}^{n}$ as follows.
Define an equivalence relation on $V(G)$ by setting $u\sim v$ if $f(u)=f(v)$.
For $u\in V(G)$ fix an enumeration of the equivalence class
$[u]=\\{v_{1},\cdots,v_{k}\\}$, where $k\leq n$, and define
$V(\delta_{u})=\delta_{f(u)}\otimes\delta_{i}$ where $v_{i}=u$. This
construction ensures that $V$ is an isometry, and that

$V^{*}(a\otimes 1)V\delta_{u}=a(f(u))\delta_{u}\qquad(u\in V(G),a\in
C(V(H))).$

The Kraus form is obtained by setting
$V(\delta_{u})=\sum_{i}b_{i}(\delta_{u})\otimes\delta_{i}$, so that
$b_{i}(\delta_{u})=\delta_{f(u)}$ if $u$ is the $i$th vertex which maps to
$f(u)$, and $0$ otherwise. For $(u,v)\in E(G)$, so that $e_{u,v}\in S_{G}$, we
see that

$b_{i}e_{u,v}b_{j}^{*}=\theta_{b_{j}(\delta_{v}),v_{i}(\delta_{u})}=e_{f(u),f(v)},$

or $0$, depending on $i,j$ (and that there is some choice of $i,j$ which gives
$e_{f(u),f(v)}$). Hence

$\overrightarrow{S_{G}}=\operatorname{lin}\\{e_{f(u),f(v)}:(u,v)\in E(G)\\}$

and so $\overrightarrow{S_{G}}\subseteq S_{H}$ if and only if $f$ is a
homomorphism.

Thus $V^{*}(a\otimes 1)V=\theta(a)$ for each $a$, and as the dilation has the
correct form, it follows that

$\overleftarrow{S_{H}}=V^{*}(S_{H}\otimes\mathcal{B}(\mathbb{C}^{n}))V.$

A calculation shows that then $\overleftarrow{S_{H}}$ is the linear span of
“blocks” of the form $\sum\\{e_{u,v}:f(u)=x,f(v)=y\\}$ as $x,y$ vary. This
indeed corresponds to a “pullback” of $H$ along $f$ to a graph on $G$, where
we expand each edge in $H$ to a clique in $G$. ∎

Stahkle shows further in [38] that given any CP morphism $C(V(H))\rightarrow
C(V(G))$, there is some graph theoretic homomorphism $G\rightarrow H$.
However, this construction is not a bijection, but rather an existence result
(which motivates Weaver to use the new terminology of “CP morphism”).
Nevertheless, for many graph theory questions, we are merely interested in the
_existence_ of any homomorphism between two graphs, and for this, the notion
of a CP morphism exactly extends this to quantum graphs (which is indeed
Stahkle’s motivation).

[20] (see Section II) and [38] (see after Proposition 9) give tentative
definitions for what subgraphs or induced subgraphs should be. Gromada in [25,
Section 2.2] looks at two notions of a “quantum subgraph”, which are defined
using the formalism of projections $e\in B\otimes B^{\operatorname{op}}$.
Firstly, $e_{1}$ is a _subgraph made by removing edges_ of $e_{2}$ if
$e_{1}\leq e_{2}$, which is clearly equivalent to containment of the related
operator bimodules. When we have $q:B_{1}\rightarrow B_{2}$ a surjective
$*$-homomorphism, then $e_{q}$ the _induced subgraph_ of $e$ along $q$ is
given by $e_{q}=(q\otimes q)(e)$. It seems tricky to give a simple description
of the associated operator bimodules. Finally, [25, Section 2.4] looks at more
general notions of “quotient”, which requires the consideration of a more
general object, a _weighted quantum graph_.

##  8 Automorphisms of Quantum graphs

As a warm-up to looking at quantum automorphisms, we look at (classical)
automorphisms of quantum graphs. Let us quickly discuss some well-known
results about automorphisms of $B$. By $\operatorname{Aut}(B,\psi)$ we mean
the automorphisms of $B$, the bijective $*$-homomorphisms $B\rightarrow B$,
which leave $\psi$ invariant. As $B$ is finite-dimensional, it is easy to
understand automorphisms of $B$. We give a quick proof of the following well-
known result, for completeness.

######  Proposition 8.1.

Let $B=\bigoplus_{k=1}^{n}\mathbb{M}_{n_{k}}$, and let $\theta:B\rightarrow B$
be an automorphism. Then $B$ restricts to a bijection from each block
$\mathbb{M}_{n_{k}}$ to a block of the same dimension, and each such
$*$-isomorphism $\mathbb{M}_{m}\rightarrow\mathbb{M}_{m}$ is of the form
$x\mapsto uxu^{*}$ for some unitary $u\in\mathbb{M}_{m}$.

###### Proof.

It is well known that a $*$-isomorphism
$\mathbb{M}_{m}\rightarrow\mathbb{M}_{m}$ is of the form stated, so we just
shown the first part of the claim. For projections $e,f$ we have by definition
that $e\leq f$ if and only if $ef=e$, and so an automorphism preserves this
partial ordering. Clearly also $\theta$ maps the centre of $B$ to itself. Thus
$\theta$ gives a bijection when restricted to the set of minimal central
projections. The minimal central projections of $B$ are exactly the units of
the matrix blocks. Fix some matrix block $\mathbb{M}_{n_{k}}$ and let the unit
be $e$. Let $(e_{ij})_{i,j=1}^{n_{k}}$ be the matrix units of this block, so
$ee_{ij}=e_{ij}e=e_{ij}$ for each $i,j$. If $\theta(e)$ is the unit of the
matrix block $\mathbb{M}_{n_{l}}$, then we see that
$\theta(e_{ij})\leq\theta(e)$ for each $i,j$, and so each $\theta(e_{ij})$ is
in this block, as required. ∎

As $\psi$ is given by the positive invertible $Q\in B$, we can work out which
automorphisms are in $\operatorname{Aut}(B,\psi)$.

######  Lemma 8.2.

Let $\theta$ be an automorphism of $B$. Then
$\theta\in\operatorname{Aut}(B,\psi)$ if and only if $\theta(Q)=Q$, and in
this case, $\theta$ and $(\sigma_{t})$ commute.

###### Proof.

We first show that $\theta\in\operatorname{Aut}(B,\psi)$ commutes with
$(\sigma_{t})$. We use the uniqueness property of $(\sigma_{t})$, see for
example [40, Theorem 1.2, Chapter VIII], which says that
$\psi\circ\sigma_{t}=\sigma_{t}$ for each $t\in\mathbb{R}$, and for every
$a,b\in B$ there is a continuous function $F=F_{a,b}$ on
$\\{z\in\mathbb{C}:0\leq{\operatorname{Im}}(z)\leq 1\\}$ which is holomorphic
on the interior, and with $F(t)=\psi(\sigma_{t}(a)b)$ and
$F(t+i)=\psi(b\sigma_{t}(a))$. These properties determine $(\sigma_{t})$
uniquely.

As $\psi\circ\theta=\psi$, also $\psi\circ\theta^{-1}=\psi$, so defining
$\sigma^{\theta}_{t}=\theta\circ\sigma_{t}\circ\theta^{-1}$, we have that
$\psi\circ\sigma^{\theta}_{t}=\psi$. Also, for $a,b\in B$, let
$F^{\prime}=F_{\theta^{-1}(a),\theta^{-1}(b)}$, and it is easily verified that
$F^{\prime}$ satisfies the condition for $(\sigma^{\theta}_{t})$. By
uniqueness, $\sigma_{t}=\sigma^{\theta}_{t}$ for all $t$, equivalently,
$\theta$ and $(\sigma_{t})$ commute.

By analytic continuation, also $\theta$ commutes with $\sigma_{-i}$, and so
$\theta(QaQ^{-1})=\theta(\sigma_{-i}(a))=\sigma_{-i}(\theta(a))=Q\theta(a)Q^{-1}$
for each $a\in B$. Thus $\theta(Q)\theta(a)\theta(Q)^{-1}=Q\theta(a)Q^{-1}$,
equivalently, $Q^{-1}\theta(Q)\theta(a)=\theta(a)Q^{-1}\theta(Q)$ for each
$a\in B$. So $Q^{-1}\theta(Q)$ is central in $B$, and so on each matrix block
of $B$ is a scalar multiple of the identity. Thus, with $Q=(Q_{k})$, there are
scalars $q_{k}$ with $\theta(Q_{k})=q_{k}Q_{k}$. As
$\psi(a)={\operatorname{Tr}}(Qa)$ and $\psi(\theta(a))=\psi(a)$, we see that
${\operatorname{Tr}}(Qa)={\operatorname{Tr}}(Q\theta(a))$ for $a\in B$. Taking
$a=Q_{k}^{z}$ (in the $k$th block, and $0$ in other blocks) yields
${\operatorname{Tr}}(Q_{k}^{z+1})=q_{k}^{z}{\operatorname{Tr}}(Q_{k}^{z+1})$
as $\theta(Q_{k}^{z})=q_{k}^{z}Q_{k}^{z}$. Setting $z=-1$ shows that
$q_{k}=1$. Thus $\theta(Q)=Q$.

Conversely, if $\theta(Q)=Q$, then notice that it follows from Proposition 8.1
that ${\operatorname{Tr}}\circ\theta={\operatorname{Tr}}$, and so
$\psi(\theta(a))={\operatorname{Tr}}(Q\theta(a))={\operatorname{Tr}}(\theta(Qa))={\operatorname{Tr}}(Qa)=\psi(a)$,
so $\theta\in\operatorname{Aut}(B,\psi)$ as claimed. ∎

######  Definition 8.3.

Given $\theta\in\operatorname{Aut}(B,\psi)$ we define
$\hat{\theta}\in\mathcal{B}(L^{2}(B))$ by

$\hat{\theta}\Lambda(b)=\Lambda(\theta(b))\qquad(b\in B).$

Notice that
$(\Lambda(\theta(b))|\Lambda(\theta(c)))=\psi(\theta(b)^{*}\theta(c))=\psi(\theta(b^{*}c))=\psi(b^{*}c)=(\Lambda(b)|\Lambda(c))$
and so $\hat{\theta}$ is an isometry. As $\theta$ preserves $\psi$ also
$\theta^{-1}$ does, and so

$(\Lambda(b)|\hat{\theta}^{*}\Lambda(c))=(\Lambda(\theta(b))|\Lambda(c))=\psi(\theta(b)^{*}c)=\psi(\theta^{-1}(\theta(b^{*})c))=\psi(b^{*}\theta^{-1}(c))=(\Lambda(b)|\widehat{\theta^{-1}}\Lambda(c)).$

Thus $\hat{\theta}^{*}=\widehat{\theta^{-1}}$, and it is now easy to see that
$\theta\mapsto\hat{\theta}$ is a unitary representation of
$\operatorname{Aut}(B,\psi)$ on $L^{2}(B)$.

Shortly $A$ will denote a quantum group, and so in this section we shall
always write $A_{G}$ for a quantum adjacency matrix. In the following, we are
deliberately a little vague as to which exact axioms we need $A_{G}$ to
satisfy.

######  Definition 8.4.

Let $A_{G}\in\mathcal{B}(L^{2}(B))$ be a quantum adjacency matrix. An
_automorphism of $A_{G}$_ is $\theta\in\operatorname{Aut}(B,\psi)$ such that
$A_{G}$ and $\hat{\theta}$ commute.

######  Remark 8.5.

Let $B=C(V)$ and let $A=A_{G}$ be the adjacency matrix of a classical (perhaps
directed) graph, so $A_{ij}=1$ exactly when there is an edge from $j$ to $i$.
Let us denote that there is an edge from $j$ to $i$ by $j\rightarrow i$. Any
automorphism $\theta$ is given by some $\sigma\in\operatorname{Sym}(V)$, and
notice that the uniform measure on $V$ is automatically preserved. Then
$L^{2}(B)\cong\mathbb{C}^{V}$ and $\hat{\theta}$ is also induced by $\sigma$,
in that $\hat{\theta}(\delta_{i})=\delta_{\sigma(i)}$. As

$A(\delta_{i})=\sum_{j}A_{ji}\delta_{j}=\sum_{i\rightarrow j}\delta_{j},$

we see that

$\displaystyle\hat{\theta}A(\delta_{i})=\sum_{\\{j:i\rightarrow
j\\}}\hat{\theta}(\delta_{j})=\sum_{\\{j:i\rightarrow
j\\}}\delta_{\sigma(j)}=\sum_{\\{k:i\rightarrow\sigma^{-1}(k)\\}}\delta_{k},\qquad
A\hat{\theta}(\delta_{i})=A(\delta_{\sigma(i)})=\sum_{\\{j:\sigma(i)\rightarrow
j\\}}\delta_{j}.$

Thus $\hat{\theta}A=A\hat{\theta}$ if and only if, for each $i$, the sets
$\\{j:i\rightarrow\sigma^{-1}(j)\\}$ and $\\{j:\sigma(i)\rightarrow j\\}$
agree. This is equivalent to the statement that $i\rightarrow k$ if and only
if $\sigma(i)\rightarrow\sigma(k)$, which we observe by setting
$k=\sigma^{-1}(j)$. That is, $\sigma$ is an automorphism of the (directed)
graph. ∎

As a final remark, we notice that it is not really necessary to work with
$L^{2}(B)$ here. Instead, we could interpret the quantum adjacency matrix
$A_{G}$ as a linear map on $B$ (compare Section 5.4), and then require that
$A_{G}$ and $\theta$ commute; this is [25, Definition 2.4] for example.

So far, we have been discussing well-known definitions. To our knowledge,
however, there is little or no discussion in the literature around what an
automorphism of an operator bimodule might be. We shall follow Theorem 5.17.
Let $A_{G}\in\mathcal{B}(L^{2}(B))$ be a quantum adjacency matrix, satisfying
axioms (1) and (2). Then $A_{G}$ corresponds to a projection $e\in B\otimes
B^{\operatorname{op}}$ with $e=\sigma(e)$ and with $e$ invariant under
$(\sigma_{t}\otimes\sigma_{t})$. Notice that when $\theta$ is an automorphism
of $B$, also $\theta$ is an automorphism of $B^{\operatorname{op}}$.

######  Proposition 8.6.

We have that $A_{G}\hat{\theta}=\hat{\theta}A_{G}$ if and only if
$(\theta\otimes\theta)(e)=e$.

###### Proof.

We can check that $e^{\prime}=(\theta\otimes\theta)(e)$ satisfies the
conditions of Theorem 5.17 and so corresponds to
$A_{G}^{\prime}\in\mathcal{B}(L^{2}(B))$ which satisfies axioms (1) and (2).
We will show that $A_{G}^{\prime}=\hat{\theta}A_{G}\hat{\theta}^{*}$, and so
$A_{G}\hat{\theta}=\hat{\theta}A_{G}$ if and only if $A_{G}^{\prime}=A_{G}$ if
and only if $e=e^{\prime}$, as claimed.

Using $\Psi_{0,1/2}:\mathcal{B}(L^{2}(B))\rightarrow B\otimes
B^{\operatorname{op}}$ we see that

$\displaystyle\hat{\theta}\theta_{\Lambda(a),\Lambda(b)}\hat{\theta}^{*}$
$\displaystyle=\theta_{\hat{\theta}\Lambda(a),\hat{\theta}\Lambda(b)}=\theta_{\Lambda(\theta(a)),\Lambda(\theta(b))}$
$\displaystyle\mapsto\theta(a)^{*}\otimes\sigma_{i/2}(\theta(b))=(\theta\otimes\theta)(a^{*}\otimes\sigma_{i/2}(b))=(\theta\otimes\theta)\Psi_{0,1/2}(\theta_{\Lambda(a),\Lambda(b)}),$

and so $\hat{\theta}A_{G}\hat{\theta}^{*}$ maps to $e^{\prime}$, showing that
$A_{G}^{\prime}=\hat{\theta}A_{G}\hat{\theta}^{*}$, as claimed. ∎

Consider now $B\subseteq\mathcal{B}(H)$ for some Hilbert space $H$. Let
$\theta_{\mathcal{B}}$ be an automorphism of $\mathcal{B}(H)$, and suppose
that $\theta_{\mathcal{B}}$ restricts to $B$ giving some
$\theta\in\operatorname{Aut}(B,\psi)$. Notice that for $a\in B,x\in
B^{\prime}$ we have that
$\theta_{\mathcal{B}}(x)\theta(a)=\theta_{\mathcal{B}}(xa)=\theta_{\mathcal{B}}(ax)=\theta(a)\theta_{\mathcal{B}}(x)$.
As $\theta$ is an automorphism, it follows that $\theta_{\mathcal{B}}(x)\in
B^{\prime}$, so $\theta_{\mathcal{B}}$ restricts to an endomorphism of
$B^{\prime}$. The same argument applies to $\theta_{\mathcal{B}}^{-1}$, and so
$\theta_{\mathcal{B}}$ restricts to an automorphism of $B^{\prime}$. As all
automorphisms of $\mathcal{B}(H)$ are inner, there is some unitary
$u_{\theta}\in\mathcal{B}(H)$ with
$\theta_{\mathcal{B}}(x)=u_{\theta}xu_{\theta}^{*}$ for $x\in\mathcal{B}(H)$.
Thus $\theta_{\mathcal{B}}$ preserves ${\operatorname{Tr}}$ on
$\mathcal{B}(H)$. Using the GNS map $\mathcal{B}(H)\rightarrow
H\otimes\overline{H}$, the automorphism $\theta_{\mathcal{B}}$ corresponds to
the unitary $u_{\theta}\otimes\overline{u_{\theta}}$ on
$H\otimes\overline{H}$.

######  Lemma 8.7.

Let $S\subseteq\mathcal{B}(H)$ be a $B^{\prime}$-bimodule satisfying the
conditions of Theorem 5.17. Then $S_{\theta}=\theta_{\mathcal{B}}(S)$ is also
a $B^{\prime}$-bimodule satisfying the conditions of Theorem 5.17. The
associated projection in $B\otimes B^{\operatorname{op}}$ is
$e_{\theta}=(\theta\otimes\theta)(e)$.

###### Proof.

As $\theta_{\mathcal{B}}$ restricts to an automorphism of $B^{\prime}$, it
follows that $B_{\theta}$ is a self-adjoint $B^{\prime}$-bimodule. From Lemma
8.2, we know that $\theta_{\mathcal{B}}(Q)=Q$, and so $QSQ^{-1}=S$ implies
that $QS_{\theta}Q^{-1}=S_{\theta}$. Let $S$ corresponds to $V\subseteq
H\otimes\overline{H}$, and similarly $S_{\theta}$ correspond to $V_{\theta}$.
As noted above, $V_{\theta}=(u_{\theta}\otimes\overline{u_{\theta}})(V)$. Let
$e\in B\otimes B^{\operatorname{op}}$ be the projection onto $V$, so that

$V_{\theta}=(u_{\theta}\otimes\overline{u_{\theta}})e(H\otimes\overline{H})=(u_{\theta}\otimes\overline{u_{\theta}})e(u_{\theta}\otimes\overline{u_{\theta}})^{*}(H\otimes\overline{H}),$

and so
$e_{\theta}=(u_{\theta}\otimes\overline{u_{\theta}})e(u_{\theta}\otimes\overline{u_{\theta}})^{*}$
is the orthogonal projection onto $V_{\theta}$. However, then
$e_{\theta}=(u_{\theta}\otimes\overline{u_{\theta}})e(u_{\theta}^{*}\otimes
u_{\theta}^{\top})=(\theta\otimes\theta)(e)$, as claimed. (Here we use that,
acting on $\overline{H}$, we have that
$\theta(b)^{\top}=(u_{\theta}bu_{\theta}^{*})^{\top}=\overline{u_{\theta}}b^{\top}u_{\theta}^{\top}$.)
∎

Thus we might define an automorphism of $S$ to be such a
$\theta_{\mathcal{B}}$ with $\theta_{\mathcal{B}}(S)=S$, equivalently, with
$(\theta\otimes\theta)(e)=e$. However, there are a number of problems here.
Firstly, different $\theta_{\mathcal{B}}$ might give the same $\theta$, and
hence the same automorphism of $S$. Secondly, we would wish there to be a
bijection between automorphisms of a quantum adjacency matrix $A_{G}$ and
automorphisms of the associated $S$, but it is not clear that every suitable
$\theta$ arises from some $\theta_{\mathcal{B}}$. Finally, when studying
operator bimodules $S$, there is no real dependence on $H$, but it does not
appear to be the case that any $\theta_{\mathcal{B}}$ on $\mathcal{B}(H)$, say
restricting to $\theta$ on $B$, with give rise to some automorphism of
$\mathcal{B}(K)$ whenever $B\subseteq\mathcal{B}(K)$.

With a special choice of $H$, we can say something. Set $H=L^{2}(B)$ and given
$\theta\in\operatorname{Aut}(B,\psi)$ form $\hat{\theta}$ as in Definition
8.3. We now define $\theta_{\mathcal{B}}(x)=\hat{\theta}x\hat{\theta}^{*}$ for
$x\in\mathcal{B}(L^{2}(B))$, an automorphism which restricts to $\theta$ on
$B$.

######  Proposition 8.8.

Let $A_{G}\in\mathcal{B}(L^{2}(B))$ be a quantum adjacency matrix, satisfying
axioms (1) and (2). Then $\theta\in\operatorname{Aut}(B,\psi)$ is an
automorphism of $A_{G}$ if and only if $\theta_{\mathcal{B}}(S)=S$, where $S$
is the $B^{\prime}$-operator bimodule associated with $A_{G}$.

###### Proof.

This follows immediately from Proposition 8.6 and Lemma 8.7 by thinking about
the projection associated to $A_{G}$ and $S$. ∎

##  9 Quantum automorphisms of Quantum graphs

We now look at how (compact) quantum groups (co)act on (quantum) graphs. For
compact quantum groups acting on classical graphs, this goes back to work of
Banica in [2]. In the subsequent years, there has been much activity, with the
quantum automorphisms of some small graphs computed, and much interest in the
question of which graphs have a non-commutative quantum automorphism group. In
this direction, let us just mention one recent paper [34] which constructs a
finite graph which has a non-commutative, yet finite-dimensional, quantum
automorphism group. More recently quantum automorphisms of quantum graphs have
been defined and studied, [31, Section V, Part C] and more completely, [7,
Section 3], compare [25, Section 4], [30, Section 2.3].

We find some of the literature in this area to be quite hard to follow, and so
we shall both motivate the definitions from the classical situation, and
develop certain aspects of compact quantum group theory to allow us to make
broader definitions than available elsewhere. We will assume the reader is
familiar with the basics of compact quantum group theory, for example, [32,
41].

We start again from the classical situation, which while well-known, we feel
provides interesting motivation. Let $G=\operatorname{Aut}(B,\psi)$ and let
$A=C(G)$ the group algebra, with coproduct

$\Delta:C(G)\rightarrow C(G)\otimes C(G)=C(G\times
G);\quad\Delta(f)(\theta_{1},\theta_{2})=f(\theta_{1}\theta_{2})\qquad(f\in
C(G),\theta_{1},\theta_{2}\in G).$

The action of $G$ on $B$ gives a coaction of $C(G)$ on $B$ by

$\alpha:B\rightarrow B\otimes C(G)\cong
C(G,B);\quad\alpha(b)=\big{(}\theta(b)\big{)}_{\theta\in G}.$

Let us check that the coaction equation holds. For $b\in B$ we identify
$(\alpha\otimes\operatorname{id})\alpha(b)\in B\otimes C(G)\otimes C(G)$ as a
member of $C(G\times G,B)$. Indeed, for arbitrary $f\in C(G,B)\cong B\otimes
C(G)$ we have that $(\alpha\otimes\operatorname{id})f$ evaluated at
$(\theta_{1},\theta_{2})$ is
$\alpha(f(\theta_{2}))(\theta_{1})=\theta_{1}(f(\theta_{2}))$, while
$(\operatorname{id}\otimes\Delta)f$ at $(\theta_{1},\theta_{2})$ is
$f(\theta_{1}\theta_{2})$. Thus

$\displaystyle(\alpha\otimes\operatorname{id})\alpha(b)(\theta_{1},\theta_{2})=\theta_{1}\big{(}\alpha(b)(\theta_{2})\big{)}=\theta_{1}(\theta_{2}(b))=(\operatorname{id}\otimes\Delta)\alpha(b)(\theta_{1},\theta_{2}),$

showing that indeed
$(\alpha\otimes\operatorname{id})\alpha=(\operatorname{id}\otimes\Delta)\alpha$.
Notice that with these conventions, we are considering _right_ coactions.

One can also verify the Podlés density condition, that
$\operatorname{lin}\\{(1\otimes a)\alpha(b):b\in B,a\in C(G)\\}$ is dense in
$B\otimes C(G)$. Even in this setting, verifying this condition does not
appear to be totally trivial to do by direct calculation, and we point the
reader to [36, Section 2] for an interesting discussion.

That $G$ is a group of automorphisms which preserve $\psi$ means that
$(\psi\otimes\operatorname{id})\alpha(b)=\psi(b)1$ for each $b\in B$. If we
regard a quantum adjacency matrix $A_{G}$ as acting on $B$, then that $G$ is a
group of automorphisms of $A_{G}$ is equivalent to
$\theta(A_{G}(b))=A_{G}(\theta(b))$ for each $b\in B$, that is,
$(A_{G}\otimes\operatorname{id})\alpha=\alpha A_{G}$. If we, as is usual,
regard $A_{G}$ as acting on $L^{2}(B)$, then we have to work harder to
interpret this at the coaction level, see below.

We now consider an arbitrary, possibly non-commutative, compact quantum group
$(A,\Delta)$.

######  Definition 9.1.

Let $(A,\Delta)$ be a compact quantum group. A _coaction_ of $(A,\Delta)$ on
$B$ is a unital $*$-homomorphism $\alpha:B\rightarrow B\otimes A$ with:

  1. 1.

$(\alpha\otimes\operatorname{id})\alpha=(\operatorname{id}\otimes\Delta)\alpha$;

  2. 2.

$\operatorname{lin}\\{(1\otimes a)\alpha(b):a\in A,b\in B\\}$ is dense in
$B\otimes A$.

A coaction $\alpha:B\rightarrow B\otimes A$ of $(A,\Delta)$ on $B$ _preserves
the state $\psi$_ when $(\psi\otimes\operatorname{id})\alpha(b)=\psi(b)1$ for
each $b\in B$.

The condition of preserving $\psi$ was first considered by Wang in [44] (in
fact, only the case $\psi={\operatorname{Tr}}$ are worked out in [44, Section
6], but as remarked there, the same ideas work in general). We write
$\operatorname{QAut}(B,\psi)$ for the maximal compact quantum group coacting
on $B$ and preserving $\psi$. This means that if $(A,\Delta)$ is any compact
quantum group coacting on $B$ and preserving $\psi$, there there is a unique
$*$-homomorphism $\operatorname{QAut}(B,\psi)\rightarrow A$ intertwining the
coproducts and the coactions. That such a maximal quantum group exists is
shown in [44, Theorem 6.1], where it is also show that
$\operatorname{QAut}(B)$ does not exist: there is no maximal compact quantum
group coacting on $B$, unless $B$ is commutative.

Let us think more about coactions. The second condition is often called the
_Podleś density_ condition after Podleś’s early work in this area. Let
$\mathcal{O}_{A}$ be the dense Hopf $*$-algebra of $A$ formed from the
coefficients of irreducible unitary corepresentations of $A$. There is a
$*$-algebra $\mathcal{O}_{\alpha}\subseteq B$ (the _Podleś subalgebra_ or
_algebraic core_ of $B$) which is dense, and such that $\alpha$ restricts to a
$*$-homomorphism
$\mathcal{O}_{\alpha}\rightarrow\mathcal{O}_{\alpha}\otimes\mathcal{O}_{A}$.
In fact, this restriction is a Hopf $*$-algebraic coaction; in particular, it
is injective. A modern reference for these results is [16].

For us, $B$ is finite-dimensional, so $\mathcal{O}_{B}=B$, and hence $\alpha$
actually maps into $B\otimes\mathcal{O}_{A}$. A modern approach to
constructing $\operatorname{QAut}(B,\psi)$ makes use of the following
observation. Let $(e_{i})_{i=1}^{n}$ be an algebraic basis of $B$ such that
$(\Lambda(e_{i}))$ is an orthonormal basis of $L^{2}(B)$. For any linear map
$\alpha:B\rightarrow B\otimes A$ there is a matrix $v=v_{ij}\in M_{n}(A)$ with
$\alpha(e_{i})=\sum_{j}e_{j}\otimes v_{ji}$ for each $i$. When $\alpha$ is a
coaction, our observations show that already $v_{ij}\in\mathcal{O}_{A}$ for
each $i,j$.

We follow here ideas of Banica, see [3, Section 4], or [4, Section 1] for the
case when $\psi$ is a trace. Given unitary matrices $v\in M_{n}(A)\cong
M_{n}\otimes A$ and $w\in M_{m}(A)\cong M_{m}\otimes A$ we define $v\boxtimes
w=v_{13}w_{23}\in M_{n}\otimes M_{m}\otimes A\cong M_{nm}(A)$, here using the
common leg-numbering notation. Define $\hom(v,w)$ to be the space of linear
maps $t:M_{n}\rightarrow M_{m}$ with $(t\otimes 1)v=w(t\otimes 1)$, an
equation we can interpret simply using matrix multiplication. These
definitions should of course be familiar to the reader who knows about
corepresentations of compact quantum groups.

With respect to the basis $(\Lambda(e_{i}))$, write the multiplication map
$m:B\otimes B\rightarrow B$ as a $n\times n^{2}$ matrix, and write the unit
map $\eta:\mathbb{C}\rightarrow B$ as a $n\times 1$ matrix. Also write the
projection $\eta\circ\eta^{*}$ as a $n\times n$ matrix. The following
proposition makes links between properties of a linear map
$\alpha:B\rightarrow B\otimes A$, and properties of $v$.

######  Proposition 9.2.

Let $A$ be a compact quantum group. Under this bijection between $\alpha$ and
$v$, the following hold:

  1. 1.

$\alpha$ is multiplicative if and only if $m\in\hom(v\boxtimes v,v)$;

  2. 2.

$\alpha$ is a unital map if and only if $\eta\in\hom(1,v)$;

  3. 3.

assuming $\alpha$ is unital, $\alpha$ is $\psi$-preserving if and only if
$\eta\circ\eta^{*}\in\hom(v,v)$;

  4. 4.

$\alpha$ satisfies the coaction equation if and only if $v$ is a
corepresentation.

Assume now that $\alpha$ is multiplicative and unital. If $v$ is unitary, then
$\alpha$ is a $*$-map. If $\alpha$ is a $*$-map, satisfies the coaction
equation, and is $\psi$-preserving, then $v$ is unitary.

###### Proof.

Most of this follows by simple direct calculation, and is rather similar to
[4, Lemma 1.2] or [25, Lemma 4.5] (which both just considers the case when
$\psi$ is a trace). However, we prove the final claim, to check that it works
for a non-trace, and to indicate that this step does not follow from direct
calculation, but rather from quite a lot of theory.

Suppose that $\alpha$ is a unital $*$-homomorphism and is $\psi$-preserving.
Then consider, for $i,j$,

$\displaystyle(\psi\otimes\operatorname{id})\alpha(e_{i}^{*}e_{j})$
$\displaystyle=(\psi\otimes\operatorname{id})(\alpha(e_{i})^{*}\alpha(e_{j}))=\sum_{s,t}\psi(e_{s}^{*}e_{t})v_{si}^{*}v_{tj}=\sum_{s,t}(\Lambda(e_{s})|\Lambda(e_{t}))v_{si}^{*}v_{tj}$
$\displaystyle=\sum_{k}v_{ki}^{*}v_{kj}=(v^{*}v)_{ij}.$

As $\alpha$ preserves $\psi$, this is equal to
$\psi(e_{i}^{*}e_{j})1=\delta_{i,j}1$ and so $v^{*}v=1$.

As we know that actually $v$ is a matrix in $M_{n}(\mathcal{O}_{A})$, Hopf
$*$-algebra theory would show that $v$ is invertible, as it has a left
inverse. An alternative way to show this is to check that $v$ is _non-
degenerate_ and then to appeal to [47, Proposition 3.2]. (We note here that we
use “non-degenerate” in the sense of [47, page 630], which is weaker than the
usual Hopf $*$-algebra definition.) Either way, we conclude that $v$ is
invertible, so necessarily also $vv^{*}=1$, so $v$ is unitary. ∎

######  Remark 9.3.

The condition that $\eta\circ\eta^{*}\in\hom(v,v)$ is often omitted, which is
perhaps surprising. In fact, it is automatic in the presence of the other
properties. Indeed, $\eta\in\hom(1,v)$ means that
$\eta\otimes\operatorname{id}=v(\eta\otimes\operatorname{id})$ so also
$\eta^{*}\otimes\operatorname{id}=(\eta^{*}\otimes\operatorname{id})v^{*}$ and
hence

$v(\eta\circ\eta^{*}\otimes\operatorname{id})v^{*}=v(\eta\otimes\operatorname{id})(\eta^{*}\otimes\operatorname{id})v^{*}=(\eta\otimes\operatorname{id})(\eta^{*}\otimes\operatorname{id})=(\eta\circ\eta^{*}\otimes\operatorname{id}).$

If $v$ is assumed unitary, this is implies that
$\eta\circ\eta^{*}\in\hom(v,v)$. ∎

It is now easy to see how to define the _quantum automorphism group_ of $B$
which preserves $\psi$, written as $\operatorname{QAut}(B,\psi)$. We define
$\operatorname{QAut}(B,\psi)$ to be the universal $C^{*}$-algebra generated by
the coefficients of an $n\times n$ unitary matrix $v=(v_{ij})$ with the
conditions

$m\in\hom(v\boxtimes v,v),\qquad\eta\in\hom(1,v).$

The coproduct on $\operatorname{QAut}(B,\psi)$ is defined to satisfy
$\Delta(v_{ij})=\sum_{k=1}^{n}v_{ik}\otimes v_{kj}$, and exists by
universality. We define $\alpha$ by $\alpha(e_{i})=\sum_{j=1}^{n}e_{j}\otimes
v_{ji}$, and the proposition shows that $\alpha$ is a coaction.

As $\operatorname{QAut}(B,\psi)$ is generated by elements of the matrix $v$,
which is by definition a (unitary) corepresentation, in fact
$\operatorname{QAut}(B,\psi)$ is a compact matrix quantum group. One point in
this definition which is far from obvious to the author is why the axioms of a
compact quantum group hold, equivalently, why the transpose of $v$ is an
invertible matrix (an alternative axiom in the case of a compact matrix
quantum group). We find it helpful to look at [5, Section 3] which explains
links with the Tannaka–Krein reconstruction theorem. Also not considered in
this construction is the Podleś density condition for $\alpha$, but this
follows easily once we know that $v^{t}$ is invertible. In fact, as we shall
need the following later, we shall prove by a more direct argument hat $v^{t}$
is invertible.

######  Lemma 9.4.

Define $X\in\mathbb{M}_{n}$ by
$X_{ij}=(\Lambda(e_{i})|\Lambda(e_{j}^{*}))=\psi(e_{i}^{*}e_{j}^{*})$. Then
$X^{-1}=\overline{X}$, and $v=(X\otimes 1)\overline{v}(\overline{X}\otimes 1)$
where $\overline{v}$ is the matrix $(v_{ij}^{*})$ (the transpose of the
adjoint). Furthermore, $\overline{v}$ is invertible with inverse
$(\overline{X}\otimes 1)v^{*}(X\otimes 1)$, and so also $v^{t}$ is invertible.

###### Proof.

As $\alpha$ is a $*$-map, and expanding using the orthonormal basis
$(\Lambda(e_{j}))$, we see that

$\displaystyle\alpha(e_{i}^{*})$
$\displaystyle=\sum_{j}(\Lambda(e_{j})|\Lambda(e_{i}^{*}))\alpha(e_{j})=\sum_{j,l}e_{l}\otimes(\Lambda(e_{j})|\Lambda(e_{i}^{*}))v_{lj}$
$\displaystyle=\alpha(e_{i})^{*}=\sum_{j}e_{j}^{*}\otimes
v_{ji}^{*}=\sum_{j,l}e_{l}\otimes(\Lambda(e_{l})|\Lambda(e_{j}^{*}))v_{ji}^{*},$

and so
$\sum_{j}(\Lambda(e_{j})|\Lambda(e_{i}^{*}))v_{lj}=\sum_{j}(\Lambda(e_{l})|\Lambda(e_{j}^{*}))v_{ji}^{*}$
for each $l$. Thus $\sum_{j}v_{lj}X_{ji}=\sum_{j}X_{lj}v_{ji}^{*}$ or
$v(X\otimes 1)=(X\otimes 1)\overline{v}$. As $\psi(b)={\operatorname{Tr}}(Qb)$
for each $b\in B$,

$(\Lambda(a^{*})|\Lambda(b))={\operatorname{Tr}}(Qab)={\operatorname{Tr}}(bQa)={\operatorname{Tr}}(QQ^{-1}bQa)=(\Lambda(Qb^{*}Q^{-1})|\Lambda(a))\qquad(a,b\in
B).$

Hence, for each $i,j$,

$\displaystyle(\overline{X}X)_{ij}$
$\displaystyle=\sum_{k}\overline{X_{ik}}X_{kj}=\sum_{k}\overline{(\Lambda(e_{i})|\Lambda(e_{k}^{*}))}(\Lambda(e_{k})|\Lambda(e_{j}^{*}))=\sum_{k}(\Lambda(Qe_{i}^{*}Q^{-1})|\Lambda(e_{k}))(\Lambda(e_{k})|\Lambda(e_{j}^{*}))$
$\displaystyle=(\Lambda(Qe_{i}^{*}Q^{-1})|\Lambda(e_{j}^{*}))=(\Lambda(e_{j})|\Lambda(e_{i}))=\delta_{i,j}.$

As $X$ is a scalar matrix, it follows that $X^{-1}=\overline{X}$, and so
$v=(X\otimes 1)\overline{v}(X^{-1}\otimes 1)=(X\otimes
1)\overline{v}(\overline{X}\otimes 1)$, as claimed. As $v^{*}v=vv^{*}=1$ we
see that $u=(\overline{X}\otimes 1)v^{*}(X\otimes 1)$ is the inverse of
$\overline{v}$, and hence $u^{*}=(X^{*}\otimes 1)v(X^{t}\otimes 1)$ is the
inverse of $v^{t}$. ∎

We shall make use of this construction, and our parenthetical comments, below.
We hope also that this quick sketch of the construction will aid the reader in
understanding links between our presentation and, for example, [7, Definition
3.7, Proposition 3.8]. See also the informative [25, Section 4].

###  9.1 Unitary implementations

In this section, let $(A,\Delta)$ be a compact quantum group with a coaction
$\alpha:B\rightarrow B\otimes A$ which preserves $\psi$. Consider $L^{2}(B)$
with GNS map $\Lambda:B\rightarrow L^{2}(B)$, and let $A$ be faithfully and
non-degenerately represented on a Hilbert space $K$. Define an operator $U$ on
$L^{2}(B)\otimes K$ by

$U(\Lambda(b)\otimes\xi)=\alpha(b)(\Lambda(1)\otimes\xi)\qquad(b\in B,\xi\in
K).$

Then

$\displaystyle\big{(}U(\Lambda(b)\otimes\xi)\big{|}U(\Lambda(d)\otimes\eta)\big{)}$
$\displaystyle=(\psi\otimes\omega_{\xi,\eta})\alpha(b^{*}d)=\psi(b^{*}d)(\xi|\eta)=\big{(}\Lambda(b)\otimes\xi\big{|}\Lambda(d)\otimes\eta\big{)},$

using that $\alpha$ is $\psi$-invariant. A similar calculation holds with
arbitrary tensors, which shows that $U$ is well-defined, and a co-isometry.
The density condition on $\alpha$ (and that $A$ acts non-degenerately on $K$)
shows that $U$ is unitary. Then

$U(b\otimes
1)U^{*}U(\Lambda(d)\otimes\xi)=U(\Lambda(bd)\otimes\xi)=\alpha(bd)(\Lambda(1)\otimes\xi)=\alpha(b)U(\Lambda(d)\otimes\xi),$

and thus $U(b\otimes 1)U^{*}=\alpha(b)$, acting on $L^{2}(B)\otimes K$.

We remark that we could instead have worked in the language of Hilbert
$C^{*}$-modules, which would remove the need to consider $K$, at the cost of
more technicalities. The following now shows how $\alpha$ commuting with
$A_{G}$ makes sense at the level of $U$.

######  Lemma 9.5.

Let $A_{0}:B\rightarrow B$ be a linear map which induces the linear map
$A_{G}:L^{2}(B)\rightarrow L^{2}(B)$, that is, $A_{G}\circ\Lambda=\Lambda\circ
A_{0}$. Then $\alpha A_{0}=(A_{0}\otimes\operatorname{id})\alpha$ if and only
if $(A_{G}\otimes 1)U=U(A_{G}\otimes 1)$.

###### Proof.

Given $b\in B,a\in A,\xi\in K$, we see that $(b\otimes
a)(\Lambda(1)\otimes\xi)=\Lambda(b)\otimes a\xi$. As $A$ acts non-degenerately
on $K$, it follows that $u\in B\otimes A$ has $u=0$ if and only if
$u(\Lambda(1)\otimes\xi)=0$ for all $\xi\in K$. We hence see that $\alpha
A_{0}=(A_{0}\otimes\operatorname{id})\alpha$ if and only if

$\displaystyle\alpha(A_{0}(b))=(A_{0}\otimes\operatorname{id})\alpha(b)$
$\displaystyle(b\in B)$ $\displaystyle\Leftrightarrow\qquad$
$\displaystyle\alpha(A_{0}(b))(\Lambda(1)\otimes\xi)=(A_{0}\otimes\operatorname{id})\alpha(b)(\Lambda(1)\otimes\xi)$
$\displaystyle(b\in B,\xi\in K)$ $\displaystyle\Leftrightarrow\qquad$
$\displaystyle U(\Lambda(A_{0}(b))\otimes\xi)=(A_{G}\otimes
1)U(\Lambda(b)\otimes\xi)$ $\displaystyle(b\in B,\xi\in K)$
$\displaystyle\Leftrightarrow\qquad$ $\displaystyle U\big{(}(A_{G}\otimes
1)(\Lambda(b)\otimes\xi)\big{)}=(A_{G}\otimes 1)U(\Lambda(b)\otimes\xi)$
$\displaystyle(b\in B,\xi\in K)$

using that $A_{G}\Lambda(d)=\Lambda(A_{0}(d))$ for $d\in B$. This is
equivalent to $U(A_{G}\otimes 1)=(A_{G}\otimes 1)U$, as claimed. ∎

We shall call $U$ the _unitary implementation of $\alpha$_, compare with the
much more general construction of [42]. Recall from Proposition 9.2 that there
is a bijection between coactions $\alpha$ and certain (unitary) matrices $v\in
M_{n}(A)$. The following clarifies the relation between $v$ and the unitary
$U$.

######  Lemma 9.6.

The unitary $U\in\mathcal{B}(L^{2}(B)\otimes K)$ is equal to the image of
$v\in M_{n}(A)\cong M_{n}\otimes A$ acting on $L^{2}(B)\otimes K$.

###### Proof.

We simply calculate that for $a\in A$,

$U\big{(}\Lambda(e_{i})\otimes\xi\big{)}=\alpha(e_{i})(\Lambda(1)\otimes\xi)=\sum_{j}\Lambda(e_{j})\otimes
v_{ji}\xi.$

Thus the matrix of $U$ equals $(v_{ij})$, acting on $L^{2}(B)\otimes K$, as
claimed. ∎

In particular, $U$ is a corepresentation (which also follows by direct
calculation).

The following definition should now be clear, given our motivation above from
the case of classical groups, and Lemma 9.5.

######  Definition 9.7.

Let $(A,\Delta)$ be a compact quantum group, and let $\alpha:B\rightarrow
B\otimes A$ be a coaction which preserves $\psi$. Let
$A_{G}\in\mathcal{B}(L^{2}(B))$ be a quantum adjacency matrix associated to
the linear map $A_{0}$ on $B$. Then $\alpha$ is _action on $A_{G}$_ when
$\alpha A_{0}=(A_{0}\otimes\operatorname{id})\alpha$, equivalently, when
$A_{G}$ and $v$ commute, where $v$ is the unitary corepresentation given by
$\alpha$.

######  Remark 9.8.

Let us consider the complete quantum graph, compare Proposition 6.1, so that
$A_{G}=\theta_{\Lambda(1),\Lambda(1)}$. Then $A_{G}=\eta\circ\eta^{*}$, and so
by Remark 9.3, as $\alpha$ is $\psi$-preserving, automatically $v$ commutes
with $A_{G}$.

Similarly, consider the empty quantum graph, compare Proposition 6.2, so that
$A_{G}=(mm^{*})^{-1}$. As $\alpha$ is a coaction, by Proposition 9.2,
$(m\otimes 1)v_{13}v_{23}=v(m\otimes 1)$. Thus also
$v_{23}^{*}v_{13}^{*}(m^{*}\otimes 1)=(m^{*}\otimes 1)v^{*}$, and so as $v$ is
unitary, also $(m^{*}\otimes 1)v=v_{13}v_{23}(m^{*}\otimes 1)$. Hence

$v(mm^{*}\otimes 1)=(m\otimes 1)v_{13}v_{23}(m^{*}\otimes 1)=(mm^{*}\otimes
1)v,$

and so also $A_{G}$ commutes with $v$.

Hence, in agreement with intuition, as soon as $\alpha$ is a $\psi$-preserving
coaction on $B$, it is automatically an action on both the empty and complete
quantum graphs. ∎

We next show that there exists a universal compact quantum group which acts on
$A_{G}$, and which we can then be justified in calling the _quantum
automorphism group of $A_{G}$_. We shall compare this definition to others in
the literature in Remark 9.10.

######  Proposition 9.9.

Consider $\operatorname{QAut}(B,\psi)$ as constructed above from a generating
unitary corepresentation $v$. Let $A$ be the quotient formed by imposing the
extra conditions that $A_{G}v=vA_{G}$. Then $\Delta$ drops to $A$ making
$(A,\Delta)$ into a quantum subgroup of $\operatorname{QAut}(B,\psi)$, which
is the quantum automorphism group of $A_{G}$: that is, universal amoungst all
compact quantum groups acting on $A_{G}$.

###### Proof.

Let $J$ be the ideal generated by the relations coming from $A_{G}v=vA_{G}$,
that is, $J$ is the ideal generated by the element

$\sum_{k=1}^{n}A^{G}_{ik}v_{kj}-\sum_{k=1}^{n}A^{G}_{kj}v_{ik}\qquad(1\leq
i,j\leq n).$

If we apply $\Delta$ to one of these elements, we obtain

$\displaystyle\sum_{k,l=1}^{n}A^{G}_{ik}v_{kl}\otimes
v_{lj}-\sum_{k,l=1}^{n}A^{G}_{kj}v_{il}\otimes v_{lk}$
$\displaystyle=\sum_{l=1}^{n}\Big{(}\sum_{k=1}^{n}A^{G}_{ik}v_{kl}-v_{ik}A^{G}_{kl}\Big{)}\otimes
v_{lj}+\sum_{l=1}^{n}\Big{(}\sum_{k=1}^{n}v_{ik}A^{G}_{kl}\Big{)}\otimes
v_{lj}-\sum_{k,l=1}^{n}A^{G}_{kj}v_{il}\otimes v_{lk}$
$\displaystyle=\sum_{l=1}^{n}\Big{(}\sum_{k=1}^{n}A^{G}_{ik}v_{kl}-v_{ik}A^{G}_{kl}\Big{)}\otimes
v_{lj}+\sum_{k,l=1}^{n}v_{ik}\otimes A^{G}_{kl}v_{lj}-v_{ik}\otimes
A^{G}_{lj}v_{kl}$
$\displaystyle=\sum_{l=1}^{n}\Big{(}\sum_{k=1}^{n}A^{G}_{ik}v_{kl}-v_{ik}A^{G}_{kl}\Big{)}\otimes
v_{lj}+\sum_{k,l=1}^{n}v_{ik}\otimes\Big{(}A^{G}_{kl}v_{lj}-A^{G}_{lj}v_{kl}\Big{)}$
$\displaystyle\in
J\otimes\operatorname{QAut}(B,\psi)+\operatorname{QAut}(B,\psi)\otimes J.$

In the first line we add in and subtract the same element, but notice that the
element in the brackets is manifestly in $J$. In the second step we re-index,
and the re-arrange in the third step so both brackets are now clearly members
of $J$.

This calculation and a simple continuity argument shows that $\Delta$ gives a
well-defined map $A\rightarrow A\otimes A$, where
$A=\operatorname{QAut}(B,\psi)/J$. By definition, $A$ coacts on $A_{G}$.

Now let $A_{1}$ be some other compact quantum group which coacts on $A_{G}$.
Then, by Proposition 9.2 for any orthonormal basis of $L^{2}(B)$ there is a
unitary corepresentation $v\in M_{n}(A_{1})$ which defines the coaction. If
necessary replacing $A_{1}$ by the $C^{*}$-subalgebra generated by the
elements of $v$, we may suppose that $A_{1}$ is a compact matrix quantum group
generated by $v$.

Consider now $\operatorname{QAut}(B,\psi)$ with generating corepresentation
$w$ say. As $A_{1}$ coacts in a $\psi$-preserving way, $v$ satisfies all the
relations which are satisfied by $w$, and so by universality, there is a
$*$-homomorphism $\theta:\operatorname{QAut}(B,\psi)\rightarrow A_{1}$ which
intertwines the coproducts and which sends $w$ to $v$. Hence $\theta$ is a
surjection. As $A_{G}v=vA_{G}$ it follows that $\theta(J)=\\{0\\}$, and so
$\theta$ drops to a morphism from $A$, which is necessarily uniquely defined.
This has hence demonstrated the required universal property for $A$. ∎

We shall write $\operatorname{QAut}(A_{G})$ for the resulting compact quantum
group.

######  Remark 9.10.

Let us compare this definition/construction with [7, Definition 3.7]. There,
$\mathcal{O}(A_{G})$ is defined to be the universal $*$-algebra generated by a
unitary matrix $u$ with the conditions that the map $\rho:B\rightarrow
B\otimes\mathcal{O}(A_{G}),e_{i}\mapsto\sum_{j}e_{j}\otimes u_{ji}$ is a
unital $*$-homomorphism with $\rho A_{0}=(A_{0}\otimes\operatorname{id})\rho$.
Here $A_{0}:B\rightarrow B$ is associated to $A_{G}$ in the usual way.
Furthermore, [7, Proposition 3.8] defines a Hopf $*$-algebra structure on
$\mathcal{O}(A_{G})$ which in particular ensures that $\rho$ satisfies the
coaction equation.

From the discussion above where we sketch the construction of
$\operatorname{QAut}(B,\psi)$, it is clear that $\mathcal{O}(A_{G})$ is a
quotient of the Hopf $*$-algebra $\mathcal{O}(\operatorname{QAut}(B,\psi))$,
where the extra relations come from requiring that $\rho
A_{0}=(A_{0}\otimes\operatorname{id})\rho$. As in Lemma 9.5, this is
equivalent to requiring that $u$ and $A_{G}$ commute. Thus this construction
does agree with Proposition 9.9.

Gromada in [25, Section 4] makes much the same definition, but with a slightly
different presentation of the relations. The links with Tannaka–Krein
reconstruction results are made explicit. However, note that all definitions
are made at the level of Hopf $*$-algebras, and so, for example, the
definition of a coaction [25, Definition 4.2] differs from ours (but is
equivalent). ∎

######  Remark 9.11.

As alluded to a number of times, much of the existing work on quantum graphs
restricts to the case when $\psi$ is a $\delta$-form (tracial, or not). There
is some motivation for this if we consider $\operatorname{QAut}(B,\psi)$, as a
result attributed to Brannan, [17, Proposition 20], describes
$\operatorname{QAut}(B,\psi)$ in terms of a free product of a decomposition of
$B$ into the coarsest direct sum $B=\bigoplus B_{i}$ where $\psi$ restricts to
a $\delta$-form on each $B_{i}$. Thus for many questions, it suffices to look
at each factor $B_{i}$ separately.

We might wonder if $\operatorname{QAut}(A_{G})$ factors similarly. This does
not appear to be the case, as there appears to be no reason why $A_{G}$ should
restrict to each factor $B_{i}$. To find a counter-example, let
$B=\mathbb{M}_{2}\oplus\mathbb{M}_{2}$ where we equip each $\mathbb{M}_{2}$
factor with differently normalised traces, say $\psi_{1},\psi_{2}$ with
$\psi=\psi_{1}\oplus\psi_{2}$. Then $\psi$ is not a $\delta$-form, but each
$\psi_{i}$ is.

Let $B$ act on $H=\mathbb{C}^{2}\oplus\mathbb{C}^{2}$ so that
$B^{\prime}\cong\ell^{\infty}_{2}$, say the span of $e_{1},e_{2}$ where
$e_{i}$ is the projection onto the $i$th $\mathbb{C}^{2}$ component. As usual,
we identify $\mathcal{B}(H)$ with
$\mathbb{M}_{2}(\mathcal{B}(\mathbb{C}^{2}))\cong\mathbb{M}_{2}(\mathbb{M}_{2})$.
Let

$S=\operatorname{lin}\\{e_{1},e_{2},x_{0},x_{0}^{*}\\},\quad\text{with}\quad
x_{0}=\begin{pmatrix}0&t\\\ 0&0\end{pmatrix},$

for some non-zero $t:\mathbb{C}^{2}\rightarrow\mathbb{C}^{2}$. Then $S$ is an
operator system, and so as $B$ has a tracial state, $S$ corresponds to a
quantum adjacency matrix $A_{G}$ say, and to a projection in $B\otimes
B^{\operatorname{op}}$, say $e_{G}$.

Towards a contradiction, suppose that $A_{G}$ commutes with the projections
onto each factor of $B$. As $A_{G}$ is self-adjoint, it follows that $A_{G}$
can be written as a sum of rank-one operators of the form
$\theta_{\Lambda(a),\Lambda(b)}$ where $a,b$ both come from the same factor of
$B$. It follows that $e_{G}$ is the span of tensors $b\otimes a^{*}$ where
$a,b$ both come from the same factor of $B$. Thus the projection of
$\mathcal{B}(L^{2}(B))$ onto $S$ is a sum of maps of the form $\theta\mapsto
b\theta a^{*}$ where $a,b$ both come from the same factor of $B$. However,

$\begin{pmatrix}b&0\\\ 0&0\end{pmatrix}x_{0}\begin{pmatrix}a^{*}&0\\\
0&0\end{pmatrix}=\begin{pmatrix}0&bt\\\
0&0\end{pmatrix}\begin{pmatrix}a^{*}&0\\\ 0&0\end{pmatrix}=0,$

and similarly if $b,a$ come from the 2nd factor. This is a contradiction, as
the projection onto $S$ should preserve $x_{0}$. ∎

Examples of quantum automorphism groups of quantum graphs are considered in
[25], and in particular in [30, Section 4], where all quantum graphs over
$B=\mathbb{M}_{2}$ are considered.

###  9.2 Quantum automorphisms of operator bimodules

As $B^{\prime}$-operator bimodules $S$ satisfying the conditions of Theorem
5.17 biject with quantum adjacency matrices $A_{G}$, we immediately obtain a
notion of an action of a quantum group on $S$. Of course, it would be more
satisfying to have a definition which only used $S$.

If we consider what happens for (non-quantum) automorphisms of $A_{G}$, say
associated to a projection $e_{G}$ and $B^{\prime}$-bimodule $S$, as in
Section 8, then we found a link between automorphisms $\theta$ of $A_{G}$, the
condition that $(\theta\otimes\theta)(e_{G})=e_{G}$, and
$\theta_{\mathcal{B}}(S)=S$, at least when $S\subseteq\mathcal{B}(L^{2}(B))$.
Unfortunately, as is well-known, for quantum groups, forming the analogue of
the “diagonal action” $\theta\otimes\theta$ is essentially impossible. We
hence focus on quantum automorphisms of $S$, but guided by the commutative
situation, we shall only consider the situation when $B$ is acting on
$L^{2}(B)$.

######  Remark 9.12.

A possible definition of $(A,\Delta)$ acting on $S$ was given in [21,
Definition 5.11], but unfortunately this does not work in general. However, it
will motivate our definition below, so we sketch the idea, and the problem.
Let $(A,\Delta)$ coact on $B$ is a $\psi$-preserving way. As in Section 9.1
form the unitary implementation $U$, which is a unitary corepresentation of
$(A,\Delta)$. We may then define

$\alpha_{U}:\mathcal{B}(L^{2}(B))\rightarrow\mathcal{B}(L^{2}(B))\otimes
A;\quad x\mapsto U(x\otimes 1)U^{*}.$

As $(\operatorname{id}\otimes\Delta)(U)=U_{12}U_{13}$, we see that

$(\operatorname{id}\otimes\Delta)(\alpha_{U}(x))=U_{12}U_{13}(x\otimes
1\otimes
1)U_{13}^{*}U_{12}^{*}=(\alpha_{U}\otimes\operatorname{id})\alpha_{U}(x),$

so $\alpha_{U}$ satisfies the coaction equation. As in fact $U\in
M_{n}(\mathcal{O}_{A})$ (see the discussion after Definition 9.1) it makes
sense to apply the counit $\epsilon$ of $\mathcal{O}_{A}$, and we see that

$(\operatorname{id}\otimes\epsilon)\alpha_{U}(x)=(\operatorname{id}\otimes\epsilon)(U)x(\operatorname{id}\otimes\epsilon)(U^{*})=x\qquad(x\in\mathcal{B}(L^{2}(B))).$

From this, the Podleś density condition follows, compare [35, Remark 2.3]. So
$\alpha_{U}$ is a coaction.

It is suggested in [21, Definition 5.11] that $(A,\Delta)$ acts on $S$ via
$\alpha$ when $\alpha_{U}(S)\subseteq S\otimes A$. In fact, [21] works with
the more general notion of _quantum metrics_ , [26], and so perhaps also in
the graph case we require $\alpha_{U}(B^{\prime})\subseteq B^{\prime}\otimes
A$ (this condition corresponding to the quantum metric at $0$). It follows
from [21, Proposition 5.14] that when $B=C(V)$ and $G$ is a classical graph,
then $\alpha$ acts on $S$ if and only if $\alpha$ acts on $A_{G}$, in the
sense of Definition 9.7.

However, suppose now $B$ is non-commutative, and consider the empty quantum
graph, compare Proposition 6.2. Suppose also that $\psi$ is a trace. Then
$A_{G}=(mm^{*})^{-1}$ and $S=B^{\prime}$. As we saw in Remark 9.8, any
coaction on $(B,\psi)$ should automatically be an action on $S$. That is, it
should be automatic that $\alpha_{U}(B^{\prime})\subseteq B^{\prime}\otimes
A$. However, there doesn’t seem to be any reason why this should be so. (This
property is claimed in [21, Remark 5.13], but we do not follow the claimed
proof. Indeed, there it seems to be claimed that $Je_{k}J=e_{k}^{*}$ where
$(e_{k})$ is a basis of $B$ with $(\Lambda(e_{k}))$ an orthonormal basis of
$L^{2}(B)$, and $J$ the modular conjugation. This relation,
$Je_{k}J=e_{k}^{*}$ would be true if $e_{k}$ were central, that is, $B$ were
commutative, but it does not hold in general.)

We shall tweak this proposed definition to more directly take account of
$B^{\prime}$. ∎

Recall the modular conjugation $J$ from Section 5, which in particular
satisfies that $JBJ=B^{\prime}$. Define
$\kappa:\mathcal{B}(L^{2}(B))\rightarrow\mathcal{B}(L^{2}(B))$ by
$\kappa(x)=Jx^{*}J$, so $\kappa$ is an anti-$*$-homomorphism,
$\kappa^{2}=\operatorname{id}$, and $\kappa(B)=B^{\prime}$. We recall that
associated to any compact quantum group $(A,\Delta)$ is both the _reduced_
form, acting on the GNS space for the Haar state of $A$, and the _universal_
form, compare [27], which is the enveloping $C^{*}$-algebra of
$\mathcal{O}_{A}$. In particular $\operatorname{QAut}(B,\psi)$ must be
universal in this sense. We shall now suppose that $(A,\Delta)$ is universal,
or more precisely, that $A$ admits a _unitary antipode_ $R$, see [27,
Proposition 7.2]. Then $R$ is an anti-$*$-homomorphism,
$R^{2}=\operatorname{id}$ and $\sigma(R\otimes R)\Delta=\Delta R$, with
$\sigma$ the tensor swap map.

######  Definition 9.13.

Given a unitary corepresentation $U$ of $(A,\Delta)$ on $L^{2}(B)$, define

$\alpha_{U}:\mathcal{B}(L^{2}(B))\rightarrow\mathcal{B}(L^{2}(B))\otimes
A;\quad x\mapsto U(x\otimes 1)U^{*},$

and define

$\alpha^{\operatorname{op}}_{U}:\mathcal{B}(L^{2}(B))\rightarrow\mathcal{B}(L^{2}(B))\otimes
A;\quad x\mapsto(\kappa\otimes R)\alpha_{U}(\kappa(x)).$

######  Lemma 9.14.

With the definition above, $\alpha^{\operatorname{op}}_{U}$ is a coaction of
$(A,\sigma\Delta)$ on $\mathcal{B}(L^{2}(B))$ and
$\alpha^{\operatorname{op}}_{U}(B^{\prime})\subseteq B^{\prime}\otimes A$.

###### Proof.

It is standard that $(A,\sigma\Delta)$ is also a compact quantum group. As
$\kappa$ and $\kappa\otimes R$ are both anti-$*$-homomorphisms, it follows
that $\alpha_{U}$ is a $*$-homomorphism. Notice that $R$ intertwines the
coproducts $\Delta$ and $\sigma\Delta$. Thus

$\displaystyle(\alpha_{U}^{\operatorname{op}}\otimes\operatorname{id})\alpha_{U}^{\operatorname{op}}(x)$
$\displaystyle=(\alpha_{U}^{\operatorname{op}}\kappa\otimes
R)\alpha_{U}(\kappa(x))=(\kappa\otimes R\otimes
R)(\alpha_{U}\otimes\operatorname{id})\alpha_{U}(\kappa(x))$
$\displaystyle=(\kappa\otimes R\otimes
R)(\operatorname{id}\otimes\Delta)\alpha_{U}(\kappa(x))=(\kappa\otimes\sigma\Delta
R)\alpha_{U}(\kappa(x))=(\operatorname{id}\otimes\sigma\Delta)\alpha_{U}^{\operatorname{op}}(x),$

and so $\alpha_{U}^{\operatorname{op}}$ satisfies the coaction equation. As
$R^{2}=\operatorname{id}$, verifying the Podleś density condition is routine.

For $x\in B^{\prime}$, we know that $b=\kappa(x)\in B$, and so
$\alpha_{U}^{\operatorname{op}}(x)=(\kappa\otimes
R)\alpha_{U}(b)\in\kappa(B)\otimes R(A)=B^{\prime}\otimes A$, as claimed. ∎

It is easy to see that $\alpha_{U}^{\operatorname{op}}(S)\subseteq S\otimes A$
if and only if $\alpha_{U}(\kappa(S))\subseteq\kappa(S)\otimes A$. In the
tracial situation, it will turn out that $\alpha$ acts on $A_{G}$, that is,
$U$ commutes with $A_{G}$, if and only if
$\alpha_{U}^{\operatorname{op}}(S)\subseteq S\otimes A$, equivalently,
$\alpha_{U}(\kappa(S))\subseteq\kappa(S)\otimes A$, see Corollary 9.19. When
$\psi$ is not a trace, things are not quite so precise, Theorem 9.18.

We start towards this proof by looking more closely at $\alpha_{U}$. We would
like to find the implementing unitary for this coaction, which first requires
us to find a natural invariant state. Set $H=L^{2}(B)$, and fix some basis
$(e_{i})$ for $B$ such that $(\Lambda(e_{i}))$ is an orthonormal basis of $H$.
To ease notation, we shall identify $\mathbb{M}_{n}$ with
$\mathbb{M}_{n}\otimes 1\subseteq M_{n}(A)$, thus allowing us, for example, to
write $XU$ for a scalar matrix $X$.

######  Lemma 9.15.

Recall the matrix $X_{ij}=(\Lambda(e_{i})|\Lambda(e_{j}^{*}))$ from Lemma 9.4,
and set $P=X^{*}X$. Then $P$ is positive and invertible, and
$x_{0}=(P^{t})^{1/2}$ agrees with $\nabla^{1/2}$ on $H=L^{2}(B)$. Define
$\tilde{\psi}(x)={\operatorname{Tr}}(P^{t}x)$ for $x\in\mathcal{B}(H)$. Then
the coaction $\alpha_{U}$ preserves $\tilde{\psi}$.

###### Proof.

From Lemma 9.4, we know that $X^{-1}=\overline{X}$, and so $P$ is positive and
invertible, with $P^{-1}=\overline{X}X^{t}$. By definition,
$x_{0}^{-2}=(P^{t})^{-1}=XX^{*}$, and we calculate that

$\displaystyle(XX^{*})_{ij}$
$\displaystyle=\sum_{k}X_{ik}\overline{X_{jk}}=\sum_{k}\psi(e_{i}^{*}e_{k}^{*})\psi(e_{k}e_{j})=\sum_{k}{\operatorname{Tr}}(Qe_{i}^{*}e_{k}^{*}){\operatorname{Tr}}(Qe_{k}e_{j})$
$\displaystyle=\sum_{k}{\operatorname{Tr}}(Qe_{k}^{*}Qe_{i}^{*}Q^{-1}){\operatorname{Tr}}(QQ^{-1}e_{j}Qe_{k})=\sum_{k}(\Lambda(e_{k})|\Lambda(Qe_{i}^{*}Q^{-1}))(\Lambda(Qe_{j}^{*}Q^{-1})|\Lambda(e_{k}))$
$\displaystyle=(\Lambda(Qe_{j}^{*}Q^{-1})|\Lambda(Qe_{i}^{*}Q^{-1})={\operatorname{Tr}}(QQ^{-1}e_{j}QQe_{i}^{*}Q^{-1})={\operatorname{Tr}}(Qe_{i}^{*}Q^{-1}e_{j}Q)$
$\displaystyle=(\Lambda(e_{i})|\Lambda(Q^{-1}e_{j}Q))=(\Lambda(e_{i})|\nabla^{-1}\Lambda(e_{j})),$

and so $x_{0}^{-2}=XX^{*}$ represents the operator $\nabla^{-1}$, so $x_{0}$
agrees with $\nabla^{1/2}$.

As $U=X\overline{U}\overline{X}$ so that $U^{t}=X^{*}U^{*}X^{t}$, we see that

$U^{t}P\overline{U}=X^{*}U^{*}X^{t}X^{*}X\overline{X}UX=X^{*}U^{*}(\overline{X}X)^{t}(X\overline{X})UX=X^{*}U^{*}UX=X^{*}X=P.$

It follows that for $x\in\mathcal{B}(H)$,

$\displaystyle(\tilde{\psi}\otimes\operatorname{id})\alpha_{U}(x)$
$\displaystyle=\sum_{i,j,k,l}\tilde{\psi}(e_{ij}xe_{kl})U_{ij}(U^{*})_{kl}=\sum_{i,j,k,l}(\Lambda(e_{j})|x\Lambda(e_{k})){\operatorname{Tr}}(P^{t}e_{il})U_{ij}U^{*}_{lk}$
$\displaystyle=\sum_{i,j,k,l}x_{jk}P_{il}(U^{t})_{ji}(\overline{U})_{lk}=\sum_{j,k}x_{jk}(U^{t}P\overline{U})_{jk}=\sum_{j,k}x_{jk}P_{jk}={\operatorname{Tr}}(P^{t}x)=\tilde{\psi}(x),$

and so $\alpha_{U}$ preserves $\tilde{\psi}$. ∎

Form now $L^{2}(\mathcal{B}(H),\tilde{\psi})$ with GNS map $\tilde{\Lambda}$,
so

$(\tilde{\Lambda}(\theta_{\xi,\eta})|\tilde{\Lambda}(\theta_{\xi^{\prime},\eta^{\prime}}))=\tilde{\psi}(\theta_{\eta,\xi}\theta_{\xi^{\prime},\eta^{\prime}})=(\eta|\eta^{\prime}){\operatorname{Tr}}(P^{t}\theta_{\xi^{\prime},\xi})=(\eta|\eta^{\prime})(x_{0}(\xi^{\prime})|x_{0}(\xi))=(\eta|\eta^{\prime})(\overline{x_{0}(\xi)}|\overline{x_{0}(\xi^{\prime})})_{\overline{H}}.$

Thus the map

$w:L^{2}(\mathcal{B}(H),\tilde{\psi})\rightarrow
H\otimes\overline{H};\quad\tilde{\Lambda}(\theta_{\xi,\eta})\mapsto\eta\otimes\overline{x_{0}\xi}$

extends by linearity to a unitary. For $x\in\mathcal{B}(H)$ we have that
$x\Lambda(\theta_{\xi,\eta})=\Lambda(\theta_{\xi,x\eta})$ and so
$wxw^{*}=x\otimes 1$. Hence $L^{2}(\mathcal{B}(H),\tilde{\psi})$ is unitarily
equivalent to $H\otimes\overline{H}$ with the usual action of
$\mathcal{B}(H)$.

Recall the map $\mathcal{B}(H)\rightarrow\mathcal{B}(\overline{H});x\mapsto
x^{\top}$ which is an anti-$*$-homomorphism. Similarly, so is the unitary
antipode $R:A\rightarrow A$. Following common usage in quantum group theory,
for $Y\in\mathcal{B}(H)\otimes A$ we write $Y^{\top\otimes R}$ for the image
of $U$ under the tensor product of $x\mapsto x^{\top}$ and $R$. One can check
that $U^{\top\otimes R}$ is also a unitary corepresentation, termed the
_contragradient corepresentation_ , compare [37, Section 3.3.3]. With respect
to the orthonormal basis $(\overline{\Lambda(e_{i})})$ of $\overline{H}$, we
have that $x\mapsto x^{\top}$ is the transpose map on matrices, and so
$U^{\top\otimes R}$ has matrix $(R(U_{ji}))_{i,j}$.

######  Lemma 9.16.

Let $\tilde{U}$ be the unitary implementation of $\alpha_{U}$ with respect to
$\tilde{\psi}$. Then $(w\otimes 1)\tilde{U}(w^{*}\otimes
1)=U_{13}U_{23}^{\top\otimes R}$.

###### Proof.

As in Section 9.1, let $A\subseteq\mathcal{B}(K)$, and then for $1\leq s,t\leq
n$ and $\xi\in K$,

$\displaystyle(w\otimes 1)\tilde{U}(w^{*}\otimes 1)$
$\displaystyle(\Lambda(e_{s})\otimes\overline{\Lambda(e_{t})}\otimes\xi)=(w\otimes
1)\tilde{U}(\tilde{\Lambda}(\theta_{x_{0}^{-1}\Lambda(e_{t}),\Lambda(e_{s})})\otimes\xi)$
$\displaystyle=(w\otimes
1)\alpha_{U}(\theta_{x_{0}^{-1}\Lambda(e_{t}),\Lambda(e_{s})})(\tilde{\Lambda}(1)\otimes\xi)$
$\displaystyle=(w\otimes
1)(U(\theta_{x_{0}^{-1}\Lambda(e_{t}),\Lambda(e_{s})}\otimes
1)U^{*})(\tilde{\Lambda}(1)\otimes\xi)$
$\displaystyle=\sum_{i,j,k,l}w\tilde{\Lambda}(e_{ij}\theta_{x_{0}^{-1}\Lambda(e_{t}),\Lambda(e_{s})}e_{kl})\otimes
U_{ij}U^{*}_{lk}\xi$
$\displaystyle=\sum_{i,k,l}(x_{0}^{-1}\Lambda(e_{t})|\Lambda(e_{k}))\Lambda(e_{i})\otimes\overline{x_{0}\Lambda(e_{l})}\otimes
U_{is}U^{*}_{lk}\xi$
$\displaystyle=\sum_{i,j,k,l}\Lambda(e_{i})\otimes\overline{\Lambda(e_{j})}\otimes(x_{0}^{-1})_{tk}(x_{0})_{lj}U_{is}U^{*}_{lk}\xi$
$\displaystyle=\sum_{i,j}\Lambda(e_{i})\otimes\overline{\Lambda(e_{j})}\otimes
U_{is}(x_{0}^{-1}U^{*}x_{0})_{tj}\xi.$

Compare this to

$\displaystyle U_{13}U_{23}^{\top\otimes
R}(\Lambda(e_{s})\otimes\overline{\Lambda(e_{t})}\otimes\xi)$
$\displaystyle=\sum_{i,j,k,l}e_{ik}\Lambda(e_{s})\otimes
e_{jl}\overline{\Lambda(e_{t})}\otimes U_{ik}R(U_{lj})\xi$
$\displaystyle=\sum_{i,j}\Lambda(e_{i})\otimes\overline{\Lambda(e_{j})}\otimes
U_{is}R(U_{tj})\xi.$

We hence need to show that $(x_{0}^{-1}U^{*}x_{0})_{tj}=R(U_{tj})$ for each
$t,j$.

To show this, we use some compact quantum group theory. Let $S$ be the
antipode on $\mathcal{O}_{A}$, which has decomposition $S=R\tau_{-i/2}$ where
$(\tau_{t})$ is the scaling group, and $\tau_{-i/2}$ is the analytic
extension. Recall the definition of the operator $\rho_{U}$ from [32, Section
1.4], and from before [32, Definition 1.4.5] recall that $\rho_{U}$ is the
unique (up to a positive scalar) operator with $((\rho_{U}^{t})^{1/2}\otimes
1)\overline{U}((\rho_{U}^{t})^{-1/2}\otimes 1)$ is unitary. From Lemma 9.4, we
know that $U=X\overline{U}X^{-1}$, and so $\rho_{U}^{1/2}$ is $|X|^{t}$, up to
a scalar. As $P=X^{*}X$, we see that $|X|^{t}=(P^{t})^{1/2}=x_{0}$. From after
[32, Remark 1.7.7] we have that
$(\operatorname{id}\otimes\tau_{i/2})(U)=(\rho_{U}^{-1/2}\otimes
1)U(\rho_{U}^{1/2}\otimes 1)$ which thus equals $x_{0}^{-1}Ux_{0}$. Hence

$R(U_{tj})=S\tau_{i/2}(U_{tj})=S((x_{0}^{-1}Ux_{0})_{tj})=(x_{0}^{-1}U^{*}x_{0})_{tj},$

as required. ∎

This previous result should be compared with [43, Corollary 2.6.3] (which
itself follows quickly from [42, Proposition 4.2]). We do not invoke this
directly, as we will need to use the exact form of the unitary $w$, and it
seems about as much work to check that $w$ really is the intertwining unitary
as it does to give the above lemma (which is also of independent interest).

Recall the functional $\psi^{\operatorname{op}}$ on $B^{\operatorname{op}}$
studied in Lemma 5.32, where we identify $L^{2}(B^{\operatorname{op}})$ with
$\overline{L^{2}(B)}$. Thus $L^{2}(B\otimes B^{\operatorname{op}})$ is
identified with $H\otimes\overline{H}$, with GNS map
$\Lambda\otimes\Lambda^{\operatorname{op}}$. As usual, we use Theorem 5.17 to
link $A_{G}$ to a projection $e_{G}=\Psi^{\prime}_{1/2,0}(A_{G})$, so
explicitly, we assume $A_{G}$ is self-adjoint satisfying axioms (1) and (2).

######  Lemma 9.17.

Let $A_{G}$ be a quantum adjacency matrix associated to projection $e_{G}\in
B\otimes B^{\operatorname{op}}$. Then:

  1. 1.

$w\tilde{\Lambda}(A_{G})=(\Lambda\otimes\Lambda^{\operatorname{op}})((\sigma_{-i}\otimes\operatorname{id})(e_{G}))$;

  2. 2.

$w\tilde{\Lambda}(aTb)=(a\otimes\sigma_{i/2}(b)^{\top})w\tilde{\Lambda}(T)$
for $T\in\mathcal{B}(H),a,b\in B$;

Let $S_{0}$ be the $B^{\prime}$-operator bimodule generated by $A_{G}$, and
set $S_{0}^{\prime}=\kappa(S_{0})$. Then:

  1. 3.

$S_{0}^{\prime}$ is the $B$-operator bimodule generated by $A_{G}$;

  2. 4.

the orthogonal projection $f$ of $L^{2}(\mathcal{B}(H),\tilde{\psi})$ onto
$\tilde{\Lambda}(S_{0}^{\prime})$ satisfies
$wfw^{*}=(J\otimes\overline{J})e_{G}(J\otimes\overline{J})$;

  3. 5.

$fw^{*}(\Lambda(1)\otimes\overline{\Lambda(1)})=\tilde{\Lambda}(A_{G})$.

###### Proof.

Let $T=\theta_{\Lambda(a),\Lambda(b)}$ for $a,b\in B$, so

$\displaystyle
w\tilde{\Lambda}(T)=\Lambda(b)\otimes\overline{x_{0}\Lambda(a)}=\Lambda(b)\otimes\overline{\nabla^{1/2}\Lambda(a)}=\Lambda(b)\otimes\overline{\Lambda(\sigma_{-i/2}(a))}=(\Lambda\otimes\Lambda^{\operatorname{op}})(b\otimes\sigma_{-i/2}(a)^{*}).$

As $\Psi^{\prime}_{1/2,0}(T)=\sigma_{i/2}(b)\otimes a^{*}$, we see that

$\displaystyle w\tilde{\Lambda}(A_{G})$
$\displaystyle=(\Lambda\otimes\Lambda^{\operatorname{op}})\big{(}(\sigma_{-i/2}\otimes\sigma_{i/2})\Psi^{\prime}_{1/2,0}(A_{G})\big{)}$
$\displaystyle=(\Lambda\otimes\Lambda^{\operatorname{op}})\big{(}(\sigma_{-i/2}\otimes\sigma_{i/2})(e_{G})\big{)}=(\Lambda\otimes\Lambda^{\operatorname{op}})((\sigma_{-i}\otimes\operatorname{id})(e_{G})),$

as $e_{G}=(\sigma_{-i/2}\otimes\sigma_{-i/2})(e_{G})$.

With $T=\theta_{\xi,\eta}$ and $a,b\in B$ we see that

$w\tilde{\Lambda}(aTb)=w\tilde{\Lambda}(\theta_{b^{*}\xi,a\eta})=a\eta\otimes\overline{\nabla^{1/2}b^{*}\xi}=a\eta\otimes\overline{(\nabla^{-1/2}b\nabla^{1/2})^{*}\nabla^{1/2}\xi}=(a\otimes\sigma_{i/2}(b)^{\top})w\tilde{\Lambda}(T),$

as claimed.

As $\kappa$ restricts to a bijection $B\rightarrow B^{\prime}$, and
$\kappa(A_{G})=JA_{G}^{*}J=A_{G}$ because $A_{G}$ is self-adjoint, and using
Proposition 5.23, it follows that $S_{0}^{\prime}$ is the $B$-operator
bimodule generated by $A_{G}$. It now follows that
$w\tilde{\Lambda}(S_{0}^{\prime})=(B\otimes
B^{\operatorname{op}})(\Lambda\otimes\Lambda^{\operatorname{op}})((\sigma_{-i}\otimes\operatorname{id})(e_{G}))$.

Set $f_{0}=(\sigma_{-i}\otimes\operatorname{id})(e_{G})\in B\otimes
B^{\operatorname{op}}$, so that

$\displaystyle(J\otimes\overline{J})(\sigma_{-i/2}\otimes\sigma_{i/2})(f_{0}^{*})(J\otimes\overline{J})$
$\displaystyle=(J\otimes\overline{J})(\sigma_{-i/2}\otimes\sigma_{i/2})(\sigma_{i}\otimes\operatorname{id})(e_{G})(J\otimes\overline{J})$
$\displaystyle=(J\otimes\overline{J})(\sigma_{i/2}\otimes\sigma_{i/2})(e_{G})(J\otimes\overline{J})$
$\displaystyle=(J\otimes\overline{J})e_{G}(J\otimes\overline{J})\in
B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}.$

However, for $u\in B\otimes B^{\operatorname{op}}$, with reference to Lemma
5.32, we see that

$(J\otimes\overline{J})(\sigma_{-i/2}\otimes\sigma_{i/2})(f_{0}^{*})(J\otimes\overline{J})(\Lambda\otimes\Lambda^{\operatorname{op}})(u)=(\Lambda\otimes\Lambda^{\operatorname{op}})(uf_{0})=u(\Lambda\otimes\Lambda^{\operatorname{op}})(f_{0}).$

Thus letting $u$ vary shows that the image of the projection
$(J\otimes\overline{J})e_{G}(J\otimes\overline{J})$ is equal to $(B\otimes
B^{\operatorname{op}})(\Lambda\otimes\Lambda^{\operatorname{op}})(f_{0})=w\tilde{\Lambda}(S_{0}^{\prime})$.
Thus necessarily $wfw^{*}=(J\otimes\overline{J})e_{G}(J\otimes\overline{J})$.
Hence

$\displaystyle wfw^{*}(\Lambda(1)\otimes\overline{\Lambda(1)})$
$\displaystyle=(J\otimes\overline{J})e_{G}(J\otimes\overline{J})(\Lambda(1)\otimes\overline{\Lambda(1)})$
$\displaystyle=(J\otimes\overline{J})(\Lambda\otimes\Lambda^{\operatorname{op}})(e_{G})=(\Lambda\otimes\Lambda^{\operatorname{op}})((\sigma_{-i/2}\otimes\sigma_{i/2})(e_{G}))$
$\displaystyle=(\Lambda\otimes\Lambda^{\operatorname{op}})((\sigma_{-i}\otimes\operatorname{id})(e_{G}))=w\tilde{\Lambda}(A_{G}).$

Thus $fw^{*}(\Lambda(1)\otimes\overline{\Lambda(1)})=\tilde{\Lambda}(A_{G})$.
∎

Again, in the following we assume that $A_{G}$ is self-adjoint satisfying
axioms (1) and (2).

######  Theorem 9.18.

Let $\alpha$ be a coaction of a compact quantum group $(A,\Delta)$ on $B$,
which preserves $\psi$, and has unitary implementation $U$. Let $A_{G}$ be a
quantum adjacency matrix, and let $S_{0}$ be the $B^{\prime}$-operator
bimodule generated by $A_{G}$. The following are equivalent:

  1. 1.

$\alpha$ acts on $A_{G}$, that is, $UA_{G}=A_{G}U$;

  2. 2.

$\alpha_{U}$ leaves $S_{0}^{\prime}=\kappa(S_{0})$ invariant, that is,
$\alpha_{U}(S_{0}^{\prime})\subseteq S_{0}^{\prime}\otimes A$;

  3. 3.

$\alpha_{U}^{\operatorname{op}}(S_{0})\subseteq S_{0}\otimes A$.

###### Proof.

We have already seen that (2) and (3) are equivalent.

Suppose that $\alpha_{U}(S_{0}^{\prime})\subseteq S_{0}^{\prime}\otimes A$,
and let $A\subseteq\mathcal{B}(K)$. By definition of the unitary
implementation $\tilde{U}$, given $s_{0}\in S_{0}^{\prime}$ and $\xi\in K$, we
see that

$\tilde{U}(\tilde{\Lambda}(s_{0}^{\prime})\otimes\xi)=\alpha_{U}(s_{0}^{\prime})(\tilde{\Lambda}(1)\otimes\xi)\in(S_{0}^{\prime}\otimes
A)(\tilde{\Lambda}(1)\otimes\xi)=\tilde{\Lambda}(S_{0}^{\prime})\otimes A\xi.$

Let $f$ be the orthogonal projection of $L^{2}(\mathcal{B}(H),\tilde{\psi})$
onto $\tilde{\Lambda}(S_{0}^{\prime})$, so this calculation shows that
$\tilde{U}(f\otimes 1)=(f\otimes 1)\tilde{U}(f\otimes 1)$. A non-trivial
result about compact quantum groups (see the proof of [29, Proposition 6.2]
for example) shows that then $\tilde{U}(f\otimes 1)=(f\otimes 1)\tilde{U}$.

From the previous lemma, it hence follows that for $\xi\in K$,

$\displaystyle\alpha_{U}(A_{G})(\tilde{\Lambda}(1)\otimes\xi)$
$\displaystyle=\tilde{U}(\tilde{\Lambda}(A_{G})\otimes\xi)=\tilde{U}(fw^{*}(\Lambda(1)\otimes\overline{\Lambda(1)})\otimes\xi)$
$\displaystyle=(f\otimes
1)\tilde{U}(w^{*}(\Lambda(1)\otimes\overline{\Lambda(1)})\otimes\xi)$
$\displaystyle=(fw^{*}\otimes 1)U_{13}U^{\top\otimes
R}_{23}(\Lambda(1)\otimes\overline{\Lambda(1)}\otimes\xi),$ (6)

in the final step using Lemma 9.16.

For $\xi\in K$, we have that
$U(\Lambda(1)\otimes\xi)=\alpha(1)(\Lambda(1)\otimes\xi)=\Lambda(1)\otimes\xi$,
from the definition of $U$. Then for $a\in B,\xi_{1}\in K$, also

$(U^{*}(\Lambda(1)\otimes\xi)|\Lambda(a)\otimes\xi_{1})=(\Lambda(1)\otimes\xi|\alpha(a)(\Lambda(1)\otimes\xi_{1}))=(\xi|(\psi\otimes\operatorname{id})\alpha(a)\xi_{1})=\psi(a)(\xi|\xi_{1}),$

as $\alpha$ is $\psi$-preserving. Thus
$U^{*}(\Lambda(1)\otimes\xi)=\Lambda(1)\otimes\xi$. Let
$U=\sum_{i}x_{i}\otimes a_{i}\in\mathcal{B}(H)\otimes A$, so that
$\sum_{i}\Lambda(x_{i}^{*})\otimes
a_{i}^{*}(\xi)=U^{*}(\Lambda(1)\otimes\xi)=\Lambda(1)\otimes\xi$. It is always
possible to choose $A\subseteq\mathcal{B}(K)$ in such a way that there is an
involution $j:K\rightarrow K$ with $R(a)=ja^{*}j$ for $a\in A$. (For example,
let $A\subseteq\mathcal{B}(K)$ be the universal representation.) Then

$U^{\top\otimes
R}(\overline{\Lambda(1)}\otimes\xi)=\sum_{i}x_{i}^{\top}\overline{\Lambda(1)}\otimes
R(a_{i})(\xi)=\sum_{i}\overline{\Lambda(x_{i}^{*})}\otimes
ja_{i}^{*}j(\xi)=\overline{\Lambda(1)}\otimes\xi.$

It hence follows that, continuing the calculation from (6),

$\displaystyle\alpha_{U}(A_{G})(\tilde{\Lambda}(1)\otimes\xi)$
$\displaystyle=(fw^{*}\otimes
1)(\Lambda(1)\otimes\overline{\Lambda(1)}\otimes\xi)=\tilde{\Lambda}(A_{G})\otimes\xi=(A_{G}\otimes
1)(\tilde{\Lambda}(1)\otimes\xi).$

As $\alpha_{U}(A_{G})\in\mathcal{B}(H)\otimes A$, and
$\tilde{\Lambda}:\mathcal{B}(H)\rightarrow L^{2}(\mathcal{B}(H))$ is
injective, and is the map $x\mapsto x\tilde{\Lambda}(1)$, the previous
calculation shows that $\alpha_{U}(A_{G})=A_{G}\otimes 1$. This is equivalent
to $U(A_{G}\otimes 1)U^{*}=A_{G}\otimes 1$, that is, that $U$ and $A_{G}$
commute. We have hence shown that (2) implies (1).

Conversely, if $A_{G}$ and $U$ commute, then $\alpha_{U}(A_{G})=A_{G}\otimes
1$. As $\alpha_{U}$ extends $\alpha$ in the sense that
$\alpha_{U}(b)=\alpha(b)$ for $b\in B\subseteq\mathcal{B}(H)$, we see that

$\alpha_{U}(aA_{G}b)=\alpha(a)\alpha_{U}(A_{G})\alpha(b)\in(B\otimes
A)(A_{G}\otimes 1)(B\otimes A)=S_{0}^{\prime}\otimes A\qquad(a,b\in B),$

and so $\alpha_{U}(S_{0}^{\prime})\subseteq S_{0}^{\prime}\otimes A$, showing
that (1) implies (2). ∎

Given the discussion before, we might have hoped instead to find that
$UA_{G}=A_{G}U$ was equivalent to $\alpha(\kappa(S))\subseteq\kappa(S)\otimes
A$, but it just seems that this is not so. However, working with $S_{0}$ and
$\kappa(S_{0})$ instead of $S$ and $\kappa(S)$ is not so unnatural in light of
Remark 5.35, because $S_{0}$ is exactly the “tracial” quantum graph which
arises from $S$ using Theorem 5.29. The dependence on the state $\psi$ arises
from the fact that we always assume that $(A,\Delta)$ coacts on $B$ in a way
which preserves $\psi$.

In the tracial case we have a more direct result.

######  Corollary 9.19.

Let $\psi$ be a trace, and let $A_{G}$ correspond to the $B^{\prime}$-operator
bimodule $S$. Then $\alpha$ acts on $A_{G}$ if and only if
$\alpha_{U}^{\operatorname{op}}$ leaves $S$ invariant.

We leave for future work an investigation of whether there is a quantum group
analogue of the ideas discussed at the end of Section 8. That is, if we have a
coaction of $(A,\Delta)$ on $\mathcal{B}(H)$, and $B\subseteq\mathcal{B}(H)$
with a quantum graph $S\subseteq\mathcal{B}(H)$, is it possible to place
conditions on the coaction so as we obtain a coaction on $S$ (whatever this
might mean, at this level of generality). For more on coactions on
$\mathcal{B}(H)$ see, for example, [18, Section 3].

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_Author’s Address:_

Matthew Daws, Jeremiah Horrocks Institute, University of Central Lancashire,
Preston, PR1 2HE, United Kingdom

_Email:_<EMAIL_ADDRESS>