# Modular Structure and Inclusions of Twisted Araki-Woods Algebras

Ricardo Correa da Silva  Department Mathematik, FAU Erlangen-Nürnberg,
<EMAIL_ADDRESS>and  Gandalf Lechner  Department Mathematik, FAU
Erlangen-Nürnberg<EMAIL_ADDRESS>
(Date: November 30, 2022)

###### Abstract.

In the general setting of twisted second quantization (including Bose/Fermi
second quantization, $S$-symmetric Fock spaces, and full Fock spaces from free
probability as special cases), von Neumann algebras on twisted Fock spaces are
analyzed. These twisted Araki-Woods algebras $\mathcal{L}_{T}(H)$ depend on
the twist operator $T$ and a standard subspace $H$ in the one-particle space.
Under a compatibility assumption on $T$ and $H$, it is proven that the Fock
vacuum is cyclic and separating for $\mathcal{L}_{T}(H)$ if and only if $T$
satisfies a standard subspace version of crossing symmetry and the Yang-Baxter
equation (braid equation). In this case, the Tomita-Takesaki modular data are
explicitly determined.

Inclusions $\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$ of twisted Araki-
Woods algebras are analyzed in two cases: If the inclusion is half-sided
modular and the twist satisfies a norm bound, it is shown to be singular. If
the inclusion of underlying standard subspaces $K\subset H$ satisfies an
$L^{2}$-nuclearity condition, $\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$
has type III relative commutant for suitable twists $T$.

Applications of these results to localization of observables in algebraic
quantum field theory are discussed.

###### Contents

  1. 1 Introduction
  2. 2 Twisted Fock spaces and Araki-Woods algebras
    1. 2.1 Twists and twisted Fock spaces
    2. 2.2 Standard subspaces and twisted Araki-Woods algebras
  3. 3 Twisted Araki-Woods algebras and standard vectors
    1. 3.1 Analytic continuations of twisted $n$-point functions
    2. 3.2 Standardness and modular properties of $(\mathcal{L}_{T}(H),\Omega)$
  4. 4 Inclusions of twisted Araki-Woods algebras
    1. 4.1 Half-sided modular inclusions
    2. 4.2 $L^{2}$-nuclearity and quasi-split inclusions
  5. 5 Applications to quantum field theory
  6. A Diagram calculus for twisted $n$-point functions

##  1\. Introduction

Second quantization von Neumann algebras play a prominent role in many areas
of mathematical physics and mathematics and exist in numerous variations: Weyl
algebras and their weak closures describe Bosonic interaction-free systems
such as Bose gases or free quantum field theories, and similarly CAR algebras
describe the corresponding Fermionic models [BR97, Pet90]. Whereas these
algebras are naturally represented on Bose/Fermi Fock spaces, there exist also
more general Fock spaces that are useful for describing generalized
statistics, including anyons [LM95a, DKLP20]. Very similar spaces – sometimes
called $S$-symmetric Fock spaces – arise from other representations of
symmetric groups on tensor powers and form convenient representation spaces
for integrable quantum field theories with prescribed two-body scattering
operator $S$, and carry families of von Neumann algebras generalizing the
CCR/CAR setting [Lec08, AL17]. Related constructions also occur in the
representation theory of Wick algebras, where the canonical (anti-)commutation
relations are deformed [JSW95, JPS01].

Another variation of second quantization algebras arises in free probability:
Here one considers representations of free group factors [Voi85] and other von
Neumann algebras [Shl97] (free Araki-Woods factors) on unsymmetrized
(“Boltzmann” or “full”) Fock spaces, and this generalizes to $q$-deformed
Araki-Woods von Neumann algebras, interpolating between the Bose $(q=1)$ and
Fermi $(q=-1)$ situations, and containing the free factors at $q=0$ as a
special case.

In this article, we adopt a general framework that includes all these examples
and goes beyond the setting of Fock spaces symmetrized by symmetric group
actions. Our main goal is to study natural families of von Neumann algebras
and their modular and inclusion properties in this setting. As our analysis is
motivated from quantum field theory, we focus on properties of the Fock vacuum
state on these algebras, and consider the structure of inclusions of such von
Neumann algebras rather than their internal structure.

The basic setup we use is due to Bożejko and Speicher [BS94] and Jørgensen,
Schmitt, and Werner [JSW95] and can briefly be described as follows: Starting
from a one-particle Hilbert space $\mathcal{H}$ and a bounded selfadjoint
operator $T$ on $\mathcal{H}\otimes\mathcal{H}$ satisfying a certain
positivity condition, one can construct a “$T$-twisted Fock space”
$\mathcal{F}_{T}(\mathcal{H})$ which specializes to the Bose, Fermi,
Boltzmann, or $S$-symmetric Fock spaces for suitable choices of $T$.

Interesting von Neumann algebras
$\mathcal{L}_{T}(H)\subset\mathcal{B}(\mathcal{F}_{T}(\mathcal{H}))$ are
generated by “field operators” $\phi_{L,T}(h)=a_{L,T}(h)+a_{L,T}^{\star}(h)$,
defined as a sum of the creation and annihilation type operators that exist on
$\mathcal{F}_{T}(\mathcal{H})$. Here $h$ ranges over a standard
subspace111Recall that a standard subspace $H$ of a complex Hilbert space
$\mathcal{H}$ is a real linear closed subspace such that $H+iH$ is dense in
$\mathcal{H}$ and $H\cap iH=\\{0\\}$, see Section 2. $H\subset\mathcal{H}$,
and the index $L$ indicates that this construction is based on operators
acting on the left. Depending on the twist $T$ and the standard subspace $H$,
the algebra $\mathcal{L}_{T}(H)$ can take various different forms.

In applications to quantum field theory, the standard subspaces serve as a
means to encode localization regions in some spacetime (see [LRT78, Foi83] for
the CCR/CAR case and [BGL02] for the concept of modular localization). Hence
one is immediately interested in inclusion and intersection properties of,
say, two von Neumann algebras $\mathcal{L}_{T}(H)$ and $\mathcal{L}_{T}(K)$.
In the context of free probability, Voiculescu’s original approach corresponds
to taking $T=0$ and $H$ as the closed real span of an orthonormal basis, in
which case $\mathcal{L}_{0}(H)$ is the free group factor of the free group on
$\operatorname{dim}\mathcal{H}$ generators. Shlyakhtenko has generalized this
setting to $T=0$ and general standard subspaces $H$, which corresponds to
choosing an orthogonal one-parameter group on a related real Hilbert space.
Also in this setting and its $q$-variations, one is interested in certain
inclusions of von Neumann algebras, for instance when studying MASAs [BM17,
BM22].

For general twist $T$, only very little is known about the structure of
$H\mapsto\mathcal{L}_{T}(H)$, which provides some of the motivation for this
article. We focus on two interrelated questions: On the one hand, we analyze
under which conditions the Fock vacuum $\Omega\in\mathcal{F}_{T}(\mathcal{H})$
is cyclic and separating for $\mathcal{L}_{T}(H)$, and what the corresponding
modular operators from Tomita-Takesaki theory are in this case. On the other
hand, we consider inclusions $\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$
arising from inclusions of standard subspaces $K\subset H$ and investigate
their relative commutants.

In Section 2, we introduce $T$-twisted Fock spaces and our von Neumann
algebras $\mathcal{L}_{T}(H)$ (Def. 2.8), called $T$-twisted Araki-Woods
algebras, as well as some instructive examples.

In Section 3 we address the standardness question of
$(\mathcal{L}_{T}(H),\Omega)$. It is easy to show that $\Omega$ is always
cyclic, but usually not separating. Our analysis is then based on a
compatibility assumption between $T$ and $H$ (namely $T$ should commute with
the modular unitaries $\Delta_{H}^{it}\otimes\Delta_{H}^{it}$, see Def. 3.1)
and two properties that $T$ might or might not have: crossing symmetry and the
Yang-Baxter equation. Crossing symmetry is a concept originating from
scattering theory; in our abstract setting it amounts to an analytic
continuation property of $T$ w.r.t. the modular group of $H$ (Def. 3.3). The
Yang-Baxter equation, on the other hand, is an algebraic equation that
expresses the essential braiding relation of the braid groups in tensor
product form.

It is shown that for compatible twists, $\Omega$ separates
$\mathcal{L}_{T}(H)$ if and only if $T$ is crossing symmetric w.r.t. $H$ and
satisfies the Yang-Baxter equation (Theorem 3.12 and Theorem 3.22). Both these
properties, the crossing symmetry and the Yang-Baxter equation, have their
origins in quantum physics and are often taken as an input in quantum field
theoretic constructions (see, for example, [LM95b, LS14, BT15, AL17, HL18,
DKLP20]). Here we do not need to assume these structures, but can rather
derive them from modular theory and actually show that they are equivalent to
the separating property of the Fock vacuum in our setting. On a technical
level, this relies on various analytic continuation arguments that are related
to the KMS condition. These arguments also involve the combinatorial structure
of $T$-twisted $n$-point functions which is best captured in a diagrammatic
form. This diagram calculus is presented in the appendix (Section A).

It turns out that $\Omega$ is cyclic and separating for $\mathcal{L}_{T}(H)$
if and only if corresponding “right” field operators $\phi_{R,T}(h)$ (3.47)
and right von Neumann algebras $\mathcal{R}_{T}(H)$ exist on
$\mathcal{F}_{T}(\mathcal{H})$ and suitably commute with the left operators.
In Proposition 3.25 we then determine the modular data
$(\mathcal{L}_{T}(H),\Omega)$ in this case, which are linked to the modular
data of $H$ via a $T$-twisted second quantization. In particular, we obtain a
left-right duality of the form

$\displaystyle\mathcal{L}_{T}(H)^{\prime}=\mathcal{R}_{T}(H^{\prime}).$ (1.1)

This generalizes several results known in special cases [EO73, LRT78, Shl97,
BJL02, BL04, Lec12].

In Section 4 we turn our attention to relative properties of the two families
of von Neumann algebras, $\mathcal{L}_{T}(H)$ and $\mathcal{R}_{T}(K)$. This
includes in particular the study of inclusions. Namely, we take an inclusion
of standard subspaces $K\subset H$ and consider the corresponding inclusion
$\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$ of von Neumann algebras and its
relative commutant $\mathcal{R}_{T}(K^{\prime})\cap\mathcal{L}_{T}(H)$. We can
determine the structure of such an inclusion in two completely different
cases, showing that $\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$ crucially
depends on $T$ and $K\subset H$.

The first case is that of a so-called half-sided modular inclusion [Wie93b,
Wie93a, AZ05]. As we recall in Section 4.1, the property of being half-sided
modular means that the small algebra arises from the large one by application
of a translation action with certain properties. It turns out that in this
situation, a recently developed criterion for determining whether a half-sided
inclusion is singular (trivial relative commutant) [LS22] is applicable in
case our twist is compatible with the inclusion and satisfies $\|T\|<1$. The
latter condition rules out familiar cases such as Bose/Fermi symmetry, for
which the inclusion is non-singular. In Theorem 4.9 we can therefore establish
many new examples of singular half-sided modular inclusions. After recent more
complicated constructions of such inclusions, using free products [LTU19] and
deformation procedures [LS22], respectively, these appear to be a much more
transparent. This shows that the relative commutant is unstable and very
sensitive to perturbations in $T$.

The results on small (trivial) relative commutants mentioned so far can be
interpreted as a consequence of the very non-commutative structure arising
from free probability (at $T=0$) respectively its “neighbourhood” (at
$\|T\|<1$). Nonetheless, such twists can also lead to very different
inclusions which might appear counter-intuitive at first. In Section 4.2, we
consider standard subspaces $K\subset H$ satisfying the $L^{2}$-nuclearity
condition of Buchholz, D’Antoni, Longo [BDL07] and use this tool and the
theory of (quasi-)split inclusions [DL84, Fid01] to give examples for the
relative commutant $\mathcal{L}_{T}(K)^{\prime}\cap\mathcal{L}_{T}(H)$ being
type III (Proposition 4.11). This section is related to work of D’Antoni,
Longo, and Radulescu [DLR01] who considered the case $T=0$, and is generalized
to other twists here. See also the paper [BM22] by Bikram and Mukherjee for
related arguments in the case of $q$-deformed Araki-Woods von Neumann
algebras.

In Section 5, we explain how our results apply to quantum field theoretic
models, in particular regarding the absence or presence of local observables
in quantum field theories on Minkowski space or the lightray, using an
abstract notion of spacetime. We discuss our results from this point of view,
review the resulting perspectives on constructive algebraic quantum field
theory, and give an outlook to ongoing research.

##  2\. Twisted Fock spaces and Araki-Woods algebras

###  2.1. Twists and twisted Fock spaces

The von Neumann algebras we are interested in are defined on twisted Fock
spaces. These Fock spaces arise as natural representation spaces of algebras
defined by a quadratic exchange relation (Wick algebras) and exist in various
versions. The version that we use here is the most general one and due to
Bożejko and Speicher [BS94] and Jørgensen, Schmitt, and Werner [JSW95].

Let $T\in\mathcal{B}(\mathcal{H}\otimes\mathcal{H})$ be an operator with
$\|T\|\leq 1$. We iteratively define operators
$R_{T,n},P_{T,n}\in\mathcal{B}(\mathcal{H}^{\otimes n})$, $n\in\mathbb{N}$,
by222Here and hereafter, we use the notation $T_{k}:=1^{\otimes(k-1)}\otimes
T\otimes 1^{\otimes(n-k-1)}\in\mathcal{B}(\mathcal{H}^{\otimes n})$.

$\displaystyle R_{T,n}$ $\displaystyle:=1+T_{1}+T_{1}T_{2}+\ldots+T_{1}\cdots
T_{n-1},$ (2.1) $\displaystyle P_{T,1}$ $\displaystyle:=1,\qquad
P_{T,n+1}:=(1\otimes P_{T,n})R_{T,n+1}.$ (2.2)

If $T$ is selfadjoint, one can show by induction in $n$ that also $P_{T,n}$ is
selfadjoint for any $n\in\mathbb{N}$. Those $T$ that lead to positive
$P_{T,n}$ will be called “twists”:

######  Definition 2.1.

A twist is an operator in

$\displaystyle{\mathcal{T}}_{\geq}$
$\displaystyle:=\\{T\in\mathcal{B}(\mathcal{H}\otimes\mathcal{H})\,\colon\,\|T\|\leq
1,\quad P_{T,n}\geq 0\;\text{ for all }n\in\mathbb{N}\\}.$ (2.3)

A strict twist is an operator in

$\displaystyle{\mathcal{T}}_{>}$
$\displaystyle:=\\{T\in{\mathcal{T}}_{\geq}\,\colon\,0\not\in\sigma(P_{T,n})\;\text{
for all }n\in\mathbb{N}\\}.$ (2.4)

We note that any twist $T$ is selfadjoint because $P_{T,2}=1+T$ is required to
be positive. To avoid misunderstanding, we emphasize that the spectral
assumption on strict twists $T\in{\mathcal{T}}_{>}$ means that for each
$n\in\mathbb{N}$, there exists an $\varepsilon_{n}>0$ such that
$P_{T,n}\geq\varepsilon_{n}1>0$.

The following theorem summarizes some known sufficient conditions for
$T\in\mathcal{B}(\mathcal{H}\otimes\mathcal{H})$ to lie in
${\mathcal{T}}_{\geq}$ or ${\mathcal{T}}_{>}$. Items a) and b) are due to
Jørgensen, Schmitt, and Werner [JSW95, Thm. 2.3.2 and Thm. 2.5.1], and item c)
is due to Bożejko and Speicher [BS94, Thm. 2.2].

######  Theorem 2.2.

Let $T=T^{*}\in\mathcal{B}(\mathcal{H}\otimes\mathcal{H})$.

  1. a)

If $\|T\|\leq\frac{1}{2}$, then333In [JSW95, Thm. 2.3.2] this conclusion is
shown under the slightly stronger assumption $\|T\|<\frac{1}{2}$, but the
arguments given there carry over to $\|T\|\leq\frac{1}{2}$.
$T\in{\mathcal{T}}_{>}$.

  2. b)

If $T\geq 0$, then $T\in{\mathcal{T}}_{>}$.

  3. c)

If $\|T\|\leq 1$ and $T$ satisfies the Yang-Baxter equation in its braid form,
i.e.

$\displaystyle T_{1}T_{2}T_{1}=T_{2}T_{1}T_{2},$ (2.5)

then $T\in{\mathcal{T}}_{\geq}$. We will refer to such twists as braided
twists. If a braided twist satisfies $\|T\|<1$, then $T\in{\mathcal{T}}_{>}$.

To construct twisted Fock spaces from a twist $T\in{\mathcal{T}}_{\geq}$, we
consider the quotient

$\displaystyle\mathcal{H}_{T,n}^{0}:=\mathcal{H}^{\otimes
n}/\operatorname{ker}P_{T,n}=\overline{{\rm Ran}P_{T,n}},\qquad
n\in\mathbb{N},$ (2.6)

where the bar denotes the closure in the canonical Hilbert norm of
$\mathcal{H}^{\otimes n}$. On this space we introduce a new (positive
definite) scalar product

$\displaystyle\langle\,\cdot\,,\,\cdot\,\rangle_{T}:=\langle\,\cdot\,,P_{T,n}\,\cdot\,\rangle,$
(2.7)

and define $\mathcal{H}_{T,n}$ as the completion of $\mathcal{H}_{T,n}^{0}$
w.r.t. the corresponding norm $\|\cdot\|_{T}$, written as
$\mathcal{H}_{T,n}=\overline{\mathcal{H}_{T,n}^{0}}^{T}=\overline{{\rm
Ran}P_{T,n}}^{T}$. In general, $\mathcal{H}_{T,n}$ is not a subspace of
$\mathcal{H}^{\otimes n}$, but, in case
$P_{T,n}|_{(\operatorname{ker}P_{T,n})^{\perp}}$ does not contain $0$ in its
spectrum, $P_{T,n}|_{(\operatorname{ker}P_{T,n})^{\perp}}$ has a bounded
inverse and the norms $\|\cdot\|$ and $\|\cdot\|_{T}$ are equivalent on
$(\operatorname{ker}P_{T,n})^{\perp}$, i.e. $\mathcal{H}_{T,n}=\overline{{\rm
Ran}P_{T,n}}\subset\mathcal{H}^{\otimes n}$. If $T\in{\mathcal{T}}_{>}$ is a
strict twist, $P_{T,n}$ is even invertible (with bounded inverse) in
$\mathcal{B}(\mathcal{H}^{\otimes n})$, i.e.
$\mathcal{H}_{T,n}=\mathcal{H}^{\otimes n}$ as vector spaces in this case.

For any twist $T\in{\mathcal{T}}_{\geq}$, the twisted Fock space is defined as
the direct sum

$\displaystyle\mathcal{F}_{T}(\mathcal{H}):=\bigoplus_{n=0}^{\infty}\mathcal{H}_{T,n},\qquad\mathcal{H}_{T,0}=\mathbb{C},$

equipped with the scalar product
$\langle\Psi,\Phi\rangle_{T}=\sum_{n=0}^{\infty}\langle\Psi_{n},P_{T,n}\Phi_{n}\rangle_{\mathcal{H}^{\otimes
n}}$, $\Psi_{n},\Phi_{n}\in\mathcal{H}_{T,n}$. We will use the notation
$\Psi=\bigoplus_{n}\Psi_{n}$ to denote the “$n$-particle components”
$\Psi_{n}\in\mathcal{H}_{T,n}$ of a vector
$\Psi\in\mathcal{F}_{T}(\mathcal{H})$, and write $\Omega=1\oplus 0\oplus
0\ldots$ for the Fock vacuum.

Although the inner product $\langle\,\cdot\,,\,\cdot\,\rangle_{T}$ is the
natural inner product in $\mathcal{F}_{T}(\mathcal{H})$, we will occasionally
also need to argue in $\mathcal{H}^{\otimes n}$ (or $\mathcal{H}^{\otimes
n}/\operatorname{ker}P_{T,n}$) using untwisted inner products
$\langle\,\cdot\,,\,\cdot\,\rangle$. Adjoints w.r.t.
$\langle\,\cdot\,,\,\cdot\,\rangle_{T}$ are denoted $A\mapsto A^{\star}$ and
adjoints w.r.t. $\langle\,\cdot\,,\,\cdot\,\rangle$ are denoted $A\mapsto
A^{*}$. The operator norm of $\mathcal{B}(\mathcal{F}_{T}(\mathcal{H}))$ is
written $\|\cdot\|_{T}$, and that of $\mathcal{B}(\mathcal{H}^{\otimes n})$ as
$\|\cdot\|$, as usual. Note that the scalar products
$\langle\,\cdot\,,\,\cdot\,\rangle$ and
$\langle\,\cdot\,,\,\cdot\,\rangle_{T}$ coincide on the zero and one-particle
vectors, i.e. on $\mathbb{C}\Omega\oplus\mathcal{H}$.

The twisted Fock spaces $\mathcal{F}_{T}(\mathcal{H})$ can look quite
different depending on the twist $T$, and we now recall some important
examples. For further discussions of these and other examples, see [BS91,
BS94, JSW95, LM95a, JPS01, LS14, DKLP20].

######  Example 2.3.

The simplest example of a strict twist is $T=0$. In this case, $P_{T,n}=1$ for
all $n\in\mathbb{N}$, and hence the $0$-twisted Fock space
$\mathcal{F}_{0}(\mathcal{H})$ over $\mathcal{H}$ coincides with the full
(Boltzmann) Fock space

$\displaystyle\mathcal{F}_{0}(\mathcal{H})=\bigoplus_{n=0}^{\infty}\mathcal{H}^{\otimes
n}$

with its canonical inner product.

######  Example 2.4.

(Involutive Yang-Baxter solutions, symmetric twists)
An important class of twists $T$ are the symmetric twists, namely the family

$\displaystyle{\mathcal{T}}_{\rm
Sym}:=\\{T\in{\mathcal{B}}(\mathcal{H}\otimes\mathcal{H})\,\colon\,T=T^{*}=T^{-1},\quad
T_{1}T_{2}T_{1}=T_{2}T_{1}T_{2}\\},$ (2.8)

of unitary (and selfadjoint) involutive solutions of the Yang-Baxter
equation444For $\operatorname{dim}\mathcal{H}<\infty$, these solutions have
been classified in [LPW19, Thm. 4.8].. These are in particular braided twists,
hence contained in ${\mathcal{T}}_{\geq}$. Any $T\in{\mathcal{T}}_{\rm Sym}$
defines a unitary representation $\rho_{T,n}$ of the symmetric group $S_{n}$
on $\mathcal{H}^{\otimes n}$ by $\rho_{T,n}(\sigma_{k}):=T_{k}$, where the
$\sigma_{k}$ are the Coxeter generators of $S_{n}$.

Looking at the definition of $R_{T,n}$ and $P_{T,n}$, one realizes that in
this case $n!^{-1}P_{T,n}$ is the projection onto the subspace ${\rm
Ran}P_{T,n}$ of $\mathcal{H}^{\otimes n}$ on which $\rho_{T,n}$ acts
trivially. Thus $T\not\in{\mathcal{T}}_{>}$ unless $T=1$.

Since the projection $n!^{-1}P_{T,n}$ acts trivially on its (closed) range,
the maps

$\displaystyle I_{n}:({\rm
Ran}P_{T,n},\langle\,\cdot\,,\,\cdot\,\rangle_{T})\to({\rm
Ran}P_{T,n},\langle\,\cdot\,,\,\cdot\,\rangle),\qquad\Psi_{n}\mapsto\sqrt{n!}\Psi_{n},$
(2.9)

define a unitary $I=\bigoplus_{n}I_{n}$ between the $T$-twisted Fock space
$\mathcal{F}_{T}(\mathcal{H})$ and another Fock space naturally associated
with a symmetric twist $T$, the $T$-symmetric Fock space. The $T$-symmetric
Fock space over $\mathcal{H}$ is defined as $\bigoplus_{n=0}^{\infty}{\rm
Ran}P_{T,n}$ with the $T$-independent scalar products inherited from
$\mathcal{H}^{\otimes n}$ by restriction. For more on $T$-symmetric Fock
spaces, see [LMR95, Lec03] and the references cited there.

The symmetric twists include in particular the flip

$\displaystyle T=F:v\otimes w\mapsto w\otimes v,$ (2.10)

giving rise (via the unitary $I$ (2.9)) to the Bose Fock space
$I\mathcal{F}_{F}(\mathcal{H})$, and the negative flip $-F$, giving rise to
the Fermi Fock space $I\mathcal{F}_{-F}(\mathcal{H})$.

Two other special examples of symmetric twists are
$T=\pm\operatorname{id}_{\mathcal{H}\otimes\mathcal{H}}$. In the case of the
positive sign, we have $P_{1,n}=n!$ and hence $I$ maps the twisted Fock space
onto the Boltzmann Fock space,
$I\mathcal{F}_{1}(\mathcal{H})=\mathcal{F}_{0}(\mathcal{H})$. In the case of
the negative sign, we have $P_{-1,n}=0$ for all $n>1$, i.e.
$\mathcal{F}_{-1}(\mathcal{H})=\mathbb{C}\oplus\mathcal{H}$.

A subset of symmetric twists are related to solutions of the Yang-Baxter
equation with (additive) spectral parameter. Let $\mathcal{K}$ be a separable
Hilbert space and $S:\mathbb{R}\to\mathcal{U}(\mathcal{K}\otimes\mathcal{K})$
a measurable function taking values in the unitaries on the tensor square
$\mathcal{K}\otimes\mathcal{K}$, and satisfying the Yang-Baxter equation with
spectral parameter555That is, the equation
$S(\theta)_{1}S(\theta+\theta^{\prime})_{2}S(\theta^{\prime})_{1}=S(\theta^{\prime})_{2}S(\theta+\theta^{\prime})_{1}S(\theta)_{2}$
holds in $\mathcal{B}(\mathcal{K}^{\otimes 3})$ for almost all
$\theta,\theta^{\prime}\in\mathbb{R}$. This condition is automatically
satisfied if $\mathcal{K}=\mathbb{C}$. as well as the symmetry condition
$S(\theta)^{*}=S(-\theta)$ for almost all $\theta\in\mathbb{R}$.

Then (the tensor square of) the Hilbert space
$\mathcal{H}=L^{2}(\mathbb{R}\to\mathcal{K},d\theta)$ carries the symmetric
twist

$\displaystyle(T_{S}\psi)(\theta_{1},\theta_{2})=S(\theta_{2}-\theta_{1})\psi(\theta_{2},\theta_{1}).$
(2.11)

These twists are often considered in the context of generalized statistics
[LM95a] and integrable quantum field theories, where $S$ has the meaning of an
elastic two-body scattering matrix – see, for example, [Smi92, LS14].

######  Example 2.5.

(Scaled Yang-Baxter solutions)
Given a symmetric twist $T\in{\mathcal{T}}_{\rm Sym}$ and $-1<q<1$, the scaled
twist $qT$ lies in ${\mathcal{T}}_{>}$ but not in ${\mathcal{T}}_{\rm Sym}$
because $\|qT\|=|q|<1$. An example that has been studied a lot in the
literature is the “$q$-Fock space” $\mathcal{F}_{qF}(\mathcal{H})$ defined by
the scaled flip $qF$, which interpolates between Bose and Fermi statistics
[FB70, BS91, BM17].

Later on we will see that in our context, braided twists are the most
interesting ones. The family of braided twists includes all finite-dimensional
contractive selfadjoint solutions $T$ to the Yang-Baxter equation. Contrary to
the situation considered in Example 2.5, these do not necessarily have
spectrum contained in a circle. Such braided twists can therefore not be
rescaled to give representations of the symmetric groups, but only define
representations of the braid groups $B_{n}$.
Moreover, there exist also genuinely infinite-dimensional braided twists such
as SO$(d,1)$-symmetric twists considered in the context of quantum field
theories on de Sitter space [HL18].
Because of the lack of general classification results on solutions of the
Yang-Baxter equation, the structure of ${\mathcal{T}}_{\geq}$ is very complex
and largely unknown in general.

Following [BS94, JSW95], we now fix an arbitrary twist
$T\in{\mathcal{T}}_{\geq}$ and describe creation and annihilation type
operators on $\mathcal{F}_{T}(\mathcal{H})$ which will be used to generate our
von Neumann algebras. Selfadjointness of $P_{T,n}$ implies

$\displaystyle P_{T,n+1}=R_{T,n+1}^{*}(1\otimes P_{T,n}),$ (2.12)

from which we read off that the left creation operators
$\Psi_{n}\mapsto\xi\otimes\Psi_{n}$, $\xi\in\mathcal{H}$,
$\Psi_{n}\in\mathcal{H}^{\otimes n}$, descend to maps

$\displaystyle a_{L,T}^{\star}(\xi):\mathcal{H}^{\otimes
n}/\operatorname{ker}P_{T,n}$
$\displaystyle\to\mathcal{H}^{\otimes(n+1)}/\operatorname{ker}P_{T,n+1},$
(2.13) $\displaystyle a_{L,T}^{\star}(\xi)[\Psi_{n}]$
$\displaystyle:=[\xi\otimes\Psi_{n}],$ (2.14)

where $[\,\cdot\,]$ denotes the quotient map, i.e. the
$\langle\,\cdot\,,\,\cdot\,\rangle$-orthogonal projection onto $\overline{{\rm
Ran}P_{T,n}}\subset\mathcal{H}^{\otimes n}$. Thus $a_{L,T}^{\star}(\xi)$
extend to densely defined operators on $\mathcal{F}_{T}(\mathcal{H})$ which
contain the quotients $\mathcal{H}^{\otimes n}/\operatorname{ker}P_{T,n}$,
$n\in\mathbb{N}$, in their domains. As shown by Bożejko and Speicher [BS94,
Thm. 3.1], these operators are bounded on each $\mathcal{H}_{T,n}$,
$n\in\mathbb{N}$. In fact, positivity and the defining formula (2.2) imply
that

$\displaystyle P_{T,n+1}^{2}$ $\displaystyle=(1\otimes
P_{T,n})R_{T,n+1}R_{T,n+1}^{*}(1\otimes P_{T,n})$ $\displaystyle\leq(1\otimes
P_{T,n})\|R_{T,n+1}^{*}\|^{2}(1\otimes P_{T,n})$ $\displaystyle\leq
c_{T,n}^{2}(1\otimes P_{T,n})^{2},$

where $c_{T,n}=\sum_{k=0}^{n}\|T\|^{k}$.

Hence $P_{T,n+1}\leq c_{T,n}(1\otimes P_{T,n})$ and for
$[\Psi_{n}]\in\mathcal{H}^{\otimes n}/\operatorname{ker}P_{T,n}$, we have

$\displaystyle\|a_{L,T}^{\star}(\xi)[\Psi_{n}]\|^{2}_{T}$
$\displaystyle=\langle\xi\otimes\Psi_{n},P_{T,n+1}\xi\otimes\Psi_{n}\rangle$
$\displaystyle\leq c_{T,n}\langle\xi\otimes\Psi_{n},(1\otimes
P_{T,n})\xi\otimes\Psi_{n}\rangle$
$\displaystyle=c_{T,n}\|\xi\|^{2}\|[\Psi_{n}]\|_{T}^{2},$

which implies the operator norm bound

$\displaystyle\|a_{L,T}^{\star}(\xi)|_{\mathcal{H}_{T,n}}\|_{T}$
$\displaystyle\leq\sqrt{c_{T,n}}\|\xi\|,\qquad
c_{T,n}=\begin{cases}\frac{1-\|T\|^{n+1}}{1-\|T\|}&\|T\|<1,\\\
n+1&\|T\|=1\end{cases}.$ (2.15)

One may calculate the adjoint $a_{L,T}(\xi)$ of $a_{L,T}^{\star}(\xi)$ (w.r.t.
$\langle\,\cdot\,,\,\cdot\,\rangle_{T}$) as

$\displaystyle a_{L,T}(\xi)[\Psi_{n}]=[a_{L}(\xi)R_{T,n}\Psi_{n}],$ (2.16)

where $a_{L}(\xi)$ is the untwisted left annihilation operator, namely the
Boltzmann Fock space operator fixed by

$\displaystyle
a_{L}(\xi)(\psi_{1}\otimes\ldots\psi_{n})=\langle\xi,\psi_{1}\rangle\cdot\psi_{2}\otimes\ldots\otimes\psi_{n}.$
(2.17)

Note that (2.16) is well-defined on $\mathcal{H}^{\otimes
n}/\operatorname{ker}P_{T,n}$ and extends to a bounded operator
$\mathcal{H}_{T,n}\to\mathcal{H}_{T,n-1}$. We will use the same symbols
$a_{L,T}(\xi)$, $a_{L,T}^{\star}(\xi)$ to denote the operators that are
defined on the dense finite particle subspace of
$\mathcal{F}_{T}(\mathcal{H})=\bigoplus_{n}\mathcal{H}_{T,n}$ consisting of
terminating direct sums.

In view of (2.15), these operators are bounded in case $\|T\|<1$, namely

$\displaystyle\|a_{L,T}^{\star}(\xi)\|_{T}\leq\frac{\|\xi\|}{\sqrt{1-\|T\|}}<\infty.$
(2.18)

######  Remark 2.6.

Let us comment on a simplified description of the operators
$a_{L,T}^{\star}(\xi)$ that is available for symmetric twists
$T\in{\mathcal{T}}_{\rm Sym}$. In this case, $\mathcal{H}_{T,n}={\rm
Ran}P_{T,n}\subset\mathcal{H}^{\otimes n}$ is the closed subspace of
$T$-symmetric vectors, i.e. the vectors $\Psi_{n}\in\mathcal{H}^{\otimes n}$
satisfying $T_{k}\Psi_{n}=\Psi_{n}$, $k=1,\ldots,n-1$. The
$\langle\,\cdot\,,\,\cdot\,\rangle$-orthogonal projection $Q_{T,n}$ onto it
(the “$T$-symmetrization”) is given by $Q_{T,n}=n!^{-1}P_{T,n}$. From this one
infers that the creation operators satisfy

$\displaystyle
Ia_{L,T}^{\star}(\xi)I^{-1}Q_{T,n}\Psi_{n}=\sqrt{n+1}Q_{T,n+1}(\xi\otimes\Psi_{n}),\qquad\Psi_{n}\in\mathcal{H}^{\otimes
n},$ (2.19)

where $I$ is the unitary (2.9). This shows that when transported to the
$T$-symmetric Fock space, we recover the familiar Zamolodchikov creation and
annihilation operators (see [Zam92] for their origin in integrable models and
[LS14] for a mathematical formulation).

In particular, these operators form representations of the CCR and CAR
algebras [BR97, Sect. 5.2.2] on the Bose and Fermi Fock space for $T=F$ and
$T=-F$, respectively.

We conclude this section with a comment on second quantization of operators on
$T$-twisted Fock spaces. Here and in the following, we write $\mathcal{D}(X)$
for the domain of an operator $X$ and we denote by $X\otimes
Y:\mathcal{D}(X)\odot\mathcal{D}(Y)\to\mathcal{H}\otimes\mathcal{H}$ the
linear operator defined by $(X\otimes Y)(x\otimes y)=X(x)\otimes Y(y)$.
Furthermore, we write $X^{\\#}$ to denote either $X$ or $X^{\star}$, i.e. an
equation of the form $f(X^{\\#})=Y^{\\#}$ means $f(X)=Y$ and
$f(X^{\star})=Y^{\star}$.

######  Lemma 2.7.

Let $T\in{\mathcal{T}}_{\geq}$ be a twist and $V:\mathcal{D}(V)\to\mathcal{H}$
a closable linear or antilinear operator such that

$\displaystyle[V\otimes V,T]=0.$ (2.20)

  1. a)

Then $\hat{V}:\bigoplus_{n=0}^{\infty}\mathcal{D}(V)^{\odot
n}/\operatorname{ker}(P_{T,n})\to\mathcal{H}_{T,n}$, given by
$\hat{V}([\Psi_{n}])=[V^{\otimes n}\Psi_{n}]$, for all
$\Psi_{n}\in\mathcal{D}(V)^{\odot n}/\operatorname{ker}(P_{T,n})$ and for all
$n\in\mathbb{N}$, where $V^{\otimes 0}:=1$ , is a well-defined closable
operator on $\mathcal{F}_{T}(\mathcal{H})$ whose closure will be denoted by
$\Gamma_{T}(V)$. In case $V$ is (anti)unitary, so is $\Gamma_{T}(V)$.

  2. b)

If $V$ is invertible, we have for any $\xi\in\mathcal{D}(V)$

$\displaystyle\overline{\Gamma_{T}(V)a_{L,T}^{\\#}(\xi)\Gamma_{T}(V^{-1})}$
$\displaystyle=a_{L,T}^{\\#}(V\xi).$ (2.21)

###### Proof.

a) Thanks to the commutation condition (2.20), one checks $T_{k}V^{\otimes
n}=V^{\otimes n}T_{k}$, $k=1,\ldots,n-1$. This implies $R_{T,n}V^{\otimes
n}=V^{\otimes n}R_{T,n}$ and inductively $P_{T,n}V^{\otimes n}=V^{\otimes
n}P_{T,n}$, i.e. $[V^{\otimes n},P_{T,n}]=0$. In particular,
$\mathcal{D}(V)^{\odot
n}/\operatorname{ker}(P_{T,n})\subset\mathcal{D}(V^{\otimes
n})/\operatorname{ker}(P_{T,n})$. It follows that $\hat{V}$ is a
$\|\cdot\|_{T}$-densely well-defined (anti)linear operator in
$\mathcal{F}_{T}(\mathcal{H})$. The exact same argument for $V^{\ast}\otimes
V^{\ast}$ proves the $\widehat{V^{\ast}}\subset\hat{V}^{\star}$ is also a
$\|\cdot\|_{T}$-densely and well-defined operator in
$\mathcal{F}_{T}(\mathcal{H})$. Hence $\hat{V}$ is a closable operator in
$\mathcal{F}_{T}(\mathcal{H})$. The fact that $\Gamma_{T}(V)$ is (anti)unitary
for (anti)unitary $V$ is easy to check from this construction.

b) can be easily checked on a core for $a_{R,T}^{\\#}(V\xi)$. ∎

###  2.2. Standard subspaces and twisted Araki-Woods algebras

We now use the twisted left field operators666These operators generalize free
quantum fields, hence the name.

$\displaystyle\phi_{L,T}(\xi)=a_{L,T}^{\star}(\xi)+a_{L,T}(\xi),\qquad\xi\in\mathcal{H},$
(2.22)

to generate von Neumann algebras. For $\|T\|<1$, these are bounded selfadjoint
operators on $\mathcal{F}_{T}(\mathcal{H})$ (2.15). If $\|T\|=1$, the bound
$\|a_{L,T}^{\\#}(\xi)|_{\mathcal{H}_{T,n}}\|_{T}\leq\sqrt{n+1}\cdot\|\xi\|$
(2.15) implies that any vector of finite particle number is an (entire)
analytic vector for $\phi_{L,T}(\xi)$, hence $\phi_{L,T}(\xi)$ is essentially
selfadjoint on the domain of finite particle number by Nelson’s Theorem and we
use the same symbol to denote its selfadjoint closure.

Before we define the von Neumann algebras we are interested in, we recall a
few facts about standard subspaces [RvD77, Lon08]: By definition, a standard
subspace of a complex Hilbert space $\mathcal{H}$ is a closed real subspace
$H\subset\mathcal{H}$ that is cyclic in the sense that
$H+iH\subset\mathcal{H}$ is dense, and separating in the sense $H\cap
iH=\\{0\\}$.

A standard subspace $H$ defines modular data through the polar decomposition
$S_{H}=J_{H}\Delta_{H}^{1/2}$ of the closed operator with domain
$\mathcal{D}_{H}:=H+iH$

$\displaystyle S_{H}:H+iH\to H+iH,\qquad h_{1}+ih_{2}\mapsto h_{1}-ih_{2}.$
(2.23)

The operator $S_{H}$ is called the Tomita operator of $H$. It satisfies an
analogue of Tomita’s theorem for von Neumann algebras, namely

$\displaystyle J_{H}H=H^{\prime},\qquad\Delta_{H}^{it}H=H,\quad
t\in\mathbb{R},$ (2.24)

where

$\displaystyle
H^{\prime}:=\\{h^{\prime}\in\mathcal{H}\colon\operatorname{Im}\langle
h,h^{\prime}\rangle=0\quad\forall h\in H\\}$ (2.25)

is the symplectic complement of $H$, which is a standard subspace as well
(with $J_{H^{\prime}}=J_{H}$, $\Delta_{H^{\prime}}=\Delta_{H}^{-1}$).

######  Definition 2.8.

Given a closed real subspace $H\subset\mathcal{H}$ and a twist
$T\in{\mathcal{T}}_{\geq}$, we define the (left) $T$-twisted Araki-Woods von
Neumann algebra

$\displaystyle\mathcal{L}_{T}(H):=\\{\exp(i\phi_{L,T}(h))\,:h\in
H\\}^{\prime\prime}\subset\mathcal{B}(\mathcal{F}_{T}(\mathcal{H})).$ (2.26)

In case the field operators are bounded, this is simply the weak closure of
the ⋆-algebra $\mathcal{P}_{L,T}(H)$ of all polynomials in $\phi_{L,T}(h)$,
$h\in H$.

If $T=F$ is the flip, $\mathcal{L}_{T}(H)$ is a second quantization von
Neumann algebra which is well understood, and closely related to the original
work of Araki and Woods [AW68]. In this case, $\mathcal{L}_{F}(H)$ is a factor
if and only if $H\cap H^{\prime}=\\{0\\}$, and $\Omega$ is cyclic and
separating if and only if $H$ is a standard subspace [LRT78, Thm. I.3.2]. For
more details on modular structure and a summary of results on the type of
$\mathcal{L}_{F}(H)$, see [FG94, Thm. 1.3].

If $T=-F$ is the negative flip, $\mathcal{L}_{-F}(H)$ is generated by a
representation of the CAR algebra, see [Foi83, BJL02] for results in this
case.

Another special case is $T=0$. These von Neumann algebras have been introduced
by Voiculescu [Voi85], initially in the case where $H$ is the closed real span
of an orthonormal basis of $\mathcal{H}$. In this case, $\mathcal{L}_{0}(H)$
is isomorphic to the group von Neumann algebra of the free group on
$\operatorname{dim}\mathcal{H}$ generators, which explains their central
importance in free probability [VDN92, Thm. 2.6.2]. Shlyakhtenko studied
$\mathcal{L}_{0}(H)$ for more general spaces $H$ [Shl97], and in particular
showed that $\mathcal{L}_{0}(H)$ is a factor for any standard subspace $H$.
When $\Delta_{H}$ is non-trivial and $\operatorname{dim}\mathcal{H}>1$, these
are full factors of type IIIλ, $0<\lambda\leq 1$ [Shl97, Thm. 6.1].

The scaled flip $T=qF$, $-1<q<1$, has been studied extensively in the
literature, as it interpolates between the CCR ($q=1$), CAR ($q=-1$), and free
($q=0$) cases. It has first been considered in the case $H=H^{\prime}$ [BS91]
and then generalized to the case of general standard subspaces by Hiai
[Hia01]. We refer to [BKS97, R03, Shl04, Sni04, Nou04, Jun06, GS14, Nel15,
BM17, BM22] for various properties (such as factoriality, type, non-
injectivity, (strong) solidity, approximation properties, MASAS) of the von
Neumann algebras $\mathcal{L}_{qF}(H)$.

In the case of general twists $T$, only very little is known about
$\mathcal{L}_{T}(H)$. The focus of our investigation below is to identify
situations in which $\Omega$ is cyclic and separating for
$\mathcal{L}_{T}(H)$, and then study properties of (specific) inclusions
$\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$ rather than the internal
structure of $\mathcal{L}_{T}(H)$. We begin with the following elementary
lemma.

######  Lemma 2.9.

Let $T\in{\mathcal{T}}_{\geq}$ and $H\subset\mathcal{H}$ be a real closed
subspace.

  1. a)

If $H$ is cyclic, $\Omega$ is cyclic for $\mathcal{P}_{L,T}(H)$ and hence for
$\mathcal{L}_{T}(H)$.

  2. b)

If $\Omega$ is separating for $\mathcal{P}_{L,T}(H)$, then $H$ is separating.

###### Proof.

a) Let us prove that $\overline{\mathcal{P}_{L,T,n}(H)\Omega}$ contains
$\mathcal{H}_{T,n}$ by induction in $n$, where $\mathcal{P}_{L,T,n}(H)$
denotes the linear space of all polynomials in the fields $\phi_{L,T}(h)$,
$h\in H$, of degree up to $n$. The case $n=0$ is trivial. For the induction
step, fix $n\in\mathbb{N}$, a vector $\Psi_{n+1}\in\mathcal{H}_{T,n+1}$, and
$\varepsilon>0$. Since we have777A remark on notation: We use square brackets
to indicate the equivalence classes in the quotient (2.6), also extended to
direct sums over several particle numbers. Indices $n$ denote projection to
the $n$-particle subspace. For example,
$[\phi_{L,T}(h_{1})\phi_{L,T}(h_{2})\Omega]_{2}=[h_{1}\otimes h_{2}]$ denotes
the $2$-particle component of the equivalence class $[h_{1}\otimes h_{2}]$.

$[\phi_{L,T}(h_{1})\cdots\phi_{L,T}(h_{n+1})\Omega]_{n+1}=[h_{1}\otimes\ldots\otimes
h_{n+1}]$

and $H$ is cyclic, it follows that we find a field polynomial
$Q_{n+1}\in\mathcal{P}_{L,T,n+1}(H)$ of degree $n+1$ such that
$\|[Q_{n+1}\Omega]_{n+1}-\Psi_{n+1}\|<\varepsilon$. By the induction
assumption, there exists $Q_{n}\in\mathcal{P}_{L,T,n}(H)$ such that

$\|Q_{n}\Omega-(Q_{n+1}\Omega-[Q_{n+1}\Omega]_{n+1})\|<\varepsilon.$

Hence $\tilde{Q}_{n+1}:=Q_{n+1}-Q_{n}\in\mathcal{P}_{L,T,n+1}(H)$ satisfies
$\|\tilde{Q}_{n+1}\Omega-\Psi_{n+1}\|\leq 2\varepsilon$.

This shows that $\Omega$ is cyclic for $\mathcal{P}_{L,T}(H)$, and by standard
arguments (only required in case $\phi_{L,T}(h)$ is unbounded), one sees that
it is also cyclic for $\mathcal{L}_{T}(H)$.

b) Let $h,\tilde{h}\in H$ and $h=i\tilde{h}\in H\cap iH$. Then
$h=\phi_{L,T}(h)\Omega=i\phi_{L,T}(\tilde{h})\Omega=i\tilde{h}$; and since
$\Omega$ separates $\mathcal{P}_{L,T}(H)$ by assumption, we find
$\phi_{L,T}(h)=i\phi_{L,T}(\tilde{h})$. Taking adjoints, this implies
$\phi_{L,T}(h)=-i\phi_{L,T}(\tilde{h})$ and hence $\phi_{L,T}(h)=0$, i.e.
$h=0$. ∎

This motivates us to restrict attention to $\mathcal{L}_{T}(H)$ with $H$ a
standard subspace. In this case $\Omega$ is cyclic for $\mathcal{L}_{T}(H)$
with no further conditions on $T$. However, in general it is not separating,
as we will see below. If $\Omega$ is separating, $(\mathcal{L}_{T}(H),\Omega)$
defines modular data $\Delta_{L,T,H}$ and $J_{L,T,H}$. When $T$ and $H$ are
clear from the context, we will use the shorter notation

$\displaystyle\Delta:=\Delta_{L,T,H},\qquad J:=J_{L,T,H},\qquad
S:=J\Delta^{1/2}.$ (2.27)

##  3\. Twisted Araki-Woods algebras and standard vectors

The aim of this section is to derive necessary and sufficient conditions on
$T$ and $H$ for $\Omega$ being cyclic and separating for $\mathcal{L}_{T}(H)$,
and compute the corresponding modular data $J$, $\Delta$. Most of our analysis
will be based on the following compatibility assumption between $T$ and $H$.

######  Definition 3.1.

Let $H\subset\mathcal{H}$ be a standard subspace. The twists compatible with
$H$ are the elements of

$\displaystyle{\mathcal{T}}_{\geq}(H)$
$\displaystyle:=\\{T\in{\mathcal{T}}_{\geq}\colon[\Delta_{H}^{it}\otimes\Delta_{H}^{it},T]=0\;\text{for
all }\,t\in\mathbb{R}\\},$ (3.1) $\displaystyle{\mathcal{T}}_{>}(H)$
$\displaystyle:={\mathcal{T}}_{\geq}(H)\cap{\mathcal{T}}_{>}.$ (3.2)

The advantage of a twist $T$ and a standard subspace $H$ being compatible is
that this ensures the existence of the unitaries $\Gamma_{T}(\Delta_{H}^{it})$
as discussed in Lemma 2.7. In the context of quantum field theory, such an
assumption occurs naturally when asking a twist given by a two-particle
S-matrix to be Poincaré invariant [LS14]. In our present abstract setup, the
main consequence of compatibility is that the one-particle restrictions of the
modular data $J,\Delta$ (2.27) of $(\mathcal{L}_{T}(H),\Omega)$ (if they
exist, i.e. if $\Omega$ is separating) coincide with the modular data of $H$,
as we will show below in Lemma 3.2.

Before this lemma, we introduce some notation. Since we often times deal with
analytic functions on strips, we denote strip regions in $\mathbb{C}$ by

$\displaystyle{\mathbb{S}}_{a}:=\\{z\in\mathbb{C}\,\colon\,\min\\{0,a\\}<\operatorname{Im}(z)<\max\\{0,a\\}\\},\qquad
a\in\mathbb{R}.$ (3.3)

For concise formulations, we also introduce the vector space

$\displaystyle{\mathbb{H}}^{\infty}_{\rm
c.bv.}({\mathbb{S}}_{a}):=\\{f:\overline{{\mathbb{S}}_{a}}\to\mathbb{C}\,\text{
continuous and bounded}\,:\,f|_{{\mathbb{S}}_{a}}\text{ is analytic}\\},$
(3.4)

which is a Banach space w.r.t.
$\|f\|_{\infty}=\sup\\{|f(z)|\,:\,z\in\overline{{\mathbb{S}}_{a}}\\}$ and can
be viewed as
$C_{b}(\overline{{\mathbb{S}}_{a}})\cap{\mathbb{H}}^{\infty}({\mathbb{S}}_{a})$.
Elements of ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{a})$ are uniquely
determined by their restriction to $\mathbb{R}$ or $\mathbb{R}+ia$, and we
will therefore identify functions on ${\mathbb{S}}_{a}$ with their boundary
values. For example, given $f:\mathbb{R}\to\mathbb{C}$ we write
$f\in{\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{a})$ to express that $f$
is the restriction of a function in ${\mathbb{H}}^{\infty}_{\rm
c.bv.}({\mathbb{S}}_{a})$ to $\mathbb{R}$.

######  Lemma 3.2.

Let $T\in{\mathcal{T}}_{\geq}(H)$, and assume that $\Omega\in\mathcal{H}$ is
separating for $\mathcal{L}_{T}(H)$. Then the modular data $J,\Delta$ of
$(\mathcal{L}_{T}(H),\Omega)$ satisfy

$\displaystyle\Delta\raisebox{-2.15277pt}{$|_{\mathcal{H}\cap\mathcal{D}(\Delta)}$}=\Delta_{H},\qquad
J\raisebox{-2.15277pt}{$|_{\mathcal{H}}$}=J_{H}.$ (3.5)

###### Proof.

As $T$ lies in ${\mathcal{T}}_{\geq}(H)$, the operators
$U(t):=\Gamma_{T}(\Delta_{H}^{it})$, $t\in\mathbb{R}$, are well-defined
unitaries on $\mathcal{F}_{T}(\mathcal{H})$ (Lemma 2.7 a)) and form a strongly
continuous one-parameter group fixing the vacuum vector $\Omega$.

For any polynomial $Q\in\mathcal{P}_{L,T}(H)$, we have
$U(t)QU(-t)\in\mathcal{P}_{L,T}(H)\subset\mathcal{L}_{T}(H)$ by Lemma 2.7 b)
and $\Delta_{H}^{it}H=H$. Hence $[U(t)QU(-t),A^{\prime}]=0$ for all
$A^{\prime}\in\mathcal{L}_{T}(H)^{\prime}$. By taking limits one concludes
that $U(t)\cdot U(-t)$ defines a ⋆-automorphism of $\mathcal{L}_{T}(H)$.
Therefore, the Tomita operator $S$ of $(\mathcal{L}_{T}(H),\Omega)$ satisfies

$SU(t)A\Omega=(U(t)AU(-t))^{\star}\Omega=U(t)A^{\star}\Omega=U(t)SA\Omega,\qquad
A\in\mathcal{L}_{T}(H),$

and therefore $U(t)$ commutes with $J$ and $\Delta$. Since $U(t)$ commutes
with $\Delta$ and $J$, the modular operator $\Delta_{H}$ commutes with
$\Delta|_{\mathcal{H}\cap\mathcal{D}(\Delta)}$ and $J|_{\mathcal{H}}$.

As $\Delta_{H}$ and $\Delta|_{\mathcal{H}\cap\mathcal{D}(\Delta)}$ commute,
there is a common core
$D\subset\mathcal{D}(\Delta_{H})\cap\mathcal{D}(\Delta)$ for these two
operators. Now, for $k_{1},h_{1}$ from this core $D$, the function defined by
$f(t)=\left\langle k_{1},U(t)h_{1}\right\rangle_{T}=\left\langle
k_{1},\Delta_{H}^{it}h_{1}\right\rangle$ belongs to
${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{-1})$ and satisfies the KMS
boundary condition $f(-i)=\left\langle
S_{H}h_{1},S_{H}k_{1}\right\rangle=\left\langle
Sh_{1},Sk_{1}\right\rangle=\langle k_{1},\Delta h_{1}\rangle$, where we have
used that the $S$ and $S_{H}$ coincide on $\mathcal{D}(S_{H})$. On the other
hand, $f(-i)=\left\langle k_{1},\Delta_{H}h_{1}\right\rangle$. Hence $\Delta
h_{1}=\Delta_{H}h_{1}$ for all $h_{1}\in D$. Hence
$\Delta_{H}=\Delta|_{\mathcal{H}\cap\mathcal{D}(\Delta)}$ and
$J|_{\mathcal{H}}=J_{H}$. ∎

For compatible twists, the standardness property of $\Omega$ turns out to be
encoded in two key properties of the twist $T$: The Yang-Baxter equation ($T$
being braided) and a “crossing symmetry”. We discuss the relation of these
properties to scattering theory later (see Remark 3.15), and first give a
mathematical formulation suitable for our setup. Note that in this definition
and various calculations below, the $T$-independent scalar products
$\langle\,\cdot\,,\,\cdot\,\rangle$ on $\mathcal{H}^{\otimes n}$ are used.

######  Definition 3.3.

(crossing symmetry)
Let $H\subset\mathcal{H}$ be a standard subspace. A bounded operator
$T\in{\mathcal{B}}(\mathcal{H}\otimes\mathcal{H})$ is called crossing-
symmetric w.r.t. $H$ if for all $\psi_{1},\ldots,\psi_{4}\in\mathcal{H}$, the
function

$\displaystyle
T^{\psi_{2},\psi_{1}}_{\psi_{3},\psi_{4}}(t):=\left\langle\psi_{2}\otimes\psi_{1},(\Delta_{H}^{it}\otimes
1)T(1\otimes\Delta_{H}^{-it})(\psi_{3}\otimes\psi_{4})\right\rangle$ (3.6)

lies in ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{1/2})$ and,
$t\in\mathbb{R}$,

$\displaystyle T^{\psi_{2},\psi_{1}}_{\psi_{3},\psi_{4}}(t+\tfrac{i}{2})$
$\displaystyle=\left\langle\psi_{1}\otimes
J_{H}\psi_{4},(1\otimes\Delta_{H}^{it})T(\Delta_{H}^{-it}\otimes
1)(J_{H}\psi_{2}\otimes\psi_{3})\right\rangle$ (3.7)
$\displaystyle=\overline{(T^{*})^{J_{H}\psi_{2},\psi_{3}}_{\psi_{1},J_{H}\psi_{4}}(t)}.$

This definition is motivated from quantum field theory and generalizes the
notion of crossing symmetry from scattering theory to a setting of standard
subspaces. We postpone the discussion of this relation to Remark 3.15 below.
For the time being, suffice it to say that from an operator-algebraic
perspective, Def. 3.3 is clearly reminiscent of the KMS / modular boundary
condition characterizing the modular group of a standard vector. As we shall
see in Thm. 3.12 a), crossing symmetry is a consequence of $\Omega$ being
separating for $\mathcal{L}_{T}(H)$.

In case $T\in{\mathcal{T}}_{\geq}(H)$ is a twist that is crossing symmetric
w.r.t. a standard subspace $H$ and compatible with $H$, we also use the
notation

$\displaystyle T(t)$ $\displaystyle:=(\Delta_{H}^{it}\otimes
1)T(1\otimes\Delta_{H}^{-it})=(1\otimes\Delta_{H}^{-it})T(\Delta_{H}^{it}\otimes
1)=T(-t)^{\ast},\quad T\in{\mathcal{T}}_{\geq}(H),$

so that the boundary condition of crossing symmetry takes the form

$\displaystyle
T^{\psi_{2},\psi_{1}}_{\psi_{3},\psi_{4}}(t+\tfrac{i}{2})=\overline{(T^{*})^{J_{H}\psi_{2},\psi_{3}}_{\psi_{1},J_{H}\psi_{4}}(t)}=T^{\psi_{1},J_{H}\psi_{4}}_{J_{H}\psi_{2},\psi_{3}}(-t).$
(3.8)

Our notion of compatibility between $T$ and $H$ only involves the modular
unitaries $\Delta_{H}^{it}$ and not the modular conjugation $J_{H}$. However,
in the presence of crossing symmetry a compatibility between $T$ and $J_{H}$
is automatic:

######  Lemma 3.4.

If a bounded selfadjoint operator $T$ is crossing symmetric and compatible
with a standard subspace $H$, then

$\displaystyle FTF=(J_{H}\otimes J_{H})T(J_{H}\otimes J_{H}),$ (3.9)

where $F$ is the tensor flip on $\mathcal{H}\otimes\mathcal{H}$.

###### Proof.

For arbitrary vectors $\psi_{k}$, the function $t\mapsto
T^{\psi_{2},\psi_{1}}_{\psi_{3},\psi_{4}}(t)$ lies in
${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{1/2})$, with boundary value
at $\mathbb{R}+\frac{i}{2}$ given by (3.8). The value of $s\mapsto
T^{\psi_{1},J_{H}\psi_{4}}_{J_{H}\psi_{2},\psi_{3}}(s)$ at $s=\frac{i}{2}$ on
the one hand coincides with $T^{\psi_{2},\psi_{1}}_{\psi_{3},\psi_{4}}(0)$,
and on the other hand coincides with
$T^{J_{H}\psi_{4},J_{H}\psi_{3}}_{J_{H}\psi_{1},J_{H}\psi_{2}}(0)$. Comparing
these expectation values yields the claim. ∎

The main results on the standardness of $(\mathcal{L}_{T}(H),\Omega)$ that we
will derive for compatible twists are:

  * •

Theorem 3.12: $\Omega$ separating $\mathcal{L}_{T}(H)$ implies that $T$ is
crossing symmetric and braided.

  * •

Theorem 3.22: $T$ crossing symmetric and braided implies that $\Omega$ is
separating for $\mathcal{L}_{T}(H)$.

  * •

Proposition 3.25: Computation of the modular data $J,\Delta$ of
$(\mathcal{L}_{T}(H),\Omega)$ in case $T$ is crossing symmetric and braided.

These results are obtained through preparatory work based on several results
related to the KMS condition [BR97, Sect. 5.3.1], which we establish in the
following technical section.

###  3.1. Analytic continuations of twisted $n$-point functions

For $\xi\in\mathcal{D}_{H}$, we consider the field operators

$\displaystyle\phi_{L,T}^{H}(\xi):=a_{L,T}^{\star}(\xi)+a_{L,T}(S_{H}\xi).$

Note that $\phi_{L,T}^{H}(\xi)$ differs from $\phi_{L,T}(\xi)$ (2.22) by the
Tomita operator $S_{H}$ in the argument of the annihilation operator. For
$\xi=h\in H$, we have $S_{H}h=h$ and both operators coincide. For general
$\xi\in\mathcal{D}_{H}$, the Tomita operator is necessary if we want
$\phi_{L,T}^{H}(\xi)$ to be affiliated with $\mathcal{L}_{T}(H)$ and $\Omega$
separating for this algebra: If $X:=a_{L,T}^{\star}(\xi)+a_{L,T}(\eta)$ is
affiliated with $\mathcal{L}_{T}(H)$ and $\Omega$ is separating, then the
Tomita operator $S$ of $(\mathcal{L}_{T}(H),\Omega)$ restricts to $S_{H}$ on
$\mathcal{H}\cap\mathcal{D}_{H}$ (Lemma 3.2), and hence we have

$\displaystyle\eta=(a_{L,T}(\xi)+a_{L,T}^{\star}(\eta))\Omega=X^{\star}\Omega=S_{H}X\Omega=S_{H}\xi.$

The expectation values of these operators will be denoted

$\displaystyle
W_{2n}^{\xi_{1},\ldots,\xi_{2n}}:=\langle\Omega,\phi_{L,T}^{H}(\xi_{1})\cdots\phi_{L,T}^{H}(\xi_{2n})\Omega\rangle_{T},\qquad\xi_{1},\ldots,\xi_{2n}\in\mathcal{D}_{H}.$
(3.10)

For an odd number of fields, these expectation values vanish. Expanding the
definitions of $\phi_{L,T}^{H}(\xi)$ and
$\langle\,\cdot\,,\,\cdot\,\rangle_{T}$, one finds that
$W_{2n}^{\xi_{1},\ldots,\xi_{2n}}$ can be written as a sum of $1\cdot
3\cdot\ldots\cdot(2n-1)$ terms of the form $\langle\Omega,A_{1}\cdots
T_{l}\cdots A_{2n})\Omega\rangle$, where the $A_{j}$ are either (untwisted)
creation operators $a_{L}^{*}(\xi_{j})$ or annihilation operators
$a_{L}(S_{H}\xi_{j})$, and $T_{l}$ denotes various insertions of twists,
coming from the $a_{L,T}(S_{H}\xi_{k})$ and
$\langle\,\cdot\,,\,\cdot\,\rangle_{T}$. The combinatorial aspects of these
terms are best captured in a diagrammatic form which was already introduced in
[BS94] in a special case. We present and further develop this diagrammatic
form in the appendix (Section A).

In the present section, we do not rely on the diagram notation in our proofs,
but still regard it as a helpful tool to keep track of the various
contributions to $W_{2n}^{\xi_{1},\ldots,\xi_{2n}}$, for instance the 15 terms
of $W_{6}^{\xi_{1},\ldots,\xi_{6}}$ that we will need below. The reader is
invited to refer to Section A as required.

We will be interested in a parameter-dependent version of
$W_{2n}^{\xi_{1},\ldots,\xi_{2n}}$, namely

$\displaystyle
W_{2n}^{\xi_{1},\ldots,\xi_{2n}}(t):=W_{2n}^{\xi_{1},\ldots,\Delta_{H}^{it}\xi_{2n}}=\langle\Omega,\phi_{L,T}^{H}(\xi_{1})\cdots\phi_{L,T}^{H}(\Delta_{H}^{it}\xi_{2n})\Omega\rangle_{T}.$
(3.11)

######  Lemma 3.5.

Let $H\subset\mathcal{H}$ be a standard subspace,
$T\in{\mathcal{T}}_{\geq}(H)$ a compatible twist, and assume that $\Omega$ is
separating for $\mathcal{L}_{T}(H)$. Then, for any $n\in\mathbb{N}$ and any
$\xi_{1},\ldots,\xi_{2n}\in\mathcal{D}_{H}$, the function
$W_{2n}^{\xi_{1},\ldots,\xi_{2n}}$ lies in ${\mathbb{H}}^{\infty}_{\rm
c.bv.}({\mathbb{S}}_{-1})$ and satisfies

$\displaystyle W_{2n}^{\xi_{1},\ldots,\xi_{2n}}(t-i)$
$\displaystyle=W_{2n}^{\Delta_{H}^{it}\xi_{2n},\xi_{1},\ldots,\xi_{2n-1}}=:(W_{2n}^{\xi_{1},\ldots,\xi_{2n}})^{\prime}(t),\qquad
t\in\mathbb{R}.$ (3.12)

###### Proof.

Since the vectors $\xi_{k}$ lie in $\mathcal{D}_{H}$, it follows that
$A:=\phi_{L,T}^{H}(\xi_{2n})$ and
$B:=\phi_{L,T}^{H}(\xi_{1})\cdots\phi_{L,T}^{H}(\xi_{2n-1})$ are closable
operators with closures affiliated to $\mathcal{L}_{T}(H)$. By assumption,
$\Omega$ is separating, so by the KMS condition,
$f(t):=\langle\Omega,B\Delta^{it}A\Omega\rangle_{T}$ has the analyticity and
boundedness properties stated in the lemma, and boundary value
$f(t-i)=\langle\Omega,A\Delta^{it}B\Omega\rangle_{T}$, $t\in\mathbb{R}$. The
claim of the lemma now follows by observing that $A\Omega$ and
$A^{\star}\Omega$ lie in the single particle space $\mathcal{H}$, on which
$\Delta^{it}$ coincides with $\Delta_{H}^{it}$ by Lemma 3.2. ∎

In the following, we will explore properties of $T$ that are consequences of
$\Omega$ being separating for $\mathcal{L}_{T}(H)$. For doing so, we need to
analyze $W_{2n}^{\xi_{1},\ldots,\xi_{2n}}$ for $n=1,2,3$. To lighten our
notation, we will often use shorthand notation and denote the functions (3.11)
and (3.12) by $W_{2n}$ and $W_{2n}^{\prime}$, respectively, leaving the
dependence on the fixed vectors $\xi_{1},\ldots,\xi_{2n}\in\mathcal{D}_{H}$
implicit. We will also refer to $W_{2n}$ as the $(2n)$-point functions because
of their similarity to correlation functions in Wightman QFT.

The most basic continuation result is the following.

######  Lemma 3.6.

Let $H\subset\mathcal{H}$ be a standard subspace and
$\xi_{1},\xi_{2}\in\mathcal{D}_{H}$. Then the function $t\mapsto\langle
S_{H}\xi_{1},\Delta_{H}^{it}\xi_{2}\rangle$ lies in
${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{-1})$ and evaluates at $t=-i$
to $\langle S_{H}\xi_{2},\xi_{1}\rangle$.

###### Proof.

In case $\Omega$ is separating, this is exactly Lemma 3.5 for $n=1$. In case
$\Omega$ is not separating, the statement follows from basic properties of
modular theory for standard subspaces:
$\langle\Delta_{H}^{1/2}S_{H}\xi_{1},\Delta_{H}^{1/2}\xi_{2}\rangle=\langle
J_{H}\xi_{1},\Delta_{H}^{1/2}\xi_{2}\rangle=\langle
S_{H}\xi_{2},\xi_{1}\rangle$. ∎

Subsequently we will often be concerned with analytic functions of the form
(3.6) or similar, namely expectation values of an operator-valued function in
tensor products between various vectors. The following lemma will be helpful
to extend such functions in their vector arguments.

We write $B(\mathbb{R},{\mathcal{B}}(\mathcal{H}))$ for the set of bounded
functions $A:\mathbb{R}\to{\mathcal{B}}(\mathcal{H})$ equipped with the
natural norm $\|A\|_{\infty}=\sup_{t\in\mathbb{R}}\\{\|A(t)\|\\}$.

######  Lemma 3.7.

Let $Z_{i}:\mathcal{D}(Z_{i})\to\mathcal{H}$, $i=1,2$ be closed antilinear
operators, $D_{i}\subset\mathcal{D}(Z_{i})$ subsets of the topological space
$\mathcal{D}(Z_{i})$ provided with the graph norm,
$E_{1},E_{2}\subset\mathcal{H}^{\otimes n}$, $\mathcal{F}\subset
B(\mathbb{R},{\mathcal{B}}(\mathcal{H}^{\otimes(n+1)}))$ subsets, and

$f:\mathcal{F}\times D_{1}\times E_{1}\times D_{2}\times
E_{2}\to{\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{\alpha}).$

If there exists a continuous map $\Theta:\mathcal{F}\to
B(\mathbb{R},{\mathcal{B}}(\mathcal{H}^{\otimes(n+1)}))$ such that

$\displaystyle f(R,\xi_{1},\Psi,\xi_{2},\Phi)(t)$
$\displaystyle=\left\langle\xi_{1}\otimes\Psi,R(t)\Phi\otimes\xi_{2}\right\rangle,$
(3.13) $\displaystyle f(R,\xi_{1},\Psi,\xi_{2},\Phi)(t+i\alpha)$
$\displaystyle=\left\langle\Psi\otimes
Z_{2}\xi_{2},\Theta(R)(t)Z_{1}\xi_{1}\otimes\Phi\right\rangle,$ (3.14)

then $f$ is separately continuous. In particular, $f$ can be uniquely extended
to

$\overline{\operatorname{span}\mathcal{F}}\times\overline{\operatorname{span}D_{1}}\times\overline{\operatorname{span}E_{1}}\times\overline{\operatorname{span}D_{2}}\times\overline{\operatorname{span}E_{2}},$

with closures taken in the respective topologies, and the extension still
satisfies (3.13) and (3.14).

###### Proof.

By linearity and antilinearity, it is obvious that we can extend $f$ to
$\operatorname{span}\mathcal{F}\times\operatorname{span}D_{1}\times\operatorname{span}E_{1}\times\operatorname{span}D_{2}\times\operatorname{span}E_{2}$.
Hence, we can suppose, $\mathcal{F}$, $D_{1},D_{2},E_{1}$, and $E_{2}$ are
vector spaces.

Fix $R$ and, for the sake of notation, let us omit the dependence on $R$ and
denote
$M=(\|\xi_{1}\|+\|Z_{1}\xi_{1}\|)(\|\xi_{2}\|+\|Z_{2}\xi_{2}\|)\|\Psi\|\|\Phi\|$.
By the three lines theorem it follows that, $z\in{\mathbb{S}}_{\alpha}$,

$\displaystyle|f(\xi_{1},\Psi,\xi_{2},\Phi)(z)|$
$\displaystyle\leq\max\left\\{\sup_{t\in\mathbb{R}}\left\\{|f(\xi_{1},\Psi,\xi_{2},\Phi)(t)|\right\\},\sup_{t\in\mathbb{R}}\left\\{\left|f(\xi_{1},\Psi,\xi_{2},\Phi)(t+i\alpha)\right|\right\\}\right\\}$
$\displaystyle\leq\left(\sup_{t\in\mathbb{R}}\|R(t)\|+\sup_{t\in\mathbb{R}}\|\Theta(R)(t)\|\right)M.$

Since $f$ depends linearly or antilinearly on all its four variables, it is
continuous in them separately in the appropriate topologies. Therefore, we can
continuously extend $f$ to
$\overline{\operatorname{span}D_{1}}\times\overline{\operatorname{span}E_{1}}\times\overline{\operatorname{span}D_{2}}\times\overline{\operatorname{span}E_{2}}$,
since ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{\alpha})$ is a Banach
space.

For the continuity in $R$, let us now omit the dependence on the fixed vectors
$\xi_{1}$, $\Psi$, $\xi_{2}$, and $\Phi$ and let $M$ be as above. Again by the
three line theorem, it follows that

$\displaystyle|f(R)(z)-f(S)(z)|$
$\displaystyle\leq\max\left\\{\sup_{t\in\mathbb{R}}\left\\{|f(R)(t)-f(S)(t)|\right\\},\sup_{t\in\mathbb{R}}\left\\{|f(R)(t+i\alpha)-f(S)(t+i\alpha)|\right\\}\right\\}$
$\displaystyle\leq\max\left\\{\sup_{t\in\mathbb{R}}\|R(t)-S(t)\|,\sup_{t\in\mathbb{R}}\|(\Theta(R)-\Theta(S))(t)\|\right\\}M,$

and the conclusion follows from the continuity of $\Theta$. ∎

For the $n$-point functions (with $n\geq 4$), it will be useful to introduce
further shorthand notation in order to increase the readability of our
formulae. We will abbreviate the vectors $\xi_{1},\ldots,\xi_{2n}$ by their
indices $1,\ldots,{2n}$, use a bar to denote the action of $S_{H}$, an index
$t$ to denote the action of $\Delta_{H}^{it}$, and symbols like $a_{k}$ to
denote $a_{L}(\xi_{k})$, the untwisted left annihilation operators (2.17). For
example,

$\displaystyle\langle\bar{1},2\rangle\langle\bar{4}\otimes\bar{3},T(5\otimes
6_{t})\rangle$ $\displaystyle=\langle S_{H}\xi_{1},\xi_{2}\rangle\langle
S_{H}\xi_{4}\otimes
S_{H}\xi_{3},T(\xi_{5}\otimes\Delta_{H}^{it}\xi_{6})\rangle,$
$\displaystyle\langle\bar{4}\otimes a_{3}T(\bar{2}\otimes\bar{1}),T(5\otimes
6_{t})\rangle$ $\displaystyle=\langle S_{H}\xi_{4}\otimes
a_{L}(\xi_{3})T(S_{H}\xi_{2}\otimes
S_{H}\xi_{1}),T(\xi_{5}\otimes\Delta_{H}^{it}\xi_{6})\rangle.$

After these preparations, we now prove results based on the analyticity of the
$4$-point function.

######  Proposition 3.8.

Let $H\subset\mathcal{H}$ be a standard subspace,
$T\in{\mathcal{T}}_{\geq}(H)$ a compatible twist, and assume that $\Omega$ is
separating for $\mathcal{L}_{T}(H)$. Let
$\varphi_{1},\varphi_{2}\in\mathcal{D}_{H}$,
$\psi_{1},\psi_{2}\in\mathcal{H}$.

  1. a)

The function

$\displaystyle
f(t):=\langle\varphi_{1}\otimes\psi_{1},T(\psi_{2}\otimes\Delta_{H}^{it}\varphi_{2})\rangle$
(3.15)

lies in ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{-1})$ and satisfies

$\displaystyle f(t-i)=\langle\psi_{1}\otimes
S_{H}\Delta_{H}^{it}\varphi_{2},T(S_{H}\varphi_{1}\otimes\psi_{2})\rangle,\qquad
t\in\mathbb{R}.$ (3.16)

  2. b)

The vector $a_{L}(\psi_{2})T(\varphi_{1}\otimes\psi_{1})$ lies in the domain
of $S_{H}$, and

$\displaystyle
S_{H}a_{L}(\psi_{2})T(\varphi_{1}\otimes\psi_{1})=a_{L}(\psi_{1})T(S_{H}\varphi_{1}\otimes\psi_{2}).$
(3.17)

###### Proof.

a) With the abbreviations introduced before, the $4$-point function reads

$\displaystyle W_{4}(t)$
$\displaystyle=\langle\bar{1},2\rangle\langle\bar{3},4_{t}\rangle+\langle\bar{2},3\rangle\langle\bar{1},4_{t}\rangle+\langle\bar{2}\otimes\bar{1},T(3\otimes
4_{t})\rangle,$ (3.18)

as follows by expanding the definitions or relying on the diagrammatic rules
explained in the appendix. Since the first two terms have the stated
analyticity, boundedness and continuity properties, and evaluate at $t=-i$ to
$\langle\bar{1},2\rangle\langle\bar{4},3\rangle$ and
$\langle\bar{2},3\rangle\langle\bar{4},1\rangle$, respectively, comparison
with

$\displaystyle
W_{4}^{\prime}(0)=\langle\bar{4},1\rangle\langle\bar{2},3\rangle+\langle\bar{1},2\rangle\langle\bar{4},3\rangle+\langle\bar{1}\otimes\bar{4},T(2\otimes
3)\rangle$ (3.19)

and Lemma 3.5 shows that $t\mapsto\langle\bar{2}\otimes\bar{1},T(3\otimes
4_{t})\rangle$ lies in ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{-1})$
and evaluates at $t=-i$ to $\langle\bar{1}\otimes\bar{4},T(2\otimes
3)\rangle$.

Up to a relabeling of vectors, this function coincides with $f$. Hence we have
shown the lemma in case $\psi_{1},\psi_{2}$ lie in $\mathcal{D}_{H}$. The
extension to $\psi_{1},\psi_{2}\in\mathcal{H}$ now follows by applying Lemma
3.7 (put $R(t):=T(1\otimes\Delta_{H}^{it})$,
$\Theta(R)(t)=(1\otimes\Delta_{H}^{-it})T$, and $Z_{1}=Z_{2}=S_{H}$).

b) We now take $\varphi_{2}$ to even be an entire analytic vector for
$\Delta_{H}$, so that $f$ is entire analytic, and satisfies at $t=-i$

$\displaystyle f(-i)=\langle\psi_{1}\otimes
S_{H}\varphi_{2},T(S_{H}\varphi_{1}\otimes\psi_{2})\rangle=\langle\varphi_{1}\otimes\psi_{1},T(\psi_{2}\otimes\Delta_{H}\varphi_{2})\rangle.$
(3.20)

This can be rewritten as

$\displaystyle\langle
a_{L}(\psi_{1})T(S_{H}\varphi_{1}\otimes\psi_{2}),S_{H}\varphi_{2}\rangle^{*}=\langle
a_{L}(\psi_{2})T(\varphi_{1}\otimes\psi_{1}),S_{H}^{*}S_{H}\varphi_{2}\rangle.$
(3.21)

This equation holds for any vector $\varphi_{2}$ which is entire analytic for
$\Delta_{H}$. As $\varphi_{2}$ ranges over this space, $S_{H}\varphi_{2}$
ranges over a core of $\Delta_{H}^{-1/2}$, and hence a core of $S_{H}^{*}$.
This implies the claim. ∎

######  Remark 3.9.

We may rephrase part b) in terms of left and right creation and annihilation
operators as follows: For any $\psi_{1},\psi_{2}\in\mathcal{H}$, the operator

$\displaystyle T_{\psi_{1},\psi_{2}}\in\mathcal{B}(\mathcal{H}),\qquad
T_{\psi_{1},\psi_{2}}\xi:=a_{L}(\psi_{2})T(\xi\otimes\psi_{1})$ (3.22)

is an endomorphism of $\mathcal{D}_{H}$, with

$\displaystyle S_{H}T_{\psi_{1},\psi_{2}}=T_{\psi_{2},\psi_{1}}S_{H}.$ (3.23)

In particular, $T_{\psi,\psi}$ (and
$T_{\psi_{1},\psi_{2}}+T_{\psi_{2},\psi_{1}}$, etc.) are endomorphisms of the
standard subspace $H$. This provides a link between endomorphisms of standard
subspaces [LW11] and crossing symmetry.

Our next analyticity result is based on the $6$-point function $W_{6}$.

######  Lemma 3.10.

Let $H\subset\mathcal{H}$ be a standard subspace,
$T\in{\mathcal{T}}_{\geq}(H)$ a compatible twist, and assume that $\Omega$ is
separating for $\mathcal{L}_{T}(H)$. Let
$\varphi_{1},\varphi_{2}\in\mathcal{D}_{H}$ and
$\Psi,\Psi^{\prime}\in\mathcal{H}\otimes\mathcal{H}$.

  1. a)

The function $t\mapsto\langle
S_{H}\varphi_{1}\otimes\Psi,T_{1}T_{2}(\Psi^{\prime}\otimes\Delta_{H}^{it}\varphi_{2})\rangle$
lies in ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{-1})$, with value at
$t=-i$ given by

$\displaystyle\langle\Psi\otimes
S_{H}\varphi_{2},T_{2}T_{1}(\varphi_{1}\otimes\Psi^{\prime})\rangle.$ (3.24)

  2. b)

The function $t\mapsto\langle
S_{H}\varphi_{1}\otimes\Psi,T_{2}T_{1}T_{2}(\Psi^{\prime}\otimes\Delta_{H}^{it}\varphi_{2})\rangle$
lies in ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{-1})$, with value at
$t=-i$ given by

$\displaystyle\langle\Psi\otimes
S_{H}\varphi_{2},T_{2}T_{1}T_{2}(\varphi_{1}\otimes\Psi^{\prime})\rangle.$
(3.25)

###### Proof.

The first step of the argument is a calculation: By expanding the definitions
of $\phi_{L,T}^{H}(\xi)$ and $\langle\,\cdot\,,\,\cdot\,\rangle_{T}$, one
finds, $t\in\mathbb{R}$,

$\displaystyle W_{6}(t)=w_{0}(t)+w_{1}(t)+w_{2}(t)+w_{3}(t),$ (3.26)

where

$\displaystyle w_{0}(t)$
$\displaystyle=\Big{(}\langle\bar{1},2\rangle\langle\bar{3},4\rangle\langle\bar{5},6_{t}\rangle+\langle\bar{2},3\rangle\langle\bar{4},5\rangle\langle\bar{1},6_{t}\rangle\Big{)}$
$\displaystyle\quad+\Big{(}\langle\bar{1},2\rangle\langle\bar{4},5\rangle\langle\bar{3},6_{t}\rangle+\langle\bar{1},4\rangle\langle\bar{2},3\rangle\langle\bar{5},6_{t}\rangle+\langle\bar{2},5\rangle\langle\bar{3},4\rangle\langle\bar{1},6_{t}\rangle\Big{)}$
$\displaystyle w_{1}(t)$
$\displaystyle=\langle\bar{1},2\rangle\langle\bar{4}\otimes\bar{3},T(5\otimes
6_{t})\rangle+\langle\bar{2},3\rangle\langle\bar{4}\otimes\bar{1},T(5\otimes
6_{t})\rangle$
$\displaystyle\quad+\langle\bar{3},4\rangle\langle\bar{2}\otimes\bar{1},T(5\otimes
6_{t})\rangle+\langle\bar{4},5\rangle\langle\bar{2}\otimes\bar{1},T(3\otimes
6_{t})\rangle$ $\displaystyle\quad+\langle\bar{2}\otimes\bar{1},T(3\otimes
4)\rangle\langle\bar{5},6_{t}\rangle+\langle\bar{3}\otimes\bar{2},T(4\otimes
5)\rangle\langle\bar{1},6_{t}\rangle,$ $\displaystyle w_{2}(t)$
$\displaystyle=\langle\bar{4}\otimes a_{3}T(\bar{2}\otimes\bar{1}),T(5\otimes
6_{t})\rangle+\langle a_{4}T(\bar{3}\otimes\bar{2})\otimes\bar{1},T(5\otimes
6_{t})\rangle$
$\displaystyle\quad+\langle\bar{2}\otimes\bar{1},T(a_{\bar{3}}T(4\otimes
5)\otimes 6_{t})\rangle,$ $\displaystyle w_{3}(t)$
$\displaystyle=\langle\bar{3}\otimes\bar{2}\otimes\bar{1},T_{2}T_{1}T_{2}(4\otimes
5\otimes 6_{t})\rangle.$

This result only becomes transparent when considering the diagram notation
(Section A). The function $w_{k}$ collects all terms corresponding to diagrams
with $k$ crossings.

Let us denote by $w_{0}^{\prime},\ldots,w_{3}^{\prime}$ the analogous
functions that we get by cyclically permuting $1,2,\ldots,6\to 6,1,\ldots,5$,
e.g. the first term of $w_{0}^{\prime}(0)$ is
$\langle\bar{6},1\rangle\langle\bar{2},3\rangle\langle\bar{4},5\rangle$.

We claim

$\displaystyle w_{k}\in{\mathbb{H}}^{\infty}_{\rm
c.bv.}({\mathbb{S}}_{-1}),\qquad w_{k}(-i)=w_{k}^{\prime}(0),\qquad
k=0,1,2,3.$ (3.27)

To prove this claim, we will investigate $w_{0},\ldots,w_{3}$ one by one.

$w_{0}$:

Each of the five terms contributing to $w_{0}(t)$ depends on $t$ via a
$2$-point function $\langle\bar{j},6_{t}\rangle$ and hence lies in
${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{-1})$ and evaluates at $t=-i$
to $\langle\bar{6},j\rangle$ (Lemma 3.6). Using this result, one sees that the
first term of $w_{0}$, evaluated at $t=-i$, coincides with the second term of
$w_{0}^{\prime}$, evaluated at $t=0$. The other terms of $w_{0}(-i)$ and
$w_{0}^{\prime}(0)$ match up similarly via cyclic permutations of the groups
of terms in round brackets. Hence (3.27) holds for $k=0$.

$w_{1}$:

For $k=1$, (3.27) follows by using Prop. 3.8 a). This result (and Lemma 3.6
for the last two terms of $w_{1}$) imply $w_{1}\in{\mathbb{H}}^{\infty}_{\rm
c.bv.}({\mathbb{S}}_{-1})$. For the boundary values, one checks that the first
term of $w_{1}(-i)$ is
$\langle\bar{1},2\rangle\langle\bar{3}\otimes\bar{6},T(4\otimes 5)\rangle$,
which coincides with the second term of $w_{1}^{\prime}(0)$. The behaviour of
the other terms is analogous: Working $\mod 6$, the $\ell$-th term of
$w_{1}(-i)$ coincides with the $(\ell+1)$st term of $w_{1}^{\prime}(0)$. Hence
(3.27) holds for $k=1$.

$w_{2}$:

For $k=2$, (3.27) follows by using Prop. 3.8: The first and third term of
$w_{2}$ are seen to lie in ${\mathbb{H}}^{\infty}_{\rm
c.bv.}({\mathbb{S}}_{-1})$ on the basis of part a) of that proposition. For
the second term, we also need part b) which ensures that
$a_{4}T(\bar{3}\otimes\bar{2})$ lies in the domain of $S_{H}$. Using these
results, we conclude $w_{2}\in{\mathbb{H}}^{\infty}_{\rm
c.bv.}({\mathbb{S}}_{-1})$ and can compute the boundary value of all terms.
The first term of $w_{2}$, $\langle\bar{4}\otimes
a_{3}T(\bar{2}\otimes\bar{1}),T(5\otimes 6_{t})\rangle$, evaluates at $t=-i$
to $\langle a_{3}T(\bar{2}\otimes\bar{1})\otimes\bar{6},T(4\otimes 5)\rangle$,
which is seen to coincide with the second term of $w_{2}^{\prime}(0)$ by
direct comparison. The second term of $w_{2}(-i)$ is

$\displaystyle\langle\bar{1}\otimes\bar{6},T(S_{H}a_{4}T(\bar{3}\otimes\bar{2})\otimes
5)\rangle$ $\displaystyle=\langle\bar{1}\otimes\bar{6},T(a_{\bar{2}}T(3\otimes
4)\otimes 5)\rangle,$

which coincides with the third term of $w_{2}^{\prime}(0)$. Similarly, the
third term of $w_{2}(-i)$ coincides with the first term of
$w_{2}^{\prime}(0)$. Hence (3.27) holds for $k=2$.

$w_{3}$:

We use (3.26). As we have already shown that $W_{6}$, $w_{0}$, $w_{1}$,
$w_{2}$ lie in ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{-1})$ and
evaluate at $t=-i$ to their primed counterparts $W_{6}^{\prime}(0)$,
$w_{0}^{\prime}(0)$, $w_{1}^{\prime}(0)$, $w_{2}^{\prime}(0)$, respectively,
we conclude that $w_{3}\in{\mathbb{H}}^{\infty}_{\rm
c.bv.}({\mathbb{S}}_{-1})$, and $w_{3}(-i)=w_{3}^{\prime}(0)$.

We now prove the two claims a) and b) made in the lemma.

The function in part a) and its claimed boundary value coincide with the
second term of $w_{2}$ and the third term of $w_{2}^{\prime}(0)$ under the
identifications $\varphi_{1}=\xi_{3}$, $\Psi=S_{H}\xi_{2}\otimes
S_{H}\xi_{1}$, $\varphi_{2}=\xi_{6}$, $\Psi^{\prime}=\xi_{4}\otimes\xi_{5}$,
respectively. Hence we have already proven a) for vectors $\Psi,\Psi^{\prime}$
that are pure tensors of vectors from $\mathcal{D}_{H}$. The general case
follows from Lemma 3.7.

The function in part b) and its is claimed boundary value coincide with
$w_{3}(t)$ and $w_{3}^{\prime}(0)$, respectively, under the identifications
$\varphi_{1}=\xi_{3}$, $\Psi=S_{H}\xi_{2}\otimes S_{H}\xi_{1}$,
$\Psi^{\prime}=\xi_{4}\otimes\xi_{5}$, $\varphi_{2}=\xi_{6}$. Similarly as in
a), this implies that b) holds for $\Psi,\Psi^{\prime}$ of tensor product
form, from which the general result follows by Lemma 3.7. ∎

We finish this section proving some analytic properties of $n$-crossing
functions from crossing-symmetry. Recall that for a twist $T$ crossing-
symmetric w.r.t. a standard subspace $H$, we had defined
$T(t)=(\Delta_{H}^{it}\otimes 1)T(1\otimes\Delta_{H}^{-it})$. Also recall the
tensor leg notation from footnote 2, e.g. $T(t)_{1}=(\Delta_{H}^{it}\otimes
1\otimes 1)(T\otimes 1)(1\otimes\Delta_{H}^{-it}\otimes 1)$ on three tensor
factors.

######  Proposition 3.11.

Let $T\in{\mathcal{T}}_{\geq}$ be crossing symmetric w.r.t. a standard
subspace $H\subset\mathcal{H}$. Then, for $\xi,\xi^{\prime}\in\mathcal{H}$,

$\displaystyle f(t):$ $\displaystyle=\langle\xi\otimes\Psi_{n},T(t)_{1}\cdots
T(t)_{n}(\Phi_{n}\otimes\xi^{\prime})\rangle$

belongs to ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{1/2})$ and
satisfies

$\displaystyle f(t+\tfrac{i}{2})=\langle\Psi_{n}\otimes
J_{H}\xi^{\prime},T(t)_{n}^{\ast}\cdots
T(t)_{1}^{\ast}\left(J_{H}\xi\otimes\Phi_{n}\right)\rangle,\qquad
t\in\mathbb{R}.$ (3.28)

###### Proof.

In order to exploit the crossing symmetry of $T$, we initially choose the
vectors $\Psi_{n}=\psi_{1}\otimes\ldots\otimes\psi_{n}$,
$\Phi_{n}=\varphi_{1}\otimes\ldots\otimes\varphi_{n}$ to be pure tensors.
Considering an orthonormal basis $(e_{k})_{k\in\mathbb{N}}$ of $\mathcal{H}$,
the tensor structure of $T(t)_{1}\cdots T(t)_{n}$ allows us to rewrite the
above scalar product as

$\displaystyle f(t)$ $\displaystyle=\sum_{k_{1},\ldots,k_{n+1}}\left\langle
e_{k_{1}},\xi\right\rangle\prod_{j=1}^{n}T^{e_{k_{j}},\psi_{j}}_{\varphi_{j},e_{k_{j+1}}}(t)\left\langle\xi^{\prime},e_{k_{n+1}}\right\rangle.$
(3.29)

Each partial sum is analytic in ${\mathbb{S}}_{1/2}$ due to the assumption of
$T$ being crossing symmetric and, to conclude that
$f\in{\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{1/2})$, it is enough to
show the partial sums of the series above are uniformly Cauchy. By defining
$Q_{N,M}$ and $Q^{\prime}_{N,M}$ to be the projections onto
$\operatorname{span}\\{e_{j}\ |\ N\leq j\leq M\\}$ and
$\operatorname{span}\\{J_{H}e_{j}\ |\ N\leq j\leq M\\}$, respectively, and
denoting $(X)_{k}=\mathbbm{1}^{\otimes(k-1)}\otimes
X\otimes\mathbbm{1}^{\otimes(n+1-k)}$ for $X\in{\mathcal{B}}(\mathcal{H})$ as
usual, one can reverse the expansion to see that

$\displaystyle f_{N,M}(t):$
$\displaystyle=\sum_{k_{1},\ldots,k_{n+1}=N}^{M}\left\langle
e_{k_{1}},\xi\right\rangle\prod_{j=2}^{n-1}T^{e_{k_{j}},\psi_{j}}_{\varphi_{j},e_{k_{j+1}}}(t)\left\langle\xi^{\prime},e_{k_{n+1}}\right\rangle$
$\displaystyle=\langle(Q_{N,M}\xi)\otimes\Psi_{n},A_{N,M}(t)(\Phi_{n}\otimes(Q_{N,M}\xi^{\prime})\rangle,$

where $A_{N,M}(t)=T(t)_{1}(Q_{N,M})_{2}T(t)_{2}\cdots(Q_{N,M})_{n}T(t)_{n}$.
This situation is similar to Lemma 3.7, because as $N,M\to\infty$, the
operator $(Q_{N,M})_{1}A_{N,M}(t)(Q_{N,M})_{n+1}$ goes uniformly (in $t$) to
zero in the strong operator topology, which is enough to guarantee that the
series is uniformly Cauchy on the real line.

Using crossing symmetry (3.7) to check the behavior of $f_{N,M}$ on the upper
boundary of the strip ${\mathbb{S}}_{1/2}$, we have

$\displaystyle f_{N,M}(t+\tfrac{i}{2})$
$\displaystyle=\sum_{k_{1},\ldots,k_{n+1}=N}^{M}\overline{\left\langle\xi,e_{k_{1}}\right\rangle}\prod_{j=1}^{n}\overline{T^{J_{H}e_{k_{j}},\varphi_{j}}_{\psi_{j},J_{H}e_{k_{j+1}}}(t)}\overline{\left\langle
e_{k_{n+1}},\xi^{\prime}\right\rangle}$
$\displaystyle=\overline{\langle(Q_{N,M}\xi)\otimes\Phi_{n},\Theta(A_{N,M})(t)^{\ast}(\Psi_{n}\otimes(Q_{N,M}\xi^{\prime}))\rangle}$
$\displaystyle=\langle\Psi_{n}\otimes(Q_{N,M}\xi^{\prime}),\Theta(A_{N,M})(t)\left((Q_{N,M}\xi)\otimes\Phi_{n}\right)\rangle,$

where
$\Theta(A_{N,M})(t)=T(t)^{\ast}_{n}(Q^{\prime}_{N,M})_{n}T(t)^{\ast}_{n-1}\cdots(Q^{\prime}_{N,M})_{2}T(t)^{\ast}_{1}$
and we choose this notation again to stress the similarity with Lemma 3.7.

For the same argument as above, this implies that the partial sums of (3.29)
converges uniformly on $\mathbb{R}+\frac{i}{2}$ to
$t\mapsto\langle\Psi_{n}\otimes J_{H}\xi,T(t)^{\ast}_{n}\cdots
T(t)^{\ast}_{1}\left(J_{H}\xi^{\prime}\otimes\Phi_{n}\right)\rangle$. Hence,
it follows by the three lines theorem that (3.29) converges uniformly on the
closure of the strip ${\mathbb{S}}_{1/2}$ and, therefore $f$ lies in
${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{1/2})$ and satisfies (3.28).
∎

###  3.2. Standardness and modular properties of $(\mathcal{L}_{T}(H),\Omega)$

We now apply the results of the previous section to study necessary and
sufficient properties of $T$ for $\Omega$ to be separating for
$\mathcal{L}_{T}(H)$.

######  Theorem 3.12.

Let $H\subset\mathcal{H}$ be a standard subspace,
$T\in{\mathcal{T}}_{\geq}(H)$ a compatible twist, and assume that $\Omega$ is
separating for $\mathcal{L}_{T}(H)$. Then

  1. a)

$T$ is crossing symmetric in the sense of Def. 3.3.

  2. b)

$T$ satisfies the Yang-Baxter equation

$\displaystyle T_{1}T_{2}T_{1}=T_{2}T_{1}T_{2},$ (3.30)

i.e. $T$ is a braided twist.

###### Proof.

a) Let $\psi_{1},\psi_{2},\varphi_{1},\varphi_{2}\in\mathcal{H}$, with
$\varphi_{1}$ and $\varphi_{2}$ entire analytic for $\Delta_{H}$. We consider
the entire analytic function $g:\mathbb{C}\times\mathbb{C}\to\mathbb{C}$,

$\displaystyle
g(t,s):=\langle\Delta_{H}^{-i\bar{s}}\varphi_{1}\otimes\psi_{1},T(\psi_{2}\otimes\Delta_{H}^{-it}\varphi_{2})\rangle,$
(3.31)

Note that the restriction of $g$ to the diagonal, $h(t):=g(t,t)$, coincides
with the function appearing in the crossing symmetry condition (3.6).

According to Prop 3.8 a) (observe that $t$ has opposite signs in $g$ and
(3.15)), we have, $t,s\in\mathbb{C}$,

$\displaystyle g(t+\tfrac{i}{2},s)$ $\displaystyle=g((t-\tfrac{i}{2})+i,s)$
$\displaystyle=\langle\psi_{1}\otimes
S_{H}\Delta_{H}^{-i(t-i/2)}\varphi_{2},T(S_{H}\Delta_{H}^{-i\bar{s}}\varphi_{1}\otimes\psi_{2})\rangle$
$\displaystyle=\langle\psi_{1}\otimes
J_{H}\Delta_{H}^{-it}\varphi_{2},T(\Delta_{H}^{-is-1/2}J_{H}\varphi_{1}\otimes\psi_{2})\rangle.$

From this it is apparent that for $t\in\mathbb{R}$

$\displaystyle h(t+\tfrac{i}{2})$
$\displaystyle=g(t+\tfrac{i}{2},t+\tfrac{i}{2})$
$\displaystyle=\langle\psi_{1}\otimes
J_{H}\Delta_{H}^{-it}\varphi_{2},T(\Delta_{H}^{-it}J_{H}\varphi_{1}\otimes\psi_{2})\rangle$
$\displaystyle=\langle\psi_{1}\otimes
J_{H}\varphi_{2},(1\otimes\Delta_{H}^{it})T(\Delta_{H}^{-it}\otimes
1)(J_{H}\varphi_{1}\otimes\psi_{2})\rangle,$

in agreement with the claimed boundary value (3.7).

By straightforward estimates, one also sees that $h$ is bounded on
$\overline{{\mathbb{S}}_{1/2}}$. Finally, Lemma 3.7 can be used to extend $h$
for general vectors $\varphi_{2},\varphi_{4}\in\mathcal{H}$.

b) The proof that $T$ is braided relies on Lemma 3.10. Let
$\xi_{1},\ldots,\xi_{6}\in\mathcal{D}_{H}$, and consider the function

$\displaystyle f(t):=\langle S_{H}\xi_{3}\otimes S_{H}\xi_{2}\otimes
S_{H}\xi_{1},T_{2}T_{1}T_{2}(\xi_{4}\otimes\xi_{5}\otimes\Delta_{H}^{it}\xi_{6})\rangle.$

By Lemma 3.10 b), $f$ analytically continues to ${\mathbb{S}}_{-1}$, and

$\displaystyle f(-i)=\langle S_{H}\xi_{2}\otimes S_{H}\xi_{1}\otimes
S_{H}\xi_{6},T_{2}T_{1}T_{2}(\xi_{3}\otimes\xi_{4}\otimes\xi_{5})\rangle.$
(3.32)

On the other hand, we may rewrite $f$ as

$\displaystyle f(t)=\langle
S_{H}\xi_{3}\otimes\Psi,T_{1}T_{2}(\xi_{4}\otimes\xi_{5}\otimes\Delta_{H}^{it}\xi_{6})\rangle,$

with $\Psi=T(S_{H}\xi_{2}\otimes S_{H}\xi_{1})$. According to Lemma 3.10 a),
this gives

$\displaystyle f(-i)$ $\displaystyle=\langle\Psi\otimes
S_{H}\xi_{6},T_{2}T_{1}(\xi_{3}\otimes\xi_{4}\otimes\xi_{5})\rangle$
$\displaystyle=\langle S_{H}\xi_{2}\otimes S_{H}\xi_{1}\otimes
S_{H}\xi_{6},T_{1}T_{2}T_{1}(\xi_{3}\otimes\xi_{4}\otimes\xi_{5})\rangle$
(3.33)

Comparing (3.32) and (3.33) now shows that matrix elements of
$T_{2}T_{1}T_{2}$ and $T_{1}T_{2}T_{1}$ between total sets of vectors in
$\mathcal{H}^{\otimes 3}$ coincide, which implies the Yang-Baxter equation
because $T$ is bounded. ∎

This result shows that $\Omega$ being separating for $\mathcal{L}_{T}(H)$ is a
strong condition on the twist. It might also explain why other classes of
twists (see part a) and b) of Thm. 2.2) have not received as much attention as
the braided case. The following examples illustrate this.

######  Example 3.13.

Let $A,B\in{\mathcal{B}}(\mathcal{H})$ be selfadjoint operators, $F$ the
tensor flip on $\mathcal{H}\otimes\mathcal{H}$ as before,
$H\subset\mathcal{H}$ a standard subspace, and

$\displaystyle T:=A\otimes B,\qquad\tilde{T}:=F(A\otimes A).$ (3.34)

Then $T$ is a twist in various cases – for example, if
$\|A\|\|B\|\leq\frac{1}{2}$ (by Thm. 2.2 a)), if $A,B$ are positive (by Thm.
2.2 b)), or if $A=qE,B=\tilde{E}$, where $E,\tilde{E}$ are commuting
orthogonal projections and $-1\leq q\leq 1$ (then $T$ solves the Yang-Baxter
equation and has norm $|q|\leq 1$, so Thm. 2.2 c) applies). Moreover, $T$ is
compatible with $H$ in case $A$ and $B$ commute with the modular unitaries
$\Delta^{it}_{H}$.

However, $T$ is braided only in the last case ($T=qE\otimes\tilde{E}$), and
satisfies crossing symmetry only if $q=0$ or $E=\tilde{E}$ is a one-
dimensional projection onto a vector that is an eigenvector of $\Delta_{H}$
and $J_{H}$. This can be proved by writing out the function $f$ (3.6) for
vectors $\xi_{1},\ldots,\xi_{4}$ analytic for $\Delta_{H}$ and comparing its
crossing boundary value (3.7), which here takes the form
$f(\frac{i}{2})=q\langle\xi_{2},J_{H}E\xi_{1}\rangle\langle
J_{H}\tilde{E}\xi_{3},\xi_{4}\rangle$ with the form
$f(\frac{i}{2})=q\langle\xi_{2},E\Delta_{H}^{-1/2}\xi_{3}\rangle\langle\tilde{E}\Delta_{H}^{1/2}\xi_{1},\xi_{4}\rangle$
that one obtains from directly continuing the modular groups $\Delta_{H}^{it}$
to $t=\frac{i}{2}$.

In comparison, $\tilde{T}$ is a braided twist if $\|A\|\leq 1$, because it is
then a selfadjoint solution to the Yang-Baxter equation. If, in addition
$[\Delta_{H}^{it},A]=0=[J_{H},A]$, then it is also crossing symmetric because

$\displaystyle
T^{\psi_{2},\psi_{1}}_{\psi_{3},\psi_{4}}(t)=\left\langle\psi_{1},A\psi_{3}\right\rangle\left\langle\psi_{2},A\psi_{4}\right\rangle=\overline{T^{J_{H}\psi_{2},\psi_{3}}_{\psi_{1},J_{H}\psi_{4}}(t)}.$

Hence, $T\in{\mathcal{T}}_{\geq}(H)$ is a compatible crossing-symmetric
braided twist in case $[\Delta_{H}^{it},A]=0=[J_{H},A]$.

The second example ($\tilde{T}$) above can be generalized to braided crossing
symmetric twists arising from symmetric twists coming from solutions of the
Yang-Baxter equation with spectral parameter. This also provides the link of
our terminology to the crossing symmetry of scattering theory.

######  Example 3.14.

Recall that on the Hilbert space $\mathcal{H}=L^{2}(\mathbb{R}\to\mathcal{K})$
(where $\mathcal{K}$ is another Hilbert space), we have symmetric twists of
the form (2.11)

$\displaystyle(T_{S}\psi)(\theta_{1},\theta_{2})=S(\theta_{2}-\theta_{1})\psi(\theta_{2},\theta_{1}),$
(3.35)

where $S:\mathbb{R}\to{\mathcal{B}}(\mathcal{K}\otimes\mathcal{K})$ is a
measurable bounded function with $S(-\theta)=S(\theta)^{*}$ almost everywhere,
satisfying the Yang-Baxter equation with spectral parameter.

Let $L\subset L^{2}(\mathbb{R},d\theta)$ denote the standard subspace with

$(\Delta_{L}^{it}\varphi)(\theta)=\varphi(\theta-2\pi
t),\qquad(J_{L}\varphi)(\theta)=\overline{\varphi(\theta)}$

(see, for example, [LL15, Sect. 4]). Let $K\subset\mathcal{K}$ be the closed
real span of an orthonormal basis of $\mathcal{K}$. Then $K$ is a standard
subspace with $K=K^{\prime}$ (such standard subspaces are called maximally
abelian in analogy to the von Neumann algebraic situation). Then the closed
real tensor product of these spaces,

$H:=L\otimes K\subset L^{2}(\mathbb{R})\otimes\mathcal{K}\cong\mathcal{H}$

is a standard subspace in $\mathcal{H}$. Since $\Delta_{K}=1$ and
$\Delta_{L}^{it}$ acts by translation, we see that $T_{S}$ (3.35) is
compatible with $H$.

To evaluate the crossing symmetry condition, consider vectors of the form
$\xi_{k}=\varphi_{k}\otimes v_{k}$, $k=1,\ldots,4$, with $\varphi_{k}\in
L^{2}(\mathbb{R},d\theta)$ and $v_{k}\in\mathcal{K}$. Then the function (3.6)
takes the form

$\displaystyle f(t)$
$\displaystyle=\int\overline{\varphi_{2}(\theta_{1})}\varphi_{4}(\theta_{1})\overline{\varphi_{1}(\theta_{2})}\varphi_{3}(\theta_{2})\langle
v_{2}\otimes v_{1},S(\theta_{2}-\theta_{1}+2\pi t)\,v_{4}\otimes
v_{3}\rangle\,d^{2}\theta.$

It has a bounded analytic continuation to the strip ${\mathbb{S}}_{1/2}$, with
upper boundary value

$\displaystyle f(t+\tfrac{i}{2})$
$\displaystyle=\int\overline{\varphi_{2}(\theta_{1})}\varphi_{4}(\theta_{1})\overline{\varphi_{1}(\theta_{2})}\varphi_{3}(\theta_{2})\langle
v_{1}\otimes J_{K}v_{4},S(\theta_{1}-\theta_{2}-2\pi t)\,v_{3}\otimes
J_{K}v_{2}\rangle\,d^{2}\theta$

This readily implies that the matrix-valued function $S$ has bounded analytic
continuation to the strip ${\mathbb{S}}_{\pi}$, with boundary value

$\displaystyle\langle v_{2}\otimes v_{1},S(t+i\pi)\,v_{4}\otimes
v_{3}\rangle=\langle v_{1}\otimes J_{K}v_{4},S(-t)\,v_{3}\otimes
J_{K}v_{2}\rangle.$ (3.36)

Hence it is clear that there are many functions
$S:\mathbb{R}\to{\mathcal{B}}(\mathcal{K}\otimes\mathcal{K})$ for which
$T_{S}$ is a symmetric twist, but $\Omega$ fails to be separating for
$\mathcal{L}_{T_{S}}(H)$.

######  Remark 3.15.

In quantum field theoretic scattering theory, crossing symmetry is a property
stating that the scattering amplitude of particles is related to the amplitude
of the corresponding antiparticles by analytic continuation [Mar69, Sect. IV].
Whereas this property has not been proven in general quantum field theory (see
[BEG65] for a proof of crossing for two-particle amplitudes, and [Miz21] for
recent work towards general crossing conditions in perturbative QFT), it is
well established – and often taken as an axiom – in integrable QFT on two-
dimensional Minkowski spacetime [Iag78, Smi92, AAR01, Sch10].

In the setting of Example 3.14, our abstract form of crossing symmetry
specializes to the crossing symmetry of scattering theory in integrable
models, with $S$ playing the role of elastic two-body scattering matrix,
$\theta$ is the rapidity and the modular conjugation $J_{K}$ corresponds to
conjugating a particle into an antiparticle [AL17]. See also [BT15, HL18] for
previous work relating standard subspaces and crossing symmetry, and [Nie98]
for a proof of the cyclic formfactor equation, related to crossing symmetry,
from modular theory.

We now proceed to show that the Yang-Baxter equation and crossing symmetry are
not only necessary, but also sufficient conditions for $\Omega$ being
separating for $\mathcal{L}_{T}(H)$. This amounts to establishing a large
commutant of $\mathcal{L}_{T}(H)$.

It is instructive to first look at the case of zero twist $T=0$, with
$\mathcal{F}_{0}(\mathcal{H})$ the full Fock space over $\mathcal{H}$. In this
case, one has in addition to the “left” creation and annihilation operators
(2.17) also “right” creation and annihilation operators, namely
($\xi,\psi_{j}\in\mathcal{H}$)

$\displaystyle a_{R}^{*}(\xi)(\psi_{1}\otimes\ldots\otimes\psi_{n})$
$\displaystyle:=\psi_{1}\otimes\ldots\otimes\psi_{n}\otimes\xi,$ (3.37)
$\displaystyle a_{R}(\xi)(\psi_{1}\otimes\ldots\otimes\psi_{n})$
$\displaystyle:=\langle\xi,\psi_{n}\rangle\cdot\psi_{1}\otimes\ldots\otimes\psi_{n-1}.$
(3.38)

It is easy to see that the left field opeartors
$\phi_{0,L}(h)=a_{L}(h)+a_{L}^{*}(h)$ and right field operators
$\phi_{0,R}(h^{\prime})=a_{R}(h^{\prime})+a_{R}^{*}(h^{\prime})$ commute if
and only if $\operatorname{Im}\langle h,h^{\prime}\rangle=0$, i.e. if $h\in H$
and $h^{\prime}\in H^{\prime}$ for some standard subspace $H$. Hence
$\mathcal{L}_{0}(H)^{\prime}$ contains the von Neumann algebra
$\mathcal{R}_{0}(H^{\prime})$ generated by right fields
$\phi_{0,R}(h^{\prime})$, $h^{\prime}\in H^{\prime}$. As $\Omega$ is also
cyclic for $\mathcal{R}_{0}(H^{\prime})$, it is separating for
$\mathcal{L}_{0}(H)$.

Furthermore, the natural unitary involution

$\displaystyle Y:\mathcal{F}_{0}(\mathcal{H})$
$\displaystyle\to\mathcal{F}_{0}(\mathcal{H})$ (3.39) $\displaystyle
Y(\psi_{1}\otimes\ldots\otimes\psi_{n})$
$\displaystyle:=\psi_{n}\otimes\ldots\otimes\psi_{1},$ (3.40)

relates the left and right operators according to
$Ya_{L}^{\\#}(\xi)Y=a^{\\#}_{R}(\xi)$, which leads to
$Y\mathcal{L}_{0}(H)Y=\mathcal{R}_{0}(H)$; this involution enters the modular
conjugation of $(\mathcal{L}_{0}(H),\Omega)$ in this case [Shl97, p. 341].

In comparison, in our general $T$-twisted setting there exist no “right”
operators. This is due to the fact that the very definition of the Hilbert
space $\mathcal{F}_{T}(\mathcal{H})$ is biased towards the left because of the
appearance of $(1\otimes P_{T,n})$ instead of $(P_{T,n}\otimes 1)$ in the
recursive definition of $P_{T,n+1}$ (2.2).

However, in the case of a braided twist, the symmetry between left and right
is restored. This observation will be a key ingredient to proving that
$\Omega$ separates $\mathcal{L}_{T}(H)$.

######  Lemma 3.16.

Let $T\in{\mathcal{T}}_{\geq}$ be a braided (but not necessarily symmetric)
twist.

  1. a)

For any $n\in\mathbb{N}$,

$\displaystyle P_{T,n+1}$ $\displaystyle=(P_{T,n}\otimes 1)\tilde{R}_{T,n+1},$
(3.41) $\displaystyle\tilde{R}_{T,n+1}$
$\displaystyle:=1+T_{n}+T_{n}T_{n-1}+\ldots+T_{n}\cdots T_{1}.$ (3.42)

  2. b)

The right creation operators

$\displaystyle
a_{R,T}^{\star}(\xi)[\Psi_{n}]=[\Psi_{n}\otimes\xi],\qquad[\Psi_{n}]\in\mathcal{H}^{\otimes
n}/\operatorname{ker}P_{T,n},$ (3.43)

are well-defined on $\mathcal{F}_{T}(\mathcal{H})$ and their adjoints
$a_{R,T}(\xi)$ are given by

$\displaystyle a_{R,T}(\xi)[\Psi_{n}]$
$\displaystyle=[a_{R}(\xi)\tilde{R}_{T,n}\Psi_{n}],$ (3.44)

where $a_{R}(\xi)$ is the untwisted right annihilation operator (3.38).

  3. c)

Let $Z:\mathcal{D}(Z)\to\mathcal{H}$ be a closed (anti)linear operator such
that

$[F(Z\otimes Z),T]=0.$

Then, $\hat{Z}_{Y}:\bigoplus_{n=0}^{\infty}\mathcal{D}(Z)^{\odot
n}/\operatorname{ker}(P_{T,n})\to\mathcal{H}_{T,n}$, given by
$\hat{Z}_{Y}([\Psi_{n}])=[YZ^{\otimes n}\Psi_{n}]$, for all
$\Psi_{n}\in\mathcal{D}(Z)^{\odot n}/\operatorname{ker}(P_{T,n})$ and for all
$n\in\mathbb{N}$, where $Z^{\otimes 0}:=1$ , is a well-defined closable
operator on $\mathcal{F}_{T}(\mathcal{H})$ whose closure will be denoted
$\Gamma_{T}^{Y}(Z)$. This operator is (anti)unitary if $Z$ is. If $Z$ is
invertible, one has for any $\xi\in\mathcal{D}(Z)$

$\displaystyle\overline{\Gamma_{T}^{Y}(Z)a_{L,T}^{\\#}(\xi)\Gamma_{T}^{Y}(Z^{-1})}$
$\displaystyle=a_{R,T}^{\\#}(Z\xi),\qquad\xi\in\mathcal{D}(Z).$ (3.45)

###### Proof.

a) We begin with a calculation in the group algebras of the symmetric groups
$S_{n}\subset S_{n+1}$, where $S_{n}$ is identified with the subgroup of
permutations of $\\{1,\ldots,n+1\\}$ leaving $n+1$ fixed. We claim

$\displaystyle\sum_{\pi\in S_{n+1}}\pi=\sum_{\rho\in
S_{n}}\rho\sum_{k=1}^{n+1}\gamma_{k},$ (3.46)

where the $\gamma_{k}$ are defined by $\gamma_{n+1}=e$ and
$\gamma_{k}=\sigma_{n}\sigma_{n-1}\cdots\sigma_{k}$ for $1\leq k\leq n$.

Indeed, it is easily checked that for any $\pi\in S_{n+1}$, one has
$\rho:=\pi\gamma_{\pi^{-1}(n+1)}^{-1}\in S_{n}$, so any $\pi\in S_{n+1}$ is of
the form $\pi=\rho\gamma_{k}$ for suitable $\rho\in S_{n}$ and
$k\in\\{1,\ldots,n+1\\}$. As $\gamma_{k^{\prime}}\gamma_{k}^{-1}$ lies in
$S_{n}$ if and only if $k=k^{\prime}$, this representation is unique. This
implies the claimed formula (3.46).

In case $T$ is involutive, we have $S_{n}$-representations given by
$\rho_{T,n}(\sigma_{k})=T_{k}$ and (3.41) follows immediately. In case $T$ is
not involutive, $P_{T,n}$ can be formulated as $P_{T,n}=\sum_{\pi\in
S_{n}}t(\pi)$, where $t:S_{n}\to\mathcal{B}(\mathcal{H}^{\otimes n})$ is the
quasi-multiplicative extension of $t(\sigma_{k}):=T_{k}$, i.e.
$t(\sigma_{i_{1}}\cdots\sigma_{i_{l}})=T_{i_{1}}\cdots T_{i_{l}}$ for every
reduced word $\sigma_{i_{1}}\cdots\sigma_{i_{l}}\in S_{n}$ [BS91]. This map is
well-defined because the $T_{k}$ satisfy all relations of $S_{n}$ except
$T_{k}^{2}=1$.

Given a reduced word $w$ representing $\rho\in S_{n}$, it is easy to check
that the word $w\sigma_{n}\sigma_{n-1}\cdots\sigma_{k}$, which represents
$\rho\gamma_{k}$, is reduced as well. Thus $t$ maps the left hand side of
(3.46) to $P_{T,n+1}$ and the right hand side of (3.46) to $(P_{T,n}\otimes
1)\tilde{R}_{T,n+1}$, which proves (3.41).

b) Considering the adjoint of equation (3.41), one gets
$P_{T,n+1}=\tilde{R}^{*}_{n+1}(P_{T,n}\otimes 1)$, from which it is then clear
that (3.43) is well-defined, namely
$\Psi_{n}\otimes\xi\in\operatorname{ker}P_{T,n+1}$ for
$\Psi_{n}\in\operatorname{ker}P_{T,n}$. We can then calculate the adjoint by

$\displaystyle\left\langle[\Psi_{n+1}],a_{R,T}^{\star}(\xi)[\Phi_{n}]\right\rangle_{T}$
$\displaystyle=\left\langle[\Psi_{n+1}],[\Phi_{n}\otimes\xi]\right\rangle_{T}$
$\displaystyle=\left\langle\Psi_{n+1},P_{T,n+1}\Phi_{n}\otimes\xi\right\rangle$
$\displaystyle=\left\langle\Psi_{n+1},\tilde{R}_{T,n+1}^{\ast}(P_{T,n}\otimes
1)\Phi_{n}\otimes\xi\right\rangle$ $\displaystyle=\left\langle
a_{R}(\xi)\tilde{R}_{T,n+1}\Psi_{n+1},P_{T,n}\Phi_{n}\right\rangle$
$\displaystyle=\left\langle[a_{R}(\xi)\tilde{R}_{T,n+1}\Psi_{n+1}],[\Phi_{n}]\right\rangle_{T},$

which implies
$a_{R,T}(\xi)[\Psi_{n+1}]=[a_{R}(\xi)\tilde{R}_{T,n+1}\Psi_{n+1}]$, as
claimed.

c) The argument is similar to that in the proof of Lemma 2.7. The new element
is the appearance of $F$ in the commutation relation $[F(Z\otimes Z),T]=0$.
With $Y_{n}:=Y|_{\mathcal{H}^{\otimes n}}$, one checks $T_{k}Y_{n}Z^{\otimes
n}=Y_{n}Z^{\otimes n}T_{n-k}$, $k=1,\ldots,n-1$. This implies
$[Y_{n}Z^{\otimes n},P_{T,n}]=0$. The remaining steps are now the same as in
Lemma 2.7, observing that the appearance of $Y$ transforms $a_{L}^{\\#}$ into
$a_{R}^{\\#}$. ∎

For braided twists $T$, we therefore also have right field operators

$\displaystyle\phi_{R,T}(\xi)=a_{R,T}^{\star}(\xi)+a_{R,T}(\xi),\qquad\xi\in\mathcal{H},$
(3.47)

and may use them to generate right versions $\mathcal{R}_{T}(H)$ of the von
Neumann algebras $\mathcal{L}_{T}(H)$ considered so far.

######  Definition 3.17.

Given a closed real subspace $H\subset\mathcal{H}$ and a braided twist
$T\in{\mathcal{T}}_{\geq}$, we define the right $T$-twisted Araki-Woods
algebra

$\displaystyle\mathcal{R}_{T}(H):=\\{\exp(i\phi_{R,T}(h))\,:h\in
H\\}^{\prime\prime}\subset\mathcal{B}(\mathcal{F}_{T}(\mathcal{H})).$ (3.48)

######  Remark 3.18.

  1. a)

In complete analogy to Lemma 2.9, one shows that $\Omega$ is cyclic for
$\mathcal{R}_{T}(H)$.

  2. b)

The right and left algebras $\mathcal{L}_{T}(H)$ and $\mathcal{R}_{T}(H)$ seem
not to be related by a natural involution in general because the unitary $Y$
(3.39) does typically not define an operator on
$\mathcal{F}_{T}(\mathcal{H})$. Only in the more specific situation of Lemma
3.16 c) such a symmetry exists.

  3. c)

In principle, the definition of $\mathcal{R}_{T}(H)$ does not need
$T\in{\mathcal{T}}_{\geq}$ to be a braided twist. Indeed, whenever
$\Psi_{n}\in\operatorname{ker}P_{T,n}$ implies
$\Psi_{n}\otimes\xi\in\operatorname{ker}P_{T,n+1}$, the operator
$a_{R,T}^{\star}(\xi)[\Psi_{n}]=[\Psi_{n}\otimes\xi]$,
$[\Psi_{n}]\in\mathcal{H}^{\otimes n}/\operatorname{ker}P_{T,n}$ is well
defined. This is for example the case if $T\in{\mathcal{T}}_{>}$. However, our
later proof that $\mathcal{L}_{T}(H)$ and $\mathcal{R}_{T}(H^{\prime})$
commute will crucially depend on $T$ being braided.

These remarks suggest that $\Omega$ being separating for $\mathcal{L}_{T}(H)$
can be proved by showing that $\mathcal{R}_{T}(H^{\prime})$ and
$\mathcal{L}_{T}(H)$ commute. Hence, we start analyzing the commutations
relations between left and right creation and annihilation operators.

######  Definition 3.19.

A twist $T\in{\mathcal{T}}_{\geq}$ is called local w.r.t. a standard subspace
$H\subset\mathcal{H}$ if $\Psi_{n}\otimes\xi\in\operatorname{ker}P_{T,n+1}$
whenever $\Psi_{n}\in\operatorname{ker}P_{T,n}$ for every $n\in\mathbb{N}$ and

$\displaystyle[\phi_{L,T}(h),\phi_{R,T}(h^{\prime})]=0,\qquad h\in
H,h^{\prime}\in H^{\prime}.$ (3.49)

######  Lemma 3.20.

Let $T\in{\mathcal{T}}_{\geq}$ and suppose
$\Psi_{n}\otimes\xi\in\operatorname{ker}P_{T,n+1}$ whenever
$\Psi_{n}\in\operatorname{ker}P_{T,n}$ for all $n\in\mathbb{N}$. Then, for
$\xi,\eta\in\mathcal{H}$, and $\Psi_{n}\in\mathcal{H}^{\otimes n}$

$\displaystyle[a_{L,T}^{\star}(\xi),a_{R,T}^{\star}(\eta)]$
$\displaystyle=0,\qquad[a_{L,T}(\xi),a_{R,T}(\eta)]=0,$ (3.50)
$\displaystyle[a_{L,T}(\xi),a_{R,T}^{\star}(\eta)]\Omega$
$\displaystyle=\left\langle\xi,\eta\right\rangle\Omega,$ (3.51)
$\displaystyle[a_{L,T}(\xi),a_{R,T}^{\star}(\eta)][\Psi_{n}]$
$\displaystyle=[a_{L}(\xi)(T_{1}\cdots T_{n})(\Psi_{n}\otimes\eta)].$ (3.52)

Furthermore if $T\in{\mathcal{T}}_{\geq}$ is a braided twist, we have in
addition

$\displaystyle[a_{R,T}(\xi),a_{L,T}^{\star}(\eta)][\Psi_{n}]$
$\displaystyle=[a_{R}(\xi)T_{n}\cdots T_{1}(\eta\otimes\Psi_{n})],$
$\displaystyle[\phi_{L,T}(\xi),\phi_{R,T}(\eta)]\Omega$
$\displaystyle=2i\operatorname{Im}\left\langle\xi,\eta\right\rangle\Omega,$
(3.53)
$\displaystyle[\phi_{L,T}(\xi),\phi_{R,T}(\eta)][\Psi_{n}]=[a_{L}(\xi)T_{1}$
$\displaystyle\cdots T_{n}(\Psi_{n}\otimes\eta)-a_{R}(\eta)T_{n}\cdots
T_{1}(\xi\otimes\Psi_{n})].$

###### Proof.

As discussed above, the condition involving kernels guarantees
$a_{R,T}^{\star}(\xi)$ to exist as an operator on
$\mathcal{F}_{T}(\mathcal{H})$. It is easy to see that the left and right
creation operators commute and, by taking adjoints, the left and right
annihilation also commute. For the mixed relation we get

$\displaystyle[a_{L,T}(\xi),a_{R,T}^{\star}(\eta)][\Psi_{n}]$
$\displaystyle=[a_{L}(\xi)R_{T,n+1}(\Psi_{n}\otimes\eta)]-[(a_{L}(\xi)R_{T,n}\Psi_{n})\otimes\eta]$
$\displaystyle=[a_{L}(\xi)T_{1}\cdots T_{n}(\Psi_{n}\otimes\eta)],$

where we have used (2.16) and $R_{T,n+1}=R_{T,n}\otimes 1+T_{1}\cdots T_{n}$.

In case $T\in{\mathcal{T}}_{\geq}$ is a braided twist, Lemma 3.16 ensures the
existence of the right creation operators and fields. One can easily calculate
the commutator acting on the vacuum directly. In addition, we have (3.44) and
$\tilde{R}_{T,n+1}=1\otimes\tilde{R}_{T,n}+T_{n}\cdots T_{1}$. Hence, it
follows that

$\displaystyle[a_{R,T}(\eta),a_{L,T}^{\star}(\xi)][\Psi_{n}]$
$\displaystyle=[a_{R}(\eta)\tilde{R}_{T,n+1}(\xi\otimes\Psi_{n})]-[\xi\otimes(a_{R}(\eta)\tilde{R}_{T,n}\Psi_{n})]$
$\displaystyle=[a_{R}(\eta)T_{n}\cdots T_{1}(\xi\otimes\Psi_{n})].$

Combining these two mixed commutators immediately yields the claimed formula
for $[\phi_{L,T}(\xi),\phi_{R,T}(\eta)][\Psi_{n}]$. ∎

One can easily see that whenever the right fields are well defined on
$\mathcal{F}_{T}(\mathcal{H})$— _i_.e., $\Psi_{n}\in\operatorname{ker}P_{T,n}$
implies $\Psi_{n}\otimes\xi\in\operatorname{ker}P_{T,n+1}$ for all
$\xi,\eta\in\mathcal{H}$, $\Psi_{n}\in\mathcal{H}^{\otimes n}$, and
$n\in\mathbb{N}$—it follows from the fields commutator in Lemma 3.20 that
$T\in{\mathcal{T}}_{\geq}$ is local if, and only if, for all
$\Psi_{n}\in\mathcal{H}^{\otimes n}$, $h\in H$ and $h^{\prime}\in H^{\prime}$,

$\displaystyle\mathcal{H}^{\otimes n}\ni a_{L}(h)T_{1}\cdots
T_{n}(\Psi_{n}\otimes h^{\prime})-a_{R}(h^{\prime})T_{n}\cdots
T_{1}(h\otimes\Psi_{n})\in\operatorname{ker}P_{T,n}.$ (3.54)

Comparing to the zero twist case, this imposes an additional constraint on an
operator $T\in{\mathcal{T}}_{\geq}$ to be local. An example worth mentioning
is $T=-\operatorname{id}_{\mathcal{H}\otimes\mathcal{H}}$, for which
$\operatorname{ker}P_{T,n}=\mathcal{H}^{\otimes n}$ for all $n>1$, so the
condition above is automatically fulfilled. More interesting is however the
situation when kernels are not the whole space. In the next result we give a
sufficient condition for a braided twist to be local, in particular, we
characterize the local braided strict twists.

######  Proposition 3.21.

For a braided twist $T\in{\mathcal{T}}_{\geq}$ to be local w.r.t. a standard
subspace $H\subset\mathcal{H}$ it is sufficient that

$\displaystyle\langle h\otimes\Psi_{n},T_{1}\cdots T_{n}(\Phi_{n}\otimes
h^{\prime})\rangle=\langle\Psi_{n}\otimes h^{\prime},T_{n}\cdots
T_{1}(h\otimes\Phi_{n})\rangle$ (3.55)

for all $\Phi_{n},\Psi_{n}\in\mathcal{H}^{\otimes n}$, $h\in H$,
$h^{\prime}\in H^{\prime}$, and all $n\in\mathbb{N}$. Furthermore, in case
$T\in{\mathcal{T}}_{>}$, (3.55) is also necessary.

###### Proof.

First notice that (3.55) is equivalent to, for all
$\Phi_{n}\in\mathcal{H}^{\otimes n}$, $h\in H$, and $h^{\prime}\in
H^{\prime}$,

$\displaystyle\mathcal{H}^{\otimes n}\ni a_{L}(h)T_{1}\cdots
T_{n}(\Phi_{n}\otimes h^{\prime})-a_{R}(h^{\prime})T_{n}\cdots
T_{1}(h\otimes\Phi_{n})=0.$

Now, Lemma 3.16 ensures the existence of the right fields and Lemma 3.20
yields

$\displaystyle[\phi_{L,T}(h),\phi_{R,T}(h^{\prime})][\Phi_{n}]=[a_{L}(h)T_{1}\cdots
T_{n}(\Phi_{n}\otimes h^{\prime})-a_{R}(h^{\prime})T_{n}\cdots
T_{1}(h\otimes\Phi_{n})]=0.$

In case $T\in{\mathcal{T}}_{>}$ is a strict twist, the kernels are trivial.
Hence one obtains also the reverse implication in this case. ∎

We are now in position for proving the converse of Theorem 3.12.

######  Theorem 3.22.

Let $H\subset\mathcal{H}$ be a standard subspace and
$T\in{\mathcal{T}}_{\geq}(H)$ be a compatible braided twist. Assume that $T$
is crossing symmetric w.r.t. $H$. Then

  1. a)

$T$ is local w.r.t. $H$, _i.e._ ,
$\mathcal{R}_{T}(H^{\prime})\subset\mathcal{L}_{T}(H)^{\prime}$;

  2. b)

$\Omega$ is cyclic and separating for $\mathcal{L}_{T}(H)$ and
$\mathcal{R}_{T}(H)$.

###### Proof.

a) We will verify (3.55). Let $h\in H$, $h^{\prime}\in H^{\prime}$,
$\Psi_{n},\Phi_{n}\in\mathcal{H}^{\otimes n}$ be arbitrary. Thanks to
Proposition 3.11, the function

$\displaystyle f(t)$ $\displaystyle:=\langle
J_{H}h\otimes\Psi_{n},T(t)_{1}\cdots T(t)_{n}(\Phi_{n}\otimes
J_{H}h^{\prime})\rangle$

has an analytic extension to the strip ${\mathbb{S}}_{1/2}$ which satisfies

$\displaystyle f(t+\tfrac{i}{2})$ $\displaystyle=\langle\Psi_{n}\otimes
h^{\prime},T(t)^{\ast}_{n}\cdots
T(t)^{\ast}_{1}(h\otimes\Phi_{n})\rangle,\qquad t\in\mathbb{R}.$

On the other hand, $T$ being compatible with $H$ (i.e.
$[\Delta_{H}^{it}\otimes\Delta_{H}^{it},T]=0$) implies that we may rewrite $f$
as

$\displaystyle f(t)$
$\displaystyle=\langle\Delta_{H}^{-it}J_{H}h\otimes\Psi_{n},T_{1}\cdots
T_{n}(\Phi_{n}\otimes\Delta_{H}^{-it}J_{H}h^{\prime})\rangle.$

Choosing $J_{H}h$ and $J_{H}h^{\prime}$ to be analytic vectors for the modular
automorphism group (this is possible because $H$ and $H^{\prime}$ are
invariant under the spectral projections $E_{(\lambda,\lambda^{-1})}$ of
$\Delta_{H}$ corresponding to spectrum in
$(\lambda,\lambda^{-1})\subset\mathbb{R}_{+}$, $0<\lambda<1$), we have

$\displaystyle f(t+\tfrac{i}{2})$
$\displaystyle=\langle\Delta_{H}^{-it-\frac{1}{2}}J_{H}h\otimes\Psi_{n},T_{1}\cdots
T_{n}(\Phi_{n}\otimes\Delta_{H}^{-it+\frac{1}{2}}J_{H}h^{\prime})\rangle$
$\displaystyle=\langle\Delta_{H}^{-it}h\otimes\Psi_{n},T_{1}\cdots
T_{n}(\Phi_{n}\otimes\Delta_{H}^{-it}h^{\prime})\rangle.$

For analytic $h,h^{\prime}$, the claim now follows by comparing the two
different expressions for $f(t+\frac{i}{2})$ at $t=0$. For general $h\in H$,
$h^{\prime}\in H^{\prime}$, the claim follows by approximation (Lemma 3.7).

b) It follows from Lemma 2.9 and its right version that $\Omega$ is cyclic for
both $\mathcal{L}_{T}(H^{\prime})$ and $\mathcal{R}_{T}(H^{\prime})$. The
inclusion $\mathcal{R}_{T}(H^{\prime})\subset\mathcal{L}_{T}(H)^{\prime}$
proven in part a) then shows that $\Omega$ is separating for
$\mathcal{L}_{T}(H)$.

Since ${\mathcal{T}}_{\geq}(H)={\mathcal{T}}_{\geq}(H^{\prime})$, part a)
implies also $\mathcal{R}_{T}(H)\subset\mathcal{L}_{T}(H^{\prime})^{\prime}$.
By taking commutants,
$\mathcal{L}_{T}(H^{\prime})\subset\mathcal{R}_{T}(H)^{\prime}$ and it follows
that $\Omega$ is separating for $\mathcal{R}_{T}(H)$. ∎

Combining Theorems 3.12 and 3.22, we obtain the following characterization.

######  Corollary 3.23.

Let $H\subset\mathcal{H}$ be a standard subspace and
$T\in{\mathcal{T}}_{\geq}(H)$ a compatible twist. The following are
equivalent:

  1. a)

$\Omega$ is separating for $\mathcal{L}_{T}(H)$.

  2. b)

$T$ is braided and crossing symmetric w.r.t. $H$.

  3. c)

$T$ is local w.r.t. $H$.

We now proceed to characterizing the modular data $J$, $\Delta$ of
$(\mathcal{L}_{T}(H),\Omega)$ (2.27) in terms of the modular data
$J_{H},\Delta_{H}$ of $H$. The main difficulty in identifying $J,\Delta$ is
the fact that a priori it is not clear that they preserve the particle number
grading of $\mathcal{F}_{T}(\mathcal{H})$.

We will be working with tensor powers of standard subspaces, and recall that
for $H\subset\mathcal{H}$ a standard subspace, $H^{\otimes n}$ is defined as
the closed real linear span of $\\{h_{1}\otimes\ldots\otimes
h_{n}\,:\,h_{1},\ldots,h_{n}\in H\\}$. Then $H^{\otimes
n}\subset\mathcal{H}^{\otimes n}$ is a standard subspace, and its modular data
are [LMR16, Prop. 2.6]

$\displaystyle S_{H^{\otimes n}}=S_{H}^{\otimes n},\qquad J_{H^{\otimes
n}}=J_{H}^{\otimes n},\qquad\Delta_{H^{\otimes n}}=\Delta_{H}^{\otimes n}.$
(3.56)

In particular, $\mathcal{D}((\Delta_{H}^{1/2})^{\otimes n})=H^{\otimes
n}+iH^{\otimes n}$.

######  Lemma 3.24.

Let $H$ be a standard subspace, $T\in{\mathcal{T}}_{\geq}(H)$ a compatible
twist, and $n\in\mathbb{N}_{\geq 2}$, $k\in\\{1,\ldots,n-1\\}$. Then
$\mathcal{D}((\Delta_{H}^{1/2})^{\otimes n})$ is an invariant subspace for
$T_{k}$.

###### Proof.

As $T$ is compatible with $H$, we have $[(\Delta_{H}^{it})^{\otimes
n},T_{k}]=0$ for all $t\in\mathbb{R}$. Hence $T_{k}$ commutes with all
spectral projections of the generator $\sum_{k=1}^{n}(\log\Delta_{H})_{k}$ of
this unitary one-parameter group, and thus leaves the domain of
$(\Delta_{H}^{1/2})^{\otimes n}$ invariant. ∎

######  Proposition 3.25.

Let $H\subset\mathcal{H}$ be a standard subspace.

  1. a)

If $T\in{\mathcal{T}}_{\geq}(H)$ is a compatible twist and $\Omega$ is
separating for $\mathcal{L}_{T}(H)$, then the Tomita operator $S$ of
$(\mathcal{L}_{T}(H),\Omega)$ is given by

$\displaystyle S=\Gamma_{T}^{Y}(S_{H}),$ (3.57)

i.e. $S[\xi_{1}\otimes\ldots\otimes\xi_{n}]=[S_{H}\xi_{n}\otimes\ldots\otimes
S_{H}\xi_{1}]$, $\xi_{k}\in\mathcal{D}_{H}$ (see Lemma 3.16 c)). The modular
conjugation and modular unitaries are given by

$\displaystyle J=\Gamma_{T}^{Y}(J_{H}),\qquad\Delta^{it}$
$\displaystyle=\Gamma_{T}(\Delta_{H}^{it}),\quad t\in\mathbb{R}.$ (3.58)

  2. b)

Conversely, suppose $T\in{\mathcal{T}}_{\geq}$ is a twist, $\Omega$ is cyclic
and separating for $\mathcal{L}_{T}$, and the Tomita operator $S$ of
$(\mathcal{L}_{T}(H),\Omega)$ satisfies
$S\raisebox{-2.15277pt}{$|_{\mathcal{H}_{T,2}}$}=\Gamma_{T}^{Y}(S_{H})\raisebox{-2.15277pt}{$|_{\mathcal{H}_{T,2}}$}$
and
$S^{\ast}\raisebox{-2.15277pt}{$|_{\mathcal{H}_{T,2}}$}=\Gamma_{T}^{Y}(S_{H}^{\ast})\raisebox{-2.15277pt}{$|_{\mathcal{H}_{T,2}}$}$.
Then $[T,\Delta^{it}_{H}\otimes\Delta^{it}_{H}]=0$ and $[T,F(J_{H}\otimes
J_{H})]=0$; in particular, $T$ is compatible with $H$.

###### Proof.

a) Compatibility of $T$ with $H$ and $\Omega$ separating $\mathcal{L}_{T}(H)$
imply that the operators on the right hand sides of (3.58) exist as
(anti)unitaries on $\mathcal{F}_{T}(\mathcal{H})$: The unitaries
$\Gamma_{T}(\Delta_{H}^{it})$ exist by Lemma 2.7. Since $T$ is crossing
symmetric w.r.t. $H$ by Thm. 3.12 a), it follows that $T$ commutes with
$F(J_{H}\otimes J_{H})$ by Lemma 3.4. Hence $\Gamma_{T}^{Y}(J_{H})$ exists as
an antiunitary involution on $\mathcal{F}_{T}(\mathcal{H})$ by Lemma 3.16 c).
Consequently,
$\Gamma_{T}^{Y}(S_{H})=\Gamma_{T}^{Y}(J_{H})\Gamma_{T}(\Delta_{H}^{1/2})$
(3.57) exists as an operator on $\mathcal{F}_{T}(\mathcal{H})$ as well.

Let us write $\mathcal{D}_{H,n}:=[H^{\otimes n}]+i[H^{\otimes
n}]\subset\mathcal{H}_{T,n}$, where the brackets $[\,\cdot\,]$ indicate the
$\|\cdot\|_{T}$-closure of the quotient by $\operatorname{ker}P_{T,n}$. This
subspace is invariant under $R_{T,n}$ by Lemma 3.24. Thus we have
$a_{L,T}(\xi)\mathcal{D}_{H,n}\subset\mathcal{D}_{H,n-1}$ for any
$\xi\in\mathcal{D}_{H}$. As
$a_{L,T}^{\star}(\xi)\mathcal{D}_{H,n}\subset\mathcal{D}_{H,n+1}$ is clear for
any $\xi\in\mathcal{D}_{H}$, it follows inductively that

$\displaystyle\phi_{L,T}^{H}(\xi_{1})\cdots\phi_{L,T}^{H}(\xi_{n})\Omega\in\hat{\mathcal{D}}_{H,n}:=\bigoplus_{k=0}^{n}\mathcal{D}_{H,k},\qquad\xi_{1},\ldots,\xi_{n}\in\mathcal{D}_{H},$
(3.59)

where $\phi_{L,T}^{H}(\xi)=a_{L,T}^{\star}(\xi)+a_{L,T}(S_{H}\xi)$. Now we
consider the Tomita operator $S$ of $(\mathcal{L}_{T}(H),\Omega)$ and prove
inductively

$\displaystyle\hat{\mathcal{D}}_{H,n}$
$\displaystyle\subset\mathcal{D}(S),\quad$ $\displaystyle
S\hat{\mathcal{D}}_{H,n}$ $\displaystyle\subset\hat{\mathcal{D}}_{H,n},\quad$
$\displaystyle S[\xi_{1}\otimes\ldots\otimes\xi_{n}]$
$\displaystyle=[S_{H}\xi_{n}\otimes\ldots\otimes S_{H}\xi_{1}],$
$\displaystyle\hat{\mathcal{D}}_{H^{\prime},n}$
$\displaystyle\subset\mathcal{D}(S^{\star}),\quad$ $\displaystyle
S^{\star}\hat{\mathcal{D}}_{H^{\prime},n}$
$\displaystyle\subset\hat{\mathcal{D}}_{H^{\prime},n},\quad$ $\displaystyle
S^{\star}[\xi_{1}^{\prime}\otimes\ldots\otimes\xi_{n}^{\prime}]$
$\displaystyle=[S_{H^{\prime}}\xi_{n}^{\prime}\otimes\ldots\otimes
S_{H^{\prime}}\xi_{1}^{\prime}],$

where $\xi_{1},\ldots,\xi_{n}\in\mathcal{D}_{H}$ and
$\xi_{1}^{\prime},\ldots,\xi_{n}^{\prime}\in\mathcal{D}_{H^{\prime}}$ are
arbitrary. This claim clearly holds for $n=1$, as
$S\xi_{1}=S\phi^{H}_{L,T}(\xi_{1})\Omega=\phi^{H}_{L,T}(\xi_{1})^{\star}\Omega=S_{H}\xi_{1}$.

For the induction step, we split off terms with highest particle number,

$\displaystyle\phi_{L,T}^{H}(\xi_{1})\cdots\phi_{L,T}^{H}(\xi_{n})\Omega$
$\displaystyle=[\xi_{1}\otimes\ldots\otimes\xi_{n}]+E_{n}^{\perp}\phi_{L,T}^{H}(\xi_{1})\cdots\phi_{L,T}^{H}(\xi_{n})\Omega,$
$\displaystyle\phi_{L,T}^{H}(S_{H}\xi_{n})\cdots\phi_{L,T}^{H}(S_{H}\xi_{1})\Omega$
$\displaystyle=[S_{H}\xi_{n}\otimes..\otimes
S_{H}\xi_{1}]+E_{n}^{\perp}\phi_{L,T}^{H}(S_{H}\xi_{n})\cdots\phi_{L,T}^{H}(S_{H}\xi_{1})\Omega.$

By induction hypothesis,
$E_{n}^{\perp}\phi_{L,T}^{H}(\xi_{1})\cdots\phi_{L,T}(\xi_{n})\Omega$ lies in
$\hat{\mathcal{D}}_{H,n-1}\subset\mathcal{D}(S)$ and is mapped by $S$ to a
vector in $\hat{\mathcal{D}}_{H,n-1}$ with vanishing $n$-particle component.
As
$\phi_{L,T}^{H}(\xi_{1})\cdots\phi_{L,T}^{H}(\xi_{n})\Omega\in\mathcal{D}(S)$
as well and is mapped to
$\phi_{L,T}^{H}(S_{H}\xi_{n})\cdots\phi_{L,T}^{H}(S_{H}\xi_{1})\Omega$, we
obtain $[\xi_{1}\otimes\ldots\otimes\xi_{n}]\in\mathcal{D}(S)$ and
$S[\xi_{1}\otimes\ldots\otimes\xi_{n}]=[S_{H}\xi_{n}\otimes\ldots\otimes
S_{H}\xi_{1}]$ as claimed. This one sees by observing that for any polynomial
$Q$ in the fields $\phi_{L,T}^{H^{\prime}}(h_{k}^{\prime})$,
$h_{k}^{\prime}\in H^{\prime}$, of degree up to $n-1$, satisfies

$\displaystyle\langle Q\Omega,S[h_{1}\otimes\ldots\otimes
h_{n-1}]\rangle_{T}=\langle[h_{1}\otimes\ldots\otimes
h_{n-1}],S^{\star}Q\Omega\rangle_{T}=0$

We now obtain $\hat{\mathcal{D}}_{H,n}\subset\mathcal{D}(S)$ by closedness of
$S$. But the above argument can be repeated in complete analogy for
$S^{\star}$ and $S_{H^{\prime}}$ instead of $S$ and $S_{H}$ (and left
exchanged with right), noting that the operators
$\phi_{L,T}^{H^{\prime}}(\xi^{\prime})$,
$\xi^{\prime}\in\mathcal{D}_{H^{\prime}}$, are affiliated with the commutant
$\mathcal{L}_{T}(H)^{\prime}\supset\mathcal{R}_{T}(H^{\prime})$ (Theorem 3.22
a)), which completes the induction.

We have shown $\Gamma_{T}^{Y}(S_{H})\subset S$ and
$\Gamma_{T}^{Y}(S_{H}^{\ast})\subset S^{\ast}$. Thus we also obtain
$\Gamma_{T}^{Y}(S_{H})^{\star}=\Gamma_{T}^{Y}(S_{H^{\prime}})\subset
S^{\star}$ and hence $\Gamma_{T}^{Y}(S_{H})=S$.

The claims about $J$ and $\Delta^{it}$ now follow by polar decomposition
(Lemma 3.16 c)).

b) For any $\eta_{1},\eta_{2}\in H+iH$ and for any
$\eta_{1}^{\prime},\eta_{2}^{\prime}\in H^{\prime}+iH^{\prime}$, we have

$\displaystyle\left\langle\eta_{1}^{\prime}\otimes\eta_{2}^{\prime},(1+T)F(S_{H}\otimes
S_{H})\eta_{1}\otimes\eta_{2}\right\rangle$
$\displaystyle=\left\langle[\eta_{1}\otimes\eta_{2}],S^{\ast}[\eta_{1}^{\prime}\otimes\eta_{2}^{\prime}]\right\rangle_{T}$
$\displaystyle=\left\langle[\eta_{1}\otimes\eta_{2}],\Gamma_{T}^{Y}(S_{H}^{\ast})[\eta_{1}^{\prime}\otimes\eta_{2}^{\prime}]\right\rangle_{T}$
$\displaystyle=\left\langle\eta_{1}\otimes\eta_{2},(1+T)F(S_{H}^{\ast}\otimes
S_{H}^{\ast})\eta_{1}^{\prime}\otimes\eta_{2}^{\prime}\right\rangle$
$\displaystyle=\left\langle\eta_{1}^{\prime}\otimes\eta_{2}^{\prime},Y(S_{H}\otimes
S_{H})(1+T)\eta_{1}\otimes\eta_{2}\right\rangle.$

Hence, $[F(S_{H}\otimes S_{H}),T]=0$ and, by the polar decomposition, we must
have that $[T,\Delta^{it}_{H}\otimes\Delta^{it}_{H}]=0=[T,F(J_{H}\otimes
J_{H})]$. ∎

This result generalizes several theorems known in special cases [EO73, LRT78,
BJL02, Shl97, BLS11, Lec12].

######  Remark 3.26.

Part b) of this theorem states that a Tomita operator of the “second quantized
form”(3.57) requires $T$ to be compatible with $H$. Such situations ($\Omega$
separating for $\mathcal{L}_{T}(H)$ despite $T$ not being compatible with $H$)
actually occur, as we explain now.

Notice that, if $H\subset\mathcal{H}$ is a standard subspace, we discussed in
Example 3.13 that for each $n\in\mathbb{N}$,

$T:=F\left(E^{\Delta_{H}}_{(\frac{1}{n},n)}\otimes
E^{\Delta_{H}}_{(\frac{1}{n},n)}\right)\in{\mathcal{T}}_{\geq}(H)$

is a braided crossing symmetric twist, where
$E^{\Delta_{H}}_{(\frac{1}{n},n)}$ are the spectral projections of
$\Delta_{H}$ on the intervals $(\frac{1}{n},n)$. Therefore, $\Omega$ is cyclic
and separating for $\mathcal{L}_{T}(H)$ thanks to Theorem 3.22. Let us now
take $K\subset H$ another standard subspace. Then, $\Omega$ is also cyclic and
separating for $\mathcal{L}_{T}(K)$. Suppose that
$T\in{\mathcal{T}}_{\geq}(K)$ for every $n\in\mathbb{N}$. Then,

$\displaystyle\Delta_{K}^{is}E^{\Delta_{H}}_{(\frac{1}{n},n)}\Delta_{K}^{-is}\otimes\Delta_{K}^{is}E^{\Delta_{H}}_{(\frac{1}{n},n)}\Delta_{K}^{-is}=E^{\Delta_{H}}_{(\frac{1}{n},n)}\otimes
E^{\Delta_{H}}_{(\frac{1}{n},n)}.$

This forces
$[E^{\Delta_{K}}_{(\frac{1}{n},n)},E^{\Delta_{H}}_{(\frac{1}{n},n)}]=0$.
Hence, the dense subspace

$D:=\bigcup_{n\in\mathbb{N}}E^{\Delta_{K}}_{(\frac{1}{n},n)}E^{\Delta_{H}}_{(\frac{1}{n},n)}\mathcal{H}\subset\mathcal{D}(\Delta_{H}^{\frac{1}{2}})\cap\mathcal{D}(\Delta_{K}^{\frac{1}{2}})$

is invariant under the action of $\Delta_{K}^{is}$ and $\Delta_{H}^{is}$ for
all $s\in\mathbb{R}$. Therefore $D$ is a common core for $\log(\Delta_{K})$
and $\log(\Delta_{H})$. On the other hand, we know that the closed operators
$S_{K},S_{H}$ form an extension $S_{K}\subset S_{H}$ because $K\subset H$. The
existence of a common core then yields $S_{K}=S_{H}$, _i.e._ , $K=H$.

From this we see that, if $\operatorname{dim}\mathcal{H}=\infty$, there are
examples of a standard subspace $K$ and of a braided twist $T$ such that
$\Omega$ is cyclic and separating for $\mathcal{L}_{T}(K)$, but $T$ is not
compatible with $K$ and consequently the Tomita operator of
$(\mathcal{L}_{T}(K),\Omega)$ is different from $\Gamma_{T}^{Y}(S_{K})$.

We also note that the properties of a twist $T$ being compatible with a
standard subspace, or being crossing symmetric w.r.t. $H$, do in general not
pass to sub standard subspaces $K\subset H$.

As a simple corollary to Proposition 3.25, we now obtain a duality between
left and right twisted Araki-Woods algebras.

######  Corollary 3.27.

Let $H\subset\mathcal{H}$ be a standard subspace and
$T\in{\mathcal{T}}_{\geq}(H)$ a compatible twist that is braided and crossing-
symmetric. Then

$\displaystyle\mathcal{L}_{T}(H)^{\prime}=\mathcal{R}_{T}(H^{\prime}).$ (3.60)

###### Proof.

We apply Lemma 3.16 c), noting that $Z=J_{H}$ satisfies $[J_{H}^{\otimes
2}F,T]=0$ by Lemma 3.4. In view of the form of the modular conjugation $J$
established in Proposition 3.25, we therefore have
$J\mathcal{L}_{T}(H)J=\mathcal{R}_{T}(J_{H}H)=\mathcal{R}_{T}(H^{\prime})$. As
$J\mathcal{L}_{T}(H)J=\mathcal{L}_{T}(H)^{\prime}$ by Tomita’s Theorem, the
claim follows. ∎

##  4\. Inclusions of twisted Araki-Woods algebras

Let $\mathcal{H}$ be a Hilbert space and $T$ a braided twist on
$\mathcal{H}\otimes\mathcal{H}$. In the previous sections we have constructed
two maps

$\displaystyle\mathcal{L}_{T},\mathcal{R}_{T}:\text{Std}_{T}(\mathcal{H})\to\text{vN}(\mathcal{F}_{T}(\mathcal{H}))$
(4.1)

from the set $\text{Std}_{T}(\mathcal{H})$ of all standard subspaces
$H\subset\mathcal{H}$ that are compatible with $T$ to the set
$\text{vN}(\mathcal{F}_{T}(\mathcal{H}))$ of von Neumann subalgebras of
$\mathcal{B}(\mathcal{F}_{T}(\mathcal{H}))$.

######  Proposition 4.1.

The maps (4.1) have the following properties.

  1. a)

$\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$ and
$\mathcal{R}_{T}(K)\subset\mathcal{R}_{T}(H)$ for standard subspaces $K\subset
H$.

  2. b)

If $T$ is crossing-symmetric and $H\in\text{\rm Std}_{T}(\mathcal{H})$, then
$\Omega$ is cyclic and separating for $\mathcal{L}_{T}(H)$ and
$\mathcal{R}_{T}(H)$, and duality holds, i.e.
$\mathcal{L}_{T}(H)^{\prime}=\mathcal{R}_{T}(H^{\prime})$.

  3. c)

Let $H\in\text{\rm Std}_{T}(\mathcal{H})$ and $U$ a unitary on $\mathcal{H}$.
If $[U\otimes U,T]=0$, then also $UH\in\text{\rm Std}_{T}(\mathcal{H})$, and

$\displaystyle\Gamma_{T}(U)\mathcal{L}_{T}(H)\Gamma_{T}(U)^{\star}$
$\displaystyle=\mathcal{L}_{T}(UH),$ (4.2)
$\displaystyle\Gamma_{T}(U)\mathcal{R}_{T}(H)\Gamma_{T}(U)^{\star}$
$\displaystyle=\mathcal{R}_{T}(UH).$ (4.3)

If $T$ is crossing symmetric w.r.t. $H$, it is crossing symmetric w.r.t. $UH$.

  4. d)

Let $H\in\text{\rm Std}_{T}(\mathcal{H})$ and $U$ an antiunitary on
$\mathcal{H}$. If $[F(U\otimes U),T]=0$, then also $UH\in\text{\rm
Std}_{T}(\mathcal{H})$, and

$\displaystyle\Gamma_{T}^{Y}(U)\mathcal{L}_{T}(H)\Gamma_{T}^{Y}(U)^{\star}$
$\displaystyle=\mathcal{R}_{T}(UH),$ (4.4)
$\displaystyle\Gamma_{T}^{Y}(U)\mathcal{R}_{T}(H)\Gamma_{T}^{Y}(U)^{\star}$
$\displaystyle=\mathcal{L}_{T}(UH).$ (4.5)

If $T$ is crossing symmetric w.r.t. $H$, it is crossing symmetric w.r.t. $UH$.

###### Proof.

Part a) is obvious from the definition of these algebras and b) is the content
of Thm. 3.22 and Cor. 3.27. The first part of c) follows by application of
Lemma 2.7, and the second part by application of Lemma 3.16 c). The last
statement follows easily by observing $\Delta_{UH}=U\Delta_{H}U^{*}$ and
$J_{UH}=UJ_{H}U^{*}$. For d), the action of $\Gamma_{T}(U)$ on
$\mathcal{L}_{T}(H)$ and $\mathcal{R}_{T}(H)$ follows from Lemma 3.16 c), so
we only need to explain the crossing symmetry w.r.t. $UH$. For arbitrary
$\psi_{1},\ldots,\psi_{4}\in\mathcal{H}$, we see that the function

$\displaystyle f(t)$
$\displaystyle:=\langle\psi_{2}\otimes\psi_{1},(\Delta_{UH}^{it}\otimes
1)T(1\otimes\Delta_{UH}^{-it})(\psi_{3}\otimes\psi_{4})\rangle$
$\displaystyle=\langle U^{*}\psi_{4}\otimes
U^{*}\psi_{3},(\Delta_{H}^{it}\otimes
1)T(1\otimes\Delta_{H}^{-it})(U^{*}\psi_{1}\otimes U^{*}\psi_{2})\rangle$

lies in ${\mathbb{H}}^{\infty}_{\rm c.bv.}({\mathbb{S}}_{1/2})$, with

$\displaystyle f(t+\tfrac{i}{2})$ $\displaystyle=\langle U^{*}\psi_{3}\otimes
J_{H}U^{*}\psi_{2},(1\otimes\Delta_{H}^{it})T(\Delta_{H}^{-it}\otimes
1)(J_{H}U^{*}\psi_{4}\otimes U^{*}\psi_{1})\rangle$
$\displaystyle=\langle\psi_{1}\otimes
J_{UH}\psi_{4},(1\otimes\Delta_{UH}^{it})T(\Delta_{UH}^{-it}\otimes
1)(J_{UH}\psi_{2}\otimes\psi_{3})\rangle.$

Hence $T$ is crossing symmetric w.r.t. $UH$.

We note that for the crossing symmetry statements in c) and d), the assumption
of $T$ being compatible with $H$ is not unnecessary. ∎

Observe that in case $T=F$ is the tensor flip, both nets coincide, i.e.

$\displaystyle\mathcal{L}_{F}(H)=\mathcal{R}_{F}(H).$ (4.6)

This easily follows from the fact that in this case, $n!^{-1}P_{F,n}$ is the
projection of $\mathcal{H}^{\otimes n}$ onto its totally symmetric subspace so
that tensor multiplication from the left and right become identical. Let us
point out that in general, the two nets are different and do not even form
inclusions.

######  Lemma 4.2.

Let $T$ be a braided twist, $H\subset K\in\text{\rm Std}_{T}(\mathcal{H})$,
and $T$ crossing symmetric w.r.t. $H$ and $K$. If
$\mathcal{L}_{T}(H)\subset\mathcal{R}_{T}(K)$, then

  1. a)

$\phi_{L,T}(\psi)=\phi_{R,T}(\psi)$ for all $\psi\in\mathcal{H}$,

  2. b)

$(1+T)(1-F)=0$.

If $T$ is a strict twist, $\mathcal{L}_{T}(H)\subset\mathcal{R}_{T}(K)$ is
impossible.

###### Proof.

We have $\phi_{R,T}(h)\Omega=h=\phi_{L,T}(h)\Omega$ for any $h\in H$. Since
the vacuum separates $\mathcal{R}_{T}(K)$ and the field operators are
affiliated, we conclude $\phi_{L,T}(h)=\phi_{R,T}(h)$, $h\in H$. The
annihilation/creation structure then implies
$a_{L,T}^{\star}(h)=a_{R,T}^{\star}(h)$, which extends to arbitrary
$h\in\mathcal{H}$.

This implies in particular

$\displaystyle[h\otimes
k]=a^{\star}_{L,T}(h)a^{\star}_{L,T}(k)\Omega=a^{\star}_{R,T}(h)a^{\star}_{R,T}(k)\Omega=[k\otimes
h]$

for any $h,k\in\mathcal{H}$, and hence $P_{T,2}=1+T$ vanishes on antisymmetric
vectors, i.e. $(1+T)(1-F)=0$. For strict twist, $(1+T)$ is invertible, which
yields a contradiction. ∎

As $\mathcal{L}_{T}(H)\neq\mathcal{R}_{T}(H)$ in general, it is interesting to
consider the relative positions of the $\mathcal{L}_{T}$\- and
$\mathcal{R}_{T}$-algebras. This relates in particular to inclusions of
twisted Araki-Woods algebras: Given an inclusion of standard subspaces
$K\subset H\subset\mathcal{H}$, we will consider the corresponding inclusion

$\displaystyle\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$ (4.7)

of von Neumann algebras.888By taking commutants, one also obtains a “right”
version $\mathcal{R}_{T}(H^{\prime})\subset\mathcal{R}_{T}(K^{\prime})$, and
in case $[T,F]=0$ (e.g., for $T=qF$), the unitary $Y$ (3.39) satisfies
$Y\mathcal{L}_{T}(H)Y^{\star}=\mathcal{R}_{T}(H)$ for every standard subspace
$H$ (Lemma 3.16 c)), so that the inclusions
$\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$ and
$\mathcal{R}_{T}(K)\subset\mathcal{R}_{T}(H)$ are isomorphic. The
investigation of the relation between the $\mathcal{L}_{T}$\- and
$\mathcal{R}_{T}$-inclusions for general $T$ is left to future work.

The structure of this inclusion depends strongly on the standard subspace
inclusion $K\subset H$ as well as the twist $T$. We here focus on the relative
commutants

$\displaystyle\mathcal{C}_{T}(K,H):=\mathcal{L}_{T}(K)^{\prime}\cap\mathcal{L}_{T}(H)=\mathcal{R}_{T}(K^{\prime})\cap\mathcal{L}_{T}(H)$
(4.8)

which contains information on the relative position of the $\mathcal{L}_{T}$\-
and $\mathcal{R}_{T}$-systems.

For general twist $T$ and standard subspaces $K\subset H$, a detailed analysis
of (4.8) is quite complicated. For instance, setting $K=H$ this question
contains the question whether $\mathcal{L}_{T}(H)$ is a factor. For the
special twists $T=qF$, this question has been completely settled only very
recently – see [KSW23] and the references therein.

In a few special cases, more is known: If $T=F$ is the tensor flip (Bose
case), then $\mathcal{C}_{F}(K,H)=\mathcal{L}_{F}(K^{\prime}\cap H)$, and
$\Omega$ is cyclic for $\mathcal{F}_{F}(\mathcal{H})$ if and only if
$K^{\prime}\cap H$ is a cyclic space. This follows from [LRT78, Thm. I.3.2] or
[EO73, Lemma 1] by mapping our setting to the Bose Fock space via the unitary
$I=\bigoplus_{n}I_{n}$. Results about the Fermionic case $T=-F$ can be found
in [Foi83, Prop. 2.5] and [BJL02, Prop. 3.4].

If $T=T_{S}\in{\mathcal{T}}_{\rm Sym}$ is a symmetric (in particular braided)
twist coming from a solution $S$ of the Yang-Baxter equation with spectral
parameter as in (2.11), and $S$ satisfies a number of conditions (including
crossing symmetry), then the relative commutant $\mathcal{C}_{T}(K,H)$ is
known to be a type III1 factor having $\Omega$ as a cyclic vector for certain
inclusions $K\subset H$ that arise in quantum field theory (see Section 5)
[AL17].

In particular, there seem to exist few results on the inclusions
$\mathcal{L}_{T}(K)\subset\mathcal{L}_{T}(H)$ in case the twist is not
unitary, for instance in case $\|T\|<1$. In this section, we investigate two
types of inclusions in which the structure of $\mathcal{C}_{T}(K,H)$ can be
decided for $\|T\|<1$.

###  4.1. Half-sided modular inclusions

Half-sided modular inclusions were introduced by Wiesbrock [Wie93b, Wie93a]
and are of prominent importance in conformal quantum field theory. They are
defined as follows.

######  Definition 4.3.

An inclusion of two von Neumann algebras $\mathcal{N}\subset\mathcal{M}$ with
a joint cyclic separating vector $\Omega$ is called half-sided modular if the
modular unitaries $\Delta^{it}$ of $(\mathcal{M},\Omega)$ satisfy

$\displaystyle\Delta^{it}\mathcal{N}\Delta^{-it}\subset\mathcal{N},\qquad
t\leq 0.$ (4.9)

For a non-trivial half-sided inclusion
$(\mathcal{N}\subset\mathcal{M},\Omega)$, there exists no conditional
expectation $\mathcal{M}\to\mathcal{N}$ (hence no meaningful notion of index),
and the inclusion is not split. Hence no standard methods for investigating
the relative commutant of a half-sided inclusion are available.

The structure of a half-sided modular inclusion is closely related to that of
a one-dimensional Borchers triple, that we also recall here [BLS11].

######  Definition 4.4.

A one-dimensional Borchers triple $(\mathcal{M},U,\Omega)$ consists of a von
Neumann algebra $\mathcal{M}$ on a Hilbert space $\mathcal{H}$, a strongly
continuous unitary one-parameter group $U$ with positive generator
${\mathrm{P}}$, and a unit vector $\Omega\in\mathcal{H}$ such that

  1. a)

$\Omega$ is cyclic and separating for $\mathcal{M}$, and $U(x)\Omega=\Omega$
for all $x\in\mathbb{R}$,

  2. b)

$U(x)\mathcal{M}U(x)^{-1}\subset\mathcal{M}$ for $x\geq 0$.

Let us recall that for a one-dimensional Borchers triple
$(\mathcal{M},U,\Omega)$, Borchers’ Theorem [Bor92] asserts that

$\displaystyle\Delta^{it}U(x)\Delta^{-it}=U(e^{-2\pi t}x),\quad
JU(x)J=U(-x),\qquad t,x\in\mathbb{R},$ (4.10)

where $J,\Delta$ are the modular data of $(\mathcal{M},\Omega)$. Thus $U$
extends to a (anti)unitary representation of the affine group of $\mathbb{R}$
(the “$ax+b$ group”).

As a consequence of (4.10), the inclusion
${\mathcal{N}}:=U(1)\mathcal{M}U(-1)\subset\mathcal{M}$ coming from a Borchers
triple is half-sided modular. Conversely, given a half-sided modular inclusion
$({\mathcal{N}}\subset\mathcal{M},\Omega)$, there exists a strongly continuous
unitary one-parameter group $U$ such that $(\mathcal{M},U,\Omega)$ is a one-
dimensional Borchers triple, and ${\mathcal{N}}=U(1)\mathcal{M}U(-1)$ [Wie93b,
AZ05]. This one-parameter group is related to the modular unitaries of
$({\mathcal{N}},\Omega)$ and $(\mathcal{M},\Omega)$ by

$\displaystyle U(e^{2\pi
t}-1)=\Delta_{\mathcal{M}}^{-it}\Delta_{\mathcal{N}}^{it}.$ (4.11)

We may therefore use the structure of one-dimensional Borchers triples to
define half-sided modular inclusions.

We will say that a half-sided inclusion has unique vacuum if the subspace of
$U$-invariant vectors is $\mathbb{C}\Omega$. In case this condition holds and
$\operatorname{dim}\mathcal{H}>1$, the von Neumann algebras $\mathcal{M}$ and
$\mathcal{N}$ are (not necessarily hyperfinite) type III1 factors [Lon79,
Wie93b].

To give examples that are of the form (4.7), we now consider a standard
subspace $H$ and a twist $T$ with a compatible translation representation.

######  Definition 4.5.

A translation group compatible with a standard subspace $H\subset\mathcal{H}$
and a twist $T\in{\mathcal{T}}_{\geq}$ is a strongly continuous one-parameter
group $U(x)$, $x\in\mathbb{R}$, such that

  1. a)

the generator ${\mathrm{P}}$ of $U(x)$ is positive,

  2. b)

$\operatorname{ker}{\mathrm{P}}=\\{0\\}$,

  3. c)

$U(x)H\subset H$, $x\geq 0$,

  4. d)

$[U(x)\otimes U(x),T]=0$ for all $x\in\mathbb{R}$.

######  Proposition 4.6.

Let $H\subset\mathcal{H}$ be a standard subspace,
$T\in{\mathcal{T}}_{\geq}(H)$ a compatible braided crossing symmetric twist
and $U$ a compatible translation group. Then
$(\mathcal{L}_{T}(H),\Gamma_{T}\circ U,\Omega)$ is a one-dimensional Borchers
triple with unique vacuum.

###### Proof.

Thanks to Def. 4.5 d) and Lemma 2.7, $x\mapsto\Gamma_{T}(U(x))$ is a unitary
one-parameter group on $\mathcal{F}_{T}(\mathcal{H})$. It is clear that it has
positive generator and fixes the Fock vacuum $\Omega$, and the latter is the
only (up to multiples) translation invariant vector because $U(x)$ has no non-
zero fixed points.

Furthermore, we have the covariance relation (cf. Lemma 2.7 b))

$\displaystyle\Gamma_{T}(U(x))\mathcal{L}_{T}(H)\Gamma_{T}(U(x))^{-1}=\mathcal{L}_{T}(U(x)H)\subset\mathcal{L}_{T}(H)\quad\text{
for }\,x\geq 0.$

As $\Omega$ is cyclic and separating for $\mathcal{L}_{T}(H)$ by Thm. 3.22,
the proof is finished. ∎

Examples of standard subspaces with compatible translation groups can be
easily constructed with the help of positive energy representations of the
Poincaré group as we shall discuss in Section 5.

To analyze the relative commutants $\mathcal{C}_{T}(K,H)$, where
$K=U(1)H\subset H$ is given by a compatible translation group $U$, we will
make use of a criterion that was recently developed in [LS22]. Given a one-
dimensional Borchers triple $(\mathcal{M},U,\Omega)$, one considers bounded
open intervals $I\subset\mathbb{R}$ and a family of von Neumann algebras
$I\mapsto\mathcal{A}(I)$ indexed by these intervals, namely

$\displaystyle\mathcal{A}((a,b)):=U(a)\mathcal{M}U(a)^{*}\cap
U(b)\mathcal{M}^{\prime}U(b)^{*},$ (4.12)

as well as the algebra at infinity,

$\displaystyle\mathcal{A}_{\infty}:=\bigcap_{I}\mathcal{A}(I)^{\prime}.$
(4.13)

Then the following holds.

######  Proposition 4.7.

[LS22] Let $(\mathcal{M},U,\Omega)$ be a one-dimensional Borchers triple with
unique vacuum. Let $J,\Delta$ denote the modular data of
$(\mathcal{M},\Omega)$, and
$\mathcal{N}:=U(1)\mathcal{M}U(1)^{-1}\subset\mathcal{M}$.

  1. a)

The relative commutants $\mathcal{A}((0,x))=\mathcal{M}\cap
U(x)\mathcal{M}^{\prime}U(x)^{*}$, $x>0$, are trivial if and only if the one-
dimensional projection $P_{\Omega}$ onto $\mathbb{C}\Omega$ lies in
$\mathcal{A}_{\infty}$.

  2. b)