# On quantum Cayley graphs

Mateusz Wasilewski  Institute of Mathematics of the Polish Academy of
Sciences, ul. Śniadeckich 8, 00–656 Warszawa, Poland<EMAIL_ADDRESS>
###### Abstract.

We clarify the correspondence between the two approaches to quantum graphs:
via quantum adjacency matrices and via quantum relations. We show how the
choice of a (possibly non-tracial) weight manifests itself on the quantum
relation side and suggest an extension of the theory of quantum graphs to the
infinite dimensional case. Then we use this framework to introduce quantum
graphs associated to discrete quantum groups, leading to a new definition of a
quantum Cayley graph.

##  1\. Introduction

Inspired by the Lovász bound for the zero-error capacity of a channel, the
authors of [DSW13] introduced the notion of a quantum graph on a matrix
algebra $M_{n}$, as an object associated to a quantum channel $\Phi:M_{n}\to
M_{n}$ – its quantum confusability graph. Independently, the author of [Wea12]
developed a theory of quantum relations, inspired by his previous joint work
with Kuperberg [KW12] on quantum metrics. Specialised to symmetric, reflexive
relations on $M_{n}$, both approaches give the same answer – operator
subsystems $V\subset M_{n}$.

Another approach has been suggested in [MRV18], where the authors introduced
the notion of a quantum adjacency matrix. They were inspired by [MR16], where
quantum graph homomorphisms have been introduced, viewed as perfect quantum
strategies of winning the graph homomorphism game. The authors of [MRV18]
developed a categorical framework in which one can talk about general (finite)
quantum sets and quantum functions. A quantum graph is then a quantum set with
the extra structure of a quantum adjacency matrix, and a quantum graph
homomorphism is a quantum function that preserves this extra structure. This
approach to quantum graphs is equivalent to the one using operator systems.
This definition has been extended in [BCE+20] to include quantum graphs
equipped with non-tracial states and new links have been found between quantum
isomorphisms between quantum graphs and monoidal equivalence of the respective
quantum automorphism groups. In [Daw22] a correspondence between the operator
systems and quantum adjacency matrices has been established in the non-tracial
case.

In this work we clarify the relation between various definitions in the non-
tracial case for general quantum graphs, without any symmetry assumptions. The
main point here is a consistent use of the KMS inner product, as opposed to
the more common GNS inner product. Using the Hilbert space techniques from
[Daw22], but with the KMS inner product replacing the GNS one, we re-establish
the one-to-one correspondence between quantum adjacency matrices on a finite
dimensional $C^{\ast}$-algebra $\operatorname{B}$ and projections in
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$. Because we do
not assume GNS symmetry, no invariance under the modular group is required.

######  Theorem A.

There is a one-to-one correspondence between quantum adjacency matrices
$A:\operatorname{B}\to\operatorname{B}$ and projections in
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$. Moreover, the
tensor flip of the projection corresponds to taking the KMS adjoint of $A$.

Another thing that has been missing is that even to define the notion of a
quantum adjacency matrix we need to have a state (or a more general positive
functional, depending on the normalisation convention) on our algebra, but
Weaver’s theory of quantum relations works for arbitrary von Neumann algebras
without any reference state: a quantum relation on
$\mathsf{M}\subset\operatorname{B}(\mathsf{H})$ is a weak∗ closed
$\mathsf{M}^{\prime}$-bimodule and there is a way of making this notion
representation independent. It turns out that, using an old result of Haagerup
from [Haa79], a weight $\psi$ on $\mathsf{M}$ gives an operator valued weight
$\psi^{-1}$ from $\operatorname{B}(\mathsf{H})$ to $\mathsf{M}^{\prime}$. In
our case, i.e. if $\mathsf{M}$ is a finite dimensional $C^{\ast}$-algebra with
a positive functional $\psi$ such that $mm^{\ast}=\operatorname{Id}$ then
$\psi^{-1}$ is a faithful normal conditional expectation, which endows
$\operatorname{B}(\mathsf{H})$ with a Hilbert $\mathsf{M}^{\prime}$-module
structure. Working with an analogue of the KMS inner product in this case we
obtain a correspondence between projections in
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ and weak∗ closed
$\operatorname{B}^{\prime}$-bimodules that interacts nicely with symmetry
conditions.

######  Theorem B.

For a finite dimensional $C^{\ast}$-algebra
$\operatorname{B}\subset\operatorname{B}(\mathsf{H})$ equipped with a positive
functional $\psi$ such that $mm^{\ast}=\operatorname{Id}$ there is a one-to-
one correspondence between projections
$P\in\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ and weak∗
closed $\operatorname{B}^{\prime}$-bimodules
$V\subset\operatorname{B}(\mathsf{H})$. Moreover, the bimodule $V^{\ast}$
corresponds to the tensor flip of $P$.

We also extended the notion of a quantum graph to an infinite direct sum of
matrix algebras. In this case the $\operatorname{B}^{\prime}$-valued inner
product on a weak∗ closed bimodule is not necessarily complete, so in general
we have to work with pre-Hilbert modules, but the genuine Hilbert modules form
a natural class from the standpoint of quantum graph theory.

######  Theorem C.

Let $\operatorname{B}:=\bigoplus_{\alpha=1}^{\infty}M_{n_{\alpha}}$ and let
$A:c_{00}-\bigoplus_{\alpha=1}^{\infty}M_{n_{\alpha}}\to\operatorname{B}$ be a
quantum adjacency matrix. Then $A$ is of bounded degree, i.e. extends to a
normal completely positive map
$\widetilde{A}:\operatorname{B}\to\operatorname{B}$ if and only if for any
faithful representation $\operatorname{B}\subset\operatorname{B}(\mathsf{H})$
the corresponding $\operatorname{B}^{\prime}$-bimodule $V$ is a (complete,
self-dual) Hilbert $\operatorname{B}^{\prime}$-module.

Next we investigate the quantum adjacency matrices on the algebras associated
to discrete quantum groups, as they are infinite direct sums of matrix
algebras possessing a natural weight, namely either of the Haar weights, and
they happen to satisfy the condition $mm^{\ast}=\operatorname{Id}$, hence fit
perfectly into the theory. To respect the group structure, we study quantum
adjacency matrices that are covariant with respect to the right action, i.e.
left convolutions.

######  Theorem D.

Let $\mathbb{\Gamma}$ be a discrete quantum group and let
$A:\ell^{\infty}(\mathbb{\Gamma})\to\ell^{\infty}(\mathbb{\Gamma})$ be given
by $Ax:=P\ast x$ for some $P\in c_{00}(\mathbb{\Gamma})$, where
$\ell^{\infty}(\mathbb{\Gamma})\simeq\ell^{\infty}-\bigoplus_{\alpha\in\operatorname{Irr}(\mathbb{G})}M_{n_{\alpha}}$
is the algebra of bounded functions on $\mathbb{\Gamma}$ and
$c_{00}(\mathbb{\Gamma}):=c_{00}-\bigoplus_{\alpha\in\operatorname{Irr}(\mathbb{G})}M_{n_{\alpha}}$
is the algebra of finitely supported functions on $\mathbb{\Gamma}$. Then $A$
is a quantum adjacency matrix if and only if $P$ is a projection. Moreover,
$A$ is GNS symmetric if $P$ is invariant under the antipode of
$\mathbb{\Gamma}$ and it is KMS symmetric if $P$ is invariant under the
_unitary_ antipode.

A projection $P\in c_{00}(\mathbb{\Gamma})$ can be viewed as a finite subset
of $\mathbb{\Gamma}$ and to define a quantum Cayley graph we need a symmetric
generating subset. Symmetry should just be the KMS symmetry of the
corresponding quantum adjacency matrix and the property of being generating is
also not difficult to generalise to the non-commutative setting. Classically
the Cayley graph is a geometric object used to study the group and its
geometry depends very little on the particular generating set. Using the
notion of a quantum metric from [KW12] we arrive at the following result.

######  Theorem E.

Let $\mathbb{\Gamma}$ be a discrete quantum group and let $P_{1},P_{2}\in
c_{00}(\mathbb{\Gamma})$ be two generating projections. Then the corresponding
quantum Cayley graphs are bi-Lipschitz equivalent.

We finish the introduction by quickly describing the contents of the paper. In
section 2 we gather some information about the KMS inner product, Hilbert
modules and quantum groups. In section 3 we discuss the correspondence between
the three approaches to quantum graphs, and also extend it to the infinite
dimensional case. Next, in section 4, we finally turn to quantum groups. We
describe the covariant quantum adjacency matrices and characterise the GNS/KMS
symmetric ones. In the last section 5 the quantum Cayley graphs are
introduced. We prove that changing the generating projection results in a bi-
Lipschitz equivalent quantum graph. We also give an easy application: if the
balls in the Cayley graph grow subexponentially then the discrete quantum
group is amenable. We finish this section with some examples.

##  2\. Preliminaries

We will work with pairs $(\operatorname{B},\psi)$, where $\operatorname{B}$ is
a finite dimensional $C^{\ast}$-algebra equipped with non-tracial positive
functional $\psi$. If $\operatorname{B}$ is just a matrix algebra $M_{n}$ then
it means that $\psi$ is of the form $\psi(x):=\operatorname{Tr}(\rho x)$,
where $\rho$ is a positive semidefinite matrix. In this case you can define an
inner product on $M_{n}$ by $\langle x,y\rangle:=\psi(x^{\ast}y)$, which we
will refer to as the GNS inner product, and if $\rho$ is invertible, then it
provides a Hilbert space structure on $\operatorname{B}$, which we will assume
from now on. There is also another way to define an inner product, for which
we need the notion of the modular group of a functional. For each
$t\in\mathbb{R}$ we define $\sigma_{t}(x):=\rho^{it}x\rho^{-it}$, i.e. a
conjugation by the unitary matrix $\rho$. We have
$\sigma_{t}\circ\sigma_{s}=\sigma_{t+s}$ and the whole collection
$(\sigma_{t})_{t\in\mathbb{R}}$ is called the modular group of $\psi$. In the
finite dimensional setting we can also define the modular group at complex
argument, namely for any $z\in\mathbb{C}$ we put
$\sigma_{z}(x):=\rho^{iz}x\rho^{-iz}$, in particular $\sigma_{-i}(x)=\rho
x\rho^{-1}$; an important and easy to check property of the modular group is
that $\psi\circ\sigma_{z}=\psi$. Even though the functional $\psi$ is not
tracial in general, i.e. $\psi(xy)\neq\psi(yx)$ for some
$x,y\in\operatorname{B}$, the modular group allows us to recover this property
to an extent. To be more precise, it enjoys the _KMS property_ , i.e.

$\psi(xy)=\psi(y\sigma_{-i}(x)),$

which follows from

$\psi(xy)=\operatorname{Tr}(\rho xy)=\operatorname{Tr}(\rho x\rho^{-1}(\rho
y))=\operatorname{Tr}(\rho y\rho x\rho^{-1})=\operatorname{Tr}(\rho
y\sigma_{-i}(x))=\psi(y\sigma_{-i}(x)).$

General finite dimensional $C^{\ast}$-algebras are direct sums of matrix
algebras, so it is easy to extend all these notions.

The Hilbert space structure will be of paramount importance to us, and it is
therefore crucial to specify it, especially since we will not use the typical
GNS inner product induced by a positive functional. Namely, we will work with
the KMS inner product, i.e. $\langle
x,y\rangle=\psi(x^{\ast}\sigma_{-\frac{i}{2}}(y))$. One can relate it to the
more common GNS inner product, yielding a more symmetric formula

(2.1)
$\psi(x^{\ast}\sigma_{-\frac{i}{2}}(y))=\psi\left((\sigma_{-\frac{i}{4}}(x))^{\ast}\sigma_{-\frac{i}{4}}(y)\right).$

The reason for using the KMS inner product is that it interacts nicely with
positivity, namely if $\psi(x)=\operatorname{Tr}(\rho x)$, then $\langle
x,y\rangle=\operatorname{Tr}((\rho^{\frac{1}{4}}x\rho^{\frac{1}{4}})^{\ast}\rho^{\frac{1}{4}}y\rho^{\frac{1}{4}})$,
i.e. we use the positivity preserving embedding
$x\mapsto\rho^{\frac{1}{4}}x\rho^{\frac{1}{4}}$. The KMS inner product also
interacts nicely with the adjoint, namely $\langle x,y\rangle=\langle
y^{\ast},x^{\ast}\rangle$, exactly like in the tracial case. Moreover, we will
want to talk about undirected graphs, which amounts to a symmetry condition on
the adjacency matrix, and GNS symmetry is too restrictive of a condition, as
the following well-known statement shows.

######  Lemma 2.1.

Suppose that $A:\operatorname{B}\to\operatorname{B}$, where $\operatorname{B}$
is a finite dimensional $C^{\ast}$-algebra, is GNS symmetric with respect to a
faithful positive functional $\psi$, i.e. $\psi((Ax)y)=\psi(x(Ay))$. Then
$A\circ\sigma_{t}=\sigma_{t}\circ A$.

###### Proof.

We have

$\displaystyle\varphi(A(\sigma_{i}(x))y)$
$\displaystyle=\varphi(\sigma_{i}(x)(Ay))=\varphi((Ay)x)$
$\displaystyle=\varphi(y(Ax))=\varphi(\sigma_{i}(Ax)y),$

hence $A\circ\sigma_{i}=\sigma_{i}\circ A$ and this is enough to conclude that
$A$ is covariant with respect to the modular group. ∎

It is in many cases much easier to perform computations with respect to the
GNS inner product, and in some instances, we can resort to that, e.g. if we
want to compute the adjoint of a map that intertwines the respective modular
groups.

######  Lemma 2.2.

Suppose that $A:(\operatorname{B},\psi)\to(\operatorname{C},\varphi)$
satisfies $A\circ\sigma_{t}^{\psi}=\sigma_{t}^{\varphi}\circ A$. Then the
adjoint of $A$ with respect to the KMS inner products is the same as the
adjoint with respect to the GNS inner products.

###### Proof.

Follows easily from formula (2.1). ∎

This lemma applies for example to the multiplication map
$m:\operatorname{B}\otimes\operatorname{B}\to\operatorname{B}$, because the
equality $m\circ(\sigma_{t}\otimes\sigma_{t})=\sigma_{t}\circ m$ just means
that $\sigma_{t}$ is a homomorphism; we will now write down the formula for
the adjoint.

######  Proposition 2.3.

Suppose that $M_{n}$ is equipped with a faithful positive functional
$\psi(x):=\alpha\operatorname{Tr}(\rho x)$, where $\alpha>0$ and $\rho\in
M_{n}$ is a positive definite matrix. Then we have

$m^{\ast}(e_{ij}\rho^{-1})=\frac{1}{\alpha}\sum_{k}e_{ik}\rho^{-1}\otimes
e_{kj}\rho^{-1}$

and $mm^{\ast}=\frac{\operatorname{Tr}(\rho^{-1})}{\alpha}$. Moreover
$m^{\ast}$ is a bimodular map.

###### Proof.

Let us compute $\langle e_{pq}\otimes
e_{rs},m^{\ast}(e_{ij}\rho^{-1})\rangle$, using the formula for $m^{\ast}$
from the statement of the proposition:

$\displaystyle\alpha\sum_{k}\operatorname{Tr}(\rho
e_{qp}e_{ik}\rho^{-1})\operatorname{Tr}(\rho e_{sr}e_{kj}\rho^{-1})$
$\displaystyle=\alpha\sum_{k}\delta_{pi}\delta_{qk}\delta_{rk}\delta_{sj}$
$\displaystyle=\alpha\delta_{pi}\delta_{qr}\delta_{sj}$

On the other hand, it should be equal to $\langle
e_{pq}e_{rs},e_{ij}\rho^{-1}\rangle$, i.e. $\delta_{qr}\langle
e_{ps},e_{ij}\rho^{-1}\rangle=\alpha\delta_{qr}\delta_{pi}\delta_{sj}$, which
is the same.

To check that $mm^{\ast}=\frac{\operatorname{Tr}(\rho^{-1})}{\alpha}$, simply
note that
$mm^{\ast}(e_{ij}\rho^{-1})=\frac{1}{\alpha}\sum_{k}e_{ik}\rho^{-1}e_{kj}\rho^{-1}$
and $\sum_{k}e_{ik}\rho^{-1}e_{kj}=\operatorname{Tr}(\rho^{-1})e_{ij}$. We
will now address the bimodularity of $m^{\ast}$. The case of the left action
is straigthforward and for the right action we will use the KMS property.
Indeed

$\displaystyle\langle a\otimes b,m^{\ast}(xy)\rangle$ $\displaystyle=\langle
ab,xy\rangle=\psi(b^{\ast}a^{\ast}xy)$
$\displaystyle=\psi(\sigma_{i}(y)b^{\ast}a^{\ast}x)=\psi((ab\sigma_{-i}(y^{\ast}))^{\ast}x)$
$\displaystyle=\langle ab\sigma_{-i}(y^{\ast}),x\rangle=\langle a\otimes
b\sigma_{-i}(y^{\ast}),m^{\ast}(x)\rangle=\langle a\otimes
b,m^{\ast}(x)y\rangle,$

where for the last equality we need to write $m^{\ast}(x)$ as a finite sum
$\sum_{i}s_{i}\otimes t_{i}$ and perform the same argument as in the second
line, but in reverse. ∎

As a consequence, the positive functionals satisfying
$mm^{\ast}=\operatorname{Id}$ are of the form
$\operatorname{Tr}(\rho^{-1})\operatorname{Tr}(\rho x)$; the extension to
direct sums of matrix algebras is immediate. For further use, we record here
the following simple lemma.

######  Lemma 2.4.

Let $\psi$ be a weight on $M_{n}$ such that $mm^{\ast}=\operatorname{Id}$ and
let $P\in M_{n}$ be a non-zero projection. Then $\psi(P)\geqslant 1$.

###### Proof.

We have $\psi(x)=\operatorname{Tr}(\rho^{-1})\operatorname{Tr}(\rho x)$. By
the Cauchy-Schwarz inequality we have

$1\leqslant(\operatorname{Tr}(P))^{2}=(\operatorname{Tr}(\rho^{-\frac{1}{2}}\rho^{\frac{1}{2}}P))^{2}\leqslant\operatorname{Tr}(\rho^{-1})\operatorname{Tr}(\rho
P^{2})=\psi(P).$

∎

We will occasionally have to work with the opposite algebra
$\operatorname{B}^{\operatorname{op}}$, which is the same vector space
equipped with the opposite multiplication, i.e.
$x^{\operatorname{op}}y^{\operatorname{op}}=(yx)^{\operatorname{op}}$, where
by $x^{op}$ we mean an element $x\in\operatorname{B}$ viewed as an element of
$\operatorname{B}^{\operatorname{op}}$. We will denote the identity map
$\operatorname{B}\ni x\mapsto
x^{\operatorname{op}}\in\operatorname{B}^{\operatorname{op}}$ by $\iota$. For
a positive functional $\psi$ on $\operatorname{B}$ we can define
$\psi^{\operatorname{op}}:\operatorname{B}^{\operatorname{op}}\to\mathbb{C}$
by $\psi^{\operatorname{op}}(x^{\operatorname{op}}):=\psi(x)$. We can relate
the modular group of this functional to the original one, namely
$\sigma_{t}^{\operatorname{op}}(x^{\operatorname{op}})=(\sigma_{-t}(x))^{\operatorname{op}}$.
Using the KMS property of $\psi$ and $\psi^{\operatorname{op}}$ we obtain

$\displaystyle=\psi(x\sigma_{-i}(y))$
$\displaystyle=\psi(yx)=\psi^{\operatorname{op}}((yx)^{\operatorname{op}})=\psi^{\operatorname{op}}(x^{\operatorname{op}}y^{\operatorname{op}})$
$\displaystyle=\psi^{\operatorname{op}}(\sigma_{i}^{\operatorname{op}}(y^{\operatorname{op}})(x^{\operatorname{op}})),$

from which it follows that
$\sigma_{i}^{\operatorname{op}}(y^{\operatorname{op}})=(\sigma_{-i}(y))^{\operatorname{op}}$
and that is enough to conclude. Most often we will work with the algebra
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$, whose natural
action on $\operatorname{B}\otimes\operatorname{B}$ will be denoted by $\\#$,
i.e. $(a\otimes b^{\operatorname{op}})\\#(c\otimes d):=ac\otimes db$.

In subsection 3.2 we will use the language of Hilbert $C^{\ast}$-modules. We
refer the reader to the book [BLM04] for all the necessary information. We
just recall here quickly the main definition.

######  Definition 2.5.

Let $A$ be a $C^{\ast}$-algebra and let $X$ be a right $A$-module. We call $X$
a pre-Hilbert module if it is equipped with a sesquilinear map (linear in the
second variable) $\langle\cdot,\cdot\rangle:X\times X\to A$ such that:

  1. (i)

$\langle x,ya\rangle=\langle x,y\rangle a$ for all $x,y\in X$ and $a\in A$;

  2. (ii)

$\langle x,y\rangle^{\ast}=\langle y,x\rangle$;

  3. (iii)

$\langle x,x\rangle\geqslant 0$ and $\langle x,x\rangle=0$ if and only if
$x=0$.

We call $X$ a Hilbert module if it is complete with respect to the norm
$\|x\|:=\sqrt{\langle x,x\rangle}$. A Hilbert module $X$ is self-dual if any
bounded right $A$-linear map $\phi:X\to A$ is of the form $\phi(x)=\langle
y,x\rangle$ for some $y\in X$.

In sections 4 and 5 we will work with compact and discrete quantum groups. For
information about compact quantum groups we refer to the excellent book
[NT13]. For information about discrete quantum groups we refer to [VD96],
although sometimes it is more convenient to treat them as general locally
compact quantum groups and use the results from [KV00]. Here we recall just
the basic definitions.

######  Definition 2.6.

A compact quantum group $\mathbf{G}$ is described by a pair $(A,\Delta)$,
where $A$ is a unital $C^{\ast}$-algebra and $\Delta:A\to
A\otimes_{\operatorname{min}}A$ is a $\ast$-homomorphism satisfying:

  1. (i)

$(\operatorname{Id}\otimes\Delta)\circ\Delta=(\Delta\otimes\operatorname{Id})\circ\Delta$
(coassociativity);

  2. (i)

the spaces
$(A\otimes\mathds{1})\Delta(A):=\operatorname{span}\\{(a\otimes\mathds{1})\Delta(b),a,b\in
A\\}$ and $(\mathds{1}\otimes A)\Delta(A)$ are dense in
$A\otimes_{\operatorname{min}}A$.

Typically the algebra $A$ will be denoted by $C(\mathbb{G})$ as it is meant to
generalize the algebra of continuous functions on a compact group.

One can prove that the Haar measure exists for compact quantum groups, and
representation theory can be developed, in particular an analogue of the
Peter-Weyl holds; $\operatorname{Irr}(\mathbb{G})$ will denote the set of
equivalence classes of irreducible representations of $\mathbb{G}$.

######  Definition 2.7.

Let
$\ell^{\infty}(\mathbb{\Gamma}):=\ell^{\infty}-\bigoplus_{\alpha\in\operatorname{Irr}(\mathbb{G})}M_{n_{\alpha}}$.
There is a unique $\ast$-homomorphism
$\Delta_{\mathbb{\Gamma}}:\ell^{\infty}(\mathbb{\Gamma})\to\ell^{\infty}(\mathbb{\Gamma})\overline{\otimes}\ell^{\infty}(\mathbb{\Gamma})$
such that
$(\Delta_{\mathbb{\Gamma}}\otimes\operatorname{Id})(W)=W_{23}W_{13}$, where
$W\in\ell^{\infty}(\mathbb{\Gamma})\overline{\otimes}L^{\infty}(\mathbb{G})$
is the unitary describing the right regular representation of $\mathbb{G}$.
One can also show existence of left and right Haar measures, thus
$(\ell^{\infty}(\mathbb{\Gamma}),\Delta_{\mathbb{\Gamma}})$ becomes a locally
compact quantum group, which is the discrete dual of $\mathbb{G}$.

##  3\. Quantum graphs

Throughout this section $\operatorname{B}$ will be a finite dimensional
$C^{\ast}$-algebra equipped with a faithful positive functional $\psi$ such
that $mm^{\ast}=\operatorname{Id}$. Thus a quantum graph structure will really
be an extra structure on the pair $(\operatorname{B},\psi)$, i.e. a _quantum
space_ , not only on the algebra $\operatorname{B}$. However, in most
situations $\psi$ will be clear from the context and will be suppressed in the
notation.

###  3.1. Quantum adjacency matrices versus projections

We will discuss a correspondence between quantum adjacency matrices
$A:\operatorname{B}\to\operatorname{B}$ and projections in
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$. This
correspondence has already been established (see [BHI+23, Proposition 2.3] and
[Daw22, Theorem 5.17]), so we will be brief but we would like to stress how
the consistent use of the KMS inner product clarifies the situation.

######  Definition 3.1.

A quantum adjacency matrix on a quantum space $(\operatorname{B},\psi)$ is a
completely positive map $A:\operatorname{B}\to\operatorname{B}$ such that
$m(A\otimes A)m^{\ast}=A$.

######  Remark 3.2.

Some authors prefer to assume that $A$ is _real_ (see [Mat22]), i.e.
$\ast$-preserving, but it is equivalent to complete positivity under the
condition $m(A\otimes A)m^{\ast}=A$. Indeed, by Proposition 3.7 the Choi
matrix of such an $A$ is an idempotent and self-adjointness of an idempotent
is equivalent its positivity.

If one mimics the tracial situation, the natural object to consider seems to
be
$\widetilde{P}:=(A\otimes\operatorname{Id})m^{\ast}(\mathds{1})\in\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$,
which is an idempotent. In the terminology of [Daw22], $\widetilde{P}$
corresponds to $A$ via the map $\Psi^{\prime}_{0,0}$, which maps the rank one
operator $|a\rangle\langle b|$ to $a\otimes
b^{\ast}\in\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$.
However, the inner product in that work is the GNS inner product and we will
see now how the situation changes if we choose to work with the KMS inner
product instead.

######  Lemma 3.3.

Suppose that $\operatorname{B}\simeq\bigoplus_{\alpha}M_{n_{\alpha}}$ is
equipped with a weight
$\psi:=\bigoplus_{\alpha}\operatorname{Tr}(\rho_{\alpha}^{-1})\operatorname{Tr}(\rho_{\alpha}\cdot)$.
Let
$\Psi^{KMS}:\operatorname{End}(\operatorname{B})\to\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$
be defined via $|a\rangle\langle b|\mapsto
a\otimes(b^{\ast})^{\operatorname{op}}$. Then for each
$A\in\operatorname{End}(\operatorname{B})$ we have

(3.1)
$\Psi^{KMS}(A)=\sum_{\alpha}\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\sum_{i,j}A\left(\rho_{\alpha}^{-\frac{1}{4}}e_{ij}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}\right)\otimes\left(\rho_{\alpha}^{-\frac{1}{4}}e_{ji}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}\right)^{\operatorname{op}}.$

###### Proof.

It suffices to check the formula (3.1) for the rank one operators
$|a\rangle\langle b|$. In this case we have ($b_{\alpha}$ will denote the
component of $b$ living in the summand $M_{n_{\alpha}}$)

$\Psi^{KMS}(|a\rangle\langle
b|)=\sum_{\alpha}\sum_{i,j}a\operatorname{Tr}\left(b^{\ast}_{\alpha}\rho_{\alpha}^{\frac{1}{4}}e_{ij}^{\alpha}\rho_{\alpha}^{\frac{1}{4}}\right)\otimes\left(\rho_{\alpha}^{-\frac{1}{4}}e_{ji}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}\right)^{\operatorname{op}}.$

The trace
$\operatorname{Tr}\left(b^{\ast}_{\alpha}\rho_{\alpha}^{\frac{1}{4}}e_{ij}^{\alpha}\rho_{\alpha}^{\frac{1}{4}}\right)$
just computes the $(j,i)$ entry of
$\rho_{\alpha}^{\frac{1}{4}}b^{\ast}_{\alpha}\rho_{\alpha}^{\frac{1}{4}}$,
hence we obtain

$\Psi^{KMS}(|a\rangle\langle
b|)=\sum_{\alpha}\sum_{i,j}a\left(\rho_{\alpha}^{\frac{1}{4}}b^{\ast}_{\alpha}\rho_{\alpha}^{\frac{1}{4}}\right)_{ji}\otimes\left(\rho_{\alpha}^{-\frac{1}{4}}e_{ji}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}\right)^{\operatorname{op}}=a\otimes\sum_{\alpha}\sum_{i,j}\left(\rho_{\alpha}^{-\frac{1}{4}}\left(\rho_{\alpha}^{\frac{1}{4}}b^{\ast}_{\alpha}\rho_{\alpha}^{\frac{1}{4}}\right)_{ji}e_{ji}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}\right)^{\operatorname{op}}.$

For any matrix $X$ the sum $\sum_{i,j}X_{ij}e_{ij}$ is clearly equal to $X$,
so the second leg reduces to $\sum_{\alpha}b^{\ast}_{\alpha}=b^{\ast}$. ∎

######  Remark 3.4.

The elements
$\frac{1}{\sqrt{\operatorname{Tr}(\rho_{\alpha}^{-1})}}\rho_{\alpha}^{-\frac{1}{4}}e_{ij}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}$
form an orthonormal basis of $\operatorname{B}$ with respect the KMS inner
product, so $\Psi^{KMS}(A)$ can be thought of as the matrix of $A$ with
respect to this orthonormal basis.

We will now relate $P:=\Psi^{KMS}(A)$ with
$\widetilde{P}=(A\otimes\iota)m^{\ast}(\mathds{1})$. Before that we need a
lemma (which we only write down in the case of a single matrix block, but the
generalization to finite dimensional $C^{\ast}$-algebras is obvious).

######  Lemma 3.5.

Suppose $\operatorname{B}=M_{n}$ equipped with a positive functional
$\psi(x):=\operatorname{Tr}(\rho^{-1})\operatorname{Tr}(\rho x)$, where
$\rho\in M_{n}$ is positive definite. Let
$\varepsilon:=m^{\ast}(\mathds{1})=\frac{1}{\operatorname{Tr}(\rho^{-1})}\sum_{i,j}e_{ij}\rho^{-1}\otimes
e_{ji}$. Then for each $z,w\in\mathbb{C}$ we have
$\varepsilon=\frac{1}{\operatorname{Tr}(\rho^{-1})}\sum_{i,j}\rho^{z}e_{ij}\rho^{w-1}\otimes\rho^{-w}e_{ij}\rho^{-z}$.

###### Proof.

Treat $\varepsilon$ as an element of
$\operatorname{B}\otimes\operatorname{B}$. Note first that
$(\sigma_{z})^{\ast}=\sigma_{-\overline{z}}$ because

$\langle
x,\sigma_{z}y\rangle=\varphi(x^{\ast}\sigma_{z}(y))=\varphi(\sigma_{-z}(x^{\ast})y)=\varphi((\sigma_{-\overline{z}}(x))^{\ast}y=\langle\sigma_{-\overline{z}}(x),y\rangle.$

By taking the adjoint of the equation
$m\circ(\sigma_{z}\otimes\sigma_{z})=\sigma_{z}\circ m$ (for all
$z\in\mathbb{C}$), we get that $m^{\ast}$ commutes with the modular group,
i.e. $(\sigma_{z}\otimes\sigma_{z})\circ m^{\ast}=m^{\ast}\circ\sigma_{z}$ for
all $z\in\mathbb{C}$. From this we get
$\varepsilon=(\sigma_{-ix}\otimes\sigma_{-ix})\varepsilon$ for all
$x\in\mathbb{C}$. Moreover, $m^{\ast}$ is a bimodular map, therefore
$m^{\ast}(\mathds{1})=m^{\ast}(\rho^{y}\mathds{1}\rho^{-y})=\rho^{y}\varepsilon\rho^{-y}$
for all $y\in\mathbb{C}$. Combining the two, we obtain

$\varepsilon=\frac{1}{\operatorname{Tr}(\rho^{-1})}\sum_{i,j}\rho^{x+y}e_{ij}\rho^{-1-x}\otimes\rho^{x}e_{ji}\rho^{-y-x}.$

Now put $z=x+y$ and $w=-x$. ∎

######  Lemma 3.6.

Let $A:\operatorname{B}\to\operatorname{B}$. We have
$\Psi^{KMS}(A)=(A\otimes\iota\circ\sigma_{-\frac{i}{2}})m^{\ast}(\mathds{1})$,
where $\iota:\operatorname{B}\to\operatorname{B}^{\operatorname{op}}$ is the
identity map $\iota(x):=x^{\operatorname{op}}$.

Since
$\iota\circ\sigma_{-\frac{i}{2}}=\sigma^{\operatorname{op}}_{\frac{i}{2}}\circ\iota$,
we get
$P=(\operatorname{Id}\otimes\sigma^{\operatorname{op}}_{\frac{i}{2}})(\widetilde{P})$.

###### Proof.

From the previous lemma we get that

$m^{\ast}(\mathds{1})=\sum_{\alpha}\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\sum_{i,j}\rho_{\alpha}^{-\frac{1}{4}}e_{ij}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}\otimes\rho_{\alpha}^{-\frac{3}{4}}e_{ji}^{\alpha}\rho_{\alpha}^{\frac{1}{4}}.$

Note that

$\rho_{\alpha}^{-\frac{3}{4}}e_{ji}^{\alpha}\rho_{\alpha}^{\frac{1}{4}}=\rho_{\alpha}^{-\frac{1}{2}}(\rho_{\alpha}^{-\frac{1}{4}}e_{ji}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}})\rho_{\alpha}^{\frac{1}{2}}=\sigma_{\frac{i}{2}}(\rho_{\alpha}^{-\frac{1}{4}}e_{ji}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}),$

which ends the proof. ∎

We are now ready to summarize the properties of $\Psi^{KMS}$.

######  Proposition 3.7.

The map
$\Psi^{KMS}:\operatorname{End}(\operatorname{B})\to\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$
establishes a one-to-one correspondence between linear maps on
$\operatorname{B}$ and elements of
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$. We have:

  1. (i)

$\Psi^{KMS}(A)$ is an idempotent if and only if $A$ satisfies $m(A\otimes
A)m^{\ast}=A$;

  2. (ii)

$\Psi^{KMS}(A)$ is self-adjoint if and only if $A$ is $\ast$-preserving;

  3. (iii)

$\Psi^{KMS}(A)$ is positive if and only if $A$ is completely positive;

  4. (iv)

$\Psi^{KMS}(A)$ is invariant under the flip map
$\Sigma:\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}\to\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$
if and only if the KMS adjoint $A^{\ast}_{KMS}$ is equal to the map
$\overline{A}(x):=(A(x^{\ast}))^{\ast}$.

In particular, if $A:\operatorname{B}\to\operatorname{B}$ is a quantum
adjacency matrix then it is self-adjoint with respect to the KMS inner product
if and only if $\Sigma(\Psi^{KMS}(A))=\Psi^{KMS}(A)$.

###### Proof.

The first bullet point follows from [BHI+23, Proposition 2.3(2)], which states
that $(A\otimes\iota)m^{\ast}(\mathds{1})$ is an idempotent iff $m(A\otimes
A)m^{\ast}=A$ and Lemma 3.6, from which we infer that $\Psi^{KMS}(A)$ and
$(A\otimes\iota)m^{\ast}(\mathds{1})$ are related by a multiplicative map (the
modular group). The only new statement is the last one. Note that
$\Psi^{KMS}(|a\rangle\langle b|)=a\otimes(b^{\ast})^{\operatorname{op}}$,
hence the flip is equal to $b^{\ast}\otimes a^{\operatorname{op}}$, which is
equal to $\Psi^{KMS}(|b^{\ast}\rangle\langle a^{\ast}|)$. Let us compute
$\overline{|a\rangle\langle b|}(c)=(a\langle
b,c^{\ast}\rangle)^{\ast}=a^{\ast}\langle c^{\ast},b\rangle=a^{\ast}\langle
b^{\ast},c\rangle$, using the properties of the KMS inner product. Using the
fact that $\left(|a\rangle\langle b|\right)^{\ast}=|b\rangle\langle a|$, we
conclude that $\Sigma(\Psi^{KMS}(A))=\Psi^{KMS}(\overline{A}^{\ast}_{KMS})$,
where $\overline{A}^{\ast}_{KMS}$ should be viewed as the transpose of $A$. ∎

We will end this subsection with some comments about recovering $A$ from
$P:=\Psi^{KMS}(A)$. By [BHI+23, Proposition 2.3 (1)] we have
$Ax=(\operatorname{Id}\otimes\psi)(P_{1}(1\otimes x))$, where
$P_{1}:=(A\otimes\operatorname{Id})m^{\ast}(\mathds{1})$ is treated is an
element of $\operatorname{B}\otimes\operatorname{B}$ such that
$\widetilde{P}=(\operatorname{Id}\otimes\iota)P_{1}$, thus we can rewrite this
formula as $Ax=(\operatorname{Id}\otimes\varphi)(1\otimes
x)\\#\widetilde{P})$. Recall from Lemma 3.6 that
$\widetilde{P}=(\operatorname{Id}\otimes\sigma^{\operatorname{op}}_{-\frac{i}{2}})(P)$.
This is to be expected, because if we have a positive element
$x\in\operatorname{C}$, where $\operatorname{C}$ is a von Neumann algebra
equipped with a normal faithful positive functional $\varphi$, and we want to
use it as a density of a positive functional, then we have to choose
$\varphi(\cdot\sigma^{\varphi}_{-\frac{i}{2}}(x))$, not $\varphi(\cdot x)$.
There is an analogous formula involving the left multiplication.

###  3.2. Projections versus bimodules

Given a representation of $\operatorname{B}$ on a Hilbert space $\mathsf{H}$
(possibly infinite dimensional), we will now establish a one-to-one
correspondence between weak∗ closed $\operatorname{B}^{\prime}$-bimodules
inside $\operatorname{B}(\mathsf{H})$ and projections in
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$. Before that we
need a quick discussion of operator valued weights. We will not go into the
details, let us just say that operator valued weights are to conditional
expectations what weights are to states, i.e. they satisfy a positivity and
bimodularity conditions but are in general unbounded. In this work we will
only see weights and bona fide conditional expectations, that is why we
decided to avoid the unnecessary technical details.

In [Haa79, Theorem 6.13] Haagerup established for a pair of von Neumann
algebras $\mathsf{N}\subset\mathsf{M}\subset\operatorname{B}(\mathsf{H})$ a
bijection between normal, faithful, semifinite (abbreviated henceforth nfs)
operator valued weights from $\mathsf{M}$ to $\mathsf{N}$ and nfs operator
valued weights from $\mathsf{N}^{\prime}$ to $\mathsf{M}^{\prime}$. In
particular, any weight $\psi$ on a von Neumann algebra $\mathsf{M}$ gives rise
to an operator valued weight $\psi^{-1}$ from $\operatorname{B}(\mathsf{H})$
to $\mathsf{M}^{\prime}$. In [Con80, Corollary 16] Connes obtained a formula
for this operator valued weight as
$\psi^{-1}(|\xi\rangle\langle\xi|)=R_{\psi}(\xi)R_{\psi}(\xi)^{\ast}$, where
$R_{\psi}(\xi):\mathsf{H}_{\psi}\to\mathsf{H}$ is the map from the GNS space
of $\psi$ given by $\Lambda(a)\mapsto a\xi$, assuming that it is bounded, that
is to say that $\xi$ is a $\psi$-bounded vector (such vectors are always dense
in $\mathsf{H}$).

We will now check that if $\operatorname{B}$ is a finite dimensional
$C^{\ast}$-algebra equipped with a positive functiona $\psi$ such that
$mm^{\ast}=\operatorname{Id}$ then $\psi^{-1}$ is a faithful conditional
expectation. First, note that it is sufficient to check for a single matrix
algebra. Indeed, if $z\in\operatorname{B}$ is a central projection (namely
$z\in\operatorname{B}\cap\operatorname{B}^{\prime}$) then (using bimodularity
of $\psi^{-1}$ over $\operatorname{B}^{\prime}$)
$\psi^{-1}(zx)=z\psi^{-1}(x)=z\psi^{-1}(x)z=\psi^{-1}(zxz)$. It follows that
if $z_{\alpha}$’s are minimal central projections in $\operatorname{B}$ and
$\psi_{\alpha}$ is the restriction of the functional $\psi$ to
$z_{\alpha}\operatorname{B}\subset\operatorname{B}(z_{\alpha}\mathsf{H})$ then
$\psi^{-1}$ can be built as a direct sum of the operator valued weights
$\psi_{\alpha}^{-1}$.

We assume now that $\operatorname{B}=M_{n}\subset\operatorname{B}(\mathsf{H})$
($\mathsf{H}$ can be infinite dimensional) equipped with a functional
$\psi(x):=\operatorname{Tr}(\rho^{-1})\operatorname{Tr}(\rho x)$, where
$\rho\in M_{n}$ is positive definite. It is well-known that
$\mathsf{H}\simeq\mathbb{C}^{n}\otimes\mathsf{K}$, where $M_{n}$ acts as
$A(v\otimes\xi):=Av\otimes\xi$. Any tensor $v\otimes\xi$ is $\psi$-bounded and
one can check that
$R_{\psi}(v\otimes\xi)^{\ast}:\mathbb{C}^{n}\otimes\mathsf{K}\to H_{\psi}$ is
given by
$R_{\psi}(v\otimes\xi)^{\ast}(w\otimes\eta)=\frac{1}{\operatorname{Tr}(\rho^{-1})}|w\rangle\langle\rho^{-1}v|\langle\xi,\eta\rangle$.
It follows that
$R_{\psi}(v\otimes\xi)R_{\psi}(v\otimes\xi)^{\ast}(w\otimes\eta)=\frac{1}{\operatorname{Tr}(\rho^{-1})}\langle\rho^{-1}v,v\rangle\langle\xi,\eta\rangle
w\otimes\xi$. In other words, if we view
$\operatorname{B}(\mathbb{C}^{n}\otimes\mathsf{K})$ as
$M_{n}\overline{\otimes}\operatorname{B}(\mathsf{K})$ and
$|v\otimes\xi\rangle\langle v\otimes\xi|$ as $|v\rangle\langle
v|\otimes|\xi\rangle\langle\xi|$ then $\psi^{-1}(A\otimes
T)=\frac{1}{\operatorname{Tr}(\rho^{-1})}\operatorname{Tr}(\rho^{-1}A)\otimes
T$, which amounts to slicing the first leg with the _state_
$\frac{1}{\operatorname{Tr}(\rho^{-1})}\operatorname{Tr}(\rho^{-1}\cdot)$ (see
also [Kos98, Remark 3.3 (iii)]). This is clearly a conditional expectation.
Let us summarize this discussion.

######  Proposition 3.8.

Let $\operatorname{B}\subset\operatorname{B}(\mathsf{H})$ be a finite
dimensional $C^{\ast}$-algebra equipped with a positive functional $\psi$ such
that $mm^{\ast}=\operatorname{Id}$. Then
$\psi^{-1}:\operatorname{B}(\mathsf{H})\to\operatorname{B}^{\prime}$ is a
normal faithful conditional expectation.

######  Remark 3.9.

Note that the statement also holds for an infinite direct sum of matrix
algebras, which will be useful in the sequel (the proof remains the same). We
cannot, however, hope for a much greater generality, as existence of a normal
conditional expectation from $\operatorname{B}(\mathsf{H})$ onto
$\operatorname{B}^{\prime}$ implies that $\operatorname{B}^{\prime}$ is atomic
[Tak03, Exercise IX.4.1(e)].

Using this conditional expectation we can equip $\operatorname{B}(\mathsf{H})$
with a structure of a pre-Hilbert $C^{\ast}$-module over
$\operatorname{B}^{\prime}$, taking the inner product as $\langle
S,T\rangle:=\psi^{-1}(S^{\ast}T)$. Following [Wea12] we will define quantum
graphs on $\operatorname{B}$ as quantum relations (dropping for now the
symmetry and reflexivity conditions).

######  Definition 3.10.

Let $\operatorname{B}\subset\operatorname{B}(\mathsf{H})$ be a finite
dimensional $C^{\ast}$-algebra equipped with a positive functional $\psi$ such
that $mm^{\ast}=\operatorname{Id}$. A quantum graph on
$(\operatorname{B},\psi)$ is a weak∗ closed
$\operatorname{B}^{\prime}$-bimodule $V\subset\operatorname{B}(\mathsf{H})$.
Using $\psi^{-1}$ to define a $\operatorname{B}^{\prime}$-valued inner
product, we can make $V$ into a pre-Hilbert
$\operatorname{B}^{\prime}$-module.

######  Proposition 3.11.

Let $V\subset\operatorname{B}(\mathsf{H})$ be a quantum graph on
$(\operatorname{B},\psi)$. Then $V$ is a self-dual Hilbert module, i.e. it is
complete for the Hilbert module norm, and any right
$\operatorname{B}^{\prime}$-linear map $T:V\to\operatorname{B}^{\prime}$ is of
the form $Tv=\langle w,v\rangle$ for some $w\in V$.

###### Proof.

We will start by checking that $\operatorname{B}(\mathsf{H})$ is a Hilbert
module. Indeed, the inclusion
$\operatorname{B}^{\prime}\subset\operatorname{B}(\mathsf{H})$ is of finite
index in the sense of [Kos98, Definition 3.4], because the index is equal to
$(\psi^{-1})^{-1}(\mathds{1})=\psi(\mathds{1})<\infty$, but also in the sense
of [BDH88, Proposition 3.3], as from the explicit formula for $\psi^{-1}$ and
the fact that we only have finitely many summands it immediately follows that
for some $K>0$ we have $x\leqslant K\psi^{-1}(x)$ for all positive
$x\in\operatorname{B}(\mathsf{H})$. As a consequence, the operator norm is
equivalent to the pre-Hilbert norm induced by $\psi^{-1}$ (because the other
inequality $\|\psi^{-1}(x)\|\leqslant\|x\|$ is clear), hence the latter is
complete. Clearly each closed submodule, such as $V$, is then also a Hilbert
$\operatorname{B}^{\prime}$-module.

To prove self-duality, note that $V$ is a dual Banach space, being a weak∗
closed subspace of $\operatorname{B}(\mathsf{H})$; moreover, the
$\operatorname{B}^{\prime}$-valued inner product is separately weak∗
continuous. It follows from [BLM04, Lemma 8.5.4] that $V$ is self-dual. It
implies that it admits an orthonormal basis, all bounded right
$\operatorname{B}^{\prime}$-modular maps on $V$ are automatically adjointable
and the algebra of adjointable operators is a von Neumann algebra [Pas73,
Theorem 3.12, Proposition 3.4, and Proposition 3.10]). ∎

######  Remark 3.12.

In all other examples appearing in this paper the existence of a predual
making the inner product separately weak∗ continuous will always be easy, so
checking self-duality will amount to proving completeness of the pre-Hilbert
norm.

We have all the necessary ingredients to associate a projection in
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ to a
$\operatorname{B}^{\prime}$-bimodule $V$; indeed, we can take an orthonormal
basis of $V$ and use it to construct a $\operatorname{B}^{\prime}$-bimodular
orthogonal projection onto $V$, which in turn we can represent using
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$. There is,
however, a slight problem, since our $\operatorname{B}^{\prime}$-valued inner
product is rather an analogue of the GNS inner product, not the KMS inner
product, so it would not interact nicely with an additional symmetry
condition; we need therefore a modular group of the operator valued weight
$\psi^{-1}$. Haagerup proved that for any normal, faithful, semifinite weight
$\psi^{\prime}$ on $\operatorname{B}^{\prime}$ the restriction of the modular
group of $\phi:=\psi^{\prime}\circ\psi^{-1}$ to $\operatorname{B}$ does not
depend on $\psi^{\prime}$ and moreover $\sigma^{\phi}_{t}=\sigma^{\psi}_{-t}$
(see [Haa79, Proposition 6.1 and Theorem 6.13]). Once we get to the explicit
correspondence between various definitions of quantum graphs, we will write
everything in terms of the modular group of $\psi$. However, for now we will
denote the modular group of the conditional expectation $\psi^{-1}$ by
$\sigma_{t}^{\psi^{-1}}$.

To an extent, one can define this modular group just by using the Hilbert
module structure of $\operatorname{B}(\mathsf{H})$; the right multiplication
operator by an element $b\in\operatorname{B}$ is adjointable, and as it is a
left $\operatorname{B}(\mathsf{H})$-modular map, it is also a right
multiplication by some element $\widetilde{b}$, which has to be equal to
$\sigma_{-i}(b^{\ast})$ by the KMS property.

This uniqueness also implies that we have a well defined action on bimodules.
Indeed, if we have two weights $\phi_{1}:=\psi^{\prime}_{1}\circ\psi^{-1}$ and
$\phi_{2}:=\psi^{\prime}_{2}\circ\psi^{-1}$ on $\operatorname{B}(\mathsf{H})$
then their respective modular groups are given using density matrices:
$\sigma^{\phi_{1}}_{t}(x)=\rho_{1}^{it}x\rho_{1}^{-it}$ and
$\sigma^{\phi_{2}}_{t}(x)=\rho_{2}^{it}x\rho_{2}^{-it}$. Since they agree on
$B$, we have $\rho_{1}^{it}\rho_{2}^{-it}\in\operatorname{B}^{\prime}$, hence
$\rho_{1}^{it}=u_{t}\rho_{2}^{it}$ for some unitary in
$\operatorname{B}^{\prime}$. As $V$ is a $\operatorname{B}^{\prime}$-bimodule,
we have $u_{t}Vu_{t}^{\ast}=V$, therefore
$\sigma_{t}^{\phi_{1}}(V)=\sigma_{t}^{\phi_{2}}(V)$.

Suppose now that we have a finite orthonormal set $X:=(x_{1},\dots,x_{d})$ in
a von Neumann algebra $\mathsf{M}$ (equipped with a state $\varphi$) with
respect to the KMS inner product, which consists of analytic elements. We
would like to find a relation between orthogonal projections with respect to
KMS and GNS inner products. The orthogonal projection onto $V$, the span of
$X$, is given by $P^{KMS}_{V}(x):=\sum_{i=1}^{d}x_{i}\langle
x_{i},x\rangle_{KMS}=\sum_{i=1}^{d}x_{i}\varphi\left(\left(\sigma_{-\frac{i}{4}}(x_{i})\right)^{\ast}\sigma_{-\frac{i}{4}}(x)\right)$.
This is equal to
$\sum_{i=1}^{d}x_{i}\langle\sigma_{-\frac{i}{4}}(x_{i}),\sigma_{-\frac{i}{4}}(x)\rangle_{GNS}$.
Thus $\sigma_{-\frac{i}{4}}(X)$ is an orthonormal set with respect to the GNS
inner product and we checked that $P^{KMS}_{V}=\sigma_{\frac{i}{4}}\circ
P^{GNS}_{\sigma_{-\frac{i}{4}}(V)}\circ\sigma_{-\frac{i}{4}}$. Therefore, if
in our setting of bimodules we can define the bimodule
$\sigma_{-\frac{1}{4}}^{\psi^{-1}}(V)$ and represent the orthogonal projection
onto it as an element of
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ then we can also
apply the conjugation by $\sigma_{\frac{i}{4}}^{\psi^{-1}}$ on the level of
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$, where the
modular group is well defined, and in this way we will obtain a representation
of the orthogonal projection onto $V$ with respect to the KMS inner product.

To check that one can define $\sigma_{-\frac{i}{4}}^{\psi^{-1}}(V)$, i.e. we
can analytically extend the action of $\sigma_{t}^{\psi^{-1}}$, recall the
discussion preceding Proposition 3.8. We have
$\operatorname{B}=\bigoplus_{\alpha}M_{n_{\alpha}}$ and any representation is
essentially a direct sum of inflations of the standard representations of
matrix algebras, i.e.
$\mathsf{H}\simeq\bigoplus_{\alpha}\mathbb{C}^{n_{\alpha}}\otimes\mathsf{K}_{\alpha}$.
The commutant $\operatorname{B}^{\prime}$ is then equal to
$\bigoplus_{\alpha}\mathds{1}_{\alpha}\otimes\operatorname{B}(\mathsf{K}_{\alpha})$.
Since the action of the modular group $\sigma_{t}^{\psi^{-1}}(V)$ does not
depend on the choice of the weight on $\operatorname{B}^{\prime}$, we might as
well assume that the weight on $\operatorname{B}(\mathsf{H})$ is given by the
density
$\bigoplus_{\alpha}\rho_{\alpha}^{-1}\otimes\operatorname{Id}_{\mathsf{K}_{\alpha}}$,
i.e. we take the trace on each $\operatorname{B}(\mathsf{K}_{\alpha})$. Then
something nontrivial happens only on the finite dimensional subspace
$\operatorname{B}(\mathbb{C}^{n_{\alpha}},\mathbb{C}^{n_{\beta}})$, thus we
may define $\sigma_{z}(V)$ for arbitrary $z\in\mathbb{C}$.

######  Lemma 3.13.

For any weak∗ closed $\operatorname{B}^{\prime}$-bimodule
$V\subset\operatorname{B}(\mathsf{H})$ and $z\in\mathbb{C}$ we can define
another weak∗-closed $\operatorname{B}^{\prime}$-bimodule
$\sigma_{z}^{\psi^{-1}}(V)$.

By [BLM04, Corollary 8.5.8] any bounded module map between self-dual Hilbert
modules over von Neumann algebras is automatically weak∗ continuous, hence the
orthogonal projection
$P_{\sigma_{-\frac{i}{4}}^{\psi^{-1}}(V)}^{GNS}:\operatorname{B}(\mathsf{H})\to\sigma_{-\frac{i}{4}}^{\psi^{-1}}(V)\subset\operatorname{B}(\mathsf{H})$
is a normal $\operatorname{B}^{\prime}$-bimodular map, therefore of the form
$P_{\sigma_{-\frac{i}{4}}^{\psi^{-1}}(V)}^{GNS}x=\sum_{k}b_{k}xc_{k}$ for some
$b_{k},c_{k}\in\operatorname{B}$ (see [Smi91, Theorem 3.1] for a slightly more
general result; there the von Neumann algebraic result we need is attributed
to Haagerup). In this case the orthogonal projection onto $V$ with respect to
the KMS inner product would be given by
$P_{V}^{KMS}(x)=\sum_{k}\sigma_{\frac{i}{4}}^{\psi^{-1}}(b_{k})x\sigma_{\frac{i}{4}}^{\psi^{-1}}(c_{k})$.
Note that the map $x\mapsto\sum_{k}b_{k}xc_{k}$ is represented by
$\sum_{k}b_{k}\otimes\left(\sigma_{\frac{i}{2}}^{\psi^{-1}}(c_{k})\right)^{\operatorname{op}}$,
because the right action needs to be twisted by the modular group so that it
becomes a $\ast$-representation. To wit, an element $b\otimes
c^{\operatorname{op}}\in\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$
acts as $x\mapsto bx\sigma_{-\frac{i}{2}}^{\psi^{-1}}(c)$, when we use the GNS
inner product. However, if we use the KMS inner product, then $b\otimes
c^{\operatorname{op}}$ needs to acts as
$x\mapsto\sigma_{\frac{i}{4}}^{\psi^{-1}}(b)x\sigma_{-\frac{i}{4}}^{\psi^{-1}}(c)$.
Therefore
$\sum_{k}b_{k}\otimes\left(\sigma_{\frac{i}{2}}^{\psi^{-1}}(c_{k})\right)^{\operatorname{op}}$
acting on $\operatorname{B}(\mathsf{H})$ with respect the KMS inner product
acts as
$x\mapsto\sum_{k}\sigma_{\frac{i}{4}}^{\psi^{-1}}(b_{k})x\sigma_{\frac{i}{4}}^{\psi^{-1}}(c_{k})$,
i.e. it is equal to $P_{V}^{KMS}$. It means that $P_{V}^{KMS}$ and
$P_{\sigma_{-\frac{i}{4}}^{\psi^{-1}}(V)}^{GNS}$ are represented by the same
element of $\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$.

######  Proposition 3.14.

The assignment
$\Phi:\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}\to\phantom{B}_{\operatorname{B}^{\prime}}\operatorname{B}_{\operatorname{B}^{\prime}}(\mathsf{H})$,
where
$\phantom{B}{}_{\operatorname{B}^{\prime}}\operatorname{B}_{\operatorname{B}^{\prime}}(\mathsf{H})$
denotes the algebra of bounded $\operatorname{B}^{\prime}$-bimodular maps,
given by the extension of $b\otimes c^{\operatorname{op}}\mapsto(x\mapsto
bx\sigma_{-\frac{i}{2}}(c))$, is an isomorphism of von Neumann algebras.

###### Proof.

$\Phi$ is a $\ast$-homomorphism, which is also surjective by the
aforementioned representation of $\operatorname{B}^{\prime}$-bimodular maps.
Let us check that it is injective. Suppose that (remember that
$\operatorname{Id}\otimes\sigma_{\frac{i}{2}}^{\psi^{-1}}:\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}\to\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$
is a bijection)
$\Phi(\sum_{k}b_{k}\otimes\sigma_{\frac{i}{2}}(c_{k})^{\operatorname{op}})=0$,
i.e. $\sum_{k}b_{k}xc_{k}=0$ for each $x\in\operatorname{B}(\mathsf{H})$. This
holds in particular for any Hilbert-Schmidt operator, identified with a vector
$\xi\in\mathsf{H}\otimes\overline{\mathsf{H}}$. The natural representation of
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ on
$\mathsf{H}\otimes\overline{\mathsf{H}}$ is clearly faithful, so injectivity
follows. ∎

######  Theorem 3.15.

Let $\operatorname{B}\subset\operatorname{B}(\mathsf{H})$ be a finite
dimensional $C^{\ast}$-algebra equipped with a positive functional $\psi$ such
that $mm^{\ast}=\operatorname{Id}$ and let
$\psi^{-1}:\operatorname{B}(\mathsf{H})\to\operatorname{B}^{\prime}$ be the
corresponding normal, fathful conditional expectation. There is a one-to-one
correspondence between weak∗ closed $\operatorname{B}^{\prime}$-bimodules
$V\subset\operatorname{B}(\mathsf{H})$ and projections
$P\in\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ obtained in
the following way. For $V\subset\operatorname{B}(\mathsf{H})$ let
$P_{\sigma_{-\frac{i}{4}}^{\psi^{-1}}(V)}$ be the
$\operatorname{B}^{\prime}$-bimodular projection onto
$\sigma_{-\frac{i}{4}}^{\psi^{-1}}(V)$ and let
$P\in\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ be its
representing projection. On the other hand for a projection
$P\in\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ let
$\widetilde{P}$ be the corresponding $\operatorname{B}^{\prime}$-bimodular
projection; then
$V=\sigma_{\frac{i}{4}}^{\psi^{-1}}(\operatorname{Im}(\widetilde{P}))$.

###### Proof.

The correspondence between projections in
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ and
$\operatorname{B}^{\prime}$-bimodular projections on
$\operatorname{B}(\mathsf{H})$ was established in Propostion 3.14. The fact
that for any weak∗ closed bimodule we can define $\sigma_{z}(V)$ for any
$z\in\mathbb{C}$ was covered in Lemma 3.13. The last ingredient is the
correspondence between bimodules and bimodular projections, which follows from
self-duality (see Proposition 3.11). ∎

######  Remark 3.16.

In Subsection 3.4 we will present a much less abstract version of this
correspondence.

###  3.3. Infinite quantum graphs

In order to work with infinite discrete quantum groups, we need to deal with
quantum graphs defined on an infinite direct sum of matrix algebras. The
theory of quantum relations developed by Weaver works equally well for general
von Neumann algebras and if $\mathsf{M}\subset\operatorname{B}(\mathsf{H})$
then a quantum relation over $\mathsf{M}$ is a weak∗-closed
$\mathsf{M}^{\prime}$-bimodule inside $\operatorname{B}(\mathsf{H})$. We will
assume that our algebra is of the form
$\operatorname{B}=\ell^{\infty}-\bigoplus_{\alpha}\operatorname{M}_{n_{\alpha}}$
and it is equipped with a weight
$\psi(x):=\bigoplus_{\alpha}\operatorname{Tr}(\rho_{\alpha}^{-1})\operatorname{Tr}(\rho_{\alpha}x_{\alpha})$;
such a weight satisfies $mm^{\ast}=\operatorname{Id}$, e.g. if we view it as a
map on $L^{2}(\operatorname{B})$, but it should be noted that $m^{\ast}$ is
not a well-defined map from $\operatorname{B}$ to
$\operatorname{B}\overline{\otimes}\operatorname{B}$. In this case Weaver’s
quantum relations correspond to projections in
$\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$,
which in turn will be in bijection with suitably defined quantum adjacency
matrices. Later in this section we will describe a certain subclass of quantum
graphs that is more suitable for describing quantum Cayley graphs.

######  Definition 3.17.

Let $I$ be a countable index set and let
$\operatorname{B}:=\ell^{\infty}-\bigoplus_{\alpha\in I}M_{n_{\alpha}}$, where
$n_{\alpha}\in\mathbb{N}$. Let $c_{00}(\operatorname{B})$ be the algebraic
direct sum $c_{00}-\bigoplus_{\alpha}M_{n_{\alpha}}$, which can be more
abstractly defined as the subalgebra generated by $\psi$-finite projections,
i.e. projections $p$ such that $\psi(p)<\infty$. A completely positive map
$A:c_{00}(\operatorname{B})\to\operatorname{B}$ is called a quantum adjacency
matrix if for each $\alpha\in I$ it satisfies

$\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\sum_{l}A(e_{il}^{\alpha}\rho_{\alpha}^{-1})A(e_{lj}^{\alpha})=A(e_{ij}^{\alpha}).$

The corresponding projection in
$\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$ is
given by
$P:=\sum_{\alpha}\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\sum_{i,j}A(\rho_{\alpha}^{-\frac{1}{4}}e_{ij}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}})\otimes\rho_{\alpha}^{-\frac{1}{4}}e_{ji}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}$.

######  Remark 3.18.

Note that in the definition of $P$ only the sum over $\alpha$ is infinite but
the summands are projections lying in disjoint blocks, so it clearly converges
in the weak/strong topology.

######  Proposition 3.19.

Let
$P\in\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$.
Then we can define a quantum adjacency matrix
$A:c_{00}(\operatorname{B})\to\operatorname{B}$ via

$A(e_{ij}^{\alpha}):=(\operatorname{Id}\otimes\psi)(P\\#(1\otimes\rho_{\alpha}^{-\frac{1}{2}}e_{ij}^{\alpha}\rho_{\alpha}^{\frac{1}{2}})).$

###### Proof.

Note that
$P\\#(1\otimes\rho_{\alpha}^{-\frac{1}{2}}e_{ij}^{\alpha}\rho_{\alpha}^{\frac{1}{2}})\in\operatorname{B}\otimes
M_{n_{\alpha}}$, so slicing with $\psi$ is certainly possible. Because there
are no analytical difficulties, one can check that $A$ is a quantum adjacency
matrix exactly the same as in the finite dimensional case. ∎

######  Proposition 3.20.

There is a one-to-one correspondence between projections
$P\in\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$
and weak∗ closed $\operatorname{B}^{\prime}$-bimodules
$V\subset\operatorname{B}(\mathsf{H})$.

###### Proof.

Let $(\mathds{1}_{\alpha})_{\alpha}$ be the minimal central projections of
$\operatorname{B}$. Then
$V_{\beta\alpha}:=\mathbf{1}_{\beta}V\mathbf{1}_{\alpha}$ is a
$\operatorname{B}^{\prime}$-bimodule, but the only nontrivial action comes
from $M_{n_{\beta}}\otimes M_{n_{\alpha}}^{\operatorname{op}}$, hence finding
the projection onto $V_{\beta\alpha}$ is done exactly the same as in the
finite dimensional case. Moreover $V$ consists precisely of those bounded
operators $T\in\operatorname{B}(\mathsf{H})$ such that
$\mathbf{1}_{\beta}T\mathbf{1}_{\alpha}\in V_{\beta\alpha}$, which is a
consequence of the condition that $V$ is weak∗ closed. To summarize, any weak∗
closed $\operatorname{B}^{\prime}$-bimodule can be equivalently described by
the collection $(V_{\beta\alpha})_{\beta,\alpha}$. Each $V_{\beta\alpha}$
corresponds to a projection $P_{\beta\alpha}\in M_{n_{\beta}}\otimes
M_{n_{\alpha}}^{\operatorname{op}}$, and such a collection
$(P_{\beta\alpha})_{\beta,\alpha}$ is exactly a projection
$P\in\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$.
∎

The established correspondence is very satisfactory because it includes all
quantum relations on $\operatorname{B}$ but the quantum adjacency matrix has
poor analytic and algebraic properties, e.g. it cannot be iterated because the
codomain does not match the domain. There is a natural subclass of quantum
graphs that is better behaved and should include all quantum Cayley graphs.
Classically, one constructs Cayley graphs from finite generating sets, hence
the resulting graphs have bounded degree; this is the class that we would like
to generalize.

######  Lemma 3.21.

Let $A=(A_{ij})_{i,j\in\mathbb{N}}$ be a $\\{0,1\\}$ matrix. It defines a
normal completely positive map $A:\ell^{\infty}\to\ell^{\infty}$ if and only
if it is of bounded degree, i.e.
$\sup_{i}|\\{j\in\mathbb{N}:a_{ij}=1\\}|<\infty$.

###### Proof.

If $A$ defines a map from $\ell^{\infty}$ to itself then we look at
$A\mathds{1}$. It is clear that
$(A\mathds{1})_{i}=\sum_{j}a_{ij}=|\\{j:a_{ij}=1\\}|$, hence
$\sup_{i}|\\{j:a_{ij}=1\\}|<\infty$.

Suppose now that $A$ is of bounded degree. Then the transpose $A^{T}$
satisfies $\sup_{j}|\\{i\in\mathbb{N}:a_{ij}=1\\}|<\infty$. It means that
$\sup_{i}\|A^{T}e_{i}\|_{1}<\infty$. Since the unit vectors $e_{i}$ are
extreme points of the unit ball of $\ell^{1}$ it follows that $A^{T}$ extends
to a bounded linear map on $\ell^{1}$. By dualizing we get that $A$ defines a
weak∗ continuous (i.e. normal) map on $\ell^{\infty}$ and (complete)
positivity is clear. ∎

######  Definition 3.22.

A normal, completely positive map $A:\operatorname{B}\to\operatorname{B}$ is
called a quantum adjacency matrix of bounded degree if for each $\alpha\in I$
it satisfies

$\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\sum_{l}A(e_{il}^{\alpha}\rho_{\alpha}^{-1})A(e_{lj}^{\alpha})=A(e_{ij}^{\alpha}).$

######  Proposition 3.23.

Quantum adjacency matrices of bounded degree
$A:\operatorname{B}\to\operatorname{B}$ correspond precisely to projections
$P\in\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$
that are _integrable with respect to the second variable_ , i.e. satisfy
$(\operatorname{Id}\otimes\psi^{\operatorname{op}})P\in\operatorname{B}$.

###### Proof.

If we start from $A$, we can use the formula
$P:=\sum_{\alpha}\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\sum_{i,j}A(\rho_{\alpha}^{-\frac{1}{4}}e_{ij}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}})\otimes(\rho_{\alpha}^{-\frac{1}{4}}e_{ji}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}})^{\operatorname{op}}$
to conclude that
$(\operatorname{Id}\otimes\psi^{\operatorname{op}})P=\sum_{\alpha}A(\mathds{1}_{\alpha})=A\mathds{1}$,
because $A$ was assumed to be normal.

Assume now that
$D:=(\operatorname{Id}\otimes\psi^{\operatorname{op}})P\in\operatorname{B}$.
Recall that in the finite dimensional case $A$ can be recovered from $P$ via
the formula
$Ax:=(\operatorname{Id}\otimes\psi)(P\\#(1\otimes\sigma_{\frac{i}{2}}(x)))$
and we just have to check that this formula still makes sense. It suffices to
show that $(\varphi\otimes\psi)(P\\#(1\otimes\sigma_{\frac{i}{2}}(x)))<\infty$
for any normal state $\varphi$ and a positive operator $x\in\operatorname{B}$.
To this end, we will denote by $x_{m}$ the projection of $x$ onto the finite
direct sum $\bigoplus_{\alpha=1}^{m}M_{n_{\alpha}}$ and show that
$\sup_{m}(\varphi\otimes\psi)(P\\#(1\otimes\sigma_{\frac{i}{2}}(x_{m})))$ is
finite; $z_{m}$ will be the central projection corresponding to
$\bigoplus_{\alpha=1}^{m}M_{n_{\alpha}}$. Because the right leg of our tensor
product is now finite dimensional, we can write $P_{m}:=P(1\otimes
z_{m})\leqslant P$ as a finite sum $P_{m}:=\sum_{i}P_{i}\otimes
Q_{i}^{\operatorname{op}}$ and our formula becomes
$\sum_{i}\varphi(P_{i})\psi(\sigma_{\frac{i}{2}}(x_{m})Q_{i})$. But $P_{m}$ is
an orthogonal projection, so we actually get
$\sum_{i,j}\varphi(P_{i}^{\ast}P_{j})\psi(\sigma_{\frac{i}{2}}(x_{m})Q_{j}Q_{i}^{\ast})$.
By the KMS property of $\psi$ we get

$\psi(\sigma_{\frac{i}{2}}(x_{m})Q_{j}Q_{i}^{\ast})=\psi((\sigma_{-\frac{i}{2}}(Q_{i}))^{\ast}x_{m}\sigma_{-\frac{i}{2}}(Q_{j})).$

It means that we apply $\varphi\otimes\psi$ to the positive operator
$R^{\ast}(1\otimes x_{m})R$, where
$R=\sum_{i}P_{i}\otimes\sigma_{-\frac{i}{2}}(Q_{i})$; it is bounded above by
$\|x_{m}\|R^{\ast}R$. Note that

$(\operatorname{Id}\otimes\psi)(R^{\ast}R)=\sum_{i,j}P_{i}^{\ast}P_{j}\psi(\sigma_{\frac{i}{2}}(Q_{i}^{\ast})\sigma_{-\frac{i}{2}}(Q_{j}))=\sum_{i,j}P_{i}^{\ast}P_{j}\psi(Q_{j}Q_{i}^{\ast})=(\operatorname{Id}\otimes\psi^{\operatorname{op}})(P_{m})\leqslant
D,$

where once again we used the KMS property. It follows that
$\varphi(A(x_{m}))\leqslant\|x_{m}\|\varphi(D)$, thus we can define $A$ for
any positive $x$; $A$ is a well-defined normal, completely positive map on
$\operatorname{B}$. ∎

There is another interesting class of infinite graphs that we can easily
generalize to the quantum setting, namely the locally finite graphs.
Classically, these are graphs whose adjacency matrix $A$ has finitely
supported rows. The rows of the adjacency matrix are images of the standard
basis vectors under the transpose $A^{T}$, i.e. locally finite graphs are
exactly the ones, for which the transpose of the adjacency matrix preserves
the subspaces of finitely supported functions. Recall from Proposition 3.7
that the transpose in the context of quantum adjacency matrices is exactly the
KMS adjoint.

######  Definition 3.24.

We say that the quantum graph defined by a quantum adjacency matrix
$A:c_{00}(\operatorname{B})\to\operatorname{B}$ is locally finite if
$A^{\ast}_{KMS}(c_{00}(\operatorname{B}))\subset c_{00}(\operatorname{B})$ or,
equivalently, the corresponding projection satisfies $x\cdot P\in
c_{00}(\operatorname{B})\otimes c_{00}(\operatorname{B}^{\operatorname{op}})$
for any $x\in c_{00}(\operatorname{B})$.

We will now check that quantum graphs of bounded degree are locally finite.

######  Proposition 3.25.

If $A:\operatorname{B}\to\operatorname{B}$ is of bounded degree then it is
locally finite.

###### Proof.

Let
$P\in\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$
be the corresponding projection. Recall that
$P=(P_{\beta\alpha})_{\beta,\alpha}$, where $P_{\beta\alpha}$ is a projection
in $M_{n_{\beta}}\otimes M_{n_{\alpha}}^{\operatorname{op}}$. Define
$P_{\beta}:=\sum_{\alpha}P_{\beta\alpha}$. Because $A$ is of bounded degree,
we clearly have $(\psi\otimes\psi^{\operatorname{op}})(P_{\beta})<\infty$. It
is easy to check that the weight $\psi\otimes\psi^{\operatorname{op}}$ on
$\operatorname{B}\otimes\operatorname{B}^{\operatorname{op}}$ satisfies
$mm^{\ast}=\operatorname{Id}$. Therefore we get
$(\psi\otimes\psi^{\operatorname{op}})(P_{\beta})\geqslant|\\{\alpha:P_{\beta\alpha}\neq
0\\}|$ by Lemma 2.4. It follows that for each $\beta$ there is only a finite
number of $\alpha$’s such that $P_{\beta\alpha}\neq 0$, which means that $A$
is locally finite. ∎

Before we describe the bimodules that give rise to quantum graphs of bounded
degree, note that from the formula (3.4) (the formula is stated in the
infinite dimensional case) it follows that the degree matrix
$D=\sum_{i}V_{i}V_{i}^{\ast}$, where $V_{i}$ is an orthonormal basis of the
$\operatorname{B}^{\prime}$-module $\sigma_{\frac{i}{4}}^{\psi^{-1}}(V)$ with
respect to the inner product induced by $\psi^{-1}$. It is standard to check
that the expression $\sum_{i}V_{i}V_{i}^{\ast}$ does not depend on the choice
of the basis. We are now ready to state the first result.

######  Proposition 3.26.

If the $\operatorname{B}^{\prime}$-bimodule
$V\subset\operatorname{B}(\mathsf{H})$ defines a quantum graph of bounded
degree then $\sigma_{\frac{i}{4}}^{\psi^{-1}}(V)$ is a self-dual Hilbert
module over $\operatorname{B}^{\prime}$.

###### Proof.

We work in a more general setup where $\mathbb{E}:\mathsf{M}\to\mathsf{N}$ is
a normal conditional expectation and $V\subset\mathsf{M}$ is a weak∗ closed
right $\mathsf{N}$-module, so $\operatorname{B}(\mathsf{H})$ is replaced by
$\mathsf{M}$, $\operatorname{B}^{\prime}$ by $\mathsf{N}$, $\psi^{-1}$ by
$\mathbb{E}$, and $\sigma_{\frac{i}{4}}^{\psi^{-1}}$ by $V$. We will follow
the proof of [BDH88, Théorème 3.5]. We want to show that if there is an
orthonormal family $(V_{j})_{j}\subset V$ such that $(V_{j})_{j}$ is an
orthonormal basis inside the self-dual completion $V_{E}$ of $V$ such that
$\sum_{j}V_{j}V_{j}^{\ast}$ converges weak∗ in $\mathsf{M}$ then $V$ is self-
dual. It is sufficient to check that $V$ is complete with respect to the norm
induced by the $\mathsf{N}$-valued inner product, because self-duality will
follow, as $V$ admits a predual that makes the inner product separately weak∗
continuous. This is equivalent to the existence of a constant $K>0$ such that
$K\|\mathbb{E}(v^{\ast}v)\|\geqslant\|v\|^{2}$ for all $v\in V$. Let
$V_{E}\otimes_{\mathsf{N}}\mathsf{M}$ denote the self-dual
$\mathsf{M}$-module, obtained as the completion of the relative tensor
product. Then $\eta:=\sum_{j}V_{j}\otimes V_{j}^{\ast}$ is a well defined
element of $V_{E}\otimes_{\mathsf{N}}\mathsf{M}$ by [BDH88, 1.8(vi)]. We now
define a completely positive map $F$ from
$\operatorname{B}_{\mathsf{N}}(V_{E})$ to $\mathsf{M}$ via
$F(T):=\langle\eta,T\cdot\eta\rangle$. For any $v,w\in V$ we have

$F(|v\rangle\langle w|)=\sum_{j,k}\langle V_{j}^{\ast},\langle
V_{j},v\rangle_{\mathsf{N}}\langle
w,V_{k}\rangle_{N}V_{k}^{\ast}\rangle_{\mathsf{M}}=\sum_{j,k}V_{j}\langle
V_{j},v\rangle_{\mathsf{N}}\langle w,V_{k}\rangle_{\mathsf{N}}V_{k}^{\ast},$

which is equal to $vw^{\ast}$ as $(V_{j})_{j}$ was chosen to be an orthonormal
basis. Since $\|F\|\leqslant\|\eta\|^{2}$, it follows that
$\|v\|^{2}=\|F(|v\rangle\langle v|)\|\leqslant\|\eta\|^{2}\|\langle
v,v\rangle\|=\|\eta\|^{2}\|\mathbb{E}(v^{\ast}v)\|$. ∎

If we want to specify a class of $\operatorname{B}^{\prime}$-bimodules that
describe quantum graphs of bounded degree, then we need a property that does
not depend on the choice of the embedding
$\operatorname{B}\subset\operatorname{B}(\mathsf{H})$. This way we arrive at
the following equivalence.

######  Proposition 3.27.

A quantum graph is of bounded degree if and only if for any faithful
representation $\operatorname{B}\subset\operatorname{B}(\mathsf{H})$ the
corresponding bimodule $\sigma_{\frac{i}{4}}^{\psi^{-1}}(V)$ is a self-dual
Hilbert module over $\operatorname{B}^{\prime}$.

###### Proof.

We proved one implication in the previous proposition. Once again we work in a
greater generality. Namely we assume that we have a faithful normal
conditional expectation $\mathbb{E}:\mathsf{M}\to\mathsf{N}$ and
$V\subset\mathsf{M}$ is a weak∗ closed self-dual $\mathsf{N}$-module. Taking
advantage of the independence of the representation, we assume moreover that
$V\overline{\otimes}\operatorname{B}(\ell^{2})$ (weak∗ closure inside
$\mathsf{M}\overline{\otimes}\operatorname{B}(\ell^{2})$) is a self-dual
module over $\mathsf{N}\overline{\otimes}\operatorname{B}(\ell^{2})$. In the
original formulation the module
$V\overline{\otimes}\operatorname{B}(\ell^{2})$ will correspond to the
inflation of the representation of $\operatorname{B}$, i.e. taking
$\operatorname{B}\subset\operatorname{B}(\mathsf{H}\otimes\ell^{2})$ instead
of $\operatorname{B}\subset\operatorname{B}(\mathsf{H})$, hence will be self-
dual by the assumption of the proposition.

Note that $V$ has two natural operator space structures, one inherited from
$\mathsf{M}$, and the other one coming from the structure of an
$\mathsf{N}$-module. The assumption that $V$ is self-dual, hence in particular
complete, implies that the two Banach space structures are isomorphic, as the
operator norm clearly dominates the Hilbert module norm, hence one can invoke
the open mapping theorem. Our assumption that
$V\overline{\otimes}\operatorname{B}(\ell^{2})$ is also self-dual implies that
even the operator space structures are isomorphic.

Let $(v_{1},\dots,v_{n})$ be an $n$-tuple in $V$ and view it as the first row
of a matrix in $M_{n}(V)$. The square of the norm of this matrix, when we use
the Hilbert module operator space structure, is equal to the norm of the
matrix $(\mathbb{E}(v_{i}^{\ast}v_{j}))$ in $M_{n}(\mathsf{N})$, which is
clearly the same as the norm in $M_{n}(\mathsf{M})$. This can be computed
using the following formula (for positive matrices):

$\|(m_{ij})\|_{M_{n}(\mathsf{M})}=\sup\\{\|\sum
a_{i}^{\ast}m_{ij}a_{j}\|_{\mathsf{M}}:a_{i}\in\mathsf{M},\|\sum
a_{i}^{\ast}a_{i}\|_{\mathsf{M}}\leqslant 1\\}.$

This formula is proved using the fact that the $C^{\ast}$-norm on
$M_{n}(\mathsf{M})$ is the same as the operator norm, where
$M_{n}(\mathsf{M})$ acts on the Hilbert $\mathsf{M}$-module $\mathsf{M}^{n}$.
It follows from the equivalence of the operator space structures that for some
$K>0$ we have the inequality

$\|\sum_{i,j}a_{i}^{\ast}v_{i}^{\ast}v_{j}a_{j}\|\leqslant
K\|\sum_{i,j}a_{i}^{\ast}\mathbb{E}(v_{i}^{\ast}v_{j})a_{j}\|.$

Once again we will follow the proof of [BDH88, Théorème 3.5]. The estimate
above shows that we can define a bounded map
$\tau:V\otimes_{\mathsf{N}}\mathsf{M}\to\mathsf{M}$ given by
$\sum_{i}v_{i}\otimes m_{i}\mapsto\sum_{i}v_{i}m_{i}$. Since it is right
$\mathsf{M}$-linear and $V\otimes_{\mathsf{N}}\mathsf{M}$ is self-dual, $\tau$
is given by an inner product with a vector $\eta\in
V\otimes_{\mathsf{N}}\mathsf{M}$. Using $\eta$, we can define a normal,
completely positive map $F:\operatorname{B}_{\mathsf{N}}(V)\to\mathsf{M}$ via
$F(T):=\langle\eta,T\cdot\eta\rangle$. Let now $(X_{k})_{k}$ be an orthonormal
basis of $V$ and define $V_{n}$ to be the self-dual submodule of $V$ generated
by $X_{1},\dots,X_{n}$. The corresponding map $\tau_{n}$ is clearly
represented by the vector $\eta_{n}:=\sum_{k=1}^{n}X_{k}\otimes X_{k}^{\ast}$.
Therefore one can compute the value of $F_{n}$ on a rank one operator
$|v\rangle\langle w|\in\operatorname{B}_{\mathsf{N}}(V_{n})$ as

$F_{n}(|v\rangle\langle w|)=\sum_{j,k=1}^{n}X_{j}\langle X_{j},w\rangle\langle
w,X_{k}\rangle X_{k}^{\ast}=vw^{\ast}.$

As $\operatorname{Id}_{V_{n}}=\sum_{k=1}^{n}|X_{k}\rangle\langle X_{k}|$, we
get $F_{n}(\operatorname{Id})=\sum_{k=1}^{n}X_{k}X_{k}^{\ast}$, therefore
$\|\sum_{k=1}^{n}X_{k}X_{k}^{\ast}\|\leqslant K$. It follows that the partial
sums $\sum_{k=1}^{n}X_{k}X_{k}^{\ast}$ form an increasing and bounded
sequence, hence the series $\sum_{k=1}^{\infty}X_{k}X_{k}^{\ast}$ converges
strongly to an operator $D$ with norm at most $K$, which is precisely the
bounded degree condition. ∎

######  Remark 3.28.

It is not clear whether one can simply assume that one of the bimodules
defining the quantum graph is self-dual, i.e. whether the equivalence of the
operator space structures follows automatically from the Banach space
equivalence. For the index of a conditional expectation this is exactly the
gap between [BDH88, Proposition 3.3] and [BDH88, Théorème 3.5]. However, this
problem has been resolved in [FK98], where the authors proved that if for a
conditional expectation $\mathbb{E}:\operatorname{A}\to\operatorname{B}$ there
exists a constant $K>0$ such that the map $K\mathbb{E}-\operatorname{Id}$ is
positive then there also exists a constant $L>0$ such that
$L\mathbb{E}-\operatorname{Id}$ is _completely_ positive. Unfortunately the
proof relies heavily on the $C^{\ast}$-algebraic structure and it is unclear
whether an analogous result holds in our situation.

###  3.4. Explicit correspondence between the three approaches

From now on we assume that
$\operatorname{B}\simeq\ell^{\infty}-\bigoplus_{\alpha}M_{n_{\alpha}}$ is
represented on $\mathsf{H}:=\bigoplus_{\alpha}\mathbb{C}^{n_{\alpha}}$. In
this case the commutant of $\operatorname{B}$ is equal to the center of
$\operatorname{B}$. Therefore each weak∗ closed
$\operatorname{B}^{\prime}$-bimodule is exactly a collection of subspaces
$V_{\alpha\beta}\subset\operatorname{B}(\mathbb{C}^{n_{\alpha}},\mathbb{C}^{n_{\beta}})$.
The conditional expectation
$\psi^{-1}:\operatorname{B}(\mathsf{H})\to\operatorname{B}^{\prime}$ is given
by
$\psi^{-1}(x)=\sum_{\alpha}\frac{\operatorname{Tr}(\rho_{\alpha}^{-1}\mathds{1}_{\alpha}x\mathds{1}_{\alpha})}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\mathds{1}_{\alpha}$,
where $\mathbb{1}_{\alpha}\in\operatorname{B}^{\prime}$ is the minimal central
projection corresponding to the summand $M_{n_{\alpha}}$. For each pair
$(\alpha,\beta)$ let
$\left(X_{i}^{\alpha\beta}\right)_{i=1}^{n_{\alpha\beta}}$ be an orthonormal
basis of $V_{\alpha\beta}$ with respect to the KMS inner product on
$\operatorname{B}(\mathsf{H})$ (where we use the density
$\bigoplus_{\alpha}\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\rho_{\alpha}^{-1}$
to define a weight on $\operatorname{B}(\mathsf{H})$), i.e.
$\frac{1}{\sqrt{\operatorname{Tr}(\rho_{\alpha}^{-1})\operatorname{Tr}(\rho_{\beta}^{-1})}}\operatorname{Tr}((X_{i}^{\alpha\beta})^{\ast}\rho_{\beta}^{-\frac{1}{2}}X_{j}^{\alpha\beta}\rho_{\alpha}^{-\frac{1}{2}})=\delta_{ij}$.
Orthogonal projection from
$\operatorname{B}(\mathbb{C}^{n_{\alpha}},\mathbb{C}^{n_{\beta}})$ onto
$V_{\alpha\beta}$ will be given by

(3.2)
$P_{V_{\alpha\beta}}(x)=\frac{1}{\sqrt{\operatorname{Tr}(\rho_{\alpha}^{-1})\operatorname{Tr}(\rho_{\beta}^{-1})}}\sum_{i}X_{i}^{\alpha\beta}\operatorname{Tr}((X_{i}^{\alpha\beta})^{\ast}\rho_{\beta}^{-\frac{1}{2}}x\rho_{\alpha}^{-\frac{1}{2}}).$

We will use the notation $e_{kl}^{\alpha\beta}$ to denote the rank one
operator $|e_{k}^{\alpha}\rangle\langle e_{l}^{\beta}|$ to rewrite it further
as

$\frac{1}{\sqrt{\operatorname{Tr}(\rho_{\alpha}^{-1})\operatorname{Tr}(\rho_{\beta}^{-1})}}\sum_{i,k,l}X_{i}^{\alpha\beta}e_{kl}^{\alpha\beta}\rho_{\beta}^{-\frac{1}{2}}x\rho_{\alpha}^{-\frac{1}{2}}(X_{i}^{\alpha\beta})^{\ast}e_{lk}^{\beta\alpha}.$

Recall from Proposition 3.14 (and the discussion preceding it) that to find
the representing element in $M_{n_{\beta}}\otimes
M_{n_{\alpha}}^{\operatorname{op}}$ we need to twist by the modular group.

######  Proposition 3.29.

Let $\left(X_{i}^{\alpha\beta}\right)_{i=1}^{n_{\alpha\beta}}$ be an
orthonormal basis of
$V_{\alpha\beta}\subset\operatorname{B}(\mathbb{C}^{n_{\alpha}},\mathbb{C}^{n_{\beta}})$
with respect to the KMS inner product induced by $\psi^{-1}$. Then the
corresponding projection $P_{\beta\alpha}\in M_{n_{\beta}}\otimes
M_{n_{\alpha}}^{\operatorname{op}}$ is given by

(3.3)
$\frac{1}{\sqrt{\operatorname{Tr}(\rho_{\alpha}^{-1})\operatorname{Tr}(\rho_{\beta}^{-1})}}\sum_{i,k,l}\rho_{\beta}^{-\frac{1}{4}}X_{i}^{\alpha\beta}e_{kl}^{\alpha\beta}\rho_{\beta}^{-\frac{1}{4}}\otimes\left(\rho_{\alpha}^{-\frac{1}{4}}(X_{i}^{\alpha\beta})^{\ast}e_{lk}^{\beta\alpha}\rho_{\alpha}^{-\frac{1}{4}}\right)^{\operatorname{op}}.$

In particular the flip $\sigma(P_{\beta\alpha})\in M_{n_{\alpha}}\otimes
M_{n_{\beta}}^{\operatorname{op}}$ gives the orthogonal projection onto
$V_{\alpha\beta}^{\ast}\subset\operatorname{B}(\mathbb{C}^{n_{\beta}},\mathbb{C}^{n_{\alpha}})$.

###### Proof.

The flip $\sigma(P_{\beta\alpha})$ is the projection corresponding to the
orthonormal set
$\left((X_{i}^{\alpha\beta})^{\ast}\right)_{i=1}^{n_{\alpha\beta}}$, which is
an orthonormal basis of $V_{\alpha\beta}^{\ast}$, because of the property
$\langle X,Y\rangle=\langle Y^{\ast},X^{\ast}\rangle$ of the KMS inner
product. ∎

To find the explicit formula for the associated adjacency matrix, we will
rewrite the formula (3.2) in a different manner

$P_{V_{\alpha\beta}}(x)=\frac{1}{\sqrt{\operatorname{Tr}(\rho_{\alpha}^{-1})\operatorname{Tr}(\rho_{\beta}^{-1})}}\sum_{i,k,l}X_{i}^{\alpha\beta}e_{kl}^{\alpha}(X_{i}^{\alpha\beta})^{\ast}\rho_{\beta}^{-\frac{1}{2}}x\rho_{\alpha}^{-\frac{1}{2}}e_{lk}^{\alpha}.$

Therefore the corresponding projection in
$\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$ will
be given by

$P_{\beta\alpha}=\frac{1}{\sqrt{\operatorname{Tr}(\rho_{\alpha}^{-1})\operatorname{Tr}(\rho_{\beta}^{-1})}}\sum_{i,k,l}\rho_{\beta}^{-\frac{1}{4}}X_{i}^{\alpha\beta}e_{kl}^{\alpha}(X_{i}^{\alpha\beta})^{\ast}\rho_{\beta}^{-\frac{1}{4}}\otimes\left(\rho_{\alpha}^{-\frac{1}{4}}e_{lk}^{\alpha}\rho_{\alpha}^{-\frac{1}{4}}\right)^{\operatorname{op}}.$

If we compare it with (3.1), we immediately obtain the following statement.

######  Proposition 3.30.

Let $\left(X_{i}^{\alpha\beta}\right)_{i=1}^{n_{\alpha\beta}}$ be an
orthonormal basis of
$V_{\alpha\beta}\subset\operatorname{B}(\mathbb{C}^{n_{\alpha}},\mathbb{C}^{n_{\beta}})$
with respect to the KMS inner product induced by $\psi^{-1}$. Then the
corresponding quantum adjacency matrix
$A:c_{00}(\operatorname{B})\to\operatorname{B}$ is given by

(3.4)
$A(e_{kl}^{\alpha})=\sum_{\beta}\sqrt{\frac{\operatorname{Tr}(\rho_{\alpha}^{-1})}{\operatorname{Tr}(\rho_{\beta}^{-1})}}\sum_{i}\rho_{\beta}^{-\frac{1}{4}}X_{i}^{\alpha\beta}\rho_{\alpha}^{\frac{1}{4}}e_{kl}^{\alpha}\rho_{\alpha}^{\frac{1}{4}}(X_{i}^{\alpha\beta})^{\ast}\rho_{\beta}^{-\frac{1}{4}}.$

######  Remark 3.31.

We could in principle get rid of the scalars
$\operatorname{Tr}(\rho_{\alpha}^{-1})$ just by normalizing our densities so
that these numbers are equal to $1$. However, in the context of quantum groups
it is more common to choose the normalization
$\operatorname{Tr}(\rho_{\alpha}^{-1})=\operatorname{Tr}(\rho_{\alpha})$, so
we decided to include a more general discussion.

##  4\. Covariant quantum adjacency matrices

Let $\mathbb{G}$ be a compact quantum group and let $\mathbb{\Gamma}$ be its
discrete dual. We want to find quantum adjacency matrices on
$\ell^{\infty}(\mathbb{\Gamma})\simeq\bigoplus_{\alpha\in\operatorname{Irr}\mathbb{G}}M_{n_{\alpha}}$
that are covariant with respect to the right action of $\mathbb{\Gamma}$ on
itself. Such operators are given by left convolutions and we just need to
figure out, which ones give rise to quantum adjacency matrices. Before that we
need to gather some information about convolutions on quantum groups.

###  4.1. Fourier transform on compact/discrete quantum groups

Here we collect some conventions about the Fourier transform on compact and
discrete quantum groups. We will always denote by $\mathbb{G}$ a compact
quantum group and by $\mathbb{\Gamma}$ its discrete dual.

First, let $\varphi$ be a bounded functional on $L^{\infty}(\mathbb{G})$. We
will define its Fourier transform to be an element
$\widehat{\varphi}\in\ell^{\infty}(\mathbb{\Gamma})\simeq\ell^{\infty}-\bigoplus_{\alpha\in\operatorname{Irr}(\mathbb{G})}\operatorname{M}_{n_{\alpha}}$
such that

$\widehat{\varphi}(\alpha):=(\operatorname{Id}\otimes\varphi)\left((u^{\alpha})^{\ast}\right).$

We define the convolution of two functionals as

$\varphi_{1}\ast\varphi_{2}:=(\varphi_{1}\otimes\varphi_{2})\circ\Delta$

With this convention it is easy to check that
$(\varphi_{1}\ast\varphi_{2})\hat{\phantom{i}}(\pi)=\widehat{\varphi_{2}}(\pi)\widehat{\varphi_{1}}(\pi)$.

Recall (see [NT13, Sections 1.3 and 1.4]) that for any
$\alpha\in\operatorname{Irr}(\mathbb{G})$ there is a unique positive matrix
$\rho_{\alpha}\in M_{n_{\alpha}}$ such that
$\operatorname{Tr}(\rho_{\alpha})=\operatorname{Tr}(\rho_{\alpha}^{-1})=:\operatorname{dim}_{q}(\alpha)$
and it implements an equivalence between the representation $\alpha$ and its
double contragredient. These matrices can be used to describe the left and
right Haar measures on the discrete dual $\mathbb{\Gamma}$:

$\displaystyle\widehat{h_{R}}(x)$
$\displaystyle=\sum_{\alpha}\operatorname{Tr}(\rho_{\alpha})\operatorname{Tr}(\rho_{\alpha}x_{\alpha})$
$\displaystyle\widehat{h_{L}}(x)$
$\displaystyle=\sum_{\alpha}\operatorname{Tr}(\rho_{\alpha})\operatorname{Tr}(\rho_{\alpha}^{-1}x_{\alpha}).$

Both $\widehat{h_{R}}$ and $\widehat{h_{L}}$ satisfy
$mm^{\ast}=\operatorname{Id}$ by Proposition 2.3.

To define the Fourier transform and convolution of elements of
$L^{\infty}(\mathbb{G})$ we will always associate to such an element $x\in
L^{\infty}(\mathbb{G})$ a linear functional $h(\cdot x)$. For further use, we
will compute now the Fourier transform of an element
$x:=\sum_{i,j=1}^{n_{\alpha}}x_{ij}u_{ij}^{\alpha}$ belonging to the span of
matrix elements of a fixed irreducible representation $\alpha$.

######  Lemma 4.1.

Let $x:=\sum_{i,j=1}^{n_{\alpha}}x_{ij}u_{ij}^{\alpha}$. We have
$\widehat{x}(\pi)=\delta_{\alpha\pi}\frac{1}{\dim_{q}(\alpha)}X^{\operatorname{T}}\rho_{\alpha}^{-1}$,
where $(X)_{ij}=x_{ij}$.

###### Proof.

We have
$\widehat{x}(\pi)=\sum_{i,j,k,l}x_{ij}h(\left(u_{lk}^{\pi}\right)^{\ast}u_{ij}^{\alpha})e_{kl}$.
Using the orthogonality relations
$h(\left(u_{lk}^{\pi}\right)^{\ast}u_{ij}^{\alpha})=\delta_{\alpha\pi}\frac{1}{\dim_{q}(\alpha)}\delta_{jk}\left(\rho_{\alpha}^{-1}\right)_{il}$,
we easily find that

$\widehat{x}(\pi)=\frac{\delta_{\alpha\pi}}{\dim_{q}(\alpha)}\sum_{i,j,l}x_{ij}\left(\rho_{\alpha}^{-1}\right)_{il}e_{jl}=\frac{\delta_{\alpha\pi}}{\dim_{q}(\alpha)}X^{\operatorname{T}}\rho_{\alpha}^{-1}.$

∎

In the special case of a single matrix element we get the formula
$(u_{ij}^{\alpha})\hat{\phantom{a}}(\pi)=\frac{\delta_{\pi\alpha}}{\dim_{q}(\alpha)}e_{ji}^{\alpha}\rho_{\alpha}^{-1}$.
Given Proposition 2.3, it suggests that we will be working with the Haar
measure $\widehat{h_{R}}$ on $\mathbb{\Gamma}$.

######  Lemma 4.2.

The Fourier transform extends to a unitary between $L^{2}(\mathbb{G})$ and
$\ell^{2}(\mathbb{\Gamma})$, where $\ell^{2}(\mathbb{\Gamma})$ is defined
using the right Haar measure on $\mathbb{\Gamma}$.

###### Proof.

Using orthogonality relations and the previous lemma one can check that
$\langle
u_{ij}^{\alpha},u_{kl}^{\pi}\rangle=\langle\widehat{u_{ij}^{\alpha}},\widehat{u_{kl}^{\pi}}\rangle$.
Therefore the Fourier transform is isometric on $Pol(\mathbb{G})$, thus it
extends to an isometry on $L^{2}(\mathbb{G})$, which is also surjective, as
the image is clearly dense. ∎

We will also need the fact that the inverse Fourier transform maps convolution
into the regular product. We will redefine the Fourier transform using the
right multiplicative unitary $W$, defined via $W(a\otimes
b)=\Delta(a)(1\otimes b)$ on $L^{\infty}(\mathbb{G})\otimes
L^{\infty}(\mathbb{G})$ and extended to $L^{2}(\mathbb{G})\otimes
L^{2}(\mathbb{G})$ by continuity; it corresponds to the right regular
representation of $\mathbb{G}$ and implements the multiplication via
$W(x\otimes\mathds{1})W^{\ast}=\Delta(x)$. It can be shown that
$W\in\ell^{\infty}(\mathbb{\Gamma})\overline{\otimes}L^{\infty}(\mathbb{G})$.
It also satisfies the pentagonal equation $W_{23}W_{12}=W_{12}W_{13}W_{23}$,
which amounts to coassociativity of the comultiplication. We can write it more
explicitly as
$W=\sum_{\alpha\in\operatorname{Irr}(\mathbb{G})}\sum_{i,j}^{n_{\alpha}}e_{ij}^{\alpha}\otimes
u_{ij}^{\alpha}$, where $e_{ij}^{\alpha}$ acts on $L^{2}(\mathbb{G})$ via
right convolution by its inverse Fourier transform. Because of this formula,
we can also compute the Fourier transform as
$\widehat{\varphi}=(\operatorname{Id}\otimes\varphi)(W^{\ast})$. We will now
use the multiplicative unitary to define the Fourier transform on
$\mathbb{\Gamma}$.

######  Definition 4.3.

Let $\psi$ be a bounded functional on $\ell^{\infty}(\mathbb{\Gamma})$ then
its Fourier transform will be defined by
$\widehat{\psi}:=(\psi\otimes\operatorname{Id})(W)\in L^{\infty}(\mathbb{G})$;
the fact that we use the same notation for Fourier transforms going both ways
should not cause any confusion. For any $x\in c_{00}(\mathbb{\Gamma})$ we can
define a functional on $\ell^{\infty}(\mathbb{\Gamma})$ by
$\widehat{h_{R}}(\cdot x)$ and define the Fourier transform of $x$ using this
embedding.

######  Lemma 4.4.

We have
$\widehat{e_{ji}^{\alpha}\rho_{\alpha}^{-1}}=\operatorname{dim}_{q}(\alpha)u_{ij}^{\alpha}$
and if
$\psi_{1},\psi_{2}\in\left(\ell^{\infty}(\mathbb{\Gamma})\right)^{\ast}$ then
$\widehat{\psi_{1}\ast\psi_{2}}=\widehat{\psi_{2}}\widehat{\psi_{1}}$.

###### Proof.

We have

$\widehat{e_{ji}^{\alpha}\rho_{\alpha}^{-1}}=\sum_{\pi,k,l}\widehat{h_{R}}(e_{kl}^{\pi}e_{ji}^{\alpha}\rho_{\alpha}^{-1})u_{kl}^{\alpha},$

so $\pi=\alpha$ and $l=j$. We can further rewrite our formula as

$\sum_{k}\operatorname{Tr}(\rho_{\alpha})\operatorname{Tr}(\rho_{\alpha}e_{ki}^{\alpha}\rho_{\alpha}^{-1})u_{kj}^{\alpha}=\operatorname{Tr}(\rho_{\alpha})u_{ij}^{\alpha}=\operatorname{dim}_{q}(\alpha)u_{ij}^{\alpha}.$

In particular, the Fourier transform on $\mathbb{\Gamma}$ is inverse to the
Fourier transform on $\mathbb{G}$ (e.g. on the level of $L^{2}$-spaces).

Let us check that the convolution is mapped into the usual product, using the
pentagonal equation. To this end, let $\psi_{1}$ and $\psi_{2}$ be two
functionals on $\ell^{\infty}(\mathbb{\Gamma})$. We have
$\widehat{\psi_{1}\ast\psi_{2}}=(\psi_{1}\ast\psi_{2}\otimes\operatorname{Id})(W)$.
By definition of the convolution, it can be rewritten as

$(\psi_{1}\otimes\psi_{2}\otimes\operatorname{Id})(\Delta_{\mathbb{\Gamma}}\otimes\operatorname{Id})(W).$

The comultiplication in $\mathbb{\Gamma}$ is defined in such a way that
$(\Delta_{\mathbb{\Gamma}}\otimes\operatorname{Id})(W)=W_{23}W_{13}$.
Therefore we have to compute
$(\psi_{1}\otimes\psi_{2}\otimes\operatorname{Id})(W_{23}W_{13})$. It is clear
that $\psi_{2}$ acts on the part $W_{23}$ and $\psi_{1}$ acts on $W_{13}$,
hence we get $\widehat{\psi_{2}}\widehat{\psi_{1}}$ as the result. ∎

We will now investigate the behaviour of the Fourier transform under the
antipode. Recall that on any locally compact quantum group the antipode $S$
admits a polar decomposition $S=\tau_{-\frac{i}{2}}R$, where
$(\tau_{t})_{t\in\mathbb{R}}$ is the so-called scaling group (hence
$\tau_{-\frac{i}{2}}$ is only defined on analytic elements) and $R$ is the
unitary antipode, which is an involutive $\ast$-antiautomorphism; we also have
$\widehat{h_{R}}\circ R=\widehat{h_{L}}$.

######  Lemma 4.5 ([KV00, Proposition 6.8 and Proposition 7.9]).

Let $\mathbb{\Gamma}$ be a locally compact quantum group. Let
$(\sigma_{t})_{t\in\mathbb{R}}$ and
$(\sigma^{{}^{\prime}}_{t})_{t\in\mathbb{R}}$ be the modular groups of the
right and left Haar measures , and let $(\tau_{t})_{t\in\mathbb{R}}$ be the
scaling group. Then the following commutation relations hold:

  1. (i)

$\Delta\circ\sigma_{t}=(\sigma_{t}\otimes\tau_{-t})\circ\Delta$;

  2. (ii)

$\Delta\circ\sigma_{t}^{{}^{\prime}}=(\tau_{t}\otimes\sigma_{t}^{{}^{\prime}})\circ\Delta$;

  3. (iii)

$\Delta\circ\tau_{t}=(\tau_{t}\otimes\tau_{t})\circ\Delta$;

  4. (iv)

$\Delta\circ\tau_{t}=(\sigma_{t}^{{}^{\prime}}\otimes\sigma_{-t}\circ\Delta$.

If $\mathbb{\Gamma}$ is discrete then $\sigma_{t}^{{}^{\prime}}=\sigma_{-t}$,
$(\sigma_{t}\otimes\sigma_{t})\circ\Delta=\Delta\circ\sigma_{t}$, and
$\tau_{t}=\sigma_{-t}$.

###### Proof.

We will only explain how to get the additional statements about discrete
quantum groups. The fact that $\sigma_{t}^{\prime}=\sigma_{-t}$ follows
immediately from the explicit formulas for the Haar measures. The modular
element of a quantum group is relating the left and right Haar measures and in
the discrete case it is equal to
$\delta:=\bigoplus_{\alpha\in\operatorname{Irr}(\mathbb{G})}\rho_{\alpha}^{2}$.
It always satisfies $\Delta(\delta)=\delta\circ\delta$, which implies that
$\Delta\circ\sigma_{t}=(\sigma_{t}\otimes\sigma_{t})\circ\Delta$. Now
$\tau_{t}=\sigma_{-t}$ follows easily from the equalities
$\Delta\circ\tau_{t}=(\sigma_{-t}\otimes\sigma_{-t})\circ\Delta$ and
$(\sigma_{-t}\otimes\sigma_{-t})\circ\Delta=\Delta\circ\sigma_{-t}$, as they
imply $\Delta\circ\tau_{t}=\Delta\circ\sigma_{-t}$ and $\Delta$ is injective.
∎

######  Lemma 4.6 ([Kah10, Lemma 3.3]).

Let $\psi\in\ell^{\infty}(\mathbb{\Gamma})^{\ast}$ be such that
$\psi^{\\#}(x):=\psi^{\ast}(S^{-1}(x))=\overline{\psi(S(x^{\ast}))}$ is also
bounded. Then $(\widehat{\psi})^{\ast}=\widehat{\psi^{\\#}}$.

###### Proof.

See the proof of [Kah10, Lemma 3.3], but note that our convention for Fourier
transform uses the adjoint of the multiplicative unitary, that is why we need
to work with $S^{-1}$ rather than $S$. We also used the formula
$(S^{-1}(x))^{\ast}=S(x^{\ast})$. ∎

We will now see, what happens when the functional comes from an element of
$c_{00}(\mathbb{\Gamma})$.

######  Lemma 4.7.

Let $x\in c_{00}(\mathbb{\Gamma})$ and $\psi:=\widehat{h_{R}}(\cdot x)$. Then
$\psi^{\\#}=\widehat{h_{R}}(\cdot S(x^{\ast})\rho^{-2})$.

###### Proof.

We have to compute

$\psi^{\\#}(y)=\overline{\widehat{h_{R}}(S(y^{\ast})^{\ast}x))}=\widehat{h_{R}}(x^{\ast}S(y^{\ast})^{\ast})=\widehat{h_{R}}(x^{\ast}S^{-1}(y)).$

Since $S=\tau_{-\frac{i}{2}}R$ and $\tau_{-\frac{i}{2}}=\sigma_{\frac{i}{2}}$,
we get

$\widehat{h_{R}}(R(\sigma_{-\frac{i}{2}}(y)R(x^{\ast}))).$

Since $\widehat{h_{R}}\circ R=\widehat{h_{L}}$ and
$\widehat{h_{L}}(y)=\widehat{h_{R}}(\rho^{-2}y)$, we obtain

$\widehat{h_{R}}(\rho^{-2}\sigma_{-\frac{i}{2}}(y)R(x^{\ast})).$

Using the invariance under the modular group, and the fact that $\rho^{-2}$ is
in the centralizer, we arrive at

$\widehat{h_{R}}(y\sigma_{\frac{i}{2}}(R(x^{\ast})\rho^{-2}).$

As $\sigma_{\frac{i}{2}}R(x^{\ast})=S(x^{\ast})$, we get our answer:

$\psi^{\\#}(y)=\widehat{h_{R}}(yS(x^{\ast})\rho^{-2}).$

∎

###  4.2. Quantum adjacency matrices as convolution operators

We will now assume that our adjacency matrix is given by a convolution against
an element $P$. To avoid technical issues, we will only convolve with elements
belonging to the algebraic direct sum of matrix algebras. This is also
justified by the fact that $P\ast\mathds{1}=\widetilde{h_{R}}(P)\mathds{1}$
and if this number is finite and $P$ is a projection (see Proposition 4.9),
then $P\in c_{00}(\mathbb{\Gamma})$ by Lemma 2.4. Suppose then that our
quantum adjacency matrix is given by the following formula:

$Ax:=P\ast x.$

We will prove a general formula for a Schur product of two convolution
operators.

######  Proposition 4.8.

Let $P_{1},P_{2}\in c_{00}(\mathbb{\Gamma})$ and define $A_{1}(x):=P_{1}\ast
x$, $A_{2}(x):=P_{2}\ast x$. Then $A:=m(A_{1}\otimes A_{2})m^{\ast}$ is given
by $Ax:=P_{1}P_{2}\ast x$.

###### Proof.

In this proof we will use the notation $\mathcal{F}$ for the Fourier transform
on $\mathbb{G}$.

The formula for $A$ is (for each $\alpha\in\operatorname{Irr}\mathbb{G}$):

$\frac{1}{\operatorname{dim}_{q}(\alpha)}\sum_{k}A_{1}(e_{ik}^{\alpha}\rho_{\alpha}^{-1})A_{2}(e_{kj}^{\alpha}\rho_{\alpha}^{-1})=A(e_{ij}^{\alpha}\rho_{\alpha}^{-1}).$

By Lemma 4.1 we can write
$e_{ij}^{\alpha}\rho_{\alpha}^{-1}=\operatorname{dim}_{q}(\alpha)\mathcal{F}(u_{ji}^{\alpha})$,
so we want

$\sum_{k}A_{1}(\mathcal{F}(u_{ki}^{\alpha}))A_{2}(\mathcal{F}(u_{jk}^{\alpha}))=A(\mathcal{F}(u_{ji}^{\alpha})).$

As the Fourier transform switches convolution to multiplication (and reverses
the order), we have $A_{k}\circ\mathcal{F}=\mathcal{F}\circ\widehat{A_{k}}$,
where $\widehat{A_{k}}y:=y\widehat{P_{k}}$, with $\widehat{P_{k}}$ being the
inverse Fourier transform of $P_{k}$. We can therefore rewrite our condition
further as

$\sum_{k}\mathcal{F}(u_{ki}^{\alpha}\widehat{P_{1}})\mathcal{F}(u_{jk}^{\alpha}\widehat{P_{2}})=\mathcal{F}(u_{ji}^{\alpha}\widehat{P}),$

and using once again the properties of the Fourier transform we finally arrive
at

$\sum_{k}u_{jk}^{\alpha}\widehat{P_{2}}\ast
u_{ki}^{\alpha}\widehat{P_{1}}=u_{ji}^{\alpha}\widehat{P}.$

As the span of $u_{ij}$’s is dense, we can rewrite this equation as (using
Sweedler’s notation)

$x_{(1)}\widehat{P_{2}}\ast x_{(2)}\widehat{P_{1}}=x\widehat{P}.$

For $x=\mathds{1}$ we get $\widehat{P_{2}}\ast\widehat{P_{1}}=\widehat{P}$,
i.e. $P=P_{1}P_{2}$. This condition turns out to be sufficient as well.
Indeed, we can compute $x_{(1)}\widehat{P_{2}}\ast x_{(2)}\widehat{P_{1}}$
acting on an element $b$ as

$\displaystyle(x_{(1)}\widehat{P_{2}}\ast x_{(2)}\widehat{P_{1}})(b)$
$\displaystyle=(h(\cdot x_{(1)}\widehat{P_{2}})\otimes h(\cdot
x_{(2)}\widehat{P_{1}})\Delta(b)$ $\displaystyle=(h\otimes
h)(\Delta(b)(x_{(1)}\otimes x_{(2)})(\widehat{P_{2}}\otimes\widehat{P_{1}}))$
$\displaystyle=(h\otimes
h)(\Delta(b)\Delta(x)(\widehat{P_{2}}\otimes\widehat{P_{1}}))$
$\displaystyle=(h\otimes
h)(\Delta(bx)(\widehat{P_{2}}\otimes\widehat{P_{1}}))$
$\displaystyle=(\widehat{P_{2}}\ast\widehat{P_{1}})(bx)=\widehat{P}(bx),$

where the last expression is clearly equal to $x\widehat{P}$ acting on $b$. ∎

######  Proposition 4.9.

The map $A:\ell^{\infty}(\mathbb{\Gamma})\to\ell^{\infty}(\mathbb{\Gamma})$
given by $Ax:=P\ast x$ with $P\in c_{00}(\mathbb{\Gamma})$ is a quantum Schur
idempotent iff $P^{2}=P$ is an idempotent. Moreover, it is completely positive
iff $P$ is a projection.

###### Proof.

We have already showed the first part. For the second part, we will check that
$(P\ast x)^{\ast}=P^{\ast}\ast x^{\ast}$. Indeed, let $P\ast x=T$, which means
that for any $y\in\ell^{\infty}(\mathbb{\Gamma})$ we have
$\widehat{h_{R}}(y_{(1)}P)\widehat{h_{R}}(y_{(2)}x)=\widehat{h_{R}}(yT)$. We
then have

$\widehat{h_{R}}(yT^{\ast})=\overline{\widehat{h_{R}}(Ty^{\ast})}=\overline{\widehat{h_{R}}(\sigma_{i}(y^{\ast})T)}.$

We can now use the defining property of $T$ and the fact that on a discrete
quantum group the comultiplication is a $\ast$-homomorphism that intertwines
the modular group and arrive at

$\widehat{h_{R}}(\sigma_{i}(y^{\ast})T)=\widehat{h_{R}}(\sigma_{i}(y_{(1)}^{\ast})P)\widehat{h_{R}}(\sigma_{i}(y_{(2)}^{\ast})x).$

After applying the complex conjugation we obtain

$\overline{\widehat{h_{R}}(\sigma_{i}(y^{\ast})T)}=\widehat{h_{R}}(P^{\ast}\sigma_{-i}(y_{(1)}))\widehat{h_{R}}(x^{\ast}\sigma_{-i}(y_{(2)})),$

which is equal to

$\widehat{h_{R}}(y_{(1)}P^{\ast})\widehat{h_{R}}(y_{(2)}x^{\ast})=(P^{\ast}\ast
x^{\ast})(y),$

thus $(P\ast x)^{\ast}=P^{\ast}\ast x^{\ast}$, hence $A$ is $\ast$-preserving
iff $P=P^{\ast}$, and, as we already mentioned in Remark 3.2, complete
positivity is equivalent to being $\ast$-preserving for quantum Schur
idempotents. ∎

We will now characterize which covariant adjacency matrices are GNS-symmetric
and which are KMS-symmetric.

######  Theorem 4.10.

Let $P=P^{\ast}\in c_{00}(\mathbb{\Gamma})$. $Ax:=P\ast x$ is GNS-symmetric
iff $P=S(P)$ and $A$ is KMS-symmetric iff $P=R(P)$.

###### Proof.

We will view the convolution operator as a multiplication operator on the dual
compact quantum group, using the Fourier transform. After taking the Fourier
transform $A$ will become a _right_ multiplication operator by $\widehat{P}$.
The adjoint of a right multiplication operator by $x$ is the right
multiplication by $\sigma_{-i}^{\mathbb{G}}(x^{\ast})$, as can be checked
using the KMS property. Therefore the GNS-symmetry amounts to
$(\widehat{P})^{\ast}=\sigma_{i}(\widehat{P})$. We know from Lemma 4.7 that
$(\widehat{P})^{\ast}$ is the Fourier transform of $S(P^{\ast})\rho^{-2}$. Let
us now check that $\sigma_{z}(\widehat{P})$ is the Fourier transform of
$\rho^{iz}P\rho^{iz}$. Indeed, it is sufficient to check the formula when
$\widehat{P}=u_{kl}^{\alpha}$, and we can moreover assume that we have chosen
a basis such that $\rho_{\alpha}$ is diagonal with diagonal entries
$\rho_{\alpha,k}$. Then
$\sigma_{z}(u_{kl}^{\alpha})=\rho_{\alpha,k}^{iz}\rho_{\alpha,l}^{iz}u_{kl}^{\alpha}$,
which is the Fourier transform of
$\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\rho_{\alpha,l}^{iz}e_{lk}^{\alpha}\rho_{\alpha,k}^{iz}\rho_{\alpha}^{-1}$,
proving the formula. We therefore arrive at the equality

$S(P^{\ast})\rho^{-2}=\rho^{-1}P\rho^{-1},$

i.e. $S(P^{\ast})=\sigma_{i}(P)$. Recall that $S^{2}=\tau_{-i}=\sigma_{i}$,
hence $S(P^{\ast})=S^{2}(P)$, i.e. $S(P)=P^{\ast}$. Since by assumption
$P=P^{\ast}$, we obtain $S(P)=P$.

For KMS-symmetry, note that it is equivalent to the GNS-symmetry of the map
$\widetilde{A}(x):=\sigma_{-\frac{i}{4}}A(\sigma_{\frac{i}{4}}(x))$. It is not
difficult to check that this map is given by
$\widetilde{A}x=\sigma_{-\frac{i}{4}}(P)\ast x$, using the fact that the
comultiplication intertwines the modular group in our case. We therefore get
$S(\sigma_{-\frac{i}{4}}(P))=(\sigma_{-\frac{i}{4}}(P))^{\ast}=\sigma_{\frac{i}{4}}(P^{\ast})$.
This leads to $S\sigma_{-\frac{i}{2}}(P)=P^{\ast}$, but
$S\sigma_{-\frac{i}{2}}=S\tau_{\frac{i}{2}}=R$, so $R(P)=P^{\ast}$. By
assumption $P=P^{\ast}$, thus $R(P)=P$. ∎

##  5\. Quantum Cayley graphs

Classically the Cayley graph is defined using a generating set of our group.
In the quantum case we will replace a finite subset of a group with a
projection $P\in c_{00}(\mathbb{\Gamma})$. Typically Cayley graphs are
undirected, so we will also assume that $P$ is invariant under the unitary
antipode $R$. Moreover, a generating set should not contain the unit, which
will easily translate to the fact that the counit $\varepsilon$ applied to $P$
should be equal to zero; less abstractly, the component of $P$ corresponding
to the trivial representation of $\mathbb{G}$ should be equal to $0$. The only
nontrivial condition is what it means for such a projection to be generating.
We will denote by $[T]$ the range projection of an operator $T$.

######  Definition 5.1.

Let $\mathbb{\Gamma}$ be a discrete quantum group and let $P\in
c_{00}(\mathbb{\Gamma})$ be a projection such that $R(P)=P$ and
$\varepsilon(P)=0$. We say that $P$ is generating if $\bigvee_{n}[P^{\ast
n}]=\mathds{1}\in\ell^{\infty}(\Gamma)$. Then we define the quantum Cayley
graph of $\mathbb{\Gamma}$ via the adjacency matrix

$Ax:=P\ast x.$

These quantum graphs are always regular with
$D:=A\mathds{1}=\widehat{h_{R}}(P)\mathds{1}$.

There is a natural special case, where we choose a central generating
projection, which leads to a more familiar notion.

######  Proposition 5.2.

Let $z\in c_{00}(\mathbb{\Gamma})$ be a central projection, i.e. it is a sum
of minimal central projections corresponding to certain irreducible
representations of $\mathbb{G}$:
$z=\bigoplus_{k=1}^{n}\mathds{1}_{\alpha_{k}}$. Let
$S=\\{\alpha_{1},\dots,\alpha_{n}\\}$. Then $z=R(z)$ iff $S=\overline{S}$,
i.e. it is closed under taking conjugate representations. The condition
$\varepsilon(z)=0$ is equivalent to the fact that $S$ does not contain the
trivial representation. Moreover, $z$ is generating iff the set $S$ is a
generating set of irreducible representations of $\mathbb{G}$.

###### Proof.

Only the statement about the criterion for being a generating set requires
proof. Let $\mathds{1}_{n_{\alpha}}$ be a minimal central projection. By Lemma
4.1 we have that
$\widehat{\mathds{1}_{n_{\alpha}}\rho_{\alpha}^{-1}}=\operatorname{Tr}(\rho_{\alpha}^{-1})\chi_{\alpha}$.
Because of the formula
$\sigma_{w}^{\mathbb{G}}(\widehat{T})=\rho^{iw}T\rho^{iw}$ established in the
proof of Theorem 4.10, we get
$\widehat{\mathds{1}_{n_{\alpha}}}=\operatorname{Tr}(\rho_{\alpha}^{-1})\sigma^{\mathbb{G}}_{-\frac{i}{2}}(\chi_{\alpha})$.
Using the Fourier transform, we can view the convolution operator by $z$ as a
multiplication operator by the character (acted on by the modular group) of
the representation $\pi:=\alpha_{1}\oplus\dots\oplus\alpha_{n}$. If $S$ is a
generating set of representations, then eventually all characters will appear
in the decomposition of the powers of the character $\chi_{\pi}$. By taking an
inverse Fourier transform, we will obtain that $z^{\ast m}$ is a linear
combination of central projections and eventually all minimal central
projections will appear in its decomposition, which shows that $z$ is
generating. If we assume that $z$ is generating we can reverse this argument
to show that $S$ is a generating set of representations. ∎

######  Example 5.3.

It is clear that if $P\in c_{00}(\mathbb{\Gamma})$ is a generating projection,
then so is its central cover $z(P)$. The converse, however is not true, as the
following simple example shows. Let $G=SU(2)$ and let $\mathbb{\Gamma}$ be its
dual. We will take as $P$ the rank one projection $e_{11}$ inside the $M_{2}$
summand corresponding to the fundamental representation $\pi$ of $SU(2)$. Its
central cover corresponds to a generating representation, hence is generating
itself; we will show that $P$ is not generating. Convolution powers of
$e_{11}$ are not difficult to compute using the Fourier transform. Indeed, the
Fourier transform of $e_{11}$ is a multiple of the matrix element $u_{11}$,
the matrix element of the fundamental representation $\pi$ at the vector
$e_{1}$. The power $u_{11}^{n}$ is the matrix element of $\pi^{\otimes n}$ at
the vector $e_{1}^{\otimes n}$. This vector happens to be a symmetric tensor,
so $u_{11}^{n}$ is a matrix element of the symmetric tensor power
$\operatorname{Sym}^{n}(\pi)$. It is well-known that this representation is
irreducible, hence the inverse Fourier transform of $u_{11}^{n}$ is a multiple
of a rank one projection inside the summand $M_{n+1}$. It is clear that $P$ is
not generating, because it misses a lot, e.g. the trivial representation.

###  5.1. Independence of the choice of the generating set

For a group $\Gamma$ the Cayley graph does depend on the choice of the
generating set but they are all bi-Lipschitz equivalent, so more or less
indistnguishable in the sense of metric/coarse geometry. Fortunately there is
a metric structure available in the quantum case that will allow us to define
the notion of bi-Lipschitz equivalence.

######  Definition 5.4 (Kuperberg-Weaver).

Let $\mathsf{M}\subset\operatorname{B}(\mathsf{H})$ be a von Neumann algebra.
A quantum metric on $\mathsf{M}$ is an increasing family of weak∗ closed
operator systems $(V_{t})_{t\geqslant 0}$ such that

  1. (i)

$V_{0}=\mathsf{M}^{\prime}$,

  2. (ii)

$V_{s}V_{t}\subset V_{s+t}$,

  3. (iii)

$V_{t}=\bigcap_{s>t}V_{s}$.

For $\mathsf{M}=\ell^{\infty}(X)$ we can intepret these operator systems
$V_{t}$ as operators of propagation at most $t$.

For _undirected_ quantum graphs the quantum metric can be built in the
following way. First of all, we can associate a weak∗ closed
$\mathsf{M}^{\prime}$-bimodule $V$ to a quantum adjacency matrix. Then we
simply define
$V_{t}:=\overline{\operatorname{span}}\\{\sum_{k=0}^{[t]}V^{k}\\}$, where we
interpret $V^{0}$ as $\mathsf{M}^{\prime}$.

######  Definition 5.5.

Let $(V_{t})_{t\geqslant 0}$ and $(W_{t})_{t\geqslant 0}$ be two quantum
metrics on $\mathsf{M}$. We say that they are bi-Lipschitz equivalent if there
exists a constant $M>0$ such that $V_{t}\subset W_{Mt}$ and $W_{t}\subset
V_{Mt}$ for all $t\geqslant 0$.

Therefore to understand the metric structure of a quantum graph, we need a
better understanding of the powers $V^{k}$. We will relate them in some way to
the powers of the adjacency matrix, so we need to be able to compose the
adjacency matrix with itself, which is possible for quantum graphs of bounded
degree111It is also possible for locally finite quantum graphs, but then the
Choi matrix associated with the power of the adjacency matrix is only
_affiliated_ with
$\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$ and
we decided to avoid these technicalities..

######  Lemma 5.6.

Let $A:\operatorname{B}\to\operatorname{B}$ be a quantum adjacency matrix of
bounded degree, let
$P\in\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$
be its associated projection, and let $V$ be the corresponding
$\operatorname{B}^{\prime}$ bimodule. Let
$A_{k}\in\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}$
be the generalized Choi matrix of $A^{k}$. Then the range projection $[A_{k}]$
corresponds to the $\operatorname{B}^{\prime}$-bimodule $V^{k}$.

###### Proof.

Since
$\operatorname{B}\overline{\otimes}\operatorname{B}^{\operatorname{op}}\simeq\ell^{\infty}-\bigoplus_{\alpha,\beta}M_{n_{\beta}}\otimes
M_{n_{\alpha}}^{\operatorname{op}}$, it suffices to check that the range
projection $[A_{k}]_{\beta\alpha}$ corresponds to
$V^{k}_{\alpha\beta}\subset\operatorname{B}(\mathbb{C}^{n_{\alpha}},\mathbb{C}^{n_{\beta}})$.

We will first work in the tracial setting, and then discuss how to modify the
proof in the general case. We are in the following situation: we have a
completely positive map $\Phi(x):=\sum_{k=1}^{n}V_{k}xV_{k}^{\ast}$ and we
would like to check that the support projection of the Choi matrix of $\Phi$
gives rise to the orthogonal projection onto the span of $V_{k}$’s inside
$\operatorname{B}(\mathbb{C}^{n_{\alpha}},\mathbb{C}^{n_{\beta}})$; indeed,
the Kraus operators of $A^{k}_{\alpha\beta}$ are the products belonging to
$V^{k}_{\alpha\beta}$ (see (3.4) and note that the sums appearing will be
finite by Proposition 3.25).

Let $T$ be the Choi matrix of $\Phi$. As a positive semidefinite matrix it can
be diagonalized and written in the form
$T=\frac{1}{n_{\alpha}}\sum_{l=1}^{d}\lambda_{l}T_{l}e_{ij}^{\alpha}T_{l}^{\ast}\otimes(e_{ji}^{\alpha})^{\operatorname{op}}$,
where $T_{l}:\mathbb{C}^{n_{\alpha}}\to\mathbb{C}^{n_{\beta}}$ satisfy
$\frac{1}{n_{\alpha}}\operatorname{Tr}(T_{k}^{\ast}T_{l})=\delta_{kl}$, $d$ is
the rank of $T$, and $\lambda_{l}$’s are the non-zero eigenvalues. The support
projection of $T$ is equal to
$[T]=\frac{1}{n_{\alpha}}\sum_{l=1}^{d}T_{l}e_{ij}^{\alpha}T_{l}^{\ast}\otimes(e_{ji}^{\alpha})^{\operatorname{op}}$,
i.e. by setting all the non-zero eigenvalues to be equal to $1$. The
corresponding subspace $V_{\alpha\beta}$ is the span of the $T_{l}$’s. Luckily
it is a well-known fact that the span of Kraus operators is independent of any
choices hence
$\operatorname{span}\\{T_{l}:l\in[d]\\}=\operatorname{span}\\{V_{k}:k\in[n]\\}$.
This finishes the proof of the lemma.

Let us comment now, what changes in the non-tracial setting. Combining Lemma
3.3 with Lemma 3.5 we see that the generalised Choi matrix of $\Phi$ is equal
to
$\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\sum_{i,j}V_{k}\rho_{\alpha}^{-\frac{1}{2}}e_{ij}^{\alpha}\rho_{\alpha}^{-\frac{1}{2}}V_{k}^{\ast}\otimes
e_{ji}^{\alpha}$. Once again we diagonalize it to obtain $(T_{1},\dots,T_{d})$
with $\frac{1}{n_{\alpha}}\operatorname{Tr}(T_{k}^{\ast}T_{l})=\delta_{kl}$.
Define
$\widetilde{T_{l}}:=\frac{\sqrt{\operatorname{Tr}(\rho_{\alpha}^{-1})}}{\sqrt{n_{\alpha}}}T_{l}\rho_{\alpha}^{\frac{1}{2}}$.
Then the generalised Choi matrix is equal to

$\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\sum_{l,i,j}\lambda_{l}\widetilde{T_{l}}\rho_{\alpha}^{-\frac{1}{2}}e_{ij}^{\alpha}\rho_{\alpha}^{-\frac{1}{2}}\widetilde{T_{l}}^{\ast}\otimes
e_{ji}^{\alpha},$

where
$\frac{1}{\operatorname{Tr}(\rho_{\alpha}^{-1})}\operatorname{Tr}(\rho_{\alpha}^{-1}\widetilde{T_{k}}^{\ast}\widetilde{T_{l}})=\delta_{kl}$,
from which it follows that the support projection is once again obtained by
setting all $\lambda_{l}$’s to $1$. ∎

######  Theorem 5.7.

Let $\mathbb{\Gamma}$ be a discrete quantum group and let
$(\mathbb{\Gamma},P_{1})$ and $(\mathbb{\Gamma},P_{2})$ be its two quantum
Cayley graphs. Then the corresponding quantum metric spaces are bi-Lipschitz
equivalent.

###### Proof.

Let $A_{1}x:=P_{1}\ast x$ and $A_{2}x:=P_{2}\ast x$. Note that the powers
$A_{1}^{k}$ and $A_{2}^{k}$ can be computed using the convolution powers
$P_{1}^{\ast k}$ and $P_{2}^{\ast k}$. From Proposition 4.8 it follows that
the support projection of the Choi matrix of $A_{i}^{k}$ is the Choi matrix of
the convolution operator by $[P_{i}^{\ast k}]$. Since both projections are
generating, there exists $M\in\mathbb{N}$ such that
$P_{1}\leqslant\bigvee_{i=1}^{M}[P_{2}^{\ast i}]$ and
$P_{2}\leqslant\bigvee_{i=1}^{M}[P_{1}^{\ast i}]$. The bi-Lipschitz
equivalence follows easily. ∎

###  5.2. Quantum Cayley graphs as quantum relations

In this subsection we will describe the $\operatorname{B}^{\prime}$-bimodule
corresponding to a covariant quantum adjacency matrix coming from a central
projection, so $A:\operatorname{B}\to\operatorname{B}$ will be given by
$Ax:=\mathds{1}_{\gamma}\ast x$; for simplicity we will assume that $\gamma$
is irreducible.

In order to find $V_{\alpha\beta}$ we need to compute the part of the
convolution $\mathbf{1}_{\gamma}\ast e_{ij}^{\alpha}$ belonging to
$M_{n_{\beta}}$. We will assume that that the $\rho$ matrices are diagonal.

######  Lemma 5.8.

We have $(\mathbf{1}_{\gamma}\ast
e_{ij}^{\alpha})_{\beta}=\frac{\operatorname{dim}_{q}(\alpha)\operatorname{dim}_{q}(\gamma)}{\operatorname{dim}_{q}(\beta)}\sum_{k=1}^{n_{\gamma}}\sum_{l=1}^{m(\beta,\alpha\otimes\gamma)}V_{kl}e_{ij}^{\alpha}V_{kl}^{\ast}$,
where $m(\beta,\alpha\otimes\gamma)$ is the multiplicity of $\beta$ inside
$\alpha\otimes\gamma$,
$V_{l}:\mathsf{H}_{\beta}\to\mathsf{H}_{\alpha}\otimes\mathsf{H}_{\gamma}$ are
morphisms, which are isometries with orthogonal ranges, and
$V_{kl}(v):=V_{l}^{\ast}(v\otimes e_{k})$.

###### Proof.

By Lemma 4.1 we have

$\widehat{\mathbf{1}_{\gamma}}=\operatorname{dim}_{q}(\gamma)\sum_{k=1}^{n_{\gamma}}\rho_{\gamma
k}u_{kk}^{\gamma}$

and

$\widehat{e_{ij}^{\alpha}}=\operatorname{dim}_{q}(\alpha)\rho_{\alpha
j}u_{ji}^{\alpha}.$

What we need now is a formula for a product of two matrix coefficients and
this is exactly [KS18, Lemma 4.1]:

$(u_{ji}^{\alpha}u_{kk}^{\gamma})_{\beta}=\sum_{l=1}^{m(\beta,\alpha\otimes\gamma)}\sum_{p,q}\langle
e_{j}\otimes e_{k},Ve_{p}\rangle u_{pq}^{\beta}\langle Ve_{q},e_{i}\otimes
e_{k}\rangle.$

∎

Inverting the Fourier transform, we obtain

$(\mathbf{1}_{\gamma}\ast
e_{ij}^{\alpha})_{\beta}=\frac{\operatorname{dim}_{q}(\alpha)\operatorname{dim}_{q}(\gamma)}{\operatorname{dim}_{q}(\beta)}\sum_{k=1}^{n_{\gamma}}\rho_{\gamma
k}\rho_{\alpha j}\sum_{l=1}^{m(\beta,\alpha\otimes\gamma)}\sum_{p,q}\langle
e_{j}\otimes e_{k},V_{l}e_{p}\rangle e_{qp}^{\beta}\rho_{\beta p}^{-1}\langle
V_{l}e_{q},e_{i}\otimes e_{k}\rangle.$

Recall that $e_{qp}^{\beta}=|e_{q}\rangle\langle e_{p}|$, so using the
equalities $\operatorname{Id}_{\beta}=\sum_{p}|e_{p}\rangle\langle e_{p}|$, we
arrive at

$(\mathbf{1}_{\gamma}\ast
e_{ij}^{\alpha})_{\beta}=\frac{\operatorname{dim}_{q}(\alpha)\operatorname{dim}_{q}(\gamma)}{\operatorname{dim}_{q}(\beta)}\sum_{k=1}^{n_{\gamma}}\rho_{\gamma
k}\rho_{\alpha
j}\sum_{l=1}^{m(\beta,\alpha\otimes\gamma)}|V_{l}^{\ast}(e_{i}\otimes
e_{k})\rangle\langle V^{\ast}(e_{j}\otimes e_{k})|\rho_{\beta}^{-1}.$

We can further rewrite it as

$(\mathbf{1}_{\gamma}\ast
e_{ij}^{\alpha})_{\beta}=\frac{\operatorname{dim}_{q}(\alpha)\operatorname{dim}_{q}(\gamma)}{\operatorname{dim}_{q}(\beta)}\sum_{k=1}^{n_{\gamma}}\rho_{\gamma
k}\sum_{l=1}^{m(\beta,\alpha\otimes\gamma)}V_{kl}e_{ij}^{\alpha}\rho_{\alpha}V_{kl}^{\ast}\rho_{\beta}^{-1}.$

The last thing to check is that $V_{kl}\rho_{\gamma
k}\rho_{\alpha}=\rho_{\beta}V_{kl}$. Recall $V_{kl}$ is a composition of two
maps, an inclusion $\iota_{k}(v):=v\otimes e_{k}$ and a morphism
$V_{l}^{\ast}:\mathsf{H}_{\alpha}\otimes\mathsf{H}_{\gamma}\to\mathsf{H}_{\beta}$.
It is clear that $\iota_{k}\rho_{\alpha}\rho_{\gamma
k}=(\rho_{\alpha}\otimes\rho_{\gamma})\iota_{k}$. It is not difficult to check
that for two representations $X$ and $Y$ and an intertwiner
$V\in\operatorname{Mor}(X,Y)$ we have $V\rho_{X}=\rho_{Y}V$ because by Schur’s
lemma it is sufficient to check it when $X$ and $Y$ are multiples of the same
irreducible representation $\alpha$ and for them the $\rho$ operator is merely
a direct sum of copies of $\rho_{\alpha}$. By [NT13, Theorem 1.4.9]
$\rho_{\alpha\otimes\gamma}=\rho_{\alpha}\otimes\rho_{\gamma}$, so we obtain
$V_{l}^{\ast}(\rho_{\alpha}\otimes\rho_{\gamma})=\rho_{\beta}V_{l}^{\ast}$. It
follows that $V_{l}^{\ast}\iota_{k}\rho_{\gamma
k}\rho_{\alpha}=\rho_{\beta}V_{l}^{\ast}\iota_{k}$.

######  Proposition 5.9.

Let $\mathbb{\Gamma}$ be a discrete quantum group and let $S$ be a generating
set of representations of $\mathbb{G}$. Then the bimodule describing the
quantum Cayley graph is given by

$V_{\alpha\beta}=\operatorname{span}\\{V_{s}^{\ast}(\iota_{s}(v)):s\in
S,V_{s}\in\operatorname{Mor}(\beta,\alpha\otimes s)\\},v\in\mathsf{H}_{s}\\}$

where
$\iota_{s}(v):\mathsf{H}_{\alpha}\to\mathsf{H}_{\alpha}\otimes\mathsf{H}_{s}$
is given by $\iota_{s}(v)(w):=w\otimes v$.

###### Proof.

From the previous lemma it follows that the span of the Kraus operators of the
convolution operator against $\mathbf{1}_{\gamma}$, or rather the block
mapping from $M_{n_{\alpha}}$ to $M_{n_{\beta}}$ is equal to

$\operatorname{span}\\{V^{\ast}(\iota(v)):V\in\operatorname{Mor}(\beta,\alpha\otimes
s)\\},v\in\mathsf{H}_{\gamma}\\}.$

From the formula 3.4 it is clear that this span is equal to
$(\sigma_{-\frac{i}{4}}^{\psi^{-1}}(V))_{\alpha\beta}$, rather than
$V_{\alpha\beta}$. However, since we convolve against a central projection,
the adjacency matrix commutes with the modular group, hence the corresponding
bimodule will be invariant under the modular group, so
$\sigma_{-\frac{i}{4}}^{\psi^{-1}}(V)=V$.

In general we convolve against a central projection corresponding to a direct
sum of irreducible representations and we obtain the result by linearity. ∎

###  5.3. A sample application: discrete quantum groups of subexponential
growth are amenable

Many algebraic properties of groups are encoded in the geometry of their
Cayley graphs. Here we would like to present an indication that the same
should hold for discrete _quantum_ groups. We will use the result from [Kye08]
to show that subexponential growth of the Cayley graph implies amenability of
the discrete quantum group (actually the result even shows that the dual
compact quantum group is coamenable, which is formally a stronger property,
but actually equivalent by [Tom06]).

######  Definition 5.10.

Let $\mathbb{\Gamma}$ be a discrete quantum group. We say that
$\mathbb{\Gamma}$ is of subexponential growth if there exists a generating
projection $P\in c_{00}(\mathbb{\Gamma})$ with $\varepsilon(P)=1$ such that
$\lim_{n\to\infty}\left(\widehat{h_{R}}([P^{\ast n}])\right)^{\frac{1}{n}}=1$.

######  Remark 5.11.

We assume that the counit applied to $P$ is equal to $1$ because we want to
measure the volume of the ball, so we want $[P^{\ast n}]$ to correspond to the
set of words of length at most $n$, not exactly $n$.

######  Proposition 5.12.

Suppose that $\mathbb{\Gamma}$ is a discrete quantum group of subexponential
growth. Then $\mathbb{\Gamma}$ is amenable.

###### Proof.

First we explain why the notion of subexponential growth is independent of the
choice of a generating projection. Take two generating projection $P$ and $Q$
and assume we have subexponential growth with respect to $P$. Call $V_{n}$ the
quantum relations corresponding to $[P^{\ast n}]$ and $W_{n}$ the ones
corresponding to $[Q^{\ast}n]$. By Theorem 5.7 there exists $M\in\mathbb{N}$
such that $V_{n}\subset W_{Mn}$ and $W_{n}\subset V_{Mn}$. Since
$\widehat{h_{R}}([P^{\ast n}])\mathds{1}$ is the degree matrix of $V_{n}$, we
obtain

$\displaystyle\widehat{h_{R}}([P^{\ast n}])$
$\displaystyle\leqslant\widehat{h_{R}}([Q^{\ast Mn}]$
$\displaystyle\widehat{h_{R}}([Q^{\ast n}])$
$\displaystyle\leqslant\widehat{h_{R}}([P^{\ast Mn}].$

It follows that if $\lim_{n\to\infty}\left(\widehat{h_{R}}([P^{\ast
n}])\right)^{\frac{1}{n}}=1$ then
$\lim_{n\to\infty}\left(\widehat{h_{R}}([Q^{\ast n}])\right)^{\frac{1}{n}}=1$.

If $P\in c_{00}(\mathbb{\Gamma})$ is generating then its central cover
$z(P)\in c_{00}(\mathbb{\Gamma})$ is generating as well, hence we can assume
that we are working with a generating set $S$ of representations, including
the trivial one (see Proposition 5.2).

According to [Kye08, Theorem 3.3 and Theorem 4.5] coamenability of
$\mathbb{G}$ is equivalent to the following Følner condition: for every
finitely supported, symmetric probability measure $\mu$ on
$\operatorname{Irr}(\mathbb{G})$ whose support contains the trivial
representation and every $\varepsilon>0$ there exists a finite subset
$F\subset\operatorname{Irr}(\mathbb{G})$ such that

$\sum_{\alpha\in\operatorname{supp}(\mu\ast\mathbf{1}_{F})}n_{\alpha}^{2}\leqslant(1+\varepsilon)\sum_{\alpha\in
F}n_{\alpha}^{2},$

where $n_{\alpha}$ is the dimension of the representation $\alpha$.

Note the inequality
$n_{\alpha}^{2}\leqslant\operatorname{Tr}(\rho_{\alpha})\operatorname{Tr}(\rho_{\alpha}^{-1})$,
from which it follows that for any central projection $z\in
c_{00}(\mathbb{\Gamma})$ corresponding to a family $T$ of irreducible
representations we have $\sum_{\alpha\in
T}n_{\alpha}^{2}\leqslant\widehat{h_{R}}(z)$.

Let $S^{k}$ denote the set of irreducible representations of $\mathbb{G}$ that
appear in $k$-fold tensor products of elements of $S$. It follows that the
sequence $a_{k}:=\sum_{\alpha\in S^{k}}n_{\alpha}^{2}$ satisfies
$\lim_{k\to\infty}(a_{k})^{\frac{1}{k}}=1$.

Fix $\mu$ as above and $\varepsilon>0$. For a large enough $k$ we will have
$\operatorname{supp}(\mu)\subset S^{k}$. Clearly
$\operatorname{supp}(\mu\ast\mathbf{1}_{S^{n}})\subset S^{k+n}$, so it
suffices to find $n$ large enough so that
$a_{k+n}\leqslant(1+\varepsilon)a_{n}$. As
$\lim_{k\to\infty}(a_{k})^{\frac{1}{k}}=1$ we have
$\lim_{n\to\infty}\frac{a_{k+n}}{a_{n}}=1$, so we are done.

∎

In the non-unimodular case (e.g. for $SU_{q}(2)$) it often happens that the
classical dimensions exhibit subexponential growth, while the quantum ones
grow exponentially, so Kyed’s can still be applied. It is however not clear at
this point how to encode such a growth condition in a natural way using the
quantum Cayley graphs.

###  5.4. Examples

Our first example will be constructed using the duals of the free unitary
quantum groups introduced in [VDW96]. Recall that if $F\in M_{n}$ is an
invertible matrix then $C(U_{F}^{+})$ is the universal $C^{\ast}$-algebra
generated by the entries of a unitary matrix $U:=(u_{ij})_{i,j\in[n]}$ such
that $F\overline{U}F^{-1}$ is also unitary, where
$\overline{U}:=(u_{ij}^{\ast})_{i,j\in[n]}$; the comultiplication is given by
the familiar formula $\Delta(u_{ij}):=\sum_{k=1}^{n}u_{ik}\otimes u_{kj}$. In
[Ban97, Théorème] the representation theory has been computed and we will need
the following two facts about it: all irreducibles are generated by the
fundamental representation $u$ and its conjugate $\overline{u}$ and the tensor
powers $u^{\otimes n}$ are irreducible.

######  Example 5.13.

Let $U_{F}^{+}$ be a free unitary quantum group. Let
$P:=\mathbf{1}_{u}+\mathbf{1}_{\overline{u}}$ be the central projection in
$\ell^{\infty}(\widehat{U_{F}^{+}})$ corresponding to the representation
$u\oplus\overline{u}$. $P$ is generating and we call the associated quantum
graph the quantum Cayley graph of $\widehat{U_{F}^{+}}$.

Here we would like to give some indication that this quantum Cayley graph
should be in fact a quantum tree. For now we do not have a notion of a tree
for quantum graphs, but we we would like to nonetheless argue why the Cayley
graph of $\widehat{U_{F}^{+}}$ should be called a tree. We can split the
quantum adjacency matrix $A$ into a sum of two convolution operators
$A_{1}(x):=\mathbf{1}_{u}\ast x$ and $A_{2}(x):=\mathbf{1}_{\overline{u}}\ast
x$. Because $u^{\otimes n}$ are $\overline{u}^{\otimes n}$ are irreducible for
each $n\in\mathbb{N}$, hence the convolution powers are easy to compute using
the Fourier transform; we have that $A_{1}^{n}(x)=\mathbf{1}_{u^{\otimes
n}}\ast x$ and analogously for $A_{2}$. Classically this would mean that the
direct graphs defined by $A_{1}$ and $A_{2}$ have the following property: for
any pair of vertices there is at most one directed path connecting them. This
property is weaker than being a tree, but provides some evidence that this
example might be a tree. Moreover, we have $m(A_{1}\otimes A_{2})m^{\ast}=0$,
which means that $A_{1}$ and $A_{2}$ have disjoint sets of edges, and $A_{2}$
is the KMS adjoint of $A_{1}$. Combining all this information, we see that
$A(x):=P\ast x$ is built as a sum of superposing two opposite orientations, so
perhaps could be an example of an unoriented quantum tree.

The next example concerns the famous compact quantum group $SU_{q}(2)$
introduced by Woronowicz. The representation theory is the same as for
$SU(2)$, so we get a two-dimensional, self-conjugate, generating
representation $\pi$.

######  Example 5.14.

Let $\pi$ be the fundamental representation of $SU_{q}(2)$ and let $P$ be the
corresponding central projection in $\ell^{\infty}(\widehat{SU_{q}(2)})$. The
associated quantum graph is the quantum Cayley graph of $\widehat{SU_{q}(2)}$.

The quantum adjacency matrix is related in this case to the Markov operator of
the random walk on the dual of $SU(2)$ considered by Biane in [Bia91]. After
all, the quantum Cayley graphs are regular, hence $A$ just needs to be
rescaled by a constant to give a random walk operator.

It is interesting to note that these quantum Cayley graphs are non-isomorphic
for all $q\in(0,1]$. Indeed, any isomorphism would be a $\ast$-isomorphism of
algebras preserving the appropriate weights and intertwining the adjacency
matrices. In particular it would preserve the quantum dimensions of
representations but these are different for different parameters.

As our last example we choose (the dual of) the free orthogonal quantum group
$O_{F}^{+}$ (see [VDW96]), where $F\in M_{n}$ is an invertible matrix such
that $F\overline{F}=c\mathds{1}$ for some $c\in\mathbb{R}$. $C(O_{F}^{+})$ is
the universal $C^{\ast}$-algebra generated by the entries of a unitary matrix
$U:=(u_{ij})_{i,j\in[n]}$ such that $U=F\overline{U}F^{-1}$. In [Ban96] it was
proved that the representation theory is the same as for $SU(2)$: the
fundamental representation is irreducible, all irreducible representations are
indexed by the natural numbers and

$n\otimes m\simeq\bigoplus_{k=|n-m|}^{n+m}k.$

In this case we will take as the generating projection the central projection
corresponding to the fundamental representation.

In [Ver05] an alternative approach to Cayley graphs of discrete quantum groups
has been developed, based on Hilbert spaces; there a notion of a tree is
defined based on the classical graph built on the set of irreducible
representations of the dual compact quantum group. The author proved (see
[Ver05, Proposition 4.5]) that one gets a tree iff this compact quantum group
is a finite free product of free unitary and orthogonal quantum groups and a
specific central projection is chosen – the one corresponding to the sum of
respective fundamental representations.

In our framework we do not yet have tools to prove that the quantum Cayley
graph of $\widehat{O_{F}^{+}}$ is a tree thus it seems necessary to understand
the precise relationship between our approach and that of Vergnioux. In
particular it would be interesting to check whether there exists a reasonable
notion of being a tree in our framework that can be detected using only
classical graphs.

###  5.5. Concluding remarks

This paper should be viewed as a first step towards understanding the quantum
graphs associated to discrete quantum groups. The motivation for this work has
been twofold: to motivate the need for studying both infinite and non-tracial
quantum graphs, but also to introduce a new geometric tool to aid in working
with quantum groups. So far we have mostly developed a general framework and
now more effort should be put into investigating the combinatorial/geometric
properties of specific examples. It might suggest what to do for general
quantum graphs, e.g. what are paths in a quantum graph. Paths are an
indispensable tool in classical theory and would likely be equally important
in our case.

Another project would be to understand the link between property (T) of a
discrete quantum group and the spectral gap of the random walk on its quantum
Cayley graph. Maybe one can even mimic the famous construction of Margulis to
obtain a nice sequence of quantum expanding graphs.

## Acknowledgements

The author is grateful to Adam Skalski and Christian Voigt for many helpful
remarks. The author was partially supported by the National Science Center,
Poland (NCN) grant no. 2021/43/D/ST1/01446. The project is co-financed by the
Polish National Agency for Academic Exchange within Polish Returns Programme.

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