# Derivations and KMS-Symmetric Quantum Markov Semigroups

Matthijs Vernooij  Delft University of Technology, Faculty EEMCS/DIAM, P.O.Box
5031, 2600 GA Delft, The Netherlands<EMAIL_ADDRESS>and  Melchior
Wirth  Institute of Science and Technology Austria (ISTA), Am Campus 1, 3400
Klosterneuburg, Austria<EMAIL_ADDRESS>
###### Abstract.

We prove that the generator of the $L^{2}$ implementation of a KMS-symmetric
quantum Markov semigroup can be expressed as the square of a derivation with
values in a Hilbert bimodule, extending earlier results by Cipriani and
Sauvageot for tracially symmetric semigroups and the second-named author for
GNS-symmetric semigroups. This result hinges on the introduction of a new
completely positive map on the algebra of bounded operators on the GNS Hilbert
space. This transformation maps symmetric Markov operators to symmetric Markov
operators and is essential to obtain the required inner product on the Hilbert
bimodule.

##  1\. Introduction

Quantum Markov semigroups are a versatile tool that has found applications not
only in quantum statistical mechanics, where they were originally introduced
in the description of certain open quantum systems [Ali76, GKS76, Kos72,
Lin76], but also in various purely mathematical fields such as noncommutative
harmonic analysis [JX07, JMP14], noncommutative probability [Bia03, CFK14],
noncommutative geometry [Arh23, CS03a, Sau96] and the structure theory of von
Neumann algebras [CS15, CS17, Pet09].

One central question from the beginning was to describe the generators of
quantum Markov semigroups. For quantum Markov semigroups acting on matrix
algebras, a characterization of their generators was given by Lindblad [Lin76]
and Gorini–Kossakowski–Sudarshan [GKS76] and later extended to generators of
uniformly continuous quantum Markov semigroups on arbitrary von Neumann
algebras by Christensen–Evans [CE79]. While partial results are known in
particular for type I factors [AZ15, Dav79, Hol95], a similarly explicit
description of unbounded generators of quantum Markov semigroups on arbitrary
von Neumann algebras seems out of reach.

Both for the modelling of open quantum systems and purely mathematical
questions in noncommutative probability, operator algebra theory, etc., one is
often not interested in arbitrary quantum Markov semigroups, but quantum
Markov semigroups that are symmetric with respect to a reference state or
weight. In quantum statistical mechanics, these describe open systems coupled
to a heat bath in thermal equilibrium. From the mathematical standpoint,
symmetry with respect to a reference state allows to extend the semigroups to
symmetric semigroups on the GNS Hilbert space, which makes the powerful tools
for self-adjoint Hilbert space operators available.

If the reference state or weight is a trace, there is an unambiguous notion of
symmetry, called tracial symmetry. The study of tracially symmetric quantum
Markov semigroups through their associated quadratic forms, so-called
Dirichlet forms, was initiated by Albeverio and Høegh-Krohn [AH77] and further
developed by Lindsay and Davies [DL92, DL93]. The (Hilbert space) generators
of tracially symmetric quantum Markov semigroups have been characterized by
Cipriani and Sauvageot [CS03] to be of the form $\delta^{\ast}\delta$, where
$\delta$ is a derivation with values in a Hilbert bimodule. This result has
lead to a far range of applications from analysis on fractals [HT13] and
metric graphs [BK19] over noncommutative geometry [CS03], noncommutative
probability [Dab10, JZ15], quantum optimal transport [Wir20, WZ21] to the
structure theory of von Neumann algebras and in particular Popa’s deformation
and rigidity theory [Pet09a, DI16, Cas21, CIW21].

Despite this success, the notion of tracial symmetry is somewhat limiting. For
one, von Neumann algebras with a type III summand do not even admit a
(faithful normal) trace. But even on semi-finite von Neumann algebras, plenty
of quantum Markov semigroups of interest are not tracially symmetric. For
example, in many models of open quantum systems one considers quantum Markov
semigroups symmetric with respect to a Gibbs state, which is only a trace in
the infinite temperature limit.

In the case when the reference state is not a trace, there are several non-
equivalent notions of symmetry, for example GNS symmetry and KMS symmetry. If
the state is of the form $\mathrm{tr}(\cdot\,\rho)$ on a type I factor, then
GNS symmetry is symmetry with respect to the inner product
$(x,y)\mapsto\mathrm{tr}(x^{\ast}y\rho)$, while KMS symmetry is symmetry with
respect to the inner product
$(x,y)\mapsto\mathrm{tr}(x^{\ast}\rho^{1/2}y\rho^{1/2})$.

GNS symmetry is the strongest one of the symmetry conditions usually
considered, and it also implies commutation with the modular group, which
makes the structure of GNS-symmetric quantum Markov semigroups particularly
nice and rich. For GNS-symmetric quantum Markov semigroups on matrix algebras,
Alicki’s theorem [Ali76] gives a characterization of their generators in the
spirit of the results of Lindblad and Gorini–Kossakowski–Sudarshan. But it can
also be recast as a representation of the generator as square of a derivation,
thus presenting an analogue of the result of Cipriani and Sauvageot mentioned
above (see [CM17]). In this process one loses the property that the left and
right action are $\ast$-homomorphisms, but this is unavoidable, as was shown
by the first-named author [Ver22].

This result of Alicki has played a central role in recent research motivated
by quantum information theory, in particular for the development of a
dynamical quantum optimal transport distance [CM17], relating
hypercontractivity and logarithmic Sobolev inequalities for quantum systems
[Bar17] and the proof of the complete modified logarithmic Sobolev inequality
for finite-dimensional GNS-symmetric quantum Markov semigroups [GR22].

Moving beyond matrix algebras, the second-named author established a version
of the Christensen–Evans theorem for generators of uniformly continuous GNS-
symmetric quantum Markov semigroups [Wir22] and a generalization of the result
of Cipriani and Sauvageot for generators of arbitrary GNS-symmetric quantum
Markov semigroups [Wir22a].

Let us describe the latter result in some more detail. If a quantum Markov
semigroup $(\Phi_{t})$ is GNS-symmetric with respect to the state (or more
generally weight) $\varphi$, then it has a GNS implementation as strongly
continuous semigroup $(T_{t})$ on the GNS Hilbert space $L^{2}(M,\varphi)$.
Let $\mathscr{E}$ denote the associated quadratic form and
$\mathfrak{A}_{\varphi}$ the maximal Tomita algebra induced by $\varphi$. It
is shown in [Wir22a] that

$\mathfrak{A}_{\mathscr{E}}=\\{a\in\mathfrak{A}_{\varphi}\mid\Delta_{\varphi}^{z}(a)\in\operatorname{dom}(\mathscr{E})\text{
for all }z\in\mathbb{C}\\}$

is a Tomita algebra and a form core for $\mathscr{E}$. Moreover, there exists
a Hilbert space $\mathscr{H}$ with a left action of
$\mathfrak{A}_{\mathscr{E}}$ that is a $\ast$-homomorphism for ♯ and a right
action that is a $\ast$-homomorphism for ♭ and a closable operator
$\delta\colon\mathfrak{A}_{\mathscr{E}}\to\mathscr{H}$ satisfying the Leibniz
rule $\delta(ab)=a\delta(b)+\delta(a)b$ such that

$\mathscr{E}(a,b)=\langle\delta(a),\delta(b)\rangle_{\mathscr{H}}$

for $a,b\in\mathfrak{A}_{\mathscr{E}}$. Moreover, $\mathscr{H}$ carries an
anti-unitary involution and a strongly continuous unitary group with certain
compatibility conditions that reflect the commutation of $(T_{t})$ with the
modular operator and modular conjugation.

It is then natural to ask whether this relation between $L^{2}$ generators of
quantum Markov semigroups and derivations can be extended to KMS-symmetric
semigroups as KMS symmetry can be seen as the more natural assumption in some
contexts. For one, every completely positive map can be decomposed as a linear
combination of KMS-symmetric ones using the Accardi–Cecchini adjoint [AC82],
while for GNS symmetry the commutation with the modular group poses an
algebraic constraint. This makes KMS-symmetric quantum Markov semigroups more
suitable for various applications, such as the characterization of the
Haagerup property in terms of KMS-symmetric quantum Markov semigroups [CS15],
while the same property for GNS-symmetric semigroups is more restrictive. But
also in quantum statistical mechanics, irreversible open quantum systems are
often modeled by quantum Markov semigroups that are only KMS-symmetric rather
than GNS-symmetric, such as the heat-bath dynamics introduced in [KB16] for
example.

The lack of commutation with the modular group poses a serious challenge. For
example, many questions regarding noncommutative $L^{p}$ spaces can be reduced
to $L^{p}$ spaces with respect to a trace by Haagerup’s reduction method
[HJX10], but commutation with the modular group is necessary for maps to be
compatible with this reduction procedure.

Even for KMS-symmetric quantum Markov semigroups on type I factors, when
explicit representations of the generator are known [FU10, AC21], it is not
obvious if the generator can be expressed in the form $\delta^{\ast}\delta$
for a derivation.

For these reasons it has not been clear if one should even expect a
Cipriani–Sauvageot-type result for KMS-symmetric quantum Markov semigroups. In
this article we show that this is indeed the case, not only on matrix
algebras, but arbitrary von Neumann algebras. For generators of uniformly
continuous quantum Markov semigroups, our main result is the following
(Theorems 4.2, 4.7 in the main part). Here $\mathscr{L}_{2}$ denotes the KMS
implementation of $\mathscr{L}$ on $L^{2}(M)$.

######  Theorem.

Let $(\Phi_{t})$ be a uniformly continuous KMS-symmetric quantum Markov
semigroup on $M$ and let $\mathscr{L}$ denote its generator. There exists a
correspondence $\mathscr{H}$, an anti-linear involution
$\mathscr{J}\colon\mathscr{H}\to\mathscr{H}$ and a bounded operator
$\delta\colon L^{2}(M)\to\mathscr{H}$ satisfying

  1. (a)

$\mathscr{J}(x\xi y)=y^{\ast}(\mathscr{J}\xi)x^{\ast}$ for all $x,y\in M$ and
$\xi\in\mathscr{H}$,

  2. (b)

$\delta(Ja)=\mathscr{J}\delta(a)$ for all $a\in L^{2}(M)$,

  3. (c)

$\delta(ab)=\pi_{l}(a)\cdot\delta(b)+\delta(a)\cdot J\pi_{r}(b)^{\ast}J$ for
all $a\in M\varphi^{1/2}$, $b\in\varphi^{1/2}M$,

  4. (d)

$\overline{\mathrm{lin}}\\{\delta(a)x\mid a\in L^{2}(M),\,x\in
M\\}=\mathscr{H}$

such that

$\mathscr{L}_{2}=\delta^{\ast}\delta.$

Moreover, there exists $\xi\in\mathscr{H}$ such that

$\delta(a)=\pi_{l}(a)\xi-\xi(J\pi_{r}(a)^{\ast}J)$

for $a\in M\varphi^{1/2}\cap\varphi^{1/2}M$.

Furthermore, a triple $(\mathscr{H},\mathscr{J},\delta)$ satisfying (a)–(d) is
uniquely determined by $\delta^{\ast}\delta$ up to isomorphism.

For KMS-symmetric quantum Markov semigroups that are not uniformly continuous
we do not have a uniqueness result, but we can still prove existence in the
following form (Theorems 5.2, 5.4 in the main part).

######  Theorem.

Let $(\Phi_{t})$ be a KMS-symmetric quantum Markov semigroup on $M$ with
generator $\mathscr{L}$. If $a\in\operatorname{dom}(\mathscr{L}_{2}^{1/2})\cap
M\varphi^{1/2}$ and
$b\in\operatorname{dom}(\mathscr{L}_{2}^{1/2})\cap\varphi^{1/2}M$, then
$ab\in\operatorname{dom}(\mathscr{L}_{2}^{1/2})$.

Moreover, there exists a Hilbert space $\mathscr{H}$ with commuting left and
right actions of $M$, an anti-unitary involution
$\mathscr{J}\colon\mathscr{H}\to\mathscr{H}$ such that

$\mathscr{J}(x\xi y)=y^{\ast}(\mathscr{J}\xi)x^{\ast}$

for $x,y\in M$ and $\xi\in\mathscr{H}$, a closed operator
$\delta\colon\operatorname{dom}(\mathscr{L}_{2}^{1/2})\to\mathscr{H}$ such
that $\mathscr{J}\delta=\delta J$ and

$\delta(ab)=\pi_{l}(a)\cdot\delta(b)+\delta(a)\cdot J\pi_{r}(a)^{\ast}J$

for $a\in\operatorname{dom}(\mathscr{L}_{2}^{1/2})\cap M\varphi^{1/2}$,
$b\in\operatorname{dom}(\mathscr{L}_{2}^{1/2})\cap\varphi^{1/2}M$, and

$\mathscr{L}_{2}=\delta^{\ast}\delta.$

To establish these results, it requires a fundamentally new tool in the form
of a quantum channel on $B(L^{2}(M))$, which we call the _$\mathscr{V}$
-transform_. Formally, the $\mathscr{V}$-transform of $T\in B(L^{2}(M))$ is
the solution $S$ of the equation

$\frac{1}{2}(\Delta^{1/4}S\Delta^{-1/4}+\Delta^{-1/4}S\Delta^{1/4})=T,$

where $\Delta$ is the modular operator.

A remarkable fact about this map is that it maps positivity-preserving maps
(with respect to the self-dual cone induced by $\varphi$) to positivity-
preserving maps. This property is key for the existence of the Hilbert space
$\mathscr{H}$ in our main results.

Let us briefly summarize the outline of this article. To make the proof
strategy transparent without the technical difficulties occuring for general
von Neumann algebras, but also to make the main results more accessible for
researchers in the quantum information theory community, we first develop the
$\mathscr{V}$-transform and prove our main result for matrix algebras in
Section 2. In Section 3 we define the $\mathscr{V}$-transform on general von
Neumann algebras and establish some of its properties in particular regarding
positvity preservation. In Section 4 we prove our main results on existence
and uniqueness of derivations associated with uniformly continuous KMS-
symmetric quantum Markov semigroups. Finally, in Section 5 we show the
existence of derivations associated with not necessarily uniformly continuous
semigroups.

### Acknowledgments

The authors are grateful to Martijn Caspers for helpful comments on a
preliminary version of this manuscript. M. V. was supported by the NWO Vidi
grant VI.Vidi.192.018 ‘Non-commutative harmonic analysis and rigidity of
operator algebras’. M. W. was funded by the Austrian Science Fund (FWF) under
the Esprit Programme [ESP 156]. For the purpose of Open Access, the authors
have applied a CC BY public copyright licence to any Author Accepted
Manuscript (AAM) version arising from this submission.

##  2\. Quantum Markov Semigroups and Derivations on Matrix Algebras

In this section we demonstrate the connection between KMS-symmetric quantum
Markov semigroups and derivations in the case of matrix algebras. We first
prove a finite-dimensional version of the main result of this article, which
allows to express generators of KMS-symmetric quantum Markov semigroups as
squares of derivations in a suitable sense (Theorem 2.4). We then use the
simple structure of bimodules over matrix algebras to give a more explicit
expression for the quadratic form associated with the generator of a KMS-
symmetric quantum Markov semigroups (Theorem 2.5). As in the general case
treated in the next sections, the crucial technical tool is the
$\mathscr{V}$-transform, which will be introduced in Subsection 2.1.

Let us start with some notation. We write $M_{n}(\mathbb{C})$ for the algebra
of $n\times n$ matrices over the complex numbers, $I_{n}$ for the identity
matrix in $M_{n}(\mathbb{C})$, and $\mathrm{id}_{n}$ for the identity map from
$M_{n}(\mathbb{C})$ to itself. The norm $\lVert\cdot\rVert$ always denotes the
operator norm, either for elements of $M_{n}(\mathbb{C})$ or for linear maps
from $M_{n}(\mathbb{C})$ to itself.

A linear map $\Phi\colon M_{n}(\mathbb{C})\to M_{n}(\mathbb{C})$ is called

  * •

_completely positive_ if, for all $A_{1},\dots,A_{n},B_{1},\dots,B_{n}\in
M_{n}(\mathbb{C})$,

$\sum_{j,k=1}^{n}B_{j}^{\ast}\Phi(A_{j}^{\ast}A_{k})B_{k}\geq 0;$

  * •

_conditionally completely negative_ if, for all
$A_{1},\dots,A_{n},B_{1},\dots,B_{n}\in M_{n}(\mathbb{C})$ with
$\sum_{j=1}^{n}A_{j}B_{j}=0$,

$\sum_{j,k=1}^{n}B_{j}^{\ast}\Phi(A_{j}^{\ast}A_{k})B_{k}\leq 0;$

  * •

_unital_ if $\Phi(I_{n})=I_{n}$.

A _quantum Markov semigroup_ is a family $(\Phi_{t})_{t\geq 0}$ of unital
completely positive maps on $M_{n}(\mathbb{C})$ such that

  * •

$\Phi_{0}=\mathrm{id}_{n}$,

  * •

$\Phi_{s}\Phi_{t}=\Phi_{s+t}$ for all $s,t\geq 0$,

  * •

$\lim_{t\to 0}\lVert\Phi_{t}-\mathrm{id}_{n}\rVert=0$.

If $(\Phi_{t})$ is a quantum Markov semigroup on $M_{n}(\mathbb{C})$, then the
limit

$\mathscr{L}=\lim_{t\to 0}\frac{1}{t}(\mathrm{id}_{n}-\Phi_{t})$

exists and is called the _generator_ of $(\Phi_{t})$. It is the unique linear
operator on $M_{n}(\mathbb{C})$ such that $e^{-t\mathscr{L}}=\Phi_{t}$ for all
$t\geq 0$.

Now fix a density matrix $\rho\in M_{n}(\mathbb{C})$, that is, a positive
matrix with trace $1$, and assume that $\rho$ is invertible. The KMS inner
product induced by $\rho$ is defined as

$\langle\cdot,\cdot\rangle_{\rho}\colon M_{n}(\mathbb{C})\times
M_{n}(\mathbb{C})\to\mathbb{C},\,(A,B)\mapsto\mathrm{tr}(A^{\ast}\rho^{1/2}B\rho^{1/2}).$

If $\Phi\colon M_{n}(\mathbb{C})\to M_{n}(\mathbb{C})$ is a linear map, we
write $\Phi^{\dagger}$ for its adjoint with respect to
$\langle\cdot,\cdot\rangle_{\rho}$ and we say that $\Phi$ is _KMS-symmetric_
if $\Phi^{\dagger}=\Phi$.

A quantum Markov semigroup $(\Phi_{t})$ is called _KMS-symmetric_ if
$\Phi_{t}$ is KMS-symmetric for all $t\geq 0$. Equivalently, $(\Phi_{t})$ is
KMS-symmetric if and only if its generator is KMS-symmetric.

The _modular group_ (or rather its analytic continuation) of $\rho$ is the
family $(\sigma_{z})_{z\in\mathbb{C}}$ of algebra homomorphisms on
$M_{n}(\mathbb{C})$ defined by

$\sigma_{z}(A)=\rho^{iz}A\rho^{-iz}$

for $A\in M_{n}(\mathbb{C})$ and $z\in\mathbb{C}$. Note that
$\sigma_{z}^{\dagger}=\sigma_{-\bar{z}}$.

If a KMS-symmetric operator on $M_{n}(\mathbb{C})$ commutes with the modular
group, then it is called GNS-symmetric (this is equivalent to the usual
definition of GNS symmetry by [CM17, Lemma 2.5]).

According to [AC21, Theorem 4.4], the generator $\mathscr{L}$ of a KMS-
symmetric quantum Markov semigroup is of the form

$\mathscr{L}(A)=(1+\sigma_{-i/2})^{-1}(\Psi(I_{n}))A+A(1+\sigma_{i/2})^{-1}(\Psi(I_{n}))-\Psi(A)$

for some KMS-symmetric completely positive map $\Psi\colon
M_{n}(\mathbb{C})\to M_{n}(\mathbb{C})$.

The main goal of this section is to show that the sesquilinear form associated
with $\mathscr{L}$ can be written as

$\langle\mathscr{L}(A),B\rangle_{\rho}=\sum_{j=1}^{N}\langle[V_{j},A],[V_{j},B]\rangle_{\rho}$

with matrices $V_{1},\dots,V_{N}\in M_{n}(\mathbb{C})$.

###  2.1. The $\mathscr{V}$-Transform

We write $B(M_{n}(\mathbb{C}))$ for the space of all linear maps from
$M_{n}(\mathbb{C})$ to itself. This space is generated by left and right
multiplication operators in the following sense. For $A\in M_{n}(\mathbb{C})$
let

$\mathbb{L}_{A},\mathbb{R}_{A}\colon M_{n}(\mathbb{C})\to
M_{n}(\mathbb{C}),\,\mathbb{L}_{A}(X)=AX,\,\mathbb{R}_{A}(X)=XA.$

By [CM17, Lemma A.1], the linear span of $\\{\mathbb{L}_{A}\mathbb{R}_{B}\mid
A,B\in M_{n}(\mathbb{C})\\}$ is $B(M_{n}(\mathbb{C}))$.

The $\mathscr{V}$-transform is a linear map on $B(M_{n}(\mathbb{C}))$, which
is most conveniently defined through its inverse. Let

$\mathscr{W}\colon B(M_{n}(\mathbb{C}))\to
B(M_{n}(\mathbb{C})),\,\Phi\mapsto\frac{1}{2}(\sigma_{i/4}\Phi\sigma_{-i/4}+\sigma_{-i/4}\Phi\sigma_{i/4}).$

In particular, if $\Phi=\mathbb{L}_{A}\mathbb{R}_{B}$, then

$\mathscr{W}(\Phi)=\frac{1}{2}(\mathbb{L}_{\sigma_{i/4}(A)}\mathbb{R}_{\sigma_{i/4}(B)}+\mathbb{L}_{\sigma_{-i/4}(A)}\mathbb{R}_{\sigma_{-i/4}(B)}).$

######  Proposition 2.1.

The map $\mathscr{W}$ is invertible with inverse given by

$\mathscr{W}^{-1}(\Phi)=2\int_{0}^{\infty}\sigma_{-i/4}e^{-r\sigma_{-i/2}}\Phi\sigma_{-i/4}e^{-r\sigma_{-i/2}}\,dr$

for $\Phi\in B(M_{n}(\mathbb{C}))$.

In particular, if $A,B\in M_{n}(\mathbb{C})$, then

$\mathscr{W}^{-1}(\mathbb{L}_{A}\mathbb{R}_{B})=2\int_{0}^{\infty}\mathbb{L}_{\sigma_{i/4}(e^{-r\sigma_{i/2}}(A))}\mathbb{R}_{\sigma_{-i/4}(e^{-r\sigma_{-i/2}}(B))}\,dr.$

###### Proof.

Since $\sigma_{-i/2}$ is an invertible operator on $M_{n}(\mathbb{C})$ and
$\mathrm{tr}(\sigma_{-i/2}(A)^{\ast}A)\geq 0$ for all $A\in
M_{n}(\mathbb{C})$, the spectrum of $\sigma_{-i/2}$ consists of strictly
positive numbers. Let $\lambda$ denote the smallest eigenvalue of
$\sigma_{-i/2}$. By the spectral theorem,

$\lVert e^{-r\sigma_{-i/2}}\rVert\leq e^{-\lambda r}$

for all $r\geq 0$. It follows that for $\Phi\in B(M_{n}(\mathbb{C}))$ we have

$\lVert\sigma_{-i/4}e^{-r\sigma_{-i/2}}\Phi\sigma_{-i/4}e^{-r\sigma_{-i/2}}\rVert\leq
e^{-2\lambda r}\lVert\sigma_{-i/4}\rVert^{2}\lVert\Phi\rVert.$

Therefore the integral

$\int_{0}^{\infty}\sigma_{-i/4}e^{-r\sigma_{-i/2}}\Phi\sigma_{-i/4}e^{-r\sigma_{-i/2}}\,dr$

converges absolutely.

Moreover,

$\displaystyle\mathscr{W}\left(2\int_{0}^{\infty}\sigma_{-i/4}e^{-r\sigma_{-i/2}}\Phi\sigma_{-i/4}e^{-r\sigma_{-i/2}}\,dr\right)$
$\displaystyle\qquad=\int_{0}^{\infty}e^{-r\sigma_{-i/2}}(\sigma_{-i/2}\Phi+\Phi\sigma_{-i/2})e^{-r\sigma_{-i/2}}\,dr$
$\displaystyle\qquad=-\int_{0}^{\infty}\frac{d}{dr}(e^{-r\sigma_{-i/2}}\Phi
e^{-r\sigma_{-i/2}})\,dr$ $\displaystyle\qquad=\Phi.$

Thus $\mathscr{W}$ is invertible and the claimed integral expression for the
inverse holds. Similar arguments yield the integral formula for
$\mathscr{W}^{-1}(\mathbb{L}_{A}\mathbb{R}_{B})$. ∎

######  Definition 2.2.

We call the inverse of $\mathscr{W}$ the $\mathscr{V}$-transform and denote it
by $\mathscr{V}$. For $\Phi\in B(M_{n}(\mathbb{C}))$ we also write
$\check{\Phi}$ for $\mathscr{V}(\Phi)$.

######  Lemma 2.3.

The $\mathscr{V}$-transform is a bijective linear map on
$B(M_{n}(\mathbb{C}))$ with the following properties.

  1. (i)

If $\Phi\in B(M_{n}(\mathbb{C}))$ is KMS-symmetric, then $\mathscr{V}(\Phi)$
is KMS-symmetric.

  2. (ii)

If $\Phi\in B(M_{n}(\mathbb{C}))$ is completely positive, then
$\mathscr{V}(\Phi)$ is completely positive.

  3. (iii)

$\mathscr{V}(\mathrm{id}_{n})=\mathrm{id}_{n}$.

###### Proof.

  1. (i)

Let $\Phi\in B(M_{n}(\mathbb{C}))$. By Proposition 2.1 we have

$\mathscr{V}(\Phi)=2\int_{0}^{\infty}\sigma_{-i/4}e^{-r\sigma_{-i/2}}\Phi\sigma_{-i/4}e^{-r\sigma_{-i/2}}\,dr.$

Since $\sigma_{-i/4}$ is KMS-symmetric, so is $e^{-r\sigma_{-i/2}}$. Thus the
KMS adjoint of $\mathscr{V}(\Phi)$ satisfies

$\displaystyle\mathscr{V}(\Phi)^{\dagger}=2\int_{0}^{\infty}\sigma_{-i/4}e^{-r\sigma_{-i/2}}\Phi^{\dagger}\sigma_{-i/4}e^{-r\sigma_{-i/2}}\,dr.$

In particular, if $\Phi$ is KMS-symmetric, so is $\mathscr{V}(\Phi)$.

  2. (ii)

If $\Phi$ is completely positive, by Kraus’ theorem there exist
$V_{1},\dots,V_{N}\in M_{n}(\mathbb{C})$ such that

$\Phi=\sum_{j=1}^{N}\mathbb{L}_{V_{j}^{\ast}}\mathbb{R}_{V_{j}}.$

From Proposition 2.1 and the identity
$\sigma_{z}(a)^{\ast}=\sigma_{\overline{z}}(a^{\ast})$ we deduce

$\mathscr{V}(\Phi)=\sum_{j=1}^{N}\int_{0}^{\infty}\mathbb{L}_{\sigma_{-i/4}(e^{-r\sigma_{-i/2}}(V_{j}))^{\ast}}\mathbb{R}_{\sigma_{-i/4}(e^{-r\sigma_{-i/2}}(V_{j}))}\,dr.$

Since maps of the form $\mathbb{L}_{A^{\ast}}\mathbb{R}_{A}$ are completely
positive and positive linear combinations and limits of completely positive
maps are again completely positive, it follows that $\mathscr{V}(\Phi)$ is
completely positive.

  3. (iii)

The identity $\mathscr{W}(\mathrm{id}_{n})=\mathrm{id}_{n}$ is immediate from
the definition, from which $\mathscr{V}(\mathrm{id}_{n})=\mathrm{id}_{n}$
follows directly. ∎

###  2.2. Derivations for KMS-Symmetric Markov Generators on Matrix Algebras

We are now in the position to prove the existence of a (twisted) derivation
that implements the Dirichlet form associated with a KMS-symmetric quantum
Markov semigroup on $M_{n}(\mathbb{C})$. We first present an abstract version
that will later be generalized to quantum Markov semigroups on arbitrary von
Neumann algebras. A more explicit version tailored for matrix algebras will be
discussed below.

######  Theorem 2.4.

Let $(\Phi_{t})$ be a KMS-symmetric quantum Markov semigroup on
$M_{n}(\mathbb{C})$ and let $\mathscr{L}$ denote its generator. There exists a
Hilbert space $\mathscr{H}$, a unital $\ast$-homorphism $\pi_{l}\colon
M_{n}(\mathbb{C})\to B(\mathscr{H})$, a unital $\ast$-antihomorphism
$\pi_{r}\colon M_{n}(\mathbb{C})\to B(\mathscr{H})$, and anti-linear isometric
involution $\mathscr{J}\colon\mathscr{H}\to\mathscr{H}$ and a linear map
$\delta\colon M_{n}(\mathbb{C})\to\mathscr{H}$ satisfying

  1. (i)

$\pi_{l}(A)\pi_{r}(B)=\pi_{r}(B)\pi_{l}(A)$ for all $A,B\in
M_{n}(\mathbb{C})$,

  2. (ii)

$\mathscr{J}(\pi_{l}(A)\pi_{r}(B)\xi)=\pi_{l}(B)^{\ast}\pi_{r}(A)^{\ast}\mathscr{J}\xi$
for all $A,B\in M_{n}(\mathbb{C})$ and $\xi\in\mathscr{H}$,

  3. (iii)

$\delta(A^{\ast})=\mathscr{J}\delta(A)$ for all $A\in M_{n}(\mathbb{C})$,

  4. (iv)

$\delta(AB)=\pi_{l}(\sigma_{-i/4}(A))\delta(B)+\pi_{r}(\sigma_{i/4}(B))\delta(A)$
and

  5. (v)

$\mathscr{H}=\mathrm{lin}\\{\pi_{l}(A)\delta(B)|A,B\in M_{n}{\mathbb{C}}\\}$

such that

(1) $\langle
A,\mathscr{L}(B)\rangle_{\rho}=\langle\delta(A),\delta(B)\rangle_{\mathscr{H}}$

for all $A,B\in M_{n}(\mathbb{C})$.

###### Proof.

First, we claim that $\check{\mathscr{L}}$ is a KMS-symmetric conditionally
completely negative map. The KMS-symmetry follows from Lemma 2.3(i). By Lemma
2.3(ii) $\check{\Phi}_{t}$ is completely positive for all $t\geq 0$. Since

$\check{\mathscr{L}}=\lim_{t\to
0}\frac{1}{t}(\mathrm{id}_{n}-\check{\Phi}_{t})$

by Lemma 2.3(iii), it follows from the definitions that $\check{\mathscr{L}}$
is conditionally completely negative, proving the claim. We also observe that

$\check{\mathscr{L}}(I_{n})=2\int_{0}^{\infty}\sigma_{-i/4}e^{-r\sigma_{-i/2}}(\mathscr{L}(\sigma_{-i/4}e^{-r\sigma_{-i/2}}(I_{n})))\,dr=0$

by Proposition 2.1.

Next, we define a sesquilinear form $\langle\cdot,\cdot\rangle_{\mathscr{H}}$
on the algebraic tensor product $M_{n}(\mathbb{C})\odot M_{n}(\mathbb{C})$ by

$\langle A_{1}\otimes B_{1},A_{2}\otimes
B_{2}\rangle_{\mathscr{H}}=-\frac{1}{2}\mathrm{tr}(B_{1}^{\ast}\rho^{1/2}\check{\mathscr{L}}(A_{1}^{\ast}A_{2})\rho^{1/2}B_{2}).$

Now consider the subspace

$N=\left\\{\sum_{j}A_{j}\otimes
B_{j}:\sum_{j}A_{j}\sigma_{-i/2}(B_{j})=0\right\\}.$

Because $\check{\mathscr{L}}$ is completely conditionally negative, we see
that for any $\sum_{j}A_{j}\otimes B_{j}\in N$ we have

$\displaystyle\sum_{jk}\langle A_{j}\otimes B_{j},A_{k}\otimes
B_{k}\rangle_{\mathscr{H}}$
$\displaystyle=-\frac{1}{2}\sum_{jk}\mathrm{tr}(B_{j}^{\ast}\rho^{1/2}\check{\mathscr{L}}(A_{j}^{\ast}A_{k})\rho^{1/2}B_{k})$
$\displaystyle=-\frac{1}{2}\sum_{jk}\mathrm{tr}(\rho^{1/2}\sigma_{-i/2}(B_{j})^{\ast}\check{\mathscr{L}}(A_{j}^{\ast}A_{k})\sigma_{-i/2}(B_{k})\rho^{1/2})$
$\displaystyle\geq 0.$

Therefore, this sesquilinear form is positive semidefinite on $N$. Now let

$\mathscr{H}=N/\\{u\in N|\langle u,u\rangle_{\mathscr{H}}=0\\}.$

Then $\langle\cdot,\cdot\rangle$ induces an inner product on $\mathscr{H}$,
which turns $\mathscr{H}$ into a Hilbert space (as $\mathscr{H}$ is finite-
dimensional). We write $\sum_{j}A_{j}\otimes_{\mathscr{H}}B_{j}$ for the image
of $\sum_{j}A_{j}\otimes B_{j}$ in $\mathscr{H}$ under the quotient map.

Define $\pi_{l}$ and $\pi_{r}$ by

$\pi_{l}(X)\sum_{j}A_{j}\otimes_{\mathscr{H}}B_{j}=\sum_{j}XA_{j}\otimes_{\mathscr{H}}B_{j},\,\pi_{r}(X)\sum_{j}A_{j}\otimes_{\mathscr{H}}B_{j}=\sum_{j}A_{j}\otimes_{\mathscr{H}}B_{j}X$

for $X\in M_{n}(\mathbb{C})$ and $\sum_{j}A_{j}\otimes B_{j}\in N$. These maps
are well defined. Indeed, they preserve $N$, and for all $u\in N$ with
$\langle u,u\rangle_{\mathscr{H}}=0$ we have

$\langle\pi_{l}(X)\pi_{r}(Y)u,\pi_{l}(X)\pi_{r}(Y)u\rangle=\langle\pi_{l}(X^{\ast}X)\pi_{r}(YY^{\ast})u,u\rangle\leq
0$

by Cauchy–Schwarz. From the definitions of $\pi_{l}$, $\pi_{r}$ and
$\langle\cdot,\cdot\rangle_{\mathscr{H}}$ we now conclude that $\pi_{l}$ is a
unital $\ast$-homomorphism and $\pi_{r}$ is a unital $\ast$-antihomomorphism.

Subsequently, we will define the anti-linear isometric involution
$\mathscr{J}$ on $\mathscr{H}$. Consider the map

$A\otimes B\mapsto-B^{\ast}\otimes A^{\ast}$

from $M_{n}(\mathbb{C})\odot M_{n}(\mathbb{C})$ to itself. This is an isometry
since $\check{\mathscr{L}}$ is KMS-symmetric. Moreover, it preserves $N$.
Consequently, it acts in a well-defined manner on the equivalence classes of
$\mathscr{H}$, and we call this map $\mathscr{J}$. It is clear that
$\mathscr{J}$ is an anti-linear involution.

Lastly, we define the map $\delta:M_{n}(\mathbb{C})\to\mathscr{H}$ by

$\delta(A)=\sigma_{-i/4}(A)\otimes_{\mathscr{H}}I_{n}-I_{n}\otimes\sigma_{i/4}(A).$

With this definition properties (i)-(iv) are immediate from the definitions.
For property (v) note that

$\sum_{j}A_{j}\otimes_{\mathscr{H}}B_{j}=\sum_{j}\pi_{l}(A_{j})\delta(\sigma_{-i/4}(B_{j}))+A_{j}\sigma_{-i/2}(B_{j})\otimes_{\mathscr{H}}I_{n}=\sum_{j}\pi_{l}(A_{j})\delta(\sigma_{-i/4}(B_{j}))$

for any $\sum_{j}A_{j}\otimes_{\mathscr{H}}B_{j}\in\mathscr{H}$.

To complete the proof of the theorem, we need to show that (1) holds. Let
$A,B\in M_{n}(\mathbb{C})$. Then we have

$\displaystyle\langle\delta(A),\delta(B)\rangle_{\mathscr{H}}=$
$\displaystyle\
\langle\sigma_{-i/4}(A)\otimes_{\mathscr{H}}I_{n}-I_{n}\otimes_{\mathscr{H}}\sigma_{i/4}(A),\sigma_{-i/4}(B)\otimes_{\mathscr{H}}I_{n}-I_{n}\otimes_{\mathscr{H}}\sigma_{i/4}(B)\rangle$
(2) $\displaystyle=$
$\displaystyle\frac{1}{2}\Big{(}\mathrm{tr}(\sigma_{-i/4}(A^{\ast})\rho^{1/2}\check{\mathscr{L}}(\sigma_{-i/4}(B))\rho^{1/2})+\mathrm{tr}(\rho^{1/2}\check{\mathscr{L}}(\sigma_{i/4}(A^{\ast}))\rho^{1/2}\sigma_{i/4}(B))$
(3)
$\displaystyle-\mathrm{tr}(\rho^{1/2}\check{\mathscr{L}}(\sigma_{i/4}(A^{\ast})\sigma_{-i/4}(B))\rho^{1/2})-\mathrm{tr}(\sigma_{-i/4}(A^{\ast})\rho^{1/2}\check{\mathscr{L}}(I_{n})\rho^{1/2}\sigma_{i/4}(B))\Big{)}.$

The two terms in line (3) are zero because $\check{\mathscr{L}}$ is KMS-
symmetric and $\check{\mathscr{L}}(I_{n})=0$. For the terms in line (2) we use
the KMS-symmetry of $\check{\mathscr{L}}$ and the fact that
$\mathrm{tr}(\sigma_{it}(C)D)=\mathrm{tr}(C\sigma_{-it}(D))$ for all
$t\in\mathbb{R}$ and $C,D\in M_{n}(\mathbb{C})$ to conclude that

$\displaystyle\langle\delta(A),\delta(B)\rangle_{\mathscr{H}}=$
$\displaystyle\
\frac{1}{2}\mathrm{tr}(A^{\ast}\rho^{1/2}\sigma_{i/4}(\check{\mathscr{L}}(\sigma_{-i/4}(B)))\rho^{1/2})+\frac{1}{2}\mathrm{tr}(\rho^{1/2}A^{\ast}\rho^{1/2}\sigma_{-i/4}(\check{\mathscr{L}}(\sigma_{i/4}(B))))$
$\displaystyle=$ $\displaystyle\
\mathrm{tr}(A^{\ast}\rho^{1/2}\mathscr{W}(\check{\mathscr{L}})(B)\rho^{1/2})$
$\displaystyle=$ $\displaystyle\ \langle A,\mathscr{L}(B)\rangle_{\rho},$

as desired. ∎

As a consequence of the previous result, we get a more explicit expression for
the quadratic form associated with the generator of a KMS-symmetric quantum
Markov semigroup on $M_{n}(\mathbb{C})$. An analogous expression can be found
in [CM17, Eq. (5.3)], [CM20, Prop. 2.5] for the special case of GNS-symmetric
quantum Markov semigroups.

######  Theorem 2.5.

If $(\Phi_{t})$ is a KMS-symmetric quantum Markov semigroup on
$M_{n}(\mathbb{C})$ with generator $\mathscr{L}$, then there exist matrices
$V_{1},\dots,V_{N}\in M_{n}(\mathbb{C})$ such that
$\\{V_{j}\\}_{j=1}^{N}=\\{V_{j}^{\ast}\\}_{j=1}^{N}$ and

$\langle
A,\mathscr{L}(B)\rangle_{\rho}=\sum_{j=1}^{N}\langle[V_{j},A],[V_{j},B]\rangle_{\rho}$

for all $A,B\in M_{n}(\mathbb{C})$.

###### Proof.

Let $\mathscr{H}$, $\pi_{l}$, $\pi_{r}$, $\mathscr{J}$ and $\delta$ be as in
the last Theorem. The map $\pi_{l}\otimes\pi_{r}$ is a unital representation
of $M_{n}(\mathbb{C})\otimes M_{n}(\mathbb{C})^{\mathrm{op}}$. It follows from
the representation theory of matrix algebras [Bla06, Proposition IV.1.2.2.]
that there exists an auxiliary Hilbert space $H$ such that $\mathscr{H}\cong
M_{n}(\mathbb{C})\otimes H$, where $M_{n}(\mathbb{C})$ is endowed with the
Hilbert–Schmidt inner product, and

$\displaystyle\pi_{l}(A)(B\otimes\xi)$ $\displaystyle=AB\otimes\xi,$
$\displaystyle\pi_{r}(A)(B\otimes\xi)$ $\displaystyle=BA\otimes\xi$

for $A,B\in M_{n}(\mathbb{C})$ and $\xi\in H$ under this identification.

Moreover, since $\mathscr{H}$ is finite-dimensional by property (v), the space
$H$ is finite-dimensional, say $H=\mathbb{C}^{N}$.

Thus $\delta(A)=\sum_{j=1}^{N}\delta_{j}(A)\otimes e_{j}$ with the canonical
orthonormal basis $(e_{j})$ on $\mathbb{C}^{N}$. It follows from property (iv)
that

$\displaystyle\rho^{-1/4}\delta_{j}(AB)\rho^{-1/4}=A\rho^{-1/4}\delta_{j}(B)\rho^{-1/4}+\rho^{-1/4}\delta_{j}(A)\rho^{-1/4}B$

for $A,B\in M_{n}(\mathbb{C})$. In other words,
$A\mapsto\rho^{-1/4}\delta_{j}(A)\rho^{-1/4}$ is a derivation. By the
derivation theorem [Kap53, Theorem 9] there exists $V_{j}\in
M_{n}(\mathbb{C})$ such that $\delta_{j}(A)=\rho^{1/4}[V_{j},A]\rho^{1/4}$.

We conclude that

$\langle
A,\mathscr{L}(B)\rangle_{\rho}=\langle\delta(A),\delta(B)\rangle_{\mathscr{H}}=\sum_{j=1}^{N}\mathrm{tr}(\delta_{j}(A)^{\ast}\delta_{j}(B))=\sum_{j=1}^{N}\langle[V_{j},A],[V_{j},B]\rangle_{\rho}.$

Since
$\langle[V_{j},A],[V_{j},B]\rangle_{\rho}=\langle[V_{j}^{\ast},B^{\ast}],[V_{j}^{\ast},A^{\ast}]\rangle_{\rho}$,
we have

$\langle\delta(A),\delta(B)\rangle_{\mathscr{H}}=\langle\mathscr{J}\delta(B),\mathscr{J}\delta(A)\rangle_{\mathscr{H}}=\sum_{j=1}^{N}\langle[V_{j},B^{\ast}],[V_{j},A^{\ast}]\rangle_{\rho}=\sum_{j=1}^{N}\langle[V_{j}^{\ast},A],[V_{j}^{\ast},B]\rangle_{\rho}.$

This shows that $V_{1},\dots,V_{N}$ can be chosen such that
$\\{V_{j}\\}_{j=1}^{N}=\\{V_{j}^{\ast}\\}_{j=1}^{N}$. ∎

######  Remark 2.6.

This theorem can also be proven without the use of Theorem 2.4. In the
appendix we include a proof of Theorem 2.5 using the $\mathscr{V}$-transform
and the structure of generators of KMS-symmetric quantum Markov semigroups
described in [AC21, Theorem 4.4].

##  3\. The $\mathscr{V}$-Transform

As we have seen in Section 2 in the case of matrix algebras, the key
ingredient to construct the derivation associated with a KMS-symmetric quantum
Markov semigroup is the $\mathscr{V}$-transform. This remains the case for
semigroups on general von Neumann algebras, but the definition and properties
of the $\mathscr{V}$-transform become much more delicate as it involves in
general unbounded operators like the analytic generator of the modular group.
In particular the fact that the $\mathscr{V}$-transform preserves completely
positive maps, which is crucial for defining an inner product, requires new
arguments in this general setting.

For technical convenience, we first define the $\mathscr{V}$-transform for
bounded operators on the Hilbert space $L^{2}(M)$, before we transfer it to
KMS-symmetric unital completely positive maps on $M$.

Throughout this section we fix a $\sigma$-finite von Neumann algebra $M$ and a
faithful normal state $\varphi$ on $M$. We also write $\varphi$ for the
element in $L^{1}(M)$ representing $\varphi$ so that $\varphi^{1/2}$ is the
unique positive vector in $L^{2}(M)$ representing $\varphi$,
$\varphi^{it}\cdot\varphi^{-it}$ is the modular group etc. We write $\Delta$
and $J$ for the modular operator and modular conjugation associated with
$\varphi$, and $L^{2}_{+}(M)$ for the standard self-dual positive cone in
$L^{2}(M)$.

###  3.1. $\mathscr{V}$-Transform of Bounded Operators on $L^{2}(M)$

In this subsection we define the $\mathscr{V}$-transform on $B(L^{2}(M))$ and
discuss some of its properties. The $\mathscr{V}$-transform in this setting is
the map

$\mathscr{V}\colon B(L^{2}(M))\to
B(L^{2}(M)),\,T\mapsto\check{T}=2\int_{0}^{\infty}\Delta^{1/4}e^{-r\Delta^{1/2}}T\Delta^{1/4}e^{-r\Delta^{1/2}}\,dr.$

We first prove that it is well-defined. Note that
$\Delta^{1/4}e^{-r\Delta^{1/2}}$ is bounded (and self-adjoint) for every
$r>0$, so that the integrand is a bounded operator and the only difficulty is
integrability.

######  Lemma 3.1.

If $T\in B(L^{2}(M))$ and $\xi,\eta\in L^{2}(M)$, then

$r\mapsto\langle\xi,\Delta^{1/4}e^{-r\Delta^{1/2}}T\Delta^{1/4}e^{-r\Delta^{1/2}}\eta\rangle$

is integrable on $(0,\infty)$, and

$2\left\lvert\int_{0}^{\infty}\langle\xi,\Delta^{1/4}e^{-r\Delta^{1/2}}T\Delta^{1/4}e^{-r\Delta^{1/2}}\eta\rangle\,dr\right\rvert\leq\lVert
T\rVert\lVert\xi\rVert\lVert\eta\rVert.$

###### Proof.

Let $E$ denote the spectral measure of $\Delta^{1/2}$. By the spectral theorem
we have

$\displaystyle
2\int_{0}^{\infty}\lvert\langle\xi,\Delta^{1/4}e^{-r\Delta^{1/2}}T\Delta^{1/4}e^{-r\Delta^{1/2}}\eta\rangle\rvert\,dr$
$\displaystyle\qquad\leq\lVert
T\rVert\int_{0}^{\infty}(\lVert\Delta^{1/4}e^{-r\Delta^{1/2}}\xi\rVert^{2}+\lVert\Delta^{1/4}e^{-r\Delta^{1/2}}\eta\rVert^{2})\,dr$
$\displaystyle\qquad=\lVert T\rVert\int_{0}^{\infty}\int_{0}^{\infty}\lambda
e^{-2\lambda
r}\,d(\langle\xi,E(\lambda)\xi\rangle+\langle\eta,E(\lambda)\eta\rangle)\,dr$
$\displaystyle\qquad=\lVert
T\rVert\int_{0}^{\infty}\lambda\int_{0}^{\infty}e^{-2\lambda
r}\,dr\,d(\langle\xi,E(\lambda)\xi\rangle+\langle\eta,E(\lambda)\eta\rangle)$
$\displaystyle\qquad=\frac{1}{2}\lVert
T\rVert(\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}).$

Thus
$r\mapsto\langle\xi,\Delta^{1/4}e^{-r\Delta^{1/2}}T\Delta^{1/4}e^{-r\Delta^{1/2}}\eta\rangle$
is integrable on $(0,\infty)$, and the desired inequality follows from the
usual rescaling trick $\xi\mapsto\alpha\xi$, $\eta\mapsto\eta/\alpha$. ∎

######  Definition 3.2.

For $T\in B(L^{2}(M))$ let $\check{T}$ denote the unique bounded linear
operator on $L^{2}(M)$ such that

$\langle\xi,\check{T}\eta\rangle=2\int_{0}^{\infty}\langle\xi,\Delta^{1/4}e^{-r\Delta^{1/2}}T\Delta^{1/4}e^{-r\Delta^{1/2}}\eta\rangle\,dr$

for all $\xi,\eta\in L^{2}(M)$.

We call the map

$\mathscr{V}\colon B(L^{2}(M))\to B(L^{2}(M)),\,T\mapsto\check{T}$

the $\mathscr{V}$-transform.

Note that if $T$ commutes with the modular operator, then $\check{T}=T$.

######  Proposition 3.3.

The $\mathscr{V}$-transform is a normal unital completely positive trace-
preserving map on $B(L^{2}(M))$.

###### Proof.

Let $E$ denote the spectral measure of $\Delta^{1/2}$. By the spectral
theorem,

$\displaystyle
2\int_{0}^{\infty}\langle\Delta^{1/4}e^{-r\Delta^{1/2}}\xi,\Delta^{1/4}e^{-r\Delta^{1/2}}\eta\rangle\,dr$
$\displaystyle=2\int_{0}^{\infty}\int_{0}^{\infty}\lambda e^{-2\lambda
r}\,d\langle\xi,E(\lambda)\eta\rangle\,dr$
$\displaystyle=2\int_{0}^{\infty}\lambda\int_{0}^{\infty}e^{-2\lambda
r}\,dr\,d\langle\xi,E(\lambda)\eta\rangle$
$\displaystyle=\langle\xi,\eta\rangle$

for all $\xi,\eta\in L^{2}(M)$. Thus $\check{1}=1$.

Since $\Delta^{1/4}e^{-r\Delta^{1/2}}$ is self-adjoint, the map
$T\mapsto\Delta^{1/4}e^{-r\Delta^{1/2}}T\Delta^{1/4}e^{-r\Delta^{1/2}}$ is
completely positive. Hence $\mathscr{V}$ is completely positive.

To prove that $\mathscr{V}$ is trace-preserving, let $T\in B(L^{2}(M))$ be
positive. By the previous part, $\mathscr{V}(T)$ is positive and we have

$\displaystyle\mathrm{tr}(\mathscr{V}(T))$
$\displaystyle=2\int_{0}^{\infty}\mathrm{tr}(\Delta^{1/4}e^{-r\Delta^{1/2}}T\Delta^{1/4}e^{-r\Delta^{1/2}})\,dr$
$\displaystyle=2\int_{0}^{\infty}\mathrm{tr}(T^{1/2}\Delta^{1/2}e^{-2r\Delta^{1/2}}T^{1/2})\,dr$
$\displaystyle=\mathrm{tr}(T^{1/2}\mathscr{V}(1)T^{1/2})$
$\displaystyle=\mathrm{tr}(T).$

Interchanging the integral and the sum defining the trace is justified by
Fubini’s theorem since the integrand is positive in each case.

To prove that the $\mathscr{V}$-transform is normal, note first that it
restricts to a bounded linear map $\mathscr{V}_{\ast}$ on the space of trace-
class operators on $L^{2}(M)$ by the previous part. Moreover, if $S,T\in
B(L^{2}(M))$ and $S$ is trace-class, then

$\mathrm{tr}(\mathscr{V}_{\ast}(S)T)=2\int_{0}^{\infty}\mathrm{tr}(\Delta^{1/4}e^{-r\Delta^{1/2}}S\Delta^{1/4}e^{-r\Delta^{1/2}}T)\,dr=\mathrm{tr}(S\mathscr{V}(T)).$

Thus $\mathscr{V}=(\mathscr{V}_{\ast})^{\ast}$, which implies that
$\mathscr{V}$ is normal. ∎

######  Lemma 3.4 (Key property).

  1. (a)

If $T\in B(L^{2}(M))$, then

$\frac{1}{2}\langle\Delta^{1/4}\xi,\check{T}\Delta^{-1/4}\eta\rangle+\frac{1}{2}\langle\Delta^{-1/4}\xi,\check{T}\Delta^{1/4}\eta\rangle=\langle\xi,T\eta\rangle$

for all
$\xi,\eta\in\operatorname{dom}(\Delta^{1/4})\cap\operatorname{dom}(\Delta^{-1/4})$.

  2. (b)

If $R\in B(L^{2}(M))$ such that

$\frac{1}{2}\langle\Delta^{1/4}\xi,R\Delta^{-1/4}\eta\rangle+\frac{1}{2}\langle\Delta^{-1/4}\xi,R\Delta^{1/4}\eta\rangle=\langle\xi,T\eta\rangle$

for all $\xi,\eta\in\bigcap_{n\in\mathbb{Z}}\operatorname{dom}(\Delta^{n})$,
then $R=\check{T}$.

###### Proof.

(a) If
$\xi,\eta\in\operatorname{dom}(\Delta^{1/4})\cap\operatorname{dom}(\Delta^{-1/4})$,
then

$\displaystyle\langle\Delta^{1/4}\xi,\check{T}\Delta^{-1/4}\eta\rangle+\langle\Delta^{-1/4}\xi,\check{T}\Delta^{1/4}\eta\rangle$
$\displaystyle\qquad=2\int_{0}^{\infty}(\langle\Delta^{1/2}e^{-r\Delta^{1/2}}\xi,Te^{-r\Delta^{1/2}}\eta\rangle+\langle
e^{-r\Delta^{1/2}}\xi,T\Delta^{1/2}e^{-r\Delta^{1/2}}\eta\rangle)\,dr$
$\displaystyle\qquad=-2\int_{0}^{\infty}\frac{d}{dr}\langle
e^{-r\Delta^{1/2}}\xi,Te^{-r\Delta^{1/2}}\eta\rangle\,dr$
$\displaystyle\qquad=2\langle\xi,T\eta\rangle.$

Here we used that since $\Delta^{1/2}$ is non-singular,
$e^{-r\Delta^{1/2}}\zeta\to 0$ as $r\to\infty$ for every $\zeta\in L^{2}(M)$.

(b) If $\xi,\eta\in\bigcap_{n\in\mathbb{Z}}\operatorname{dom}(\Delta^{n})$,
then
$\Delta^{1/4}e^{-r\Delta^{1/2}}\xi,\Delta^{1/4}e^{-r\Delta^{1/2}}\eta\in\operatorname{dom}(\Delta^{1/4})\cap\operatorname{dom}(\Delta^{-1/4})$
and hence

$\displaystyle\langle\xi,\check{T}\eta\rangle$
$\displaystyle=2\int_{0}^{\infty}\langle\Delta^{1/4}e^{-r\Delta^{1/2}}\xi,T\Delta^{1/4}e^{-r\Delta^{1/2}}\eta\rangle\,dr$
$\displaystyle=\int_{0}^{\infty}(\langle\Delta^{1/2}e^{-r\Delta^{1/2}}\xi,Re^{-r\Delta^{1/2}}\eta\rangle+\langle
e^{-r\Delta^{1/2}}\xi,R\Delta^{1/2}e^{-r\Delta^{1/2}}\eta\rangle)\,dr.$

From here we conclude
$\langle\xi,\check{T}\eta\rangle=\langle\xi,R\eta\rangle$ as above. Since
$\bigcap_{n\in\mathbb{Z}}\operatorname{dom}(\Delta^{n})$ is dense in
$L^{2}(M)$, the equality $R=\check{T}$ follows. ∎

Informally, the identity from the previous lemma reads

$\frac{1}{2}\Delta^{1/4}\check{T}\Delta^{-1/4}+\frac{1}{2}\Delta^{-1/4}\check{T}\Delta^{1/4}=T.$

######  Proposition 3.5.

Let $T\in B(L^{2}(M))$.

  1. (a)

$\check{T}=\int_{0}^{\infty}\Delta^{-1/4}e^{-r\Delta^{-1/2}}T\Delta^{-1/4}e^{-r\Delta^{-1/2}}\,dr$
in the weak operator topology.

  2. (b)

If $JT=TJ$, then $J\check{T}=\check{T}J$.

  3. (c)

If $T\varphi^{1/2}=\varphi^{1/2}$, then
$\check{T}\varphi^{1/2}=\varphi^{1/2}$.

###### Proof.

  1. (a)

Let
$R=\int_{0}^{\infty}\Delta^{-1/4}e^{-r\Delta^{-1/2}}T\Delta^{-1/4}e^{-r\Delta^{1/2}}\,dr$.
The existence is justified by the same arguments as for $\check{T}$. Replacing
$\Delta$ by $\Delta^{-1}$ in Lemma 3.4 (a) we obtain

$\displaystyle\frac{1}{2}\langle\Delta^{1/4}\xi,R\Delta^{-1/4}\eta\rangle+\frac{1}{2}\langle\Delta^{-1/4}\xi,R\Delta^{1/4}\eta\rangle=\langle\xi,T\eta\rangle$

for
$\xi,\eta\in\operatorname{dom}(\Delta^{1/4})\cap\operatorname{dom}(\Delta^{-1/4})$.
Now $R=\check{T}$ follows from Lemma 3.4 (b).

  2. (b)

One has $J\Delta^{1/4}=\Delta^{-1/4}J$, and by the spectral theorem also
$Je^{-r\Delta^{1/2}}=e^{-r\Delta^{-1/2}}J$. Thus

$\displaystyle J\check{T}$
$\displaystyle=\int_{0}^{\infty}\Delta^{-1/4}e^{-r\Delta^{-1/2}}JT\Delta^{1/4}e^{-r\Delta^{1/2}}\,dr$
$\displaystyle=\int_{0}^{\infty}\Delta^{-1/4}e^{-r\Delta^{-1/2}}TJ\Delta^{1/4}e^{-r\Delta^{1/2}}\,dr$
$\displaystyle=\int_{0}^{\infty}\Delta^{-1/4}e^{-r\Delta^{-1/2}}T\Delta^{-1/4}e^{-r\Delta^{-1/2}}\,dr\,J.$

Now $J\check{T}=\check{T}J$ follows from (a).

  3. (c)

This is immediate from the definition of $\check{T}$.∎

We are particularly interested in the action of the $\mathscr{V}$-transform on
Markov operators. Let us first recall the definition.

We call an operator $T\in B(L^{2}(M))$ _positivity-preserving_ if
$T(L^{2}_{+}(M))\subset L^{2}_{+}(M)$, and _completely positivity-preserving_
if the amplification $T\otimes\mathrm{id}_{M_{n}(\mathbb{C})}$ is positivity-
preserving on $L^{2}(M)\otimes M_{n}(\mathbb{C})\cong L^{2}(M_{n}(M))$ for
every $n\in\mathbb{N}$. We use the terms “positivity-preserving” and
“completely positivity-preserving” instead of positive and completely positive
to avoid confusion with the concept of positive operators on a Hilbert space.

######  Definition 3.6.

An operator $T\in B(L^{2}(M))$ is a _Markov operator_ if it is completely
positivity-preserving and $T\varphi^{1/2}=\varphi^{1/2}$.

To prove that the $\mathscr{V}$-transform also preserves Markov operators, we
need a series of lemmas. Recall that the cones $P^{\sharp}$ and $P^{\flat}$
are defined as

$P^{\sharp}=\overline{\\{x\varphi^{1/2}\mid x\in
M_{+}\\}},\;P^{\flat}=\overline{\\{\varphi^{1/2}x\mid x\in M_{+}\\}},$

and further recall that $P^{\sharp}$ and $P^{\flat}$ are dual cones, that is,
$\xi\in P^{\sharp}$ if and only if $\langle\xi,\eta\rangle\geq 0$ for all
$\eta\in P^{\flat}$, and vice versa.

######  Lemma 3.7.

Let $T\in B(L^{2}(M))$ be positivity-preserving. If $\xi\in P^{\sharp}$ (resp.
$\xi\in P^{\flat}$) and $\eta\in P^{\flat}$ (resp. $\eta\in P^{\sharp}$), then

$\operatorname{Re}\langle\xi,\check{T}\eta\rangle\geq 0.$

###### Proof.

Let $x,y\in M_{+}$. By Lemma 3.4 (a) we have

$\frac{1}{2}\langle
x\varphi^{1/2},\check{T}(\varphi^{1/2}y)\rangle+\frac{1}{2}\langle\varphi^{1/2}x,\check{T}(y\varphi^{1/2})\rangle=\langle\varphi^{1/4}x\varphi^{1/4},T(\varphi^{1/4}y\varphi^{1/4})\rangle\geq
0.$

Since $T$ is positivity-preserving, it commutes with $J$. Hence so does
$\check{T}$ by Proposition 3.5 (b). If we apply this to the first summand from
the previous equation, we obtain

$\displaystyle\langle x\varphi^{1/2},\check{T}(\varphi^{1/2}y)\rangle$
$\displaystyle=\langle
J\check{T}(\varphi^{1/2}y),J(x\varphi^{1/2})\rangle=\langle\check{T}(y\varphi^{1/2}),\varphi^{1/2}x\rangle.$

Therefore

$\operatorname{Re}\langle x\varphi^{1/2},\check{T}(\varphi^{1/2}y)\rangle\geq
0.$

This settles the claim for $\xi=x\varphi^{1/2}$ and $\eta=\varphi^{1/2}y$. For
arbitrary $\xi\in P^{\sharp}$ and $\eta\in P^{\flat}$ the inequality follows
by approximation. The proof for $\xi\in P^{\flat}$ and $\eta\in P^{\sharp}$ is
analogous. ∎

######  Lemma 3.8.

If $R\in B(L^{2}(M))$ such that

$\displaystyle\operatorname{Re}\langle\xi,R\eta\rangle$ $\displaystyle\geq
0,\qquad\xi\in P^{\sharp},\eta\in P^{\flat},$
$\displaystyle\operatorname{Re}\langle\xi,R\eta\rangle$ $\displaystyle\geq
0,\qquad\xi\in P^{\flat},\eta\in P^{\sharp},$

then $\operatorname{Re}\langle\xi,R\eta\rangle\geq 0$ for all $\xi,\eta\in
L^{2}_{+}(M)$.

###### Proof.

Let $S=\\{z\in\mathbb{C}\mid 0<\operatorname{Re}z<1/2\\}$ and

$f\colon\overline{S}\to\mathbb{C},\,f(z)=e^{-\langle
J\Delta^{z}(x\varphi^{1/2}),R\Delta^{z}(y\varphi^{1/2})\rangle}$

for $x,y\in M_{+}$. The function $f$ is continuous on $\overline{S}$,
holomorphic on $S$ and satisfies

$\displaystyle\lvert f(z)\rvert$ $\displaystyle=e^{-\operatorname{Re}\langle
J\Delta^{z}(x\varphi^{1/2}),R\Delta^{z}(y\varphi^{1/2})\rangle}$
$\displaystyle\leq e^{\lVert J\Delta^{z}(x\varphi^{1/2})\rVert\lVert
R\Delta^{z}(y\varphi^{1/2})\rVert}$ $\displaystyle\leq e^{\lVert
R\rVert\lVert\Delta^{\operatorname{Re}z}(x\varphi^{1/2})\rVert\lVert\Delta^{\operatorname{Re}z}(y\varphi^{1/2})\rVert}.$

By the spectral theorem,

$\lVert\Delta^{\operatorname{Re}z}(x\varphi^{1/2})\rVert^{2}\leq\lVert
x\varphi^{1/2}\rVert^{2}+\lVert\Delta^{1/2}(x\varphi^{1/2})\rVert^{2}$

and likewise for $y\varphi^{1/2}$. Thus $f$ is bounded on $\overline{S}$.

If $\operatorname{Re}z=0$, then
$\Delta^{z}(y\varphi^{1/2})=\sigma^{\varphi}_{\operatorname{Im}z}(y)\varphi^{1/2}\in
P^{\sharp}$ and
$J\Delta^{z}(x\varphi^{1/2})=\varphi^{1/2}\sigma^{\varphi}_{\operatorname{Im}z}(x)\in
P^{\flat}$. Thus $\lvert f(z)\rvert\leq 1$ by assumption. Similarly, $\lvert
f(z)\rvert\leq 1$ if $\operatorname{Re}z=1/2$.

It follows from the Phragmen–Lindelöf principle that $\lvert f(1/4)\rvert\leq
1$, which means

$\operatorname{Re}\langle\Delta^{1/4}(x\varphi^{1/2}),R\Delta^{1/4}(y\varphi^{1/2})=\operatorname{Re}\langle
J\Delta^{1/4}(x\varphi^{1/2}),R\Delta^{1/4}(y\varphi^{1/2})\rangle\rangle\geq
0.$

This settles the claim for $\xi=\Delta^{1/4}(x\varphi^{1/2})$,
$\eta=\Delta^{1/4}(y\varphi^{1/2})$ with $x,y\in M_{+}$. For arbitrary
$\xi,\eta\in L^{2}_{+}(M)$ the claim follows by approximation. ∎

######  Lemma 3.9.

If $R\in B(L^{2}(M))$ such that $\operatorname{Re}\langle\xi,R\eta\rangle\geq
0$ for all $\xi,\eta\in L^{2}_{+}(M)$ and $JR=RJ$, then $R$ is positivity-
preserving.

###### Proof.

For $\xi,\eta\in L^{2}_{+}(M)$ we have $J\xi=\xi$, $J\eta=\eta$, and hence

$\langle R\eta,\xi\rangle=\langle J\xi,JR\eta\rangle=\langle\xi,R\eta\rangle.$

Therefore $\langle\xi,R\eta\rangle$ is real and thus positive by assumption.
As $L^{2}_{+}(M)$ is self-dual, the claim follows. ∎

######  Proposition 3.10.

The $\mathscr{V}$-transform maps symmetric Markov operators to symmetric
Markov operators.

###### Proof.

If $T\in B(L^{2}(M))$ is positivity-preserving, it follows from the previous
three lemmas that $\check{T}$ is positivity-preserving as well. If $T$ is
completely positivity-preserving, the same argument applied to the
amplifications $T\otimes\mathrm{id}_{M_{n}(\mathbb{C})}$ shows that
$\check{T}$ is completely positivity-preserving. That
$T\varphi^{1/2}=\varphi^{1/2}$ implies $\check{T}\varphi^{1/2}=\varphi^{1/2}$
was established in Proposition 3.5 (c). ∎

###  3.2. $\mathscr{V}$-Transform of KMS-Symmetric Operators on $M$

Formally, the $\mathscr{V}$-transform of a bounded linear operator $\Phi$ on
$M$ should be given by

$2\int_{0}^{\infty}\sigma_{-i/4}e^{-r\sigma_{-i/2}}\Phi\sigma_{-i/4}e^{-r\sigma_{-i/2}}\,dr,$

where we simply replaced the modular operator in the definition of
$\mathscr{V}$ by the analytic generator of the modular group on $M$. However,
it seems hard to make this formula rigorous. It is not even clear if
$\sigma_{-i/2}$ generates a semigroup on $M$ in a suitable sense.

Instead, we take a different approach that relies on the correspondence
between certain operators in $B(M)$ and operators in $B(L^{2}(M))$. This way
we can only define the $\mathscr{V}$-transform for KMS-symmetric unital
completely positive maps and generators of KMS-symmetric quantum Markov
semigroups, but this suffices for our purposes.

Let us first recall the definition of a KMS-symmetric map in this setting. A
linear map $\Phi\colon M\to M$ is called _KMS-symmetric_ (with respect to
$\varphi$) if

$\langle J\Phi(x)^{\ast}J\varphi^{1/2},y\varphi^{1/2}\rangle=\langle
Jx^{\ast}J\varphi^{1/2},\Phi(y)\varphi^{1/2}\rangle$

for all $x,y\in M$. Using the trace-like functional $\mathrm{tr}$ on the
Haagerup $L^{1}$ space, this can compactly be rewritten as

$\mathrm{tr}(\Phi(x)^{\ast}\varphi^{1/2}y\varphi^{1/2})=\mathrm{tr}(x^{\ast}\varphi^{1/2}\Phi(y)\varphi^{1/2})$

in analogy with the definition for matrix algebras.

There is the following correspondence between symmetric completely positivity-
preserving maps on $L^{2}(M)$ and KMS-symmetric completely positive maps on
$M$ [Cip97, Section 2]: If $\Phi$ is KMS-symmetric unital completely positive
map on $M$, then there exists a unique bounded linear operator $T$ on
$L^{2}(M)$ such that

$T(\varphi^{1/4}x\varphi^{1/4})=\varphi^{1/4}\Phi(x)\varphi^{1/4}$

for all $x\in M$. This operator is a symmetric Markov operator, which we
denote by $\Phi^{(2)}$.

Conversely, if $T$ is a symmetric Markov operator on $L^{2}(M)$, then there
exists a KMS-symmetric unital completely positive map $\Phi$ on $M$ such that
$\Phi^{(2)}=T$.

If $\Phi$ is a KMS-symmetric unital completely positive map on $M$, then
$\mathscr{V}(\Phi^{(2)})$ is a symmetric Markov operator by Proposition 3.10.
This justifies the following definition.

######  Definition 3.11.

Let $\Phi$ be a KMS-symmetric unital completely positive map on $M$. Its
$\mathscr{V}$-transform $\check{\Phi}$ is the unique KMS-symmetric unital
completely positive map $\Psi$ on $M$ such that

$\mathscr{V}(\Phi^{(2)})=\Psi^{(2)}.$

We also write $\mathscr{V}(\Phi)$ for $\check{\Phi}$.

The main object of interest of this article are semigroups of unital
completely positive maps. While the $\mathscr{V}$-transform preserves
unitality and complete positivity, it does not, in general, preserve the
semigroup property. We will discuss in the following how one can still
$\mathscr{V}$-transform generators of a class of such semigroups in a way that
preserves complete positivity of the generated semigroup. We start by
recalling some definitions.

A _quantum Markov semigroup_ is a family $(\Phi_{t})_{t\geq 0}$ of normal
unital completely positive maps on $M$ such that

  * •

$\Phi_{0}=\mathrm{id}_{M}$,

  * •

$\Phi_{s+t}=\Phi_{s}\Phi_{t}$ for all $s,t\geq 0$,

  * •

$\Phi_{t}\to\mathrm{id}_{M}$ in the pointwise weak∗ topology as $t\to 0$.

If $\Phi_{t}\to\mathrm{id}_{M}$ in operator norm as $t\to 0$, then
$(\Phi_{t})$ is called _uniformly continuous_.

The generator of $(\Phi_{t})$ is the weak∗ closed and densely defined operator
$\mathscr{L}$ given by

$\displaystyle\operatorname{dom}(\mathscr{L})$ $\displaystyle=\left\\{x\in
M:\lim_{t\to 0}\frac{1}{t}(x-\Phi_{t}(x))\text{ exists in the weak*
topology}\right\\}$ $\displaystyle\mathscr{L}(x)$ $\displaystyle=\lim_{t\to
0}\frac{1}{t}(x-\Phi_{t}(x))\text{ in the weak* topology}.$

The semigroup $(\Phi_{t})$ is uniformly continuous if and only if the
generator is bounded. The generators of uniformly continuous quantum Markov
semigroups can be characterized as follows [EL77a, Theorem 14.7]: A bounded
operator $\mathscr{L}$ on $M$ is the generator of a uniformly continuous
quantum Markov semigroup if and only it is normal, $\mathscr{L}(1)=0$ and
$\mathscr{L}$ is _conditionally negative definite_ , that is,

$\sum_{j=1}^{n}y_{j}^{\ast}\mathscr{L}(x_{j}^{\ast}x_{k})y_{k}\leq 0$

whenever $x_{1},\dots,x_{n},y_{1},\dots,y_{n}\in M$ with
$\sum_{j=1}^{n}x_{j}y_{j}=0$.

Let $(\Phi_{t})$ be a uniformly continuous KMS-symmetric Markov semigroup on
$M$ and $(T_{t})$ the associated symmetric Markov semigroup on
$L^{2}(M,\varphi)$. Let $\mathscr{L}$ and $\mathscr{L}_{2}$ denote the
generator of $(\Phi_{t})$ on $M$ and $(T_{t})$ on $L^{2}(M,\varphi)$,
respectively. By the uniform continuity assumption, both are bounded linear
operators. Thus we can form the $\mathscr{V}$-transform of $\mathscr{L}_{2}$,
and the continuity of the $\mathscr{V}$-transform implies

$\check{\mathscr{L}}_{2}=\lim_{t\to 0}\frac{1}{t}(1-\check{T}_{t})$

in operator norm. However, since $(\check{T}_{t})$ is not a semigroup in
general, this does not imply directly that $\check{\mathscr{L}}_{2}$ generates
a symmetric Markov semigroup. The lack of semigroup property can be taken care
of by a suitable rescaling of the time parameter. More precisely, we have the
following result.

######  Proposition 3.12.

If $(T_{t})$ is a symmetric Markov semigroup on $L^{2}(M,\varphi)$, then the
semigroup generated by $\check{\mathscr{L}}_{2}$ satisfies

$e^{-t\check{\mathscr{L}}_{2}}\xi=\lim_{n\to\infty}\check{T}_{t/n}^{n}\xi$

for all $t\geq 0$ and $\xi\in L^{2}(M,\varphi)$. In particular,
$(e^{-t\check{\mathscr{L}}_{2}})$ is a symmetric Markov semigroup.

###### Proof.

Since the $\mathscr{V}$-transform is a contraction, we have
$\lVert\check{T}_{t}\rVert\leq\lVert T_{t}\rVert\leq 1$ for all $t\geq 0$. It
follows from the Chernoff product formula [Engel, Nagel, Theorem III.5.2] that
$e^{-t\check{\mathscr{L}}_{2}}\xi=\lim_{n\to\infty}\check{T}_{t/n}^{n}\xi$ for
all $t\geq 0$ and $\xi\in L^{2}(M,\varphi)$. Finally, as $\check{T}_{s}$ is a
symmetric Markov operator for all $s\geq 0$, so is
$e^{-t\check{\mathscr{L}}_{2}}$ for all $t\geq 0$. ∎

As a consequence of the previous theorem, there exists a unique KMS-symmetric
quantum Markov semigroup $(\Psi_{t})$ on $M$ such that

$\varphi^{1/4}\Psi_{t}(x)\varphi^{1/4}=e^{-t\check{\mathscr{L}}_{2}}(\varphi^{1/4}x\varphi^{1/4})$

for all $x\in M$ and $t\geq 0$. Moreover, $(\Psi_{t})$ is uniformly
continuous. This justifies the following definition.

######  Definition 3.13.

If $\mathscr{L}$ is the generator of a uniformly continuous KMS-symmetric
Markov semigroup on $M$, its $\mathscr{V}$-transform $\check{\mathscr{L}}$ is
the unique normal linear operator on $M$ such that

$\varphi^{1/4}\check{\mathscr{L}}(x)\varphi^{1/4}=\check{\mathscr{L}}_{2}(\varphi^{1/4}x\varphi^{1/4})$

for all $x\in M$.

By the discussion above, $\check{\mathscr{L}}$ is again the generator of a
KMS-symmetric quantum Markov semigroup on $M$. Note moreover that since
$\mathscr{L}(1)=0$, the only completely positive generator of a quantum Markov
semigroup is $\mathscr{L}=0$. Therefore there is no conflict between our
definitions of the $\mathscr{V}$-transform of completely positive maps and
Markov generators.

##  4\. Derivations for Uniformly Continuous Quantum Markov Semigroups

In this section we study the correspondence between uniformly KMS-symmetric
quantum Markov semigroups and certain twisted derivations with values in
bimodules. We show that – like in the case of matrix algebras – every
uniformly continuous KMS-symmetric quantum Markov semigroup gives rise to a
derivation (Theorem 4.2) and that this derivation is unique (Theorem 4.7).

Throughout this section let $M$ be a von Neumann algebra and $\varphi$ a
faithful normal state on $M$. KMS symmetry is always understood with respect
to this state $\varphi$ and multiplication of elements in $L^{2}(M)$ is
understood as the multiplication induced by the full left Hilbert algebra
associated with $\varphi$. For a left-bounded vector $a\in L^{2}(M)$ we write
$\pi_{l}(a)$ for the bounded operator of left multiplication by $a$. Likewise,
if $a\in L^{2}(M)$ is right-bounded, we write $\pi_{r}(a)$ for the right
multiplication operator.

###  4.1. Existence and Innerness

After we established the existence and properties of the
$\mathscr{V}$-transform in the previous section, the proof for the existence
of a derivation associated with a uniformly continuous KMS-symmetric quantum
Markov semigroup follows the same strategy as in the finite-dimensional case.

To establish that the derivation is inner, we need the following result, which
is an easy consequence of the Christensen–Evans theorem [CE79]. Recall that a
bounded linear map between $C^{\ast}$-algebras is called _decomposable_ if it
is a linear combination of completely positive maps [Haa85].

######  Proposition 4.1.

Every generator of a uniformly continuous quantum Markov semigroup on $M$ is
the difference of two normal completely positive maps. In particular, it is
decomposable.

###### Proof.

Let $\mathscr{L}$ be a normal conditionally negative definite map on $M$. By
[CE79, Theorem 3.1] there exists $k\in M$ and a completely positive map
$\Phi\colon M\to M$ such that

$\mathscr{L}(x)=k^{\ast}x+xk-\Phi(x)$

for all $x\in M$. Since $\mathscr{L}$ is normal, so is $\Phi$. It follows from
[Pis20, Proposition 6.10] that $\mathscr{L}$ is the difference of two normal
completely positive maps. ∎

For the following result, let us recall the notion of correspondences (see
[Con94, Chapter 5, Appendix B] for example). An _$M$ -$M$-correspondence_ or
simply _correspondence_ in our case is a Hilbert space $\mathscr{H}$ endowed
with normal unital $\ast$-homomorphisms $\pi_{l}^{\mathscr{H}}\colon M\to
B(\mathscr{H})$, $\pi_{r}^{\mathscr{H}}\colon M^{\mathrm{op}}\to
B(\mathscr{H})$ such that $\pi_{l}^{\mathscr{H}}(M)$ and
$\pi_{r}^{\mathscr{H}}(M^{\mathrm{op}})$ commute. We write $x\cdot\xi\cdot y$
or simply $x\xi y$ for
$\pi_{l}^{\mathscr{H}}(x)\pi_{r}^{\mathscr{H}}(y^{\mathrm{op}})\xi$.

Every correspondence gives rise to a representation of the _binormal tensor
product_ $M\otimes_{\mathrm{bin}}M^{\mathrm{op}}$, which is defined as follows
(see [EL77]). A linear functional $\omega$ from the algebraic tensor product
$M\odot M^{\mathrm{op}}$ to $\mathbb{C}$ is called a _binormal state_ if
$\omega(u^{\ast}u)\geq 0$ for all $u\in M\odot M^{\mathrm{op}}$, $\omega(1)=1$
and the maps $x\mapsto\omega(x\otimes y_{0}^{\mathrm{op}})$,
$y^{\mathrm{op}}\mapsto\omega(x_{0}\otimes y^{\mathrm{op}})$ are weak∗
continuous for every $x_{0},y_{0}\in M$. For $u$ in the algebraic tensor
product $M\odot M^{\mathrm{op}}$ let

$\lVert u\rVert_{\mathrm{bin}}=\sup\\{\omega(u^{\ast}u)^{1/2}\mid\omega\text{
binormal state on }M\odot M^{\mathrm{op}}\\}.$

The binormal tensor product $M\otimes_{\mathrm{bin}}M^{\mathrm{op}}$ is the
completion of $M\odot M^{\mathrm{op}}$ with respect to the norm
$\lVert\cdot\rVert_{\mathrm{op}}$.

######  Theorem 4.2.

Let $(\Phi_{t})$ be a uniformly continuous KMS-symmetric quantum Markov
semigroup on $M$ and let $\mathscr{L}$ denote its generator. There exists a
correspondence $\mathscr{H}$, an anti-unitary involution
$\mathscr{J}\colon\mathscr{H}\to\mathscr{H}$ and a bounded operator
$\delta\colon L^{2}(M)\to\mathscr{H}$ satisfying

  1. (a)

$\mathscr{J}(x\xi y)=y^{\ast}(\mathscr{J}\xi)x^{\ast}$ for all $x,y\in M$ and
$\xi\in\mathscr{H}$,

  2. (b)

$\delta(Ja)=\mathscr{J}\delta(a)$ for all $a\in L^{2}(M)$,

  3. (c)

$\delta(ab)=\pi_{l}(a)\cdot\delta(b)+\delta(a)\cdot J\pi_{r}(b)^{\ast}J$ for
all $a\in M\varphi^{1/2}$, $b\in\varphi^{1/2}M$,

  4. (d)

$\overline{\mathrm{lin}}\\{\delta(a)x\mid a\in L^{2}(M),\,x\in
M\\}=\mathscr{H}$

such that

$\mathscr{L}_{2}=\delta^{\ast}\delta.$

Moreover, there exists $\xi_{0}\in\mathscr{H}$ such that

$\delta(a)=\pi_{l}(a)\cdot\xi_{0}-\xi_{0}\cdot J\pi_{r}(a)^{\ast}J$

for $a\in M\varphi^{1/2}\cap\varphi^{1/2}M$.

###### Proof.

Let

$\omega\colon M\odot M^{\mathrm{op}}\to\mathbb{C},\,x\otimes
y^{\mathrm{op}}\mapsto-\frac{1}{2}\langle\varphi^{1/2},\check{\mathscr{L}}(x)\varphi^{1/2}y\rangle.$

We first show that $\omega$ extends continuously to
$M\otimes_{\mathrm{bin}}M^{\mathrm{op}}$.

Since $\check{\mathscr{L}}$ is the generator of a uniformly continuous quantum
Markov semigroup, by Proposition 4.1 there exist normal completely positive
maps $\Psi_{1},\Psi_{2}$ on $M$ such that
$\check{\mathscr{L}}=\Psi_{2}-\Psi_{1}$. For $j\in\\{1,2\\}$, the maps

$\omega_{j}\colon M\odot M^{\mathrm{op}}\to\mathbb{C},\,x\otimes
y^{\mathrm{op}}\mapsto\frac{1}{2}\langle\varphi^{1/2},\Psi_{j}(x)\varphi^{1/2}y\rangle$

are positive and separately weak∗ continuous. Hence they extend to positive
functionals on $M\otimes_{\mathrm{bin}}M^{\mathrm{op}}$ by definition of
$\lVert\cdot\rVert_{\mathrm{bin}}$. Thus $\omega$ extends to a bounded linear
map on $M\otimes_{\mathrm{bin}}M^{\mathrm{op}}$. We continue to denote the
extension of $\omega$ to $M\otimes_{\mathrm{bin}}M^{\mathrm{op}}$ by $\omega$.

Let

$q\colon M\odot M^{\mathrm{op}}\to L^{2}(M),\,x\otimes y^{\mathrm{op}}\mapsto
x\varphi^{1/2}y.$

Clearly, the kernel of $q$ is a left ideal of $M\odot M^{\mathrm{op}}$. Let
$I$ denote its closure, which is a closed left ideal of
$M\otimes_{\mathrm{bin}}M^{\mathrm{op}}$.

If $u=\sum_{j=1}^{n}x_{j}\otimes y_{j}^{\mathrm{op}}\in\ker q$, then

$\displaystyle\omega(u^{\ast}u)$
$\displaystyle=-\frac{1}{2}\sum_{j,k=1}^{n}\langle\varphi^{1/2}y_{j},\check{\mathscr{L}}(x_{j}^{\ast}x_{k})\varphi^{1/2}y_{k}\rangle$
$\displaystyle=\lim_{t\to
0}\frac{1}{2t}\left(\sum_{j,k=1}^{n}\langle\varphi^{1/2}y_{j},\check{\Phi}_{t}(x_{j}^{\ast}x_{k})\varphi^{1/2}y_{k}\rangle-\lVert
q(u)\rVert^{2}\right)$ $\displaystyle=\lim_{t\to
0}\frac{1}{2t}\sum_{j,k=1}^{n}\langle\varphi^{1/2}y_{j},\check{\Phi}_{t}(x_{j}^{\ast}x_{k})\varphi^{1/2}y_{k}\rangle.$

The last expression is positive since $\check{\Phi}_{t}$ is completely
positive for all $t\geq 0$. By continuity, it follows that
$\omega(u^{\ast}u)\geq 0$ for all $u\in I$.

Let $\mathscr{H}$ be the GNS Hilbert space associated with $\omega|_{I}$ and
$\pi_{\omega}$ the GNS representation of
$M\otimes_{\mathrm{bin}}M^{\mathrm{op}}$ on $\mathscr{H}$, that is,
$\mathscr{H}$ is the completion of $I$ with respect to the inner product
$(x,y)\mapsto\omega(x^{\ast}y)$ and
$\pi_{\omega}(x)[y]_{\mathscr{H}}=[xy]_{\mathscr{H}}$, were
$[\cdot]_{\mathscr{H}}$ denotes the canonical map $I\to\mathscr{H}$. As
$\omega$ is separately weak∗ continuous, it follows easily that the actions
$x\mapsto\pi_{\omega}(x\otimes 1)$ and
$y^{\mathrm{op}}\mapsto\pi_{\omega}(1\otimes y^{\mathrm{op}})$ are normal.
These actions make $\mathscr{H}$ into an $M$-$M$-correspondence.

Moreover, let $(e_{\lambda})$ be a right approximate identity for $I$
consisting of positive contractions. Since
$\lVert[e_{\lambda}]_{\mathscr{H}}\rVert=\omega(e_{\lambda}^{2})^{1/2}\leq\lVert\omega\rVert^{1/2}$,
we may assume additionally that $[e_{\lambda}]_{\mathscr{H}}$ converges weakly
to some vector $\xi_{\omega}\in\mathscr{H}$. If $u\in I$, then

$[u]_{\mathscr{H}}=[\lim_{\lambda}ue_{\lambda}]=\lim_{\lambda}\pi_{\omega}(u)[e_{\lambda}]_{\mathscr{H}}=\pi_{\omega}(u)\xi_{\omega}.$

In particular, $\xi_{\omega}$ is a cyclic vector for $\pi_{\omega}$.

To define $\mathscr{J}$, first define an anti-linear map $\mathscr{J}_{0}$ on
$M\odot M^{\mathrm{op}}$ by $\mathscr{J}_{0}(x\otimes
y^{\mathrm{op}})=y^{\ast}\otimes(x^{\ast})^{\mathrm{op}}$. A direct
computation shows $q\mathscr{J}_{0}=Jq$. In particular, $\mathscr{J}_{0}$
leaves $\ker q$ invariant.

Furthermore, if $x_{1},x_{2},y_{1},y_{2}\in M$, then

$\displaystyle\omega(\mathscr{J}_{0}(x_{1}\otimes
y_{1}^{\mathrm{op}})^{\ast}\mathscr{J}_{0}(x_{2}\otimes y_{2}^{\mathrm{op}}))$
$\displaystyle=-\frac{1}{2}\langle\varphi^{1/2},\check{\mathscr{L}}(y_{1}y_{2}^{\ast})\varphi^{1/2}x_{2}^{\ast}x_{1}\rangle$
$\displaystyle=-\frac{1}{2}\langle\varphi^{1/2}x_{1}^{\ast}x_{2},\check{\mathscr{L}}(y_{1}y_{2}^{\ast})\varphi^{1/2}\rangle$
$\displaystyle=-\frac{1}{2}\langle\varphi^{1/2}\check{\mathscr{L}}(x_{1}^{\ast}x_{2}),y_{1}y_{2}^{\ast}\varphi^{1/2}\rangle$
$\displaystyle=-\frac{1}{2}\langle\varphi^{1/2},y_{1}y_{2}^{\ast}\varphi^{1/2}\check{\mathscr{L}}(x_{2}^{\ast}x_{1})\rangle$
$\displaystyle=-\frac{1}{2}\langle
J(y_{1}y_{2}^{\ast}\varphi^{1/2}\check{\mathscr{L}}(x_{2}^{\ast}x_{1})),J\varphi^{1/2}\rangle$
$\displaystyle=-\frac{1}{2}\langle\check{\mathscr{L}}(x_{1}^{\ast}x_{2})\varphi^{1/2}y_{2}y_{1}^{\ast},\varphi^{1/2}\rangle$
$\displaystyle=-\frac{1}{2}\langle\varphi^{1/2},\check{\mathscr{L}}(x_{2}^{\ast}x_{1})\varphi^{1/2}y_{1}y_{2}^{\ast}\rangle$
$\displaystyle=\omega((x_{2}\otimes y_{2}^{\mathrm{op}})^{\ast}(x_{1}\otimes
y_{1}^{\mathrm{op}})).$

Therefore the map

$[I]_{\mathscr{H}}\to[I]_{\mathscr{H}},\,[x\otimes
y^{\mathrm{op}}]_{\mathscr{H}}\mapsto[y^{\ast}\otimes(x^{\ast})^{\mathrm{op}}]_{\mathscr{H}}$

extends to an isometric anti-linear operator $\mathscr{J}$ on $\mathscr{H}$.
Obviously, $\mathscr{J}$ is an involution and property (a) follows directly
from the definition.

Finally let us construct $\delta$. For $a\in M\varphi^{1/2}\cap\varphi^{1/2}M$
let

$\delta(a)=\pi_{\omega}(\pi_{l}(a)\otimes
1-1\otimes(J\pi_{r}(a)^{\ast}J)^{\mathrm{op}})\xi_{\omega}.$

Note that

$q(\pi_{l}(a)\otimes
1-1\otimes(J\pi_{r}(a)^{\ast}J)^{\mathrm{op}})=\pi_{l}(a)\varphi^{1/2}-\pi_{r}(a)\varphi^{1/2}=0$

so that $\pi_{l}(a)\otimes 1-1\otimes(J\pi_{r}(a)^{\ast}J)^{\mathrm{op}}\in I$
and $\delta(a)=[\pi_{l}(a)\otimes
1-1\otimes(J\pi_{r}(a)^{\ast}J)^{\mathrm{op}}]_{\mathscr{H}}$.

We have

$\displaystyle\lVert\delta(a)\rVert_{\mathscr{H}}^{2}$
$\displaystyle=\lVert[\pi_{l}(a)\otimes
1-1\otimes(J\pi_{r}(a)^{\ast}J)^{\mathrm{op}}]_{\mathscr{H}}\rVert^{2}$
$\displaystyle=\omega(\lvert\pi_{l}(a)\otimes
1-1\otimes(J\pi_{r}(a)^{\ast}J)^{\mathrm{op}}\rvert^{2})$
$\displaystyle=\omega(\pi_{l}(a^{\sharp}a)\otimes
1)+\omega(1\otimes(J\pi_{r}(aa^{\flat})J)^{\mathrm{op}})$
$\displaystyle\quad-\omega(\pi_{l}(a^{\sharp})\otimes(J\pi_{r}(a)^{\ast}J)^{\mathrm{op}})-\omega(\pi_{l}(a)\otimes(J\pi_{r}(a^{\flat})^{\ast}J)^{\mathrm{op}})$
$\displaystyle=-\frac{1}{2}\langle\varphi^{1/2},\check{\mathscr{L}}_{2}(\Delta^{1/4}(a^{\sharp}a))\rangle-\frac{1}{2}\langle\Delta^{1/4}J(aa^{\flat}),\check{\mathscr{L}}_{2}\varphi^{1/2}\rangle$
$\displaystyle\quad+\frac{1}{2}\langle\Delta^{1/4}Ja,\check{\mathscr{L}}_{2}\Delta^{1/4}(a^{\sharp})\rangle+\frac{1}{2}\langle\Delta^{1/4}Ja^{\flat},\check{\mathscr{L}}_{2}\Delta^{1/4}a\rangle$
$\displaystyle\overset{(1)}{=}\frac{1}{2}\langle
J\Delta^{-1/4}a,\check{\mathscr{L}}_{2}J\Delta^{1/4}a\rangle+\frac{1}{2}\langle\Delta^{-1/4}a,\check{\mathscr{L}}_{2}\Delta^{1/4}a\rangle$
$\displaystyle\overset{(2)}{=}\frac{1}{2}\langle\Delta^{1/4}a,\check{\mathscr{L}}_{2}\Delta^{-1/4}a\rangle+\frac{1}{2}\langle\Delta^{-1/4}a,\check{\mathscr{L}}_{2}\Delta^{1/4}a\rangle$
$\displaystyle\overset{(3)}{=}\langle a,\mathscr{L}_{2}(a)\rangle.$

Here we used the symmetry of $\check{\mathscr{L}}_{2}$ and
$\check{\mathscr{L}}_{2}\varphi^{1/2}=0$ for (1), the symmetry of
$\check{\mathscr{L}}_{2}$ and
$\check{\mathscr{L}}_{2}J=J\check{\mathscr{L}}_{2}$ for (2) and the key
property from Lemma 3.4 for (3).

Therefore the map $\delta$ extends to a bounded linear operator from
$L^{2}(M)$ to $\mathscr{H}$, and this extension, still denoted by $\delta$,
satisfies $\delta^{\ast}\delta=\mathscr{L}_{2}$. Clearly,

$\delta(ab)=\pi_{l}(a)\cdot\delta(b)+\delta(a)\cdot J\pi_{r}(b)^{\ast}J$

for $a,b\in M\varphi^{1/2}\cap\varphi^{1/2}M$. If we only have $a\in
M\varphi^{1/2}$ and $b\in\varphi^{1/2}M$, a standard approximation argument
[Haa75, Lemma 1.3] shows that this identity continues to hold, which settles
property (c).

Property (b) is clear from the definition if $a\in
M\varphi^{1/2}\cap\varphi^{1/2}M$, and can be extended to $a\in L^{2}(M)$
again by approximation.

Finally, if $u=\sum_{j=1}^{n}x_{j}\otimes y_{j}^{\mathrm{op}}\in\ker q$ and
$x_{j}$ is analytic for $\sigma^{\varphi}$ for all $j\in\\{1,\dots,n\\}$, then

$\displaystyle u$ $\displaystyle=\sum_{j=1}^{n}x_{j}\otimes
y_{j}^{\mathrm{op}}$ $\displaystyle=\sum_{j=1}^{n}(x_{j}\otimes
y_{j}^{\mathrm{op}}-1\otimes(\sigma^{\varphi}_{i/2}(x_{j})y_{j})^{\mathrm{op}}),$

where we used
$0=q(u)=\varphi^{1/2}\sum_{j=1}^{n}\sigma^{\varphi}_{i/2}(x_{j})y_{j}$. Thus

$[u]_{\mathscr{H}}=\sum_{j=1}^{n}\delta(x_{j}\varphi^{1/2})y_{j}.$

Since $\pi_{\omega}$ is normal and $[\ker q]_{\mathscr{H}}$ is dense in
$\mathscr{H}$ by definition, property (d) follows. ∎

######  Remark 4.3.

If one only wants to show the existence of the derivation $\delta$, one could
work directly with the GNS representation of $\omega$ on $\ker q$ without
passing to the binormal tensor product. Hence Proposition 4.1 and thus the
Christensen–Evans theorem is only needed to show that the derivation is inner.

###  4.2. Uniqueness

We show next that the triple $(\mathscr{H},\mathscr{J},\delta)$ constructed in
Theorem 4.2 is uniquely determined by the semigroup $(\Phi_{t})$ up to
isomorphism. Let us first introduce some terminology for triples of this kind.

######  Definition 4.4.

We call a pair $(\mathscr{H},\mathscr{J})$ consisting of an
$M$-$M$-correspondence and an anti-unitary involution
$\mathscr{J}\colon\mathscr{H}\to\mathscr{H}$ a _self-dual
$M$-$M$-correspondence_ if

$\mathscr{J}(x\xi y)=y^{\ast}(\mathscr{J}\xi)x^{\ast}$

for all $x,y\in M$ and $\xi\in\mathscr{H}$.

We call a triple $(\mathscr{H},\mathscr{J},\delta)$ consisting of a self-dual
$M$-$M$-correspondence $(\mathscr{H},\mathscr{J})$ and a closed operator
$\delta\colon\operatorname{dom}(\delta)\subset L^{2}(M)\to\mathscr{H}$ a
_first-order differential calculus_ if

  1. (a)

$J\operatorname{dom}(\delta)=\operatorname{dom}(\delta)$ and
$\delta(Ja)=\mathscr{J}\delta(a)$ for all $a\in L^{2}(M)$,

  2. (b)

Whenever $a\in\operatorname{dom}(\delta)\cap M\varphi^{1/2}$,
$b\in\operatorname{dom}(\delta)\cap\varphi^{1/2}M$, then
$ab\in\operatorname{dom}(\delta)$ and
$\delta(ab)=\pi_{l}(a)\delta(b)+\delta(a)J\pi_{r}(b)^{\ast}J$,

  3. (c)

$\overline{\mathrm{lin}}\\{\delta(a)x\mid a\in L^{2}(M),\,x\in
M\\}=\mathscr{H}$.

With this definition, Theorem 4.2 says that for every bounded Markov generator
$\mathscr{L}_{2}$ on $L^{2}(M)$ there exists a first-order differential
calculus $(\mathscr{H},\mathscr{J},\delta)$ such that
$\mathscr{L}_{2}=\delta^{\ast}\delta$. In this subsection we will show that
$(\mathscr{H},\mathscr{J},\delta)$ is uniquely determined by
$\mathscr{L}_{2}$.

To lighten the notation, we write $a\xi$ for $\pi_{l}(a)\xi$ if $a\in
M\varphi^{1/2}$ and $\xi b$ for $\xi\cdot J\pi_{r}(b)^{\ast}J$ if $b\in
M\varphi^{1/2}$ for the remainder of this subsection.

The first step towards uniqueness is a purely algebraic consequence of the
properties (a) and (b) of a first-order differential calculus.

######  Lemma 4.5.

If $(\mathscr{H},\mathscr{J},\delta)$ is a first-order differential calculus
and $\delta$ is bounded, then

$\displaystyle\langle\delta(\Delta^{1/4}$ $\displaystyle
a)\Delta^{1/4}b,\delta(\Delta^{-1/4}c)\rangle_{\mathscr{H}}+\langle\delta(\Delta^{-1/4}a)\Delta^{-1/4}b,\delta(\Delta^{1/4}c)\rangle_{\mathscr{H}}$
$\displaystyle=$
$\displaystyle\langle\delta(\Delta^{-1/4}(ab)),\delta(\Delta^{1/4}c)\rangle_{\mathscr{H}}+\langle\delta(\Delta^{1/4}a),\delta(\Delta^{-1/4}cJ(\Delta^{-1/4}b))\rangle_{\mathscr{H}}$
$\displaystyle-\langle\delta(J(\Delta^{1/4}c)\Delta^{1/4}a),\delta(J(\Delta^{-1/4}b))\rangle_{\mathscr{H}}$

for all $a,b,c\in M_{\varphi}^{a}\varphi^{1/2}$, where $M^{a}_{\varphi}$
denotes the set of all entire analytic elements for $\sigma^{\varphi}$.

###### Proof.

As $a,b,c\in M_{\varphi}^{a}\varphi^{1/2}$, these elements lie in
$\bigcap_{n\in\mathbb{Z}}\operatorname{dom}(\Delta^{n})$, and arbitrary powers
of the modular operator map them to left- and right-bounded vectors. In
particular, all expressions in the claimed equation are well-defined.

Using properties (a) and (b) of a first-order differential calculus, we can do
the following computations. Let $x,y,z\in M_{\varphi}^{a}\varphi^{1/2}$. Then
we have

$\displaystyle\langle\delta(x)y,\delta(z)\rangle_{\mathscr{H}}$
$\displaystyle=\langle\delta(x)y,\delta(z)\rangle_{\mathscr{H}}+\langle
x\delta(y),\delta(z)\rangle_{\mathscr{H}}-\langle\mathscr{J}(\delta(z)),\mathscr{J}(x\delta(y))\rangle_{\mathscr{H}}$
$\displaystyle=\langle\delta(xy),\delta(z)\rangle_{\mathscr{H}}-\langle\delta(Jz),\delta(Jy)Jx\rangle_{\mathscr{H}}$
$\displaystyle=\langle\delta(xy),\delta(z)\rangle_{\mathscr{H}}-\langle\delta(Jz)\Delta^{1/2}x,\delta(Jy)\rangle_{\mathscr{H}}$

and

$\displaystyle\langle\delta(x)y,\delta(z)\rangle_{\mathscr{H}}$
$\displaystyle=\langle\delta(x),\delta(z)J\Delta^{-1/2}y\rangle_{\mathscr{H}}+\langle\delta(x),z\delta(J\Delta^{-1/2}y)\rangle_{\mathscr{H}}-\langle\delta(x),z\delta(J\Delta^{-1/2}y)\rangle_{\mathscr{H}}$
$\displaystyle=\langle\delta(x),\delta(zJ\Delta^{-1/2}y)\rangle_{\mathscr{H}}-\langle\delta(x),z\delta(J\Delta^{-1/2}y)\rangle_{\mathscr{H}}$
$\displaystyle=\langle\delta(x),\delta(zJ\Delta^{-1/2}y)\rangle_{\mathscr{H}}-\langle
J(\Delta^{1/2}z)\delta(x),\delta(J\Delta^{-1/2}y)\rangle_{\mathscr{H}}.$

If we take $x=\Delta^{-1/4}a$, $y=\Delta^{-1/4}b$, $z=\Delta^{1/4}c$ in the
first identity and $x=\Delta^{1/4}a$, $y=\Delta^{1/4}b$, $z=\Delta^{-1/4}c$ in
the second identity and add them up, we obtain

$\displaystyle\langle\delta(\Delta^{-1/4}a)\Delta^{-1/4}b,\delta(\Delta^{1/4}c)\rangle_{\mathscr{H}}+\langle\delta(\Delta^{1/4}a)\Delta^{1/4}b,\delta(\Delta^{-1/4}c)\rangle_{\mathscr{H}}$
$\displaystyle=\langle\delta(\Delta^{-1/4}(ab)),\delta(\Delta^{1/4}c)\rangle_{\mathscr{H}}-\langle\delta(J\Delta^{1/4}c)\Delta^{1/4}a,\delta(J\Delta^{-1/4}b)\rangle_{\mathscr{H}}$
$\displaystyle\quad+\langle\delta(\Delta^{1/4}a),\delta((\Delta^{-1/4}c)J\Delta^{-1/4}b)\rangle_{\mathscr{H}}-\langle(J\Delta^{1/4}c)\delta(\Delta^{1/4}a),\delta(J\Delta^{-1/4}b)\rangle_{\mathscr{H}}$
$\displaystyle=\langle\delta(\Delta^{-1/4}(ab)),\delta(\Delta^{1/4}c)\rangle_{\mathscr{H}}+\langle\delta(\Delta^{1/4}a),\delta((\Delta^{-1/4}c)J\Delta^{-1/4}b)\rangle_{\mathscr{H}}$
$\displaystyle\quad-\langle\delta((J\Delta^{1/4}c)\Delta^{1/4}a),\delta(J\Delta^{-1/4}b)\rangle_{\mathscr{H}},$

where we used again property (b) of a first-order differential calculus in the
last step. ∎

The significance of this result is that the right side depends only on the
inner product of elements from the range of $\delta$ and not the bimodule
generated by the range. If the modular operator $\Delta$ is trivial, that is,
$\varphi$ is a trace, one can conclude uniqueness of the derivation directly
from this lemma. In the general case, substantially more work is needed. In
particular, there are analytical difficulties that are absent in the case of
tracially symmetric (or more generally GNS-symmetric) quantum Markov
semigroups.

One tool we use are spectral subspaces of the analytic generator of the
modular group. We recall the definition and some of their properties here. See
[CZ76] for more details.

For $0<\lambda_{1}<\lambda_{2}$ let

$M[\lambda_{1},\lambda_{2}]=\left\\{x\in\bigcap_{t\in\mathbb{R}}\operatorname{dom}(\sigma^{\varphi}_{it}):\varlimsup_{t\to\infty}\lVert\sigma^{\varphi}_{it}(x)\rVert^{1/t}\leq\frac{1}{\lambda_{1}},\,\varlimsup_{t\to\infty}\lVert\sigma^{\varphi}_{-it}(x)\rVert^{1/t}\leq\lambda_{2}\right\\}.$

This is a norm closed subspace of $M$, invariant under $\sigma^{\varphi}$, and
the spectrum of the restriction of $\sigma^{\varphi}_{-i}$ to
$M[\lambda_{1},\lambda_{2}]$ is contained in $[\lambda_{1},\lambda_{2}]$ (see
[CZ76, (iii)–(v), p. 351]). Moreover, the union
$\bigcup_{0<\lambda_{1}\leq\lambda_{2}<\infty}M[\lambda_{1},\lambda_{2}]$ is
weak∗ dense in $M$ [CZ76, (vi), p. 356]. Additionally, $\sigma^{\varphi}_{t}$
is given by $e^{itH}$ for some $H\in B(M[\lambda_{1},\lambda_{2}])$ with
$\mathrm{sp}(H)\subset[-\ln(\lambda_{1}),\ln(\lambda_{2})]$ [CZ76, Theorem
5.2, p. 349].

######  Lemma 4.6.

Let $0<\lambda_{1}<\lambda_{2}$. Define $X$ to be the completion of
$(\mathds{1}_{[\lambda_{1},\lambda_{2}]}(\Delta)L^{2}(M))\odot
M[\lambda_{1},\lambda_{2}]$ with the projective cross norm. Then the bounded
operator $T:X\to X$ defined on pure tensors by

$T(\eta\otimes x)=\Delta^{1/4}(\eta)\otimes\sigma^{\varphi}_{-i/4}(x)$

is well defined and $\mathrm{sp}(T)\subset(0,\infty)$.

###### Proof.

The spectrum of $\Delta^{1/4}$ restricted to
$\mathds{1}_{[\lambda_{1},\lambda_{2}]}(\Delta)L^{2}(M)$ is contained in
$[\lambda_{1}^{1/4},\lambda_{2}^{1/4}]$ by definition. Since the restriction
of $\sigma^{\varphi}_{-i/4}$ to $M[\lambda_{1},\lambda_{2}]$ is $e^{1/4H}$ for
some $H\in B(M[\lambda_{1},\lambda_{2}])$ with
$\mathrm{sp}(H)\subset[-\ln(\lambda_{1}),\ln(\lambda_{2})]$, we know that the
spectrum of the restriction of $\sigma^{\varphi}_{-i/4}$ to
$M[\lambda_{1},\lambda_{2}]$ is also contained in
$[\lambda_{1}^{1/4},\lambda_{2}^{1/4}]$. Then $\Delta^{1/4}\otimes\mathrm{id}$
and $\mathrm{id}\otimes\sigma^{\varphi}_{-i/4}$ are well defined and have
spectra contained in $[\lambda_{1}^{1/4},\lambda_{2}^{1/4}]$, so
$\Delta^{1/4}\otimes\sigma^{\varphi}_{-i/4}$ has spectrum contained in
$(0,\infty)$ [Sch69, p. 96]. ∎

######  Theorem 4.7.

Let $(\mathscr{H}_{1},\mathscr{J}_{1},\delta_{1})$ and
$(\mathscr{H}_{2},\mathscr{J}_{2},\delta_{2})$ be first order differential
calculi for $M$ such that $\delta_{1}$ and $\delta_{2}$ are bounded and
$\delta_{1}^{\ast}\delta_{1}=\delta_{2}^{\ast}\delta_{2}$. Then there exists a
unitary bimodule map $\Theta:\mathscr{H}_{1}\to\mathscr{H}_{2}$ intertwining
$\mathscr{J}_{1}$ and $\mathscr{J}_{2}$ such that
$\Theta(\delta_{1}(a))=\delta_{2}(a)$ for all $a\in L^{2}(M)$.

###### Proof.

The unitary bimodule map $\Theta$ will be given by

$\Theta(\delta_{1}(a)b)=\delta_{2}(a)b$

on elements of the form $\delta_{1}(a)b$ with $a,b\in L^{2}(M)$. The difficult
part of the proof is to show that this map is isometric; the other properties
will follow naturally.

Let $0<\lambda_{1}<\lambda_{2}$ be arbitrary. Let $X$ and $T$ be as in Lemma
4.6. Note that $T$ is invertible since $0\notin\mathrm{sp}(T)$. On
$(\mathds{1}_{[\lambda_{1},\lambda_{2}]}(\Delta)L^{2}(M))\odot
M[\lambda_{1},\lambda_{2}]\subset X$ we can define the maps $q_{1}$ and
$q_{2}$ to $\mathscr{H}_{1}$ and $\mathscr{H}_{2}$, respectively, by

$q_{1}(\eta\otimes x)=\delta_{1}(\eta)x\text{ and }q_{2}(\eta\otimes
x)=\delta_{2}(\eta)x.$

Because $\delta_{1}$ and $\delta_{2}$ are bounded and right multiplication is
bounded in the operator norm, we can boundedly extend $q_{1}$ and $q_{2}$ to
$X$.

Using Lemma 4.5 we can now show that for all $x,y\in X$ we have

(4) $\displaystyle\langle
q_{1}(T(x)),q_{1}(T^{-1}(y))\rangle_{\mathscr{H}_{1}}$ $\displaystyle+\langle
q_{1}(T^{-1}(x)),q_{1}(T(y))\rangle_{\mathscr{H}_{1}}$ $\displaystyle=$
$\displaystyle\langle
q_{2}(T(x)),q_{2}(T^{-1}(y))\rangle_{\mathscr{H}_{2}}+\langle
q_{2}(T^{-1}(x)),q_{2}(T(y))\rangle_{\mathscr{H}_{2}}.$

Indeed, for all $j\in\\{1,2\\}$,
$\eta,\xi\in(\mathds{1}_{[\lambda_{1},\lambda_{2}]}(\Delta)L^{2}(M))$ and
$a,b\in M[\lambda_{1},\lambda_{2}]$ we have

$\displaystyle\langle q_{j}(T(\eta\otimes a)),q_{j}(T^{-1}(\xi\otimes
b))\rangle_{\mathscr{H}_{j}}+\langle q_{j}(T^{-1}(\eta\otimes
a)),q_{j}(T(\xi\otimes b))\rangle_{\mathscr{H}_{j}}$ $\displaystyle=$
$\displaystyle\langle\delta_{j}(\Delta^{1/4}\eta)\sigma^{\varphi}_{-i/4}(a),\delta_{j}(\Delta^{-1/4}\xi)\sigma^{\varphi}_{i/4}(b)\rangle_{\mathscr{H}_{j}}+\langle\delta_{j}(\Delta^{-1/4}\eta)\sigma^{\varphi}_{i/4}(a),\delta_{j}(\Delta^{1/4}\xi)\sigma^{\varphi}_{-i/4}(b)\rangle_{\mathscr{H}_{j}}$
$\displaystyle=$
$\displaystyle\langle\delta_{j}(\Delta^{1/4}\eta)\sigma^{\varphi}_{-i/4}(ab^{\ast}),\delta_{j}(\Delta^{-1/4}\xi)\rangle_{\mathscr{H}_{j}}+\langle\delta_{j}(\Delta^{-1/4}\eta)\sigma^{\varphi}_{i/4}(ab^{\ast}),\delta_{j}(\Delta^{1/4}\xi))\rangle_{\mathscr{H}_{j}}$
$\displaystyle=$
$\displaystyle\langle\delta_{j}(\Delta^{-1/4}(\eta\varphi^{1/2}ab^{\ast})),\delta_{j}(\Delta^{1/4}\xi)\rangle_{\mathscr{H}_{j}}+\langle\delta_{j}(\Delta^{1/4}\eta),\delta_{j}(\Delta^{-1/4}\xi
J(\Delta^{-1/4}(\varphi^{1/2}ab^{\ast})))\rangle_{\mathscr{H}_{j}}$
$\displaystyle-\langle\delta_{j}(J(\Delta^{1/4}\xi)\Delta^{1/4}\eta),\delta_{j}(J(\Delta^{-1/4}\varphi^{1/2}ab^{\ast}))\rangle_{\mathscr{H}_{j}},$

where the last step follows from Lemma 4.5 and the fact that $va=v\cdot
J\pi_{r}(\varphi^{1/2}a)^{\ast}J$ for $v\in\mathscr{H}_{j}$ and $a\in M$.
Since we have for all $\eta,\xi\in L^{2}(M)$ that

$\langle\delta_{1}(\eta),\delta_{1}(\xi)\rangle_{\mathscr{H}_{1}}=\langle\eta,\delta_{1}^{\ast}\delta_{1}(\xi)\rangle=\langle\eta,\delta_{2}^{\ast}\delta_{2}(\xi)\rangle=\langle\delta_{2}(\eta),\delta_{2}(\xi)\rangle_{\mathscr{H}_{2}},$

we can now conclude that

$\displaystyle\langle q_{1}(T(\eta\otimes a))$
$\displaystyle,q_{1}(T^{-1}(\xi\otimes b))\rangle_{\mathscr{H}_{1}}+\langle
q_{1}(T^{-1}(\eta\otimes a)),q_{1}(T(\xi\otimes b))\rangle_{\mathscr{H}_{1}}$
$\displaystyle=\langle q_{2}(T(\eta\otimes a)),q_{2}(T^{-1}(\xi\otimes
b))\rangle_{\mathscr{H}_{2}}+\langle q_{2}(T^{-1}(\eta\otimes
a)),q_{2}(T(\xi\otimes b))\rangle_{\mathscr{H}_{2}}.$

By linearity and density of
$(\mathds{1}_{[\lambda_{1},\lambda_{2}]}(\Delta)L^{2}(M))\odot
M[\lambda_{1},\lambda_{2}]$ in $X$ we find that (4) holds.

The next part of the proof is to show that (4) implies that

$\langle q_{1}(x),q_{1}(y)\rangle_{\mathscr{H}_{1}}=\langle
q_{2}(x),q_{2}(y)\rangle_{\mathscr{H}_{2}}$

for $x,y\in X$. For this, we consider the operator $Te^{-sT^{2}}$ for $s>0$,
defined by holomorphic functional calculus. We start with the observation that

(5) $\displaystyle\begin{split}&\langle
q_{j}(T(Te^{-sT^{2}}(x))),q_{j}(T^{-1}(Te^{-sT^{2}}(y)))\rangle_{\mathscr{H}_{j}}+\langle
q_{j}(T^{-1}(Te^{-sT^{2}}(x))),q_{j}(T(Te^{-sT^{2}}(y)))\rangle_{\mathscr{H}_{j}}\\\
&\quad=-\frac{d}{ds}\langle
q_{j}(e^{-sT^{2}}(x)),q_{j}(e^{-sT^{2}}(y))\rangle_{\mathscr{H}_{j}}\end{split}$

for $j\in\\{1,2\\}$ and $x,y\in X$. Since $T$ is bounded and
$\mathrm{sp}(T)\subset(0,\infty)$, we know that $\lim_{s\to\infty}\lVert
e^{-sT^{2}}\rVert=0$ [Nee22, Theorem 6.24]. Consequently, we have for
$j\in\\{1,2\\}$ and $x,y\in X$ that

$-\lim_{r\to\infty}\int_{0}^{r}\frac{d}{ds}\langle
q_{j}(e^{-sT^{2}}(x)),q_{j}(e^{-sT^{2}}(y))\rangle_{\mathscr{H}_{j}}\,ds=\langle
q_{j}(x),q_{j}(y)\rangle_{\mathscr{H}_{j}}.$

Since the integrand of the above integral is equal for $j=1$ and $j=2$ by (4)
and (LABEL:eq:uniqueness_integrand), we deduce that

$\langle q_{1}(x),q_{1}(y)\rangle_{\mathscr{H}_{1}}=\langle
q_{2}(x),q_{2}(y)\rangle_{\mathscr{H}_{2}}$

for all $x,y\in X$. Therefore, we have for all
$\eta,\xi\in(\mathds{1}_{[\lambda_{1},\lambda_{2}]}(\Delta)L^{2}(M))$ and
$a,b\in M[\lambda_{1},\lambda_{2}]$ that

$\langle\delta_{1}(\eta)a,\delta_{1}(\xi)b\rangle_{\mathscr{H}_{1}}=\langle\delta_{2}(\eta)a,\delta_{2}(\xi)b\rangle_{\mathscr{H}_{2}}.$

So far we have shown that $\Theta$ preserves the inner product on certain
subsets of $\mathscr{H}_{1}$, and the goal is to extend this to all of
$\mathscr{H}_{1}$. This takes a few steps. First, note that $\Theta$ preserves
the inner product on all of

$\mathrm{lin}\\{\delta_{1}(\eta)a|\lambda_{1},\lambda_{2}>0,\eta\in(\mathds{1}_{[\lambda_{1},\lambda_{2}]}(\Delta)L^{2}(M)),a\in
M[\lambda_{1},\lambda_{2}]\\},$

since for each
$\eta_{1},\eta_{2}\in\bigcup_{\lambda_{1},\lambda_{2}>0}(\mathds{1}_{[\lambda_{1},\lambda_{2}]}(\Delta)L^{2}(M))$
and
$a_{1},a_{2}\in\bigcup_{\lambda_{1},\lambda_{2}>0}M[\lambda_{1},\lambda_{2}]$
we can find $\lambda_{1}^{\prime},\lambda_{2}^{\prime}>0$ such that
$\eta_{1},\eta_{2}\in(\mathds{1}_{[\lambda_{1}^{\prime},\lambda_{2}^{\prime}]}(\Delta)L^{2}(M))$
and $a_{1},a_{2}\in M[\lambda_{1}^{\prime},\lambda_{2}^{\prime}]$. Next, since
$\bigcup_{\lambda_{1},\lambda_{2}>0}(\mathds{1}_{[\lambda_{1},\lambda_{2}]}(\Delta)L^{2}(M))$
is dense in $L^{2}(M)$ and $\delta_{1}$ is bounded, we can extend this to
$\mathrm{lin}\\{\delta_{1}(\eta)a|\eta\in
L^{2}(M),a\in\bigcup_{\lambda_{1},\lambda_{2}>0}M[\lambda_{1},\lambda_{2}]\\}$
and subsequently to $\mathrm{lin}\\{\delta_{1}(\eta)a|\eta\in L^{2}(M),a\in
M\\}$ because $\bigcup_{\lambda_{1},\lambda_{2}>0}M[\lambda_{1},\lambda_{2}]$
is weak* dense in $M$. Lastly, by property (c) of a first-order differential
calculus we conclude that $\Theta$ is isometric on all of $\mathscr{H}_{1}$.

We will finish the proof by discussing the other desired properties of
$\Theta$. By property (c) of a first-order differential calculus,
$\mathrm{lin}\\{\delta_{2}(\eta)a|\eta\in L^{2}(M),a\in M\\}$ is dense in
$\mathscr{H}_{2}$. Because the image of an isometric map is closed, we know
that $\Theta$ is surjective and therefore that it is a linear isometric
isomorphism. By property (b) of a first-order differential calculus it is a
unitary bimodule map, and it is clear that it intertwines $\mathscr{J}_{1}$
and $\mathscr{J}_{2}$. ∎

##  5\. Derivations for Quantum Markov Semigroups with Unbounded Generators

In this section we study derivations for quantum Markov semigroups that are
not necessarily uniformly continuous. In this case the generator can be
unbounded, and it is convenient to work with the associated quadratic forms on
$L^{2}(M)$, which we call quantum Dirichlet forms. We show that the bounded
vectors in the form domain form an algebra (Theorem 5.2), which gives a
suitable domain for a derivation. We then show that there exists a (possibly
unbounded) first-order differential calculus associated with our given quantum
Dirichlet form (Theorem 5.4).

We keep the notation from the previous section. In particular, $M$ is a von
Neumann algebra and $\varphi$ is a normal faithful state on $M$.

Let us recall some basic definitions concerning quadratic forms on Hilbert
spaces. A quadratic form on a Hilbert space $H$ is a map $q\colon
H\to[0,\infty]$ such that

  * •

$q(\lambda\xi)=\lvert\lambda\rvert^{2}q(\xi)$

  * •

$q(\xi+\eta)+q(\xi-\eta)=2q(\xi)+2q(\eta)$

for all $\xi,\eta\in H$ and $\lambda\in\mathbb{C}$. The form $q$ is called
closed if it is lower semicontinuous and densely defined if
$\operatorname{dom}(q)=\\{\xi\in H\mid q(\xi)<\infty\\}$ is dense.

If $q$ is a quadratic form, it gives rise to a bilinear form on
$\operatorname{dom}(q)\times\operatorname{dom}(q)$ by the polarization
identity. Vice versa, the diagonal of a bilinear form extended by $\infty$ to
the complement of its domain is a quadratic form. We will use both viewpoints
interchangeably and denote both objects by the same symbol.

The generator of a closed densely defined quadratic form $q$ is the positive
self-adjoint operator $L$ given by

$\displaystyle\operatorname{dom}(L)$
$\displaystyle=\\{\xi\in\operatorname{dom}(q)\mid\exists\eta\in
H\,\forall\zeta\in\operatorname{dom}(q)\colon
q(\xi,\zeta)=\langle\eta,\zeta\rangle\\},$ $\displaystyle L\xi$
$\displaystyle=\eta.$

Conversely, if $L$ is a positive self-adjoint operator, then

$q\colon H\to[0,\infty],\,q(\xi)=\begin{cases}\lVert
L^{1/2}\xi\rVert^{2}&\text{if }\xi\in\operatorname{dom}(L^{1/2}),\\\
\infty&\text{otherwise}\end{cases}$

is a closed densely defined quadratic form with generator $q$.

To describe the quadratic forms associated with symmetric Markov semigroups,
we need the following piece of notation. For $a\in L^{2}(M)$ let
$a\wedge\varphi^{1/2}$ be the projection onto the closed convex cone
$\varphi^{1/2}-L^{2}_{+}(M)$.

######  Definition 5.1.

A closed densely defined quadratic form $\mathscr{E}\colon
L^{2}(M)\to[0,\infty]$ is called _conservative Dirichlet form_ if

  * •

$\mathscr{E}(Ja)=\mathscr{E}(a)$ for all $a\in L^{2}(M)$,

  * •

$\mathscr{E}(a\wedge\varphi^{1/2})\leq\mathscr{E}(a)$ for all $a\in L^{2}(M)$,

  * •

$\mathscr{E}(\varphi^{1/2})=0$.

The form $\mathscr{E}$ is called conservative completely Dirichlet form or
_quantum Dirichlet form_ if for every $n\in\mathbb{N}$ the quadratic form

$\mathscr{E}^{(n)}\colon
L^{2}(M_{n}(M))\to[0,\infty],\,\mathscr{E}^{(n)}([a_{jk}])=\sum_{j,k=1}^{n}\mathscr{E}(a_{jk})$

is a conservative Dirichlet form.

There is a one-to-one correspondence between quantum Dirichlet forms on
$L^{2}(M)$ and KMS-symmetric quantum Markov semigroups on $M$ (see [Cip97,
Theorem 4.11], [GL95, Theorem 5.7]): If $(\Phi_{t})$ is a KMS-symmetric
quantum Markov semigroup on $M$ and $\mathscr{L}_{2}$ the KMS implementation
of its generator on $L^{2}(M)$, then the quadratic form associated with
$\mathscr{L}_{2}$ is a quantum Dirichlet form. Vice versa, every quantum
Dirichlet form arises this way.

######  Theorem 5.2.

Let $M$ be a von Neumann algebra, $\varphi$ a normal faithful state on $M$ and
$\mathscr{E}$ a quantum Dirichlet form on $L^{2}(M)$. If
$a\in\operatorname{dom}(\mathscr{E})\cap M\varphi^{1/2}$,
$b\in\operatorname{dom}(\mathscr{E})\cap\varphi^{1/2}M$, then
$ab\in\operatorname{dom}(\mathscr{E})$ and

$\mathscr{E}(ab)^{1/2}\leq\lVert\pi_{l}(a)\rVert\mathscr{E}(b)^{1/2}+\mathscr{E}(a)^{1/2}\lVert\pi_{r}(a)\rVert.$

In particular, $\operatorname{dom}(\mathscr{E})\cap
M\varphi^{1/2}\cap\varphi^{1/2}M$ with the involution $J$ is a $\ast$-algebra.

###### Proof.

Let $(T_{t})$ be the strongly continuous semigroup associated with
$\mathscr{E}$ and let $\mathscr{E}_{t}(a)=\frac{1}{t}\langle
a,a-T_{t}a\rangle$ for $a\in L^{2}(M)$. By the spectral theorem,
$\mathscr{E}_{t}(a)\nearrow\mathscr{E}(a)$ as $t\searrow 0$.

Let $(\Phi_{t})$ be the KMS-symmetric quantum Markov semigroup on $M$
associated with $(T_{t})$. Since $\Phi_{t}$ is completely positive,
$\frac{1}{t}(I-\Phi_{t})$ is conditionally completely negative. Thus
$\frac{1}{t}(I-\Phi_{t})$ generates a quantum Markov semigroup on $M$, which
is clearly KMS-symmetric, and the associated symmetric Markov semigroup on
$L^{2}(M)$ has generator $\frac{1}{t}(I-T_{t})$.

By Theorem 4.2 there exists a bounded first-order differential calculus
$(\mathscr{H}_{t},\mathscr{J}_{t},\delta_{t})$ such that
$\mathscr{E}_{t}(a)=\lVert\delta_{t}(a)\rVert_{\mathscr{H}_{t}}^{2}$ for $a\in
L^{2}(M)$. Thus, if $a\in\operatorname{dom}(\mathscr{E})\cap M\varphi^{1/2}$
and $b\in\operatorname{dom}(\mathscr{E})\cap\varphi^{1/2}M$, then

$\displaystyle\mathscr{E}_{t}(ab)^{1/2}$
$\displaystyle=\lVert\delta_{t}(ab)\rVert_{\mathscr{H}_{t}}$
$\displaystyle=\lVert\pi_{l}(a)\delta_{t}(b)+\delta_{t}(a)J\pi_{r}(b)^{\ast}J\rVert_{\mathscr{H}_{t}}$
$\displaystyle=\lVert\pi_{l}(a)\rVert\lVert\delta_{t}(b)\rVert_{\mathscr{H}_{t}}+\lVert\delta(a)\rVert_{\mathscr{H}_{t}}\lVert\pi_{r}(b)\rVert$
$\displaystyle=\lVert\pi_{l}(a)\rVert\mathscr{E}_{t}(b)^{1/2}+\mathscr{E}(a)^{1/2}\lVert\pi_{r}(b)\rVert.$

The claim follows by taking the limit $t\searrow 0$ of both sides. ∎

######  Remark 5.3.

In contrast to the case of GNS-symmetric quantum Markov semigroups [Wir22a,
Theorem 6.3] we could not show that $\operatorname{dom}(\mathscr{E})\cap
M\varphi^{1/2}\cap\varphi^{1/2}M$ is a form core for $\mathscr{E}$. The space
$\operatorname{dom}(\mathscr{E})\cap\varphi^{1/4}M\varphi^{1/4}$ is always a
form core, but we do not expect it to be an algebra in general.

######  Theorem 5.4.

Let $M$ be a von Neumann algebra and $\varphi$ a faithful normal state on $M$.
If $\mathscr{E}$ is a quantum Dirichlet form on $L^{2}(M)$, then there exists
a Hilbert space $\mathscr{H}$ with commuting left and right actions of $M$, an
anti-unitary involution $\mathscr{J}\colon\mathscr{H}\to\mathscr{H}$ such that

$\mathscr{J}(x\xi y)=y^{\ast}(\mathscr{J}\xi)x^{\ast}$

for $x,y\in M$ and $\xi\in\mathscr{H}$, a closed operator
$\delta\colon\operatorname{dom}(\mathscr{E})\to\mathscr{H}$ such that
$\mathscr{J}\delta=\delta J$ and

$\delta(ab)=\pi_{l}(a)\cdot\delta(b)+\delta(a)\cdot J\pi_{r}(a)^{\ast}J$

for $a\in\operatorname{dom}(\mathscr{E})\cap M\varphi^{1/2}$,
$b\in\operatorname{dom}(\mathscr{E})\cap\varphi^{1/2}M$, and

$\mathscr{E}(a,b)=\langle\delta(a),\delta(b)\rangle_{\mathscr{H}}$

for $a,b\in\operatorname{dom}(\mathscr{E})$.

######  Remark 5.5.

By commuting left and right actions of $M$ on $\mathscr{H}$ we mean unital
$\ast$-homomorphisms $\pi_{l}^{\mathscr{H}}\colon M\to B(\mathscr{H})$,
$\pi_{r}^{\mathscr{H}}\colon M^{\mathrm{op}}\to B(\mathscr{H})$ with commuting
images. As usual, we write $x\xi y$ for
$\pi_{l}^{\mathscr{H}}(x)\pi_{r}^{\mathscr{H}}(y^{\mathrm{op}})\xi$.

We do not claim that the actions of $M$ on $\mathscr{H}$ are normal so that
$\mathscr{H}$ is in general not a correspondence. This is not only an artefact
of our proof or a feature of KMS symmetry specifically, but happens even for
symmetric Markov semigroups on commutative von Neumann algebras. For GNS-
symmetric quantum Markov semigroups, the normality of the action is linked to
the existence of the carré du champ (see [Wir22a, Section 7]).

###### Proof.

Let $(T_{t})$ be the strongly continuous semigroup associated with
$\mathscr{E}$. As in the proof of Theorem 5.2, we consider the quantum
Dirichlet form $\mathscr{E}_{t}$ given by
$\mathscr{E}_{t}(a)=\frac{1}{t}\langle a,a-T_{t}a\rangle$ and the associated
first-order differential calculus
$(\mathscr{H}_{t},\mathscr{J}_{t},\delta_{t})$. We will use an ultraproduct
construction to define $(\mathscr{H},\mathscr{J},\delta)$ for $\mathscr{E}$.

Choose $\omega\in\beta\mathbb{N}\setminus\mathbb{N}$ and a null sequence
$(t_{n})$ in $(0,\infty)$ and let $\mathscr{H}$ be the ultraproduct
$\prod_{n\to\omega}\mathscr{H}_{t_{n}}$. We write $[\xi_{n}]$ for the
equivalence class of $(\xi_{n})$ in $\mathscr{H}$.

We can define commuting left and right actions of $M$ on $\mathscr{H}$ by

$\displaystyle x[\xi_{n}]y=[x\xi_{n}y].$

As noted in the previous remark, these actions are not necessarily normal.

Moreover, let

$\mathscr{J}\colon\mathscr{H}\to\mathscr{H},\,\mathscr{J}[\xi_{n}]=[\mathscr{J}_{t_{n}}\xi_{n}].$

It is easy to verify that $\mathscr{J}$ is an anti-unitary involution such
that $\mathscr{J}(x\xi y)=y^{\ast}(\mathscr{J}\xi)x^{\ast}$ for $x,y\in M$ and
$\xi\in\mathscr{H}$.

Finally, if $a\in\operatorname{dom}(\mathscr{E})$, let
$\delta(a)=[\delta_{t_{n}}(a)]$. Since

$\lVert\delta_{t_{n}}(a)\rVert_{\mathscr{H}_{t_{n}}}^{2}=\mathscr{E}_{t_{n}}(a)\leq\mathscr{E}(a),$

the map $\delta\colon\operatorname{dom}(\mathscr{E})\to\mathscr{H}$ is well-
defined. Furthermore,

$\langle\delta(a),\delta(b)\rangle_{\mathscr{H}}=\lim_{n\to\omega}\langle\delta_{t_{n}}(a),\delta_{t_{n}}(b)\rangle_{\mathscr{H}_{t_{n}}}=\lim_{n\to\omega}\mathscr{E}_{t_{n}}(a,b)=\mathscr{E}(a,b)$

The operator $\delta$ is closed because $\mathscr{E}$ is closed. The other
properties of $\delta$ can be checked componentwise. ∎

######  Remark 5.6.

At this point we do not know if the triple $(\mathscr{H},\mathscr{J},\delta)$
is uniquely determined. Given the known results for tracially symmetric or
more generally GNS-symmetric quantum Markov semigroups, it seems reasonable to
suspect that the left and right action are not uniquely determined on $M$, but
only on some $\ast$-subalgebra.

In the case of tracially symmetric or more generally GNS-symmetric quantum
Markov semigroups it can be useful to realize the bimodule $\mathscr{H}$
inside $L^{2}(\hat{M})$ for some von Neumann algebra $\hat{M}$ containing $M$
with (faithful normal) expectation (see [JRS, Wir22a]). We will show that in
certain cases such a von Neumann algebra $\hat{M}$ can also be found for KMS-
symmetric semigroups, although the construction seems less canonical.

Clearly, a necessary condition for the existence is that $M$ acts normally on
$\mathscr{H}$. Since we do not have a criterion for this to happen in terms of
the semigroup, we formulate the result purely in terms of correspondences.

######  Proposition 5.7.

Let $M$ be a von Neumann algebra, $\varphi$ a faithful normal state on $M$ and
$(\mathscr{H},\mathscr{J})$ a self-dual $M$-$M$-correspondence. If $M$ is
semi-finite or $M$ has finite-dimensional center and $\mathscr{H}$ has finite
index, then there exists a von Neumann algebra $\hat{M}$ containing $M$, a
faithful normal conditional expectation $E\colon\hat{M}\to M$, and an
isometric bimodule map $V\colon\mathscr{H}\to L^{2}(\hat{M})$ such that
$V\mathscr{J}=J_{\varphi\circ E}V$,

###### Proof.

We will show that in either case there is a strongly continuous unitary group
$(\mathscr{U}_{t})$ on $\mathscr{H}$ such that
$\mathscr{U}_{t}^{\mathscr{H}}(x\xi
y)=\sigma_{t}^{\varphi}(x)(\mathscr{U}_{t}^{\mathscr{H}}\xi)\sigma_{t}^{\varphi}(y)$
for $x,y\in M$, $\xi\in\mathscr{H}$, $t\in\mathbb{R}$ and
$\mathscr{J}\mathscr{U}^{\mathscr{H}}_{t}=\mathscr{U}_{t}^{\mathscr{H}}\mathscr{J}$
for $t\in\mathbb{R}$. This makes
$(\mathscr{H},\mathscr{J},(\mathscr{U}_{t}^{\mathscr{H}}))$ into a Tomita
correspondence in the terminology of [Wir22a], and the result follows from
[Wir22a, Section 4.2].

First assume that $M$ is semi-finite and let $\tau$ be a normal semi-finite
faithful trace on $M$. There exists $\rho\in L^{1}_{+}(M,\tau)$ such that
$\varphi=\tau(\,\cdot\,\rho)$. Let
$\mathscr{U}_{t}^{\mathscr{H}}\xi=\rho^{it}\xi\rho^{-it}$. Since the actions
of $M$ on $\mathscr{H}$ are normal, $(\mathscr{U}_{t}^{\mathscr{H}})$ is a
strongly continuous unitary group. Moreover,

$\mathscr{U}_{t}^{\mathscr{H}}(x\xi y)=\rho^{it}x\xi
y\rho^{-it}=\sigma^{\varphi}_{t}(x)\rho^{it}\xi\rho^{-it}\sigma^{\varphi}_{t}(y)=\sigma^{\varphi}_{t}(x)(\mathscr{U}_{t}^{\mathscr{H}}\xi)\sigma^{\varphi}_{t}(y)$

and

$\mathscr{U}_{t}^{\mathscr{H}}\mathscr{J}\xi=\rho^{it}(\mathscr{J}\xi)\rho^{-it}=\mathscr{J}(\rho^{it}\xi\rho^{-it})=\mathscr{J}\mathscr{U}_{t}^{\mathscr{H}}\xi$

for $x,y\in M$, $\xi\in\mathscr{H}$ and $t\in\mathbb{R}$. This settles the
claim for semi-finite von Neumann algebras.

Now assume that $M$ has finite-dimensional center and $\mathscr{H}$ has finite
index. In this case we use Longo’s construction from [Lon18], which we briefly
recall.

Write $\pi_{l}^{\mathscr{H}}$ and $\pi_{r}^{\mathscr{H}}$ for the left and
right action of $M$ on $\mathscr{H}$ and let
$\varepsilon\colon\pi_{r}^{\mathscr{H}}(M)^{\prime}\to\pi_{l}^{\mathscr{H}}(M)$
be the minimal conditional expectation. Define $\Delta_{\mathscr{H}}$ to be
the spatial derivative of
$\varphi\circ(\pi_{l}^{\mathscr{H}})^{-1}\circ\varepsilon$ with respect to
$\varphi\circ(\pi_{r}^{\mathscr{H}})^{-1}$, that is,

$\Delta_{\mathscr{H}}=\frac{d(\varphi\circ(\pi_{l}^{\mathscr{H}})^{-1}\circ\varepsilon)}{d(\varphi\circ(\pi_{r}^{\mathscr{H}})^{-1})}.$

Here $(\pi_{l}^{\mathscr{H}})^{-1}$ and $(\pi_{r}^{\mathscr{H}})^{-1}$ are to
be understood as follows: There exist central projections $p,q\in M$ such that
$\ker\pi_{l}^{\mathscr{H}}=(1-p)M$, $\ker\pi_{r}^{\mathscr{H}}=(1-q)M$. Then
$\pi_{l}^{\mathscr{H}}$ restricted to $pM$ (resp. $\pi_{r}^{\mathscr{H}}$
restricted to $qM$) is injective, and we write $(\pi_{l}^{\mathscr{H}})^{-1}$
(resp. $(\pi_{r}^{\mathscr{H}})^{-1})$ for the inverse of this restriction.

By definition of the spatial derivative, $\Delta_{\mathscr{H}}$ is a non-
singular positive self-adjoint operator on $\mathscr{H}$.

Since $M$ has finite-dimensional center, there are central projections
$e_{1},\dots,e_{n}\in M$ such that $M\cap
M^{\prime}=\bigoplus_{j}\mathbb{C}e_{j}$. Let
$\mathscr{H}_{ij}=e_{i}\mathscr{H}e_{j}$. This is an
$e_{i}M$-$e_{j}M$-correspondence and
$\mathscr{H}=\bigoplus_{i,j}\mathscr{H}_{ij}$ as Hilbert spaces. Let
$d_{ij}=\sqrt{\mathrm{Ind}(\mathscr{H}_{ij})}$ and

$D_{\mathscr{H}}=\sum_{i,j}d_{ij}\pi_{l}^{\mathscr{H}}(e_{i})\pi_{r}^{\mathscr{H}}(e_{j}^{\mathrm{op}}).$

Finally let
$\mathscr{U}_{t}^{\mathscr{H}}=\Delta_{\mathscr{H}}^{it}D_{\mathscr{H}}^{it}$.
The unitary group $(\mathscr{U}_{t})$ satisfies

$\mathscr{U}_{t}(x\xi
y)=\sigma^{\varphi}_{t}(x)(\mathscr{U}_{t}\xi)\sigma^{\varphi}_{t}(y)$

for $x,y\in M$, $\xi\in\mathscr{H}$ and $t\in\mathbb{R}$ by [Lon18, Theorem
2.3].

Moreover, let

$T\colon\mathscr{H}\to\overline{\mathscr{H}},\,\xi\mapsto\overline{\mathscr{J}\xi}.$

By definition of $\mathscr{J}$, the map $T$ is a unitary bimodule map. It
follows from [Lon18, Theorem 2.3] that

$\overline{\mathscr{J}\mathscr{U}_{t}^{\mathscr{H}}\xi}=T\mathscr{U}_{t}^{\mathscr{H}}\xi=\mathscr{U}_{t}^{\bar{\mathscr{H}}}T\xi=\mathscr{U}_{t}^{\bar{\mathscr{H}}}\,\overline{\mathscr{J}\xi}=\overline{\mathscr{U}_{t}^{\mathscr{H}}\mathscr{J}\xi}$

for $\xi\in\mathscr{H}$ and $t\in\mathbb{R}$. Thus $\mathscr{J}$ commutes with
$(\mathscr{U}_{t}^{\mathscr{H}})$. ∎

##  Appendix A Alternative proof of Theorem 2.5

The proof of Theorem 2.5 shows that this result can be easily obtained from
Theorem 2.4, and therefore that one can prove it in a basis-independent
fashion. On the other hand, it gives very little intuition for the $V_{j}$
mentioned in the theorem. To provide a bit more feeling for the $V_{j}$, we
include an alternative proof of the result.

Recall that by [AC21, Theorem 4.4] the generator $\mathscr{L}$ of a KMS-
symmetric quantum Markov semigroup on $M_{n}(\mathbb{C})$ can be written as

($\ast$)
$\mathscr{L}(A)=(1+\sigma_{-i/2})^{-1}(\Psi(I_{n}))A+A(1+\sigma_{i/2})^{-1}(\Psi(I_{n}))-\Psi(A)$

with a KMS-symmetric completely positive map $\Phi\colon M_{n}(\mathbb{C})\to
M_{n}(\mathbb{C})$.

###### Alternative proof of Theorem 2.5.

Let $\Psi$ be a KMS-symmetric completely positive map such that ($\ast$ ‣ A)
holds, i.e.

$\mathscr{L}(A)=(1+\sigma_{-i/2})^{-1}(\Psi(I_{n}))A+A(1+\sigma_{i/2})^{-1}(\Psi(I_{n}))-\Psi(A).$

By Lemma 2.3(ii), $\check{\Psi}$ is also completely positive and KMS-
symmetric. Next, we define the completely positive map $\Xi$ by

$\Xi(A)=\rho^{1/4}\check{\Psi}(\rho^{-1/4}A\rho^{-1/4})\rho^{1/4}$

for all $A\in M_{n}{\mathbb{C}}$. Now for all $A,B\in M_{n}(\mathbb{C})$ we
have

$\mathrm{tr}(A\Xi(B))=\mathrm{tr}(\rho^{-1/4}A\rho^{-1/4}\rho^{1/2}\check{\Psi}(\rho^{-1/4}B\rho^{-1/4})\rho^{1/2})=\mathrm{tr}(\Xi(A)B)$

by the KMS-symmetry of $\check{\Psi}$.

Let $V_{1},\dots,V_{N}$ be the Kraus representation of $\Xi$, meaning that

$\Xi(A)=\sum_{j=1}^{N}V_{j}^{\ast}AV_{j}$

for all $A\in M_{n}(\mathbb{C})$. Since
$\mathrm{tr}(A\Xi(B))=\mathrm{tr}(\Xi(A)B)$, we can assume without loss of
generality that for each $j$ there exists an index $j^{\ast}$ such that
$V_{j}^{\ast}=V_{j^{\ast}}$ and $(j^{\ast})^{\ast}=j$. Note that
$\sigma_{-i/4}(V_{1}),\dots,\sigma_{-i/4}(V_{N})$ is a Kraus representation of
$\check{\Psi}$. Calculating $\sum_{j}\langle[V_{j},A],[V_{j},B]\rangle$ then
gives

$\displaystyle\sum_{j}\langle[V_{j},A],[V_{j},B]\rangle_{\rho}=$
$\displaystyle\sum_{j}\mathrm{tr}(\rho^{1/2}(A^{\ast}V_{j}^{\ast}-V_{j}^{\ast}A^{\ast})\rho^{1/2}(V_{j}B-BV_{j}))$
$\displaystyle=$
$\displaystyle\sum_{j}\mathrm{tr}(\rho^{1/2}A^{\ast}\rho^{1/2}(\sigma_{i/2}(V_{j}^{\ast})V_{j}B+BV_{j}\sigma_{-i/2}(V_{j})$
$\displaystyle\quad-\sigma_{i/2}(V_{j}^{\ast})BV_{j}-V_{j}B\sigma_{-i/2}(V_{j}^{\ast})))$
$\displaystyle=$
$\displaystyle\sum_{j}\mathrm{tr}(\rho^{1/2}A^{\ast}\rho^{1/2}(\sigma_{i/2}(V_{j}^{\ast})V_{j}B+BV_{j}^{\ast}\sigma_{-i/2}(V_{j})$
$\displaystyle\quad-\sigma_{i/2}(V_{j}^{\ast})BV_{j}-V_{j}^{\ast}B\sigma_{-i/2}(V_{j}))).$

We observe that

$\displaystyle\sum_{j}\sigma_{i/2}(V_{j}^{\ast})BV_{j}+V_{j}^{\ast}B\sigma_{-i/2}(V_{j})=$
$\displaystyle\sum_{j}\sigma_{i/4}(\sigma_{-i/4}(V_{j})^{\ast})B\sigma_{i/4}(\sigma_{-i/4}(V_{j}))$
$\displaystyle-\sigma_{-i/4}(\sigma_{-i/4}(V_{j})^{\ast})B\sigma_{-i/4}(\sigma_{-i/4}(V_{j}))$
$\displaystyle=$ $\displaystyle\mathscr{W}(\check{\Psi})(B)$

and consequently that

$(1+\sigma_{i/2})(\sum_{j}V_{j}^{\ast}\sigma_{-i/2}(V_{j}))=\sum_{j}\sigma_{i/2}(V_{j}^{\ast})V_{j}+V_{j}^{\ast}\sigma_{-i/2}(V_{j})=\mathscr{W}(\check{\Psi})(I_{n}).$

Therefore we have

$\displaystyle\sum_{j}V_{j}^{\ast}\sigma_{-i/2}(V_{j})$
$\displaystyle=(1+\sigma_{i/2})^{-1}(\Psi(I_{n})),$
$\displaystyle\sum_{j}\sigma_{i/2}(V_{j}^{\ast})V_{j}$
$\displaystyle=(1+\sigma_{-i/2})^{-1}(\Psi(I_{n})).$

Coming back to our original expression we find

$\displaystyle\sum\nolimits_{j}\langle[V_{j},A],[V_{j},B]\rangle_{\rho}$
$\displaystyle=\mathrm{tr}(\rho^{1/2}A^{\ast}\rho^{1/2}((1+\sigma_{-i/2})^{-1}(\Psi(I_{n}))B$
$\displaystyle\quad+B(1+\sigma_{i/2})^{-1}(\Psi(I_{n}))-\Psi(B)))$
$\displaystyle=\langle\mathscr{L}(A),B\rangle_{\rho}.\qed$

## References

  * [AC21] Érik Amorim and Eric A. Carlen  “Complete positivity and self-adjointness”  In _Linear Algebra Appl._ 611, 2021, pp. 389–439  DOI: 10.1016/j.laa.2020.10.038
  * [AC82] Luigi Accardi and Carlo Cecchini  “Conditional expectations in von Neumann algebras and a theorem of Takesaki”  In _J. Functional Analysis_ 45.2, 1982, pp. 245–273  DOI: 10.1016/0022-1236(82)90022-2
  * [AH77] Sergio Albeverio and Raphael Høegh-Krohn  “Dirichlet forms and Markov semigroups on $C^{*}$-algebras”  In _Comm. Math. Phys._ 56.2, 1977, pp. 173–187
  * [Ali76] Robert Alicki  “On the detailed balance condition for non-Hamiltonian systems”  In _Rep. Math. Phys._ 10.2, 1976, pp. 249–258  DOI: 10.1016/0034-4877(76)90046-X
  * [Arh23] Cédric Arhancet  “Spectral triples, Coulhon-Varopoulos dimension and heat kernel estimates”, 2023  arXiv:2209.12263
  * [AZ15] George Androulakis and Matthew Ziemke  “Generators of quantum Markov semigroups”  In _J. Math. Phys._ 56.8, 2015, pp. 08351216  DOI: 10.1063/1.4928936
  * [Bar17] Ivan Bardet  “Estimating the decoherence time using non-commutative Functional Inequalities”  arXiv, 2017  DOI: 10.48550/ARXIV.1710.01039
  * [Bia03] Philippe Biane  “Logarithmic Sobolev inequalities, matrix models and free entropy” International Workshop on Operator Algebra and Operator Theory (Linfen, 2001)  In _Acta Math. Sin. (Engl. Ser.)_ 19.3, 2003, pp. 497–506  DOI: 10.1007/s10114-003-0271-5
  * [BK19] Fabrice Baudoin and Daniel J. Kelleher  “Differential one-forms on Dirichlet spaces and Bakry-Émery estimates on metric graphs”  In _Trans. Amer. Math. Soc._ 371.5, 2019, pp. 3145–3178  DOI: 10.1090/tran/7362
  * [Bla06] Bruce Blackadar  “Operator algebras” Theory of $C^{*}$-algebras and von Neumann algebras, Operator Algebras and Non-commutative Geometry, III 122, Encyclopaedia of Mathematical Sciences  Springer-Verlag, Berlin, 2006, pp. xx+517  DOI: 10.1007/3-540-28517-2
  * [Cas21] Martijn Caspers  “Gradient forms and strong solidity of free quantum groups”  In _Math. Ann._ 379.1-2, 2021, pp. 271–324  DOI: 10.1007/s00208-020-02109-y
  * [CE79] Erik Christensen and David E. Evans  “Cohomology of operator algebras and quantum dynamical semigroups”  In _J. London Math. Soc. (2)_ 20.2, 1979, pp. 358–368  DOI: 10.1112/jlms/s2-20.2.358
  * [CFK14] Fabio Cipriani, Uwe Franz and Anna Kula  “Symmetries of Lévy processes on compact quantum groups, their Markov semigroups and potential theory”  In _J. Funct. Anal._ 266.5, 2014, pp. 2789–2844  DOI: 10.1016/j.jfa.2013.11.026
  * [Cip97] Fabio Cipriani  “Dirichlet forms and Markovian semigroups on standard forms of von Neumann algebras”  In _J. Funct. Anal._ 147.2, 1997, pp. 259–300  DOI: 10.1006/jfan.1996.3063
  * [CIW21] Martijn Caspers, Yusuke Isono and Mateusz Wasilewski  “$L_{2}$-cohomology, derivations, and quantum Markov semi-groups on $q$-Gaussian algebras”  In _Int. Math. Res. Not. IMRN_ , 2021, pp. 6405–6441  DOI: 10.1093/imrn/rnaa044
  * [CM17] Eric A. Carlen and Jan Maas  “Gradient flow and entropy inequalities for quantum Markov semigroups with detailed balance”  In _J. Funct. Anal._ 273.5, 2017, pp. 1810–1869  DOI: 10.1016/j.jfa.2017.05.003
  * [CM20] Eric A. Carlen and Jan Maas  “Non-commutative Calculus, Optimal Transport and Functional Inequalities in Dissipative Quantum Systems”  In _J. Stat. Phys._ 178.2, 2020, pp. 319–378  DOI: 10.1007/s10955-019-02434-w
  * [Con94] Alain Connes  “Noncommutative geometry”  Academic Press, Inc., San Diego, CA, 1994, pp. xiv+661
  * [CS03] Fabio Cipriani and Jean-Luc Sauvageot  “Derivations as square roots of Dirichlet forms”  In _J. Funct. Anal._ 201.1, 2003, pp. 78–120  DOI: 10.1016/S0022-1236(03)00085-5
  * [CS03a] Fabio Cipriani and Jean-Luc Sauvageot  “Noncommutative potential theory and the sign of the curvature operator in Riemannian geometry”  In _Geom. Funct. Anal._ 13.3, 2003, pp. 521–545  DOI: 10.1007/s00039-003-0421-z
  * [CS15] Martijn Caspers and Adam Skalski  “The Haagerup approximation property for von Neumann algebras via quantum Markov semigroups and Dirichlet forms”  In _Comm. Math. Phys._ 336.3, 2015, pp. 1637–1664  DOI: 10.1007/s00220-015-2302-3
  * [CS17] Fabio Cipriani and Jean-Luc Sauvageot  “Amenability and subexponential spectral growth rate of Dirichlet forms on von Neumann algebras”  In _Adv. Math._ 322, 2017, pp. 308–340  DOI: 10.1016/j.aim.2017.10.017
  * [CZ76] Ioana Ciorănescu and László Zsidó  “Analytic generators for one-parameter groups”  In _Tohoku Math. J. (2)_ 28.3, 1976, pp. 327–362  DOI: 10.2748/tmj/1178240775
  * [Dab10] Yoann Dabrowski  “A non-commutative Path Space approach to stationary free Stochastic Differential Equations”  arXiv, 2010  arXiv:1006.4351 [math.OA]
  * [Dav79] E.Brian Davies  “Generators of dynamical semigroups”  In _Journal of Functional Analysis_ 34.3, 1979, pp. 421–432  DOI: https://doi.org/10.1016/0022-1236(79)90085-5
  * [DI16] Yoann Dabrowski and Adrian Ioana  “Unbounded derivations, free dilations, and indecomposability results for $\rm II_{1}$ factors”  In _Trans. Amer. Math. Soc._ 368.7, 2016, pp. 4525–4560  DOI: 10.1090/tran/6470
  * [DL92] E. Davies and J. Lindsay  “Noncommutative symmetric Markov semigroups”  In _Math. Z._ 210.3, 1992, pp. 379–411  DOI: 10.1007/BF02571804
  * [DL93] E. Davies and J. Lindsay  “Superderivations and symmetric Markov semigroups”  In _Comm. Math. Phys._ 157.2, 1993, pp. 359–370  URL: http://projecteuclid.org.libraryproxy.ista.ac.at/euclid.cmp/1104253944
  * [EL77] Edward G. Effros and E. Lance  “Tensor products of operator algebras”  In _Adv. Math._ 25.1, 1977, pp. 1–34  DOI: 10.1016/0001-8708(77)90085-8
  * [EL77a] David E. Evans and John T. Lewis  “Dilations of irreversible evolutions in algebraic quantum theory”  In _Communications Dublin Inst. Advanced Studies. Ser. A_ , 1977, pp. v+104
  * [FU10] Franco Fagnola and Veronica Umanità  “Generators of KMS symmetric Markov semigroups on $B(h)$ symmetry and quantum detailed balance”  In _Comm. Math. Phys._ 298.2, 2010, pp. 523–547  DOI: 10.1007/s00220-010-1011-1
  * [GKS76] Vittorio Gorini, Andrzej Kossakowski and E.. Sudarshan  “Completely positive dynamical semigroups of $N$-level systems”  In _J. Mathematical Phys._ 17.5, 1976, pp. 821–825  DOI: 10.1063/1.522979
  * [GL95] Stanisław Goldstein and J. Lindsay  “KMS-symmetric Markov semigroups”  In _Math. Z._ 219.4, 1995, pp. 591–608  DOI: 10.1007/BF02572383
  * [GR22] Li Gao and Cambyse Rouzé  “Complete entropic inequalities for quantum Markov chains”  In _Arch. Ration. Mech. Anal._ 245.1, 2022, pp. 183–238  DOI: 10.1007/s00205-022-01785-1
  * [Haa75] Uffe Haagerup  “The standard form of von Neumann algebras”  In _Math. Scand._ 37.2, 1975, pp. 271–283  DOI: 10.7146/math.scand.a-11606
  * [Haa85] Uffe Haagerup  “Injectivity and decomposition of completely bounded maps”  In _Operator algebras and their connections with topology and ergodic theory (Buşteni, 1983)_ 1132, Lecture Notes in Math.  Springer, Berlin, 1985, pp. 170–222  DOI: 10.1007/BFb0074885
  * [HJX10] Uffe Haagerup, Marius Junge and Quanhua Xu  “A reduction method for noncommutative $L_{p}$-spaces and applications”  In _Trans. Amer. Math. Soc._ 362.4, 2010, pp. 2125–2165  DOI: 10.1090/S0002-9947-09-04935-6
  * [Hol95] Alexander S. Holevo  “On the structure of covariant dynamical semigroups”  In _J. Funct. Anal._ 131.2, 1995, pp. 255–278  DOI: 10.1006/jfan.1995.1089
  * [HT13] Michael Hinz and Alexander Teplyaev  “Vector analysis on fractals and applications”  In _Fractal geometry and dynamical systems in pure and applied mathematics. II. Fractals in applied mathematics_ 601, Contemp. Math.  Amer. Math. Soc., Providence, RI, 2013, pp. 147–163  DOI: 10.1090/conm/601/11960
  * [JMP14] Marius Junge, Tao Mei and Javier Parcet  “An invitation to harmonic analysis associated with semigroups of operators”  In _Harmonic analysis and partial differential equations_ 612, Contemp. Math.  Amer. Math. Soc., Providence, RI, 2014, pp. 107–122  DOI: 10.1090/conm/612/12227
  * [JRS] Marius Junge, Eric Ricard and Dimitri Shlyakhtenko  “Noncommutative diffusion semigroups and free probability” Unpublished preprint
  * [JX07] Marius Junge and Quanhua Xu  “Noncommutative maximal ergodic theorems”  In _J. Amer. Math. Soc._ 20.2, 2007, pp. 385–439  DOI: 10.1090/S0894-0347-06-00533-9
  * [JZ15] Marius Junge and Qiang Zeng  “Noncommutative martingale deviation and Poincaré type inequalities with applications”  In _Probab. Theory Related Fields_ 161.3-4, 2015, pp. 449–507  DOI: 10.1007/s00440-014-0552-1
  * [Kap53] Irving Kaplansky  “Modules over operator algebras”  In _Amer. J. Math._ 75, 1953, pp. 839–858  DOI: 10.2307/2372552
  * [KB16] Michael J. Kastoryano and Fernando G… Brandão  “Quantum Gibbs samplers: the commuting case”  In _Comm. Math. Phys._ 344.3, 2016, pp. 915–957  DOI: 10.1007/s00220-016-2641-8
  * [Kos72] Andrzej Kossakowski  “On quantum statistical mechanics of non-Hamiltonian systems”  In _Reports on Mathematical Physics_ 3.4, 1972, pp. 247–274  DOI: https://doi.org/10.1016/0034-4877(72)90010-9
  * [Lin76] Göran Lindblad  “On the generators of quantum dynamical semigroups”  In _Comm. Math. Phys._ 48.2, 1976, pp. 119–130  URL: http://projecteuclid.org/euclid.cmp/1103899849
  * [Lon18] Roberto Longo  “On Landauer’s principle and bound for infinite systems”  In _Comm. Math. Phys._ 363.2, 2018, pp. 531–560  DOI: 10.1007/s00220-018-3116-x
  * [Nee22] Jan Neerven  “Functional Analysis”, Cambridge Studies in Advanced Mathematics  Cambridge University Press, 2022  DOI: 10.1017/9781009232487
  * [Pet09] Jesse Peterson  “$L^{2}$-rigidity in von Neumann algebras”  In _Invent. Math._ 175.2, 2009, pp. 417–433  DOI: 10.1007/s00222-008-0154-6
  * [Pet09a] Jesse Peterson  “A 1-cohomology characterization of property (T) in von Neumann algebras”  In _Pacific J. Math._ 243.1, 2009, pp. 181–199  DOI: 10.2140/pjm.2009.243.181
  * [Pis20] Gilles Pisier  “Tensor products of $C^{*}$-algebras and operator spaces—the Connes-Kirchberg problem” 96, London Mathematical Society Student Texts  Cambridge University Press, Cambridge, 2020, pp. x+484  DOI: 10.1017/9781108782081
  * [Sau96] Jean-Luc Sauvageot  “Semi-groupe de la chaleur transverse sur la $C^{*}$-algèbre d’un feuilletage riemannien”  In _J. Funct. Anal._ 142.2, 1996, pp. 511–538  DOI: 10.1006/jfan.1996.0158
  * [Sch69] Martin Schechter  “On the spectra of operators on tensor products”  In _Journal of Functional Analysis_ 4.1, 1969, pp. 95–99  DOI: https://doi.org/10.1016/0022-1236(69)90024-X
  * [Ver22] Matthijs Vernooij  “On the existence of derivations as square roots of generators of state-symmetric quantum Markov semigroups”, 2022  arXiv:2203.12307
  * [Wir20] Melchior Wirth  “A Noncommutative Transport Metric and Symmetric Quantum Markov Semigroups as Gradient Flows of the Entropy”  arXiv, 2020  DOI: 10.48550/ARXIV.1808.05419
  * [Wir22] Melchior Wirth  “Christensen–Evans theorem and extensions of GNS-symmetric quantum Markov semigroups”  arXiv, 2022  DOI: 10.48550/ARXIV.2203.00341
  * [Wir22a] Melchior Wirth  “The Differential Structure of Generators of GNS-symmetric Quantum Markov Semigroups”  arXiv, 2022  DOI: 10.48550/ARXIV.2207.09247
  * [WZ21] Melchior Wirth and Haonan Zhang  “Complete gradient estimates of quantum Markov semigroups”  In _Comm. Math. Phys._ 387.2, 2021, pp. 761–791  DOI: 10.1007/s00220-021-04199-4