Bound State Operators and Wedge-Locality in Integrable Quantum Field Theories

Bound State Operators and Wedge-Locality
in Integrable Quantum Field Theories

Yoh TANIMOTO †‡

Y. Tanimoto

† Graduate School of Mathematical Sciences, The University of Tokyo,
† 3-8-1 Komaba Meguro-ku Tokyo 153-8914, Japan<EMAIL_ADDRESS>http://www.mat.uniroma2.it/~tanimoto/

‡ Institut für Theoretische Physik, Göttingen University,
‡ Friedrich-Hund-Platz 1, 37077 Göttingen, Germany

Received February 19, 2016, in final form October 10, 2016; Published online
October 19, 2016

We consider scalar two-dimensional quantum field theories with a factorizing
$S$-matrix which has poles in the physical strip. In our previous work, we
introduced the bound state operators and constructed candidate operators for
observables in wedges. Under some additional assumptions on the $S$-matrix, we
show that, in order to obtain their strong commutativity, it is enough to
prove the essential self-adjointness of the sum of the left and right bound
state operators. This essential self-adjointness is shown up to the two-
particle component.

Haag–Kastler net; integrable models; wedge; von Neumann algebras; Hardy space;
self-adjointness

81T05; 81T40; 81U40

##  1 Introduction

The goal of this series of papers [6, 7, 9, 35] is to construct more
interacting two-dimensional integrable quantum field theories in the
Haag–Kastler framework. The novelty of the recent progresses is that one first
constructs observables localized in infinitely-extended wedge regions, then
obtains compactly localized observables in an indirect way. In the preceding
paper [6], we constructed candidates for wedge-local observables, however,
they have subtle domain properties and their strong commutativity remained
open. In this paper, we provide a possible way to settle this question, which
is a crucial step towards construction of Haag–Kastler nets.

The main strategy was explained in [6] and we only briefly summarize it. See
also [21] for an overview of the program. The operator-algebraic framework for
quantum field theory (QFT), or a Haag–Kastler net [16], concerns the
collection of algebras of local observables. As in any mathematical framework
for relativistic QFT, it is a very difficult problem to construct interacting
examples of Haag–Kastler net. For certain integrable models with prescribed
$S$-matrix, Schroer proposed [31] to construct certain wedge-local
observables, then obtain compactly localized observables as the intersection
of two wedge-algebras. Some constructions of wedge-local algebras have been
obtained for a class of $S$-matrices [18, 22, 34], as well as operator-
algebraic constructions or deformations [1, 2, 5, 12, 20, 33, 34]. The final
step to show the existence of local observables has been completed in certain
cases, by showing modular nuclearity [4, 19], or by split property [34].

In [18], Lechner constructed wedge-local fields for scalar factorizing
$S$-matrices without poles in the physical strip. We attempted to extend this
construction to the cases with poles in [6], however, certain spectral
properties remained unsolved. More precisely, our candidates for wedge-local
fields were a slight modification of the fields of [18], but contained a very
singular term, the bound state operator, which has been introduced exactly in
order to treat the poles. The strong commutativity of these candidates
remained unclear. We studied the one-particle component of these operators in
detail in [35] and found convenient self-adjoint extensions. In the present
work, we propose a strategy to prove the strong commutativity between the
candidate operators.

Self-adjointness of unbounded operators is in general a hard problem. There
are several criteria (e.g., see [26]), but we saw in [6, 35] that most of them
do not apply to our situation: the reason is that the domain of self-
adjointness depends very much on the observables and one cannot find either a
single domain or a single dominating operator. We try to solve this problem by
a front attack, namely, we should find an appropriate common domain for any
pair of wedge-observables and show their strong commutativity. Thanks to the
Driessler–Fröhlich theorem, the problem of strong commutativity is reduced to
the (essential) self-adjointness of the sum of the left and right bound state
operators. We show this self-adjointness up to the two-particle component. It
is worth mentioning that, because of all these technicalities, the candidate
observable ${\widetilde{\phi}}$ is no longer an operator-valued distribution.

This paper is organized as follows. In Section 2 we recall our approach to
integrable QFT with bound states. Especially, we extend the domain of the
bound state operators introduced in [6] using the convenient extension found
in [35]. Then in Section 3 it is shown that the strong commutativity of wedge-
observable candidates follows from the essential self-adjointness of the sum
bound state operator. We do this by combining the Driessler–Fröhlich theorem.
With this choice of domain, the covariance of the candidate operators also
follows, which allows one to proceed to Borchers triples. In Section 4, we
show the essential self-adjointness of the one and two particle components of
the sum bound state operator. We also present some observations on
$n$-particle components which demonstrates how the poles and zeros in the
$S$-matrix affect the analytic properties of the observables.

##  2 Preliminaries

###  2.1 Wedge-local nets and Borchers triples

Let us summarize the construction of [6]. Our goal is to construct interacting
Haag–Kastler nets in two dimensions. As an intermediate step, we need to
construct nets of observables localized in wedges, or wedge-local nets (of von
Neumann algebras) for short. This notion is equivalent to that of Borchers
triples, which we will use in this work. We refer to [6, Section 2.1] for an
overview of the construction program through Borchers triples.

A (two-dimensional) Borchers triple $({\mathcal{M}},U,\Omega)$ consists of a
von Neumann algebra ${\mathcal{M}}$, a unitary representation of the proper
orthochronous Poincaré group ${{\mathcal{P}}^{\uparrow}_{+}}$ and a vector
$\Omega$ which satisfy the following conditions (see [6, Section 2.1]):

  * •

$\Omega$ is cyclic and separating for ${\mathcal{M}}$;

  * •

the restriction of $U$ to the translation subgroup ${\mathbb{R}}^{2}$ has the
joint spectrum included in $\overline{V_{+}}=\\{(a_{0},a_{1})\colon
a_{0}\geq|a_{1}|\\}$;

  * •

${\mathcal{M}}$ is covariant with respect to $U$, namely,
$\operatorname{Ad}U(g){\mathcal{M}}\subset{\mathcal{M}}$ for
$g\in{{\mathcal{P}}^{\uparrow}_{+}}$ which preserves the standard right-wedge
$W_{\mathrm{R}}$.

In [6], we constructed for test functions $f$, $g$ supported in
$W_{\mathrm{L}}$, $W_{\mathrm{R}}$ respectively, unbounded operators
${\widetilde{\phi}}(f)$, ${\widetilde{\phi}}^{\prime}(g)$ which commute weakly
on a dense domain. We also showed that, if ${\widetilde{\phi}}(f)$ and
${\widetilde{\phi}}^{\prime}(g)$ commute strongly, then $\Omega$ is cyclic and
separating for
${\mathcal{M}}:=\big{\\{}e^{i{\widetilde{\phi}}^{\prime}(g)}\colon\operatorname{supp}g\big{\\}}^{\prime\prime}$
(strong commutativity is important for the separating property). Furthermore,
if we can show the covariance of ${\widetilde{\phi}}$ with respect to $U$,
then the covariance of the Borchers triple follows and the construction is
complete. We will investigate strict locality, namely whether this Borchers
triple defines a Haag–Kastler net, in a separate paper [8] (actually, for this
purpose the definition of ${\widetilde{\phi}}$ must be slightly extended as we
will do in Section 2.4). Therefore, our most urgent problem is the strong
commutativity.

We have ${\widetilde{\phi}}(f)=\phi(f)+\chi(f)$, where $\phi(f)$ is very well
under control, while $\chi(f)$ is a new sort of operator and we investigated
its one-particle component $\chi_{1}(f)$ in detail. We review the construction
of these operators below.

###  2.2 Factorizing $\boldsymbol{S}$-matrix with poles

As an input, we fix a two-particle scattering function $S$, which is at first
a meromorphic function on ${\mathbb{R}}+i(0,\pi)$, the region we call the
physical strip, and later extended to ${\mathbb{C}}$ which satisfies the
following properties:

  1. (S1)

Unitarity. $S(\theta)^{-1}=\overline{S(\theta)}$ for $\theta\in{\mathbb{R}}$.

  2. (S2)

Hermitian analyticity. $S(-\theta)=S(\theta)^{-1}$ for
$\theta\in{\mathbb{R}}$.

  3. (S3)

Crossing symmetry. $S$ has a meromorphic extension in ${\mathbb{C}}$ and
satisfies $S(\zeta)=S(\pi i-\zeta)$.

  4. (S4)

Bootstrap equation. $S(\zeta)=S\big{(}\zeta+\frac{\pi
i}{3}\big{)}S\big{(}\zeta-\frac{\pi i}{3}\big{)}$.

  5. (S5)

Positive residue. $S$ has a simple pole at $\frac{2\pi i}{3}$ and
$R:={\underset{\zeta=\frac{2\pi i}{3}}{\operatorname{Res}}S(\zeta)\in
i{\mathbb{R}}_{+}}$. Except this and the pole at $\frac{\pi i}{3}$, there is
no pole in the physical strip ${\mathbb{R}}+i(0,\pi)$ and $S(\zeta)$ is
bounded in the complement of the union of neighborhoods of these poles in the
physical strip.

  6. (S6)

Value at zero. $S(0)=-1$.

  7. (S7)

A bound on the real part. $0\leq\operatorname{Re}S\big{(}\theta+\frac{2\pi
i}{3}\big{)}$ $\big{(}{=}\operatorname{Re}S\big{(}\theta+\frac{\pi
i}{3}\big{)}\big{)}$.

  8. (S8)

A bound on the modulus. $\big{|}S\big{(}\theta+\frac{\pi
i}{6}\big{)}\big{|}\leq 1$.

  9. (S9)

Regularity. $S$ has only finitely many zeros in the physical strip and there
is $\kappa>0$ such that
$\|S\|_{\kappa}:=\sup\\{|S(\zeta)|\colon\zeta\in{\mathbb{R}}+i(-\kappa,\kappa)\\}<\infty$.

We showed in [6] that (S6) follows from (S1)–(S5). It turns out crucial in the
consideration of domains of our operators. We also classified all functions
which satisfy these conditions [6, Appendix A]. We show that (S8) follows from
(S1)–(S5) in Appendix A.2.

(S7) and (S9) are new requirements. The property (S7) had not appeared either
in the literature on form factor program or in our previous works [6, 7]. Yet,
it is necessary in order to apply our methods, for some properties of the sum
bound state operators. The last assumption (S9) will be necessary also in the
proof of modular nuclearity (cf. [19]), and we use it here already for wedge-
observables, in Lemma E.5. With the finiteness of zeros in the physical strip,
the finiteness of $\|S\|_{\kappa}$ is equivalent to the absence of the
singular part in $S$ (cf. [23, Appendix A]). From the condition (S6), there
must be at least one zero on the interval $i(0,\frac{\pi}{3})$, hence
$\kappa<\frac{\pi}{3}$, as we saw in [6].

We check in Appendix A that there are concrete examples of $S$ which satisfy
all these requirements. In the following, we mark explicitly the points where
(S7) is needed.

###  2.3 The wedge-local observables

Some materials here are parallel to those of [18]. Assume that our model has
only a single species of particle with mass $m>0$. We fix a two-particle
scattering function $S$, which satisfies the properties explained in Section
2.2.

Our Hilbert space is as follows. The one-particle space ${\mathcal{H}}_{1}$ is
simply $L^{2}({\mathbb{R}})$ with the Lebesgue measure. We consider
$n$-particle space ${\mathcal{H}}_{1}^{\otimes n}$. An element of
${\mathcal{H}}_{1}^{\otimes n}$ can be considered as a function in
$L^{2}({\mathbb{R}}^{n})$. On each of these spaces, there is the
representation of the symmetric group $\mathfrak{S}_{n}$:

$\displaystyle(D_{n}(\tau_{k})\Psi_{n})(\theta_{1},\dots,\theta_{n}):=S(\theta_{k+1}-\theta_{k})\Psi_{n}(\theta_{1},\dots,\theta_{k+1},\theta_{k},\dots,\theta_{n}),$

where $\tau_{k}$ is the transposition of $k$ and $k+1$. This indeed extends to
a unitary representation of $\mathfrak{S}_{n}$. We say that
$\Psi_{n}\in{\mathcal{H}}_{1}^{\otimes n}$ is $S$-symmetric if
$D_{n}(\sigma)\Psi_{n}=\Psi_{n}$ for $\sigma\in\mathfrak{S}_{n}$. This notion
is equivalent to that for each $k$

$\displaystyle\Psi_{n}(\theta_{1},\dots,\theta_{n})=S(\theta_{k+1}-\theta_{k})\Psi_{n}(\theta_{1},\dots,\theta_{k+1},\theta_{k},\dots,\theta_{n}).$

We denote the orthogonal projection onto the $S$-symmetric functions by
$P_{n}$ and the space of $S$-symmetric functions by
${\mathcal{H}}_{n}=P_{n}{\mathcal{H}}_{1}^{\otimes n}$. The physical Fock
space is the direct sum

$\displaystyle{\mathcal{H}}=\bigoplus_{n}{\mathcal{H}}_{n},$

where ${\mathcal{H}}_{0}={\mathbb{C}}$. Let us also introduce an auxiliary
Hilbert space, the unsymmetrized Fock space, which is
${\mathcal{H}}^{\Sigma}=\bigoplus_{n}{\mathcal{H}}_{1}^{\otimes n}$. By
putting $P_{S}=\bigoplus_{n}P_{n}$, we have
${\mathcal{H}}=P_{S}{\mathcal{H}}^{\Sigma}$.

For $\psi\in{\mathcal{H}}_{1}$, the unsymmetrized creation and annihilation
operators are defined on the algebraic direct sum of
${\mathcal{H}}_{1}^{\otimes n}$ by

$\displaystyle\big{(}a^{\dagger}(\psi)\Psi\big{)}_{n}(\theta_{1},\dots,\theta_{n})=\sqrt{n}\psi(\theta_{1})\Psi_{n-1}(\theta_{2},\dots,\theta_{n}),\qquad
a(\psi)=\big{(}a^{\dagger}(\psi)\big{)}^{*}.$

By our convention $a^{\dagger}(\psi)$ is linear in $\psi$, while $a(\psi)$ is
antilinear. The $S$-symmetrized creation and annihilation operators are
$z^{\dagger}(\psi)=P_{S}a^{\dagger}(\psi)P_{S}$, $z(\psi)=P_{S}a(\psi)P_{S}$.

The one-particle CPT operator $J_{1}$ is given by
$(J_{1}\psi)(\theta)=\overline{\psi(\theta)}$, and the $n$-particle component
$J_{n}$ is

$\displaystyle(J_{n}\Psi_{n})(\theta_{1},\dots,\theta_{n})=\overline{\Psi_{n}(\theta_{n},\dots,\theta_{1})}.$

It is easy to see that $J_{n}$ commutes with $P_{n}$, hence its preserves the
subspace ${\mathcal{H}}_{n}$. The full CPT operator is $J=\bigoplus_{n}J_{n}$,
where $J_{0}$ is the complex conjugation on ${\mathcal{H}}_{0}={\mathbb{C}}$.

For a test function $f$ in ${\mathbb{R}}^{2}$, we define

$\displaystyle f^{\pm}(\theta)=\int dx\,f(x)e^{\pm ip(\theta)\cdot x},\qquad
p(\theta)=m\left(\begin{matrix}\cosh\theta\\\ \sinh\theta\end{matrix}\right).$

As $f$ is compactly supported, its Fourier transform decays fast, and so do
$f^{\pm}$, therefore, they belong to ${\mathcal{H}}_{1}$. On the level of test
functions, we define the CPT transformation by $f_{j}(x)=\overline{f(-x)}$.

If $S$ had no pole, then Lechner took the field
$\phi(f)=z^{\dagger}(f^{+})+z(J_{1}f^{-})$ and the reflected field
$\phi^{\prime}(g)=z^{\prime\dagger}(g^{+})+z^{\prime}(J_{1}g^{+})$, where
$z^{\prime\dagger}(\psi)=Jz^{\dagger}(J_{1}\psi)J$ and
$z^{\prime}(\psi)=Jz(J_{1}\psi)J$ and proved that they were wedge-local [18]:
namely, if $\operatorname{supp}f\subset W_{\mathrm{L}}$ and
$\operatorname{supp}g\subset W_{\mathrm{R}}$, then $\phi(f)$ and
$\phi^{\prime}(g)$ strongly commute.

#### Extension to Hardy space vectors

Actually, it is possible and also natural to consider $\phi$, $z^{\dagger}$
and $z$ as a function of $f^{+}$, and in this sense we can extend them to any
$\xi\in{H^{2}(\SS_{0,\pi})}\cap{H^{\infty}(\SS_{0,\pi})}$. For $a\leq 0\leq
b$, recall the definition of the Hardy space:

$\displaystyle
H^{2}(\SS_{a,b})=\left\\{\xi\in{\mathcal{H}}_{1}\colon\xi(\theta)\mbox{ has an
}L^{2}\mbox{-bounded analytic continuation to }\SS_{a,b}\right\\}.$

In other words, a function $\xi\in H^{2}(\SS_{a,b})$ is an analytic function
on $\SS_{a,b}$ whose $L^{2}$-norms on horizontal lines $\xi(\,\cdot+i\lambda)$
are bounded uniformly for $\lambda\in(a,b)$. It is well-known that for
$\lambda\to a,b$ the limits $\xi(\,\cdot+i\lambda)$ exist in the $L^{2}$-sense
[32, Corollary III.2.10]. As $0\in[a,b]$, we can idenfity
${\mathcal{H}}_{1}=L^{2}({\mathbb{R}})$ as a subspace of $H^{2}(\SS_{a,b})$ by
considering the value $\xi(\theta)$, $\theta\in{\mathbb{R}}$. We denote the
$L^{2}$-norm by $\|\xi\|$, while the Hardy space norm is
$\|\xi\|_{2}:=\sup\limits_{\lambda\in(a,b)}\|\xi(\,\cdot+i\lambda)\|$.

Similarly, $H^{\infty}(\SS_{a,b})$ is the space of bounded analytic functions
on $\SS_{a,b}$ and we define the norm
$\|\xi\|_{\infty}:=\sup\limits_{\zeta\in\SS_{a,b}}|\xi(\zeta)|$.

The operators $z^{\dagger}(\xi)$ and $z(\xi)$ are defined for vectors
$\xi\in{\mathcal{H}}_{1}$, hence with the understanding that
${\mathcal{H}}_{1}=L^{2}({\mathbb{R}})\supset{H^{2}(\SS_{0,\pi})}$ by taking
the boundary value at $\operatorname{Im}\zeta=0$, $z^{\dagger}$ and $z$ are
also defined on ${H^{2}(\SS_{0,\pi})}$. The reality condition
$f(a)=\overline{f(a)}$ on a test function $f$ translates to an element
$\xi\in{H^{2}(\SS_{0,\pi})}$ as $\xi(\zeta+i\pi)=\overline{\xi(\zeta)}$. It is
straightforward to see that the proof of wedge-locality for $S$ without pole
[18] works for such $\xi$’s: the Cauchy theorem works for $H^{2}$-functions
(see, e.g., [6, Lemma B.2]).

As we are interested in cases where $S$ has poles, we introduce the bound
state operator $\chi(\xi)$ (cf. in [6] we considered it as associated with
$f^{+}$, instead of $\xi$). We associate an operator $\chi(\xi)$ to
$\xi\in{H^{2}(\SS_{0,\pi})}\cap{H^{\infty}(\SS_{0,\pi})}$ which satisfies
$\overline{\xi(\theta)}=\xi(\theta+\pi i)$ by

$\displaystyle(\chi_{1}(\xi)\Psi)(\theta):=\sqrt{2\pi|R|}\xi\left(\theta+\frac{\pi
i}{3}\right)\Psi\left(\theta-\frac{\pi i}{3}\right),$
$\displaystyle\chi_{n}(\xi):=nP_{n}\left(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\right)P_{n},\qquad\chi(\xi):=\bigoplus_{n}\chi_{n}(\xi),$

where $\chi_{0}(\xi)=0$ and $R=\underset{\zeta=\frac{2\pi
i}{3}}{\operatorname{Res}}S(\zeta)$. Of course, $\chi_{1}(\xi)$ cannot be
defined on arbitrary vectors, as it involves an analytic continuation of
$\Psi$. In Section 2.4, we will specify a domain of self-adjointness of
$\chi_{1}(\xi)$ and define $\chi_{n}(\xi)$ accordingly. It follows that
$\chi(\xi)$ is symmetric as in [6, Proposition 3.1]. For
$\eta\in{H^{2}(\SS_{-\pi,0})}\cap{H^{\infty}(\SS_{-\pi,0})}$,
$\overline{\eta(\theta)}=\eta(\theta-\pi i)$ we introduce also

$\displaystyle(\chi^{\prime}_{1}(\eta)\Psi)(\theta):=\sqrt{2\pi|R|}\eta\left(\theta-\frac{\pi
i}{3}\right)\Psi\left(\theta+\frac{\pi i}{3}\right),$
$\displaystyle\chi_{n}(\eta):=nP_{n}\left({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta)\right)P_{n},\qquad\chi^{\prime}(\eta):=\bigoplus_{n}\chi^{\prime}_{n}(\eta).$

It holds that $\chi^{\prime}_{1}(\eta)=J_{1}\chi_{1}(J_{1}\eta)J_{1}$,
$\chi^{\prime}_{n}(\eta)=J_{n}\chi_{n}(J_{1}\eta)J_{n}$ and
$\chi^{\prime}(\eta)=J\chi(J_{1}\eta)J$.

Our candidate operators for wedge-observables are
${\widetilde{\phi}}(\xi):=\phi(\xi)+\chi(\xi)$ and
${\widetilde{\phi}}^{\prime}(\eta):=\phi^{\prime}(\eta)+\chi^{\prime}(\eta)$.
We will show that they weakly commute on the intersection of their domains and
propose a way to prove their strong commutativity.

###  2.4 The bound state operator

We studied the one-particle component $\chi_{1}(\xi)$ in detail in [35]:
actually, we studied the domain ${H^{2}(\SS_{-\pi,0})}$ rather than
${H^{2}\big{(}\SS_{-\frac{\pi}{3},0}\big{)}}$, but the adaptation is
straightforward. The important lessons obtained there are that
$\chi_{1}(\xi)$, when defined on
${H^{2}\big{(}\SS_{-\frac{\pi}{3},0}\big{)}}$, does not always have a nice
self-adjoint extension and even if it does, its domain varies as $\xi$ varies.
This is due to the fact that the deficiency subspaces of $\chi_{1}(\xi)$ is
closely related with the zeros and the decay rate of $\xi$, and
$\chi_{1}(\xi)$ does not have any self-adjoint extension if $\xi$ has an odd
number of zeros and does not decay sufficiently fast as
$\operatorname{Re}\zeta\to\pm\infty$. And even if $\xi$ has an even number of
zeros, it is not obvious which extension is the suited one for our purpose of
finding quantum observables. Therefore, we restrict ourselves to the certain
subclass of functions.

Recall that one can define the analytic continuation operator111We use the
symbol $\Delta$, because if we could prove the strong commutativity we will be
discussing in this paper, then $\Delta_{1}$ will turn out to be the one-
particle component of the modular operator of the wedge-algebra with respect
to the vacuum [8].

$\displaystyle\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{6}}}\big{)}={H^{2}\big{(}\SS_{-\frac{\pi}{3},0}\big{)}},\qquad\big{(}{\Delta_{1}^{\frac{1}{6}}}\Psi\big{)}(\theta)=\Psi\left(\theta-\frac{\pi
i}{3}\right),$

namely, any vector $\Psi\in{H^{2}\big{(}\SS_{-\frac{\pi}{3},0}\big{)}}$ has an
$L^{2}$-boundary value $\Psi(\theta)$ and $\Psi(\theta-\frac{\pi i}{3})$ and
it is considered as an element of $L^{2}({\mathbb{R}})$ by taking
$\Psi(\theta)$, while ${\Delta_{1}^{\frac{1}{6}}}\Psi$ is defined to be the
other boundary value $\Psi(\theta-\frac{\pi i}{3})$. This is self-adjoint,
indeed unitarily equivalent to the multiplication by the exponential function
through Fourier transform [35, Appendix A].

A property which we will use repeatedly is the “Cauchy theorem” in the
following form: since $\Delta^{\frac{t}{2\pi}}$ is self-adjoint on
$H^{2}(\SS_{-t,0})$, we have
$\langle\Phi,\Delta^{\frac{t}{2\pi}}\Psi\rangle=\langle\Delta^{\frac{t}{2\pi}}\Phi,\Psi\rangle$
for $\Phi,\Psi\in H^{2}(\SS_{-t,0})$. By noting that $\overline{\Phi}\in
H^{2}(\SS_{0,t})$, where $\overline{\Phi}(z):=\overline{\Phi(\overline{z})}$,
we can write this as

$\displaystyle\langle\Phi,\Delta^{\frac{t}{2\pi}}\Psi\rangle=\int
d\theta\,\overline{\Phi(\theta)}\Psi(\theta-it)=\int
d\theta\,\overline{\Phi}(\theta)\Psi(\theta-it)=\int
d\theta\,\overline{\Phi}(\theta+it)\Psi(\theta)$
$\displaystyle\hphantom{\langle\Phi,\Delta^{\frac{t}{2\pi}}\Psi\rangle}{}=\int
d\theta\,\overline{\Phi(\theta-t)}\Psi(\theta)=\langle\Delta^{\frac{t}{2\pi}}\Phi,\Psi\rangle,$

namely, we can shift the argument in the integral. An analogous formula is
valid even if the functions have more variables, by considering the operator
$\Delta^{\frac{t}{2\pi}}\otimes{\mathbbm{1}}$ with the identity operator
${\mathbbm{1}}$ on the complementary Hilbert space. To apply this, it is
important to check that the involved vectors belong to the domain of the
operator $\Delta^{\frac{t}{2\pi}}\otimes{\mathbbm{1}}$.

We learned in [35, Section 5.1] that, if $\xi$ is the square of another
analytic function $\xi=\underline{\xi}^{2}$,
$\underline{\xi}\in{H^{2}(\SS_{0,\pi})}\cap{H^{\infty}(\SS_{0,\pi})}$ and if
$\underline{\xi}$ satisfies the reality condition
$\underline{\xi}(\zeta+i\pi)=\overline{\underline{\xi}(\zeta)}$ (which implies
that $\underline{\xi}(\zeta+\frac{\pi
i}{3})=\overline{\underline{\xi}(\zeta+\frac{2\pi i}{3})}$), then
$\chi_{1}(\xi)$ has a canonical self-adjoint extension, which we denote by the
same symbol for simplicity. More precisely, for $\xi$ as above, we can find a
function $\xi_{0}\in H^{\infty}(\SS_{\frac{\pi}{3},\frac{2\pi}{3}})$ such that
$|\xi_{0}(\theta+\frac{2\pi i}{3})|=1$ (almost everywhere) and

$\displaystyle\xi_{0}\left(\theta+\frac{\pi
i}{3}\right)\overline{\xi_{0}\left(\theta+\frac{2\pi
i}{3}\right)}=\sqrt{2\pi|R|}\xi\left(\theta+\frac{\pi i}{3}\right).$ (2.1)

Now, the multiplication operator $M_{\xi_{0}(\cdot\,+\frac{2\pi i}{3})}$,
which is unitary, corresponds to the unitary operator of [35, Theorem 5.1]
(with the notation replacing $\xi$ with $f$ and the scaling between the
domains ${H^{\infty}\big{(}\SS_{-\frac{\pi}{3},0}\big{)}}$ and
${H^{\infty}(\SS_{-\pi,0})}$). This unitary operator
$M_{\xi_{0}(\cdot\,+\frac{2\pi i}{3})}$ implements the unitary equivalence
between ${\Delta_{1}^{\frac{1}{6}}}$ and our bound state operator

$\displaystyle\operatorname{Dom}(\chi_{1}(\xi)):=M_{\xi_{0}(\cdot\,+\frac{2\pi
i}{3})}^{*}{H^{2}\big{(}\SS_{-\frac{\pi}{3},0}\big{)}},$
$\displaystyle(\chi_{1}(\xi)\Psi)(\theta):=\sqrt{2\pi|R|}\xi\left(\theta+\frac{\pi
i}{3}\right)\Psi\left(\theta-\frac{\pi i}{3}\right),$

or in other words, $\chi_{1}(\xi)=M_{\xi_{0}(\cdot\,+\frac{2\pi
i}{3})}^{*}{\Delta_{1}^{\frac{1}{6}}}M_{\xi_{0}(\cdot\,+\frac{2\pi i}{3})}$,
which follows straightforwardly from the property (2.1). The point is that
$\xi_{0}(\theta+\frac{2\pi
i}{3})\Psi(\theta)\in{H^{2}\big{(}\SS_{-\frac{\pi}{3},0}\big{)}}$ and the
analytic continuation $\theta\to\theta+\frac{\pi i}{3}$ does not see any pole
or non-$L^{2}$ behavior on this domain.

The $n$-particle component $\chi_{n}(\xi)$ is defined by
$\chi_{n}(\xi):=P_{n}(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})P_{n}$.
This means that
$\operatorname{Dom}(\chi_{n}(\xi))=\\{\Psi_{n}\in{\mathcal{H}}_{n}\colon\Psi_{n}\in\operatorname{Dom}(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\\}$.
As the expression of $\chi_{1}(\eta)$ is same as [6] (although the domain is
larger), we have also the following alternative expression (we can do this
because the continuation $\theta\to\theta+\frac{\pi i}{3}$ is always
accompanied by the multiplication by $\xi_{0}(\theta+\frac{2\pi i}{3})$, hence
one does not encounter non-$L^{2}$ problem by definition):

$\displaystyle(\chi_{n}(f)\Psi_{n})(\theta_{1},\dots,\theta_{n})$ (2.2)
$\displaystyle=\sqrt{2\pi|R|}\sum_{1\leq k\leq n}\prod_{1\leq j\leq
k-1}S\left(\theta_{k}-\theta_{j}+\frac{\pi
i}{3}\right)\xi\left(\theta_{k}+\frac{\pi
i}{3}\right)\Psi_{n}\left(\theta_{1},\dots,\theta_{k}-\frac{\pi
i}{3},\dots,\theta_{n}\right).$

We can rephrase all this as follows: the domain of $\chi_{1}(\xi)$ consists of
functions $\Psi_{1}$ such that $\xi_{0}(\theta+\frac{2\pi
i}{3})\Psi_{1}(\theta)$ is the boundary value of an element in
${H^{2}\big{(}\SS_{-\frac{\pi}{3},0}\big{)}}$. The domain of $\chi_{n}(\xi)$
consists of functions $\Psi_{n}$ such that $P_{n}\Psi_{n}$ belongs to the
domain of
$\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}$. In
particular, $\xi_{0}(\theta_{1}+\frac{2\pi
i}{3})(P_{n}\Psi_{n})(\theta_{1},\dots,\theta_{n})$ can be viewed as an
$L^{2}({\mathbb{R}}^{n-1})$-valued analytic function in $\theta_{1}$. By
$S$-symmetry, $\xi_{0}(\theta_{k}+\frac{2\pi
i}{3})(P_{n}\Psi_{n})(\theta_{1},\dots,\theta_{n})$ is also analytic in other
variables $\theta_{k}$, but as the $S$-factor has poles, hence it belong to
${\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\underset{k\text{-th}}{\Delta_{1}^{\frac{\kappa}{2\pi}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}$,
where the boundedness of $S$ is assured in ${\mathbb{R}}+i(-\kappa,\kappa)$ as
(9).

Similarly, for each $\eta=\underline{\eta}^{2}$ where
$\underline{\eta}\in{H^{2}(\SS_{-\pi,0})}\cap{H^{\infty}(\SS_{-\pi,0})}$ such
that $\underline{\eta}(\theta-i\pi)=\overline{\underline{\eta}(\theta)}$,
there is $\eta_{0}\in H^{\infty}(\SS_{-\frac{2\pi}{3},-\frac{\pi}{3}})$ such
that $|\eta_{0}(\theta-\frac{2\pi i}{3})|=1$ and

$\displaystyle\eta_{0}\left(\theta-\frac{\pi
i}{3}\right)\overline{\eta_{0}\left(\theta-\frac{2\pi
i}{3}\right)}=\sqrt{2\pi|R|}\eta\left(\theta-\frac{\pi i}{3}\right).$

We define

$\displaystyle\operatorname{Dom}(\chi^{\prime}_{1}(\eta))=M_{\eta_{0}(\cdot\,-\frac{2\pi
i}{3})}^{*}{H^{2}\big{(}\SS_{0,\frac{\pi}{3}}\big{)}},$
$\displaystyle(\chi^{\prime}_{1}(\eta)\Psi)(\theta):=\sqrt{2\pi|R|}\eta\left(\theta-\frac{\pi
i}{3}\right)\Psi\left(\theta+\frac{\pi i}{3}\right),$

where $M_{\eta_{0}(\cdot\,-\frac{2\pi i}{3})}$ is unitary, and this is
equivalent to $\chi^{\prime}_{1}(\eta)=M_{\eta_{0}(\cdot\,-\frac{2\pi
i}{3})}^{*}{\Delta_{1}^{-\frac{1}{6}}}M_{\eta_{0}(\cdot\,-\frac{2\pi i}{3})}$.
This operator $\chi^{\prime}(\eta)$ has an alternative expression as $\chi(f)$
does:

$\displaystyle(\chi^{\prime}_{n}(\eta)\Psi_{n})(\theta_{1},\dots,\theta_{n})=\sqrt{2\pi|R|}\sum_{1\leq
k\leq n}\prod_{n-k+2\leq j\leq n}\\!S\left(\theta_{j}-\theta_{n-k+1}+\frac{\pi
i}{3}\right)\eta\left(\theta_{n-k+1}-\frac{\pi i}{3}\right)$
$\displaystyle\hphantom{(\chi^{\prime}_{n}(\eta)\Psi_{n})(\theta_{1},\dots,\theta_{n})=}{}\times\Psi_{n}\left(\theta_{1},\dots,\theta_{n-k+1}+\frac{\pi
i}{3},\dots,\theta_{n}\right).$

##  3 Towards strong commutativity

In this Section we show that, in order to construct the Borchers triple
associated with $S$ with poles, it is enough to prove that
$\chi(\xi)+\chi^{\prime}(\eta)$ is essentially self-adjoint.

###  3.1 A general criterion

This Section is technically independent from QFT.

For unbounded operators, strong commutativity is a hard problem. It may fail
even for a two self-adjoint operators which commute on a common invariant core
[25, Section 10], [26, Section VIII.5, Example 1]. We also exhibit more
examples of weakly-commuting but not strongly-commuting operators in Appendix
B. A sufficient condition which is used very widely in the context of QFT
(e.g., [2, 10, 15]) is the Driessler–Fröhlich theorem [11]. The theorem says
roughly that, if there is a positive self-adjoint operator $T$ (“Hamiltonian”)
which can nicely bound the weakly commuting symmetric operators $A$, $B$, and
their commutators with $T$, then $A$ and $B$ are actually self-adjoint and
commute strongly (see Appendix C for the precise statements).

The trouble in our situation is that the physical Hamiltonian is not strong
enough to estimate our candidates which contain the bound state operator
$\chi(\xi)$: the Hamiltonian is the multiplication operator, while $\chi(\xi)$
contains an analytic continuation. We have no idea for any other operator
which dominates $\chi(\xi)$ “nicely”, and anyway there cannot be a single
operator which bounds all $\chi(\xi)$’s with different $\xi$’s, as
$\chi_{1}(\xi)$’s have already different domains of self-adjointness. In such
a case, the question of strong commutativity is considerably difficult (cf.
[10]).

Yet, the Driessler–Fröhlich theorem can be used to reduce the problem of
strong commutativity to self-adjointness of a certain positive operator.

######  Proposition 3.1.

Let $A$, $B$ and $T_{0}$ be symmetric operators. Assume that
$T:=A+B+T_{0}\geq{\mathbbm{1}}$ and is essentially self-adjoint on its natural
domain
$\operatorname{Dom}(T):=\operatorname{Dom}(A)\cap\operatorname{Dom}(B)\cap\operatorname{Dom}(T_{0})$,
and there are ${\mathscr{D}}\subset\operatorname{Dom}(T)$ which is a core of
$T$ and $c_{1},c_{2},c_{3}>0$ such that

  1. (1)

$\|A\psi\|\leq c_{1}\|T\psi\|$ and $\|B\psi\|\leq c_{1}\|T\psi\|$ for all
$\psi\in{\mathscr{D}}$,

  2. (2)

for all $\varphi,\psi\in{\mathscr{D}}$,

     * •

$|\langle T_{0}\psi,A\varphi\rangle-\langle A\psi,T_{0}\varphi\rangle|\leq
c_{2}\big{\|}T^{\frac{1}{2}}\psi\big{\|}\cdot\big{\|}T^{\frac{1}{2}}\varphi\big{\|}$,

     * •

$|\langle T_{0}\psi,B\varphi\rangle-\langle B\psi,T_{0}\varphi\rangle|\leq
c_{2}\big{\|}T^{\frac{1}{2}}\psi\big{\|}\cdot\big{\|}T^{\frac{1}{2}}\varphi\big{\|}$,

  3. (3)

for all $\varphi,\psi\in{\mathscr{D}}$,

     * •

$|\langle T_{0}\psi,A\varphi\rangle-\langle A\psi,T_{0}\varphi\rangle|\leq
c_{3}\|\psi\|\cdot\|T\varphi\|$,

     * •

$|\langle T_{0}\psi,B\varphi\rangle-\langle B\psi,T_{0}\varphi\rangle|\leq
c_{3}\|\psi\|\cdot\|T\varphi\|$,

  4. (4)

$\langle A\psi,B\varphi\rangle=\langle B\psi,A\varphi\rangle$ for all
$\psi,\varphi\in{\mathscr{D}}$.

Then $A$ and $B$ are essentially self-adjoint on any core of $T$ and they
strongly commute.

###### Proof.

As we have $\langle A\psi,B\varphi\rangle=\langle B\psi,A\varphi\rangle$ by
assumption and $T=A+B+T_{0}$ by definition, it follows that for all
$\varphi,\psi\in{\mathscr{D}}$

  * •

$|\langle T\psi,A\varphi\rangle-\langle A\psi,T\varphi\rangle|\leq
c_{2}\big{\|}T^{\frac{1}{2}}\psi\big{\|}\cdot\big{\|}T^{\frac{1}{2}}\varphi\big{\|}$,

  * •

$|\langle T\psi,B\varphi\rangle-\langle B\psi,T\varphi\rangle|\leq
c_{2}\big{\|}T^{\frac{1}{2}}\psi\big{\|}\cdot\big{\|}T^{\frac{1}{2}}\varphi\big{\|}$,

  * •

$|\langle T\psi,A\varphi\rangle-\langle A\psi,T\varphi\rangle|\leq
c_{3}\|\psi\|\cdot\|T\varphi\|$,

  * •

$|\langle T\psi,B\varphi\rangle-\langle B\psi,T\varphi\rangle|\leq
c_{3}\|\psi\|\cdot\|T\varphi\|$.

Hence we can apply Theorem C.1, by taking $\overline{T}$ as the positive self-
adjoint operator. ∎

The point of this proposition is that, instead of using a single “Hamiltonian”
for all cases, we take the operator $T=A+B+T_{0}$ which depends on $A$ and
$B$. Of course, as weak commutativity does not imply strong commutativity, the
essential self-adjointness hypothesis of $T$ and the estimate of $A$, $B$ by
$T$ are crucial.

We have the following case in mind: $A={\widetilde{\phi}}(\xi)$,
$B={\widetilde{\phi}}^{\prime}(\eta)$, $T_{0}=c(N+{\mathbbm{1}})$, where $N$
is the number operator on the $S$-symmetric Fock space and
$c\in{\mathbb{R}}_{+}$. Furthermore, we will see in Section 3.3 that
$T=A+B+T_{0}$ is a Kato–Rellich perturbation of the operator
$\chi(\xi)+\chi^{\prime}(\eta)+c(N+{\mathbbm{1}})$. This last operator
preserves each $n$-particle subspace, and the question of (essential) self-
adjointness is reduced to each $n$.

This self-adjointness is far from obvious. By a slight modification, the self-
adjointness of a similar operator easily fails (see Appendix D).

###  3.2 Proof of weak commutativity on the intersection domain

We show that ${\widetilde{\phi}}(\xi)$ and ${\widetilde{\phi}}^{\prime}(\eta)$
commute weakly on
$\operatorname{Dom}({\widetilde{\phi}}(\xi))\cap\operatorname{Dom}({\widetilde{\phi}}^{\prime}(g))$.
Thanks to our choice of the extension (Section 2.4), this will be a
straightforward modification of the proof of [6], with the help of some
techniques in complex analysis of several variables in Appendix E.

######  Proposition 3.2.

Let $\underline{\xi}\in{H^{\infty}(\SS_{0,\pi})}$, $\underline{\xi}(\theta+\pi
i)=\overline{\underline{\xi}(\theta)}$ and
$\underline{\eta}\in{H^{\infty}(\SS_{-\pi,0})}$, $\underline{\eta}(\theta-\pi
i)=\overline{\underline{\eta}(\theta)}$ for $\theta\in{\mathbb{R}}$ and assume
that $\xi=\underline{\xi}^{2}\in{H^{2}(\SS_{0,\pi})}$,
$\eta=\underline{\eta}^{2}\in{H^{2}(\SS_{-\pi,0})}$. Then, for each
$\Psi,\Phi\in\operatorname{Dom}({\widetilde{\phi}}(\xi))\cap\operatorname{Dom}({\widetilde{\phi}}^{\prime}(\eta))$,
it holds that

$\displaystyle\big{\langle}{\widetilde{\phi}}(\xi)\Phi,{\widetilde{\phi}}^{\prime}(\eta)\Psi\big{\rangle}=\big{\langle}{\widetilde{\phi}}^{\prime}(\eta)\Phi,{\widetilde{\phi}}(\xi)\Psi\big{\rangle}.$

###### Proof.

We may assume that $\Phi$, $\Psi$ have finitely many non-zero components and
since we are considering the algebraic direct sum
$\underline{\bigoplus}_{n}\operatorname{Dom}(\chi_{n}(\xi))\subset\operatorname{Dom}(\phi(\xi))$,
$\underline{\bigoplus}_{n}\operatorname{Dom}(\chi_{n}(\eta))\subset\operatorname{Dom}(\phi^{\prime}(\eta))$,
therefore, we can compute the action of the operators as the sum:

$\displaystyle{\widetilde{\phi}}(\xi)=\phi(\xi)+\chi(\xi)=z^{\dagger}(\xi)+\chi(\xi)+z(\xi),$
$\displaystyle{\widetilde{\phi}}^{\prime}(\eta)=\phi^{\prime}(\eta)+\chi^{\prime}(\eta)=z^{\prime\dagger}(\eta)+\chi^{\prime}(\eta)+z^{\prime}(\eta).$

We may also assume that $\Phi$ and $\Psi$ are already $S$-symmetric, although
we use the unsymmetrized Fock space ${\mathcal{H}}^{\Sigma}$ in the
intermediate steps. We are going to follow the steps of our previous result
and compute the commutator term by term. Since the most of the computations
are same as the previous ones, we will be brief and indicate only the points
where care is needed, and refer the reader to the proof of [6, Theorem 3.4].

The commutator $[\chi(\xi),z^{\prime}(\eta)]$. We can compute this commutator
as operators as before [6]. Although we have an extended domain for
$\chi(\xi)$, the expression (2.2) is valid as we explained before. By
computing the operators term by term, we get

$\displaystyle([\chi(\xi),z^{\prime}(\eta)]\Psi_{n})(\theta_{1},\dots,\theta_{n-1})$
$\displaystyle=-\sqrt{2\pi|R|n}\int
d\theta\,\overline{\eta(\theta)}\prod_{1\leq j\leq
n-1}S\left(\theta-\theta_{j}+\frac{\pi i}{3}\right)\xi\left(\theta+\frac{\pi
i}{3}\right)\Psi_{n}\left(\theta_{1},\dots,\theta_{n-1},\theta-\frac{\pi
i}{3}\right)$ $\displaystyle=-\sqrt{2\pi|R|n}\int d\theta\,\eta(\theta-\pi
i)\prod_{1\leq j\leq n-1}S(\theta-\theta_{j})\xi\left(\theta+\frac{\pi
i}{3}\right)\Psi_{n}\left(\theta-\frac{\pi
i}{3},\theta_{1},\dots,\theta_{n-1}\right),$

where in the last equation we used also the symmetry property
$\overline{\eta(\theta)}=\eta(\theta-i\pi)$ and the $S$-symmetry of
$\Psi_{n}$.

Let us look at the integrand without the $S$-factor and shift the first
variable $\theta$ by $\frac{\pi i}{3}$: $\eta(\theta-\frac{2\pi
i}{3})\xi\left(\theta+\frac{2\pi
i}{3}\right)\Psi_{n}\left(\theta,\theta_{1},\dots,\theta_{n-1}\right)$. Recall
that:

  * •

$\xi\in{H^{\infty}(\SS_{0,\pi})}$, $\eta\in{H^{\infty}(\SS_{-\pi,0})}$,

  * •

$\overline{\xi_{0}\left(\theta+\frac{\pi
i}{3}\right)}\xi_{0}\left(\theta+\frac{2\pi
i}{3}\right)=\sqrt{2\pi|R|}\xi\left(\theta+\frac{2\pi i}{3}\right)$,
$\xi_{0}\in H^{\infty}(\SS_{\frac{\pi}{3},\frac{2\pi}{3}})$,

  * •

$\overline{\eta_{0}\left(\theta-\frac{\pi
i}{3}\right)}\eta_{0}\left(\theta-\frac{2\pi
i}{3}\right)=\sqrt{2\pi|R|}\eta\left(\theta-\frac{2\pi i}{3}\right)$,
$\eta_{0}\in H^{\infty}\big{(}\SS_{-\frac{2\pi}{3},-\frac{\pi}{3}}\big{)}$,

  * •

$\xi_{0}\big{(}\theta+\frac{2\pi
i}{3}\big{)}\Psi_{n}(\theta,\theta_{1},\dots,\theta_{n-1})\in\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$,

  * •

$\eta_{0}\big{(}\theta-\frac{2\pi
i}{3}\big{)}\Psi_{n}(\theta,\theta_{1},\dots,\theta_{n-1})\in\operatorname{Dom}\big{(}\Delta_{1}^{-\frac{\kappa}{2\pi}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$.

We consider the function

$\displaystyle\eta\left(\theta-\frac{2\pi
i}{3}\right)\xi\left(\theta+\frac{2\pi
i}{3}\right)\prod_{j=1}^{n-1}\xi_{0}\left(\theta_{j}+\frac{2\pi
i}{3}\right)\cdot\Psi_{n}\left(\theta,\theta_{1},\dots,\theta_{n-1}\right),$

which belongs to

$\displaystyle\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}\bigcap\operatorname{Dom}\big{(}\Delta_{1}^{-\frac{\kappa}{2\pi}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$
$\displaystyle\qquad{}\bigcap_{j=1}^{n-1}\operatorname{Dom}\big{(}{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\underset{j+1\textrm{-th}}{\Delta_{1}^{\frac{\kappa}{2\pi}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$

by $S$-symmetry of $\Psi$. Furthermore, again from the $S$-symmetry of
$\Psi_{n}$ and from the property (S6), it has zeros at any point where two of
the variables coincide: $\theta=\theta_{j}$. By applying Lemma E.7, we obtain
that

$\displaystyle\prod_{1\leq j\leq n-1}S\left(\theta-\theta_{j}+\frac{\pi
i}{3}\right)\cdot\eta\left(\theta-\frac{2\pi
i}{3}\right)\xi\left(\theta+\frac{2\pi i}{3}\right)$
$\displaystyle\qquad{}\times\prod_{j=2}^{n}\xi_{0}\left(\theta_{j}+\frac{2\pi
i}{3}\right)\cdot\Psi_{n}\left(\theta,\theta_{1},\dots,\theta_{n-1}\right)$

still belongs to the same domain. Now we can remove extra $\xi_{0}$’s with
variables $\theta_{j}$, which does not affect the domain of
${\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}$
and get that

$\displaystyle\prod_{1\leq j\leq n-1}S\left(\theta-\theta_{j}+\frac{\pi
i}{3}\right)\cdot\eta\left(\theta-\frac{2\pi
i}{3}\right)\xi\left(\theta+\frac{2\pi
i}{3}\right)\Psi_{n}\left(\theta,\theta_{1},\dots,\theta_{n-1}\right)$

belongs to
$\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$.
Therefore, we can apply the Cauchy theorem (on Hilbert-space valued functions)
and

$\displaystyle([\chi(\xi),z^{\prime}(\eta)]\Psi_{n})(\theta_{1},\dots,\theta_{n-1})=-\sqrt{2\pi|R|n}\int
d\theta\,\eta\left(\theta-\frac{2\pi i}{3}\right)\prod_{1\leq j\leq
n-1}S\left(\theta-\theta_{j}+\frac{\pi i}{3}\right)$
$\displaystyle\hphantom{([\chi(\xi),z^{\prime}(\eta)]\Psi_{n})(\theta_{1},\dots,\theta_{n-1})=-\sqrt{2\pi|R|n}\int}{}\times\xi\left(\theta+\frac{2\pi
i}{3}\right)\Psi_{n}\left(\theta,\theta_{1},\dots,\theta_{n-1}\right).$

The commutator $[z(\xi),\chi^{\prime}(\eta)]$. This can be computed in a
similar way as the previous one:

$\displaystyle[(z(\xi),\chi^{\prime}(\eta)]\Psi_{n})(\theta_{1},\dots,\theta_{n-1})=\sqrt{2\pi|R|n}\int
d\theta\,\xi\left(\theta+\frac{2\pi i}{3}\right)\prod_{1\leq j\leq
n-1}S\left(\theta-\theta_{j}+\frac{\pi i}{3}\right)$
$\displaystyle\hphantom{[(z(\xi),\chi^{\prime}(\eta)]\Psi_{n})(\theta_{1},\dots,\theta_{n-1})=\sqrt{2\pi|R|n}\int}{}\times\eta\left(\theta-\frac{2\pi
i}{3}\right)\Psi_{n}\left(\theta,\theta_{1},\dots,\theta_{n-1}\right),$

which coincides with the result of the commutator
$[\chi(f),z^{\prime}(J_{1}g^{-})]$ up to a sign, therefore they cancel each
other.

The commutators $[z^{\dagger}(\xi),\chi^{\prime}(\eta)]$ and
$[\chi(\xi),z^{\prime\dagger}(\eta)]$. One can show that these two commutators
cancel each other by taking the adjoints and repeat the computations as in the
commutators before.

The commutator $[\phi(\xi),\phi^{\prime}(\eta)]$. This term has been
essentially computed in [18]. The only difference is that $\xi$ does not come
from the Fourier transform of a test function $f$ supported in
$W_{\mathrm{L}}$, but the properties needed in the computations are that
$\xi\in{H^{2}(\SS_{0,\pi})}$ (which allows one to apply the Cauchy theorem)
and the symmetry property $\overline{\xi(\theta)}=\xi(\theta+i\pi)$ (which
assures that the terms appearing in the commutator is the boundary values of
the same analytic function), and the corresponding properties of $\eta$. Here
is no effect of the domain and the result is essentially taken from [18,
before Proposition 2]:

$\displaystyle([\phi(\xi),\phi^{\prime}(\eta)]\Psi_{n})(\theta_{1},\dots,\theta_{n})$
$\displaystyle\qquad{}=-2\pi
i\Bigg{(}\sum_{k=1}^{n}R\eta\left(\theta_{k}-\frac{\pi
i}{3}\right)\xi\left(\theta_{k}+\frac{2\pi i}{3}\right)\prod_{j\neq
k}S\left(\theta_{k}-\theta_{j}+\frac{2\pi i}{3}\right)$
$\displaystyle\qquad\qquad\qquad{}-\sum_{k=1}^{n}R\eta\left(\theta_{k}-\frac{2\pi
i}{3}\right)\xi\left(\theta_{k}+\frac{\pi i}{3}\right)\prod_{j\neq
k}S\left(\theta_{k}-\theta_{j}+\frac{\pi
i}{3}\right)\Bigg{)}\Psi_{n}(\theta_{1},\dots,\theta_{n}).$

The commutator $[\chi(\xi),\chi^{\prime}(\eta)]$. We have the expressions
(2.2) and those for $\chi_{n}^{\prime}(\eta)$. This allows one to expand the
commutator into the sum of $n^{2}$ terms as in [6] by a straightforward
computation. Of them, there are $2n(n-1)$ terms which come from the actions of
$\chi(\xi)$, $\chi^{\prime}(\eta)$ on different variables in the sense of
(2.2), and cancel each other. For that, we only need that $\chi_{n}(\xi)$ is
the projection onto $P_{n}{\mathcal{H}}^{\otimes n}$ of the operator
$\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}$ acting on
the first component (and the corresponding property of
$\chi^{\prime}_{n}(\eta)$ and the $S$-symmetry of the vectors). The extended
domains do not affect this.

The remaining terms are those where $\chi_{1}(\xi)$ and
$\chi^{\prime}_{1}(\eta)$ act on the same variable in the sense of (2.2):

$\displaystyle\langle
D_{n}(\rho^{\prime}_{n-k+1})({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))\Phi,\;D_{n}(\rho_{k})(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi\rangle$
$\displaystyle=2\pi|R|\int d\theta_{1}\cdots
d\theta_{n}\,\prod_{l=1}^{k-1}S\left(\theta_{k}-\theta_{l}+\frac{\pi
i}{3}\right)\xi\left(\theta_{k}+\frac{\pi
i}{3}\right)\Psi_{n}\left(\theta_{1},\dots,\theta_{k}-\frac{\pi
i}{3},\dots,\theta_{n}\right)$
$\displaystyle\hphantom{=2\pi|R|\int}{}\times\overline{\prod_{l=k+1}^{n}S\left(\theta_{l}-\theta_{k}+\frac{\pi
i}{3}\right)\eta\left(\theta_{k}-\frac{\pi
i}{3}\right)\Phi_{n}\left(\theta_{1},\dots,\theta_{k}+\frac{\pi
i}{3},\dots,\theta_{n}\right)}$ $\displaystyle=2\pi|R|\int d\theta_{1}\cdots
d\theta_{n}\,\prod_{l\neq k}S\left(\theta_{k}-\theta_{l}+\frac{2\pi
i}{3}-i\kappa\right)\xi\left(\theta_{k}+\frac{2\pi
i}{3}-i\kappa\right)\overline{\eta\left(\theta_{k}-\frac{2\pi
i}{3}+i\kappa\right)}$
$\displaystyle\hphantom{=2\pi|R|\int}{}\times\Psi_{n}\left(\theta_{1},\dots,\theta_{k}-i\kappa,\dots,\theta_{n}\right)\overline{\Phi_{n}\left(\theta_{1},\dots,\theta_{k}+i\kappa,\dots,\theta_{n}\right)},$

where $\rho_{k}$, $\rho_{n-k+1}^{\prime}$ are the cyclic permutations

$\displaystyle\rho_{k}\colon\
(1,2,\dots,n)\mapsto(k,1,2,\dots,k-1,k+1,\dots,n),$
$\displaystyle\rho_{n-k+1}^{\prime}\colon\
(1,2,\dots,n)\mapsto(1,2,\dots,k-1,k+1,\dots,n,k),$

and in the second equality we used that $S$ has no pole except $\frac{\pi
i}{3}$, $\frac{2\pi i}{3}$ and applied the (Hilbert space-valued) Cauchy
theorem: the pole at $\frac{\pi i}{3}$ makes no problem by the fact that
$\Psi_{n}\in\operatorname{Dom}(\chi_{n}(\xi))$,
$\Phi_{n}\in\operatorname{Dom}(\chi_{n}^{\prime}(\eta))$.

Now, the question is the pole of $S$ at $\frac{2\pi i}{3}$. Let us forget at
the moment the $S$-factors and look at the further shifted integrand:

$\displaystyle\xi\left(\theta_{k}+\frac{2\pi
i}{3}\right)\Psi_{n}\left(\theta_{1},\dots,\theta_{n}\right)\overline{\eta\left(\theta_{k}-\frac{2\pi
i}{3}\right)\Phi_{n}\left(\theta_{1},\dots,\theta_{n}\right)}$
$\displaystyle=\frac{1}{\sqrt{2\pi|R|}}\xi\left(\theta_{k}+\frac{2\pi
i}{3}\right)\eta_{0}\left(\theta_{k}-\frac{\pi
i}{3}\right)\Psi_{n}\left(\theta_{1},\dots,\theta_{n}\right)\cdot\overline{\eta_{0}\left(\theta_{k}-\frac{2\pi
i}{3}\right)\Phi_{n}\left(\theta_{1},\dots,\theta_{n}\right)},$

where we used that $\overline{\eta_{0}\left(\theta_{k}-\frac{\pi
i}{3}\right)}\eta_{0}\left(\theta_{k}-\frac{2\pi
i}{3}\right)=\sqrt{2\pi|R|}\eta\left(\theta_{k}-\frac{2\pi i}{3}\right)$.

Let us observe that

  * •

$\eta_{0}\big{(}\theta_{k}-\frac{2\pi
i}{3}\big{)}\Phi_{n}(\theta_{1},\dots,\theta_{n})\in\operatorname{Dom}\big{(}{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\underset{k\textrm{-th}}{\Delta_{1}^{-\frac{\kappa}{2\pi}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$,

  * •

$\xi\big{(}\theta_{k}+\frac{2\pi
i}{3}\big{)}\eta_{0}\big{(}\theta_{k}-\frac{\pi
i}{3}\big{)}\Psi_{n}(\theta_{1},\dots,\theta_{n})\in\operatorname{Dom}\big{(}{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\underset{k\textrm{-th}}{\big{(}\Delta_{1}^{\frac{\kappa}{2\pi}}+\Delta_{1}^{-\frac{\kappa}{2\pi}}\big{)}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$,
since $\xi\in{H^{\infty}(\SS_{0,\pi})}$ and $\eta_{0}(\theta_{k}-\frac{\pi
i}{3})=\sqrt{2\pi|R|}\eta\big{(}\theta_{k}-\frac{\pi
i}{3}\big{)}\eta_{0}\big{(}\theta_{k}-\frac{2\pi i}{3}\big{)}$, $\eta\in
H^{\infty}(\SS_{-\pi,0})$.

We use the same trick as before: consider the function

$\displaystyle\xi\left(\theta_{k}+\frac{2\pi
i}{3}\right)\eta_{0}\left(\theta_{k}-\frac{\pi i}{3}\right)\prod_{l\neq
k}\xi_{0}\left(\theta_{l}+\frac{2\pi
i}{3}\right)\cdot\Psi_{n}\left(\theta_{1},\dots,\theta_{n}\right),$

which belongs to

$\displaystyle\operatorname{Dom}\big{(}{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\underset{k\textrm{-th}}{\big{(}\Delta_{1}^{\frac{\kappa}{2\pi}}+\Delta_{1}^{-\frac{\kappa}{2\pi}}\big{)}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$
$\displaystyle\qquad{}\bigcap_{l\neq
k}\operatorname{Dom}\big{(}{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\underset{l\textrm{-th}}{\Delta_{1}^{\frac{\kappa}{2\pi}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$

by $S$-symmetry of $\Psi$. Furthermore, again by $S$-symmetry, for $k<l$ we
have

$\displaystyle\Psi(\theta_{1},\dots,\theta_{n})=\prod_{k<j<l}S(\theta_{l}-\theta_{j})S(\theta_{j}-\theta_{k})\cdot
S(\theta_{l}-\theta_{k})\Psi(\theta_{1},\dots,\theta_{l},\dots,\theta_{k},\dots,\theta_{n}).$

From this expression, it is clear that $\Psi$ has a zero at
$\theta_{l}-\theta_{k}=0$, $k<l$. A parallel argument holds for $l<k$.

By applying Lemma E.7, we obtain that

$\displaystyle\prod_{l\neq k}S\left(\theta_{k}-\theta_{l}+\frac{2\pi
i}{3}\right)\cdot\xi\left(\theta_{k}+\frac{2\pi
i}{3}\right)\eta_{0}\left(\theta_{k}-\frac{\pi i}{3}\right)\prod_{l\neq
k}\xi_{0}\left(\theta_{l}+\frac{2\pi
i}{3}\right)\cdot\Psi_{n}\left(\theta_{1},\dots,\theta_{n}\right)$

still belongs to the same domain. We remove extra $\xi_{0}$’s with variables
$\theta_{l}$ and get that

$\displaystyle\prod_{l\neq k}S\left(\theta_{k}-\theta_{l}+\frac{2\pi
i}{3}\right)\cdot\xi\left(\theta_{k}+\frac{2\pi
i}{3}\right)\eta_{0}\left(\theta_{k}-\frac{\pi
i}{3}\right)\Psi_{n}\left(\theta_{1},\dots,\theta_{n}\right)$

belongs to
$\operatorname{Dom}\big{(}{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\underset{k\textrm{-th}}{\Delta_{1}^{\frac{\kappa}{2\pi}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}$.

Therefore, we can apply the Cauchy theorem and

$\displaystyle\langle
D_{n}(\rho^{\prime}_{n-k+1})({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))\Phi,D_{n}(\rho_{k})(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi\rangle$
$\displaystyle\qquad{}=2\pi|R|\int d\theta_{1}\cdots d\theta_{n}\prod_{l\neq
k}S\left(\theta_{k}-\theta_{l}+\frac{2\pi
i}{3}\right)\xi\left(\theta_{k}+\frac{2\pi
i}{3}\right)\Psi_{n}\left(\theta_{1},\dots,\theta_{n}\right)$
$\displaystyle\qquad\hphantom{=2\pi|R|\int}{}\times\overline{\eta\left(\theta_{k}-\frac{2\pi
i}{3}\right)\Phi_{n}\left(\theta_{1},\dots,\theta_{n}\right)}$
$\displaystyle\qquad{}=2\pi|R|\int d\theta_{1}\cdots d\theta_{n}\prod_{l\neq
k}S\left(\theta_{k}-\theta_{l}+\frac{2\pi
i}{3}\right)\xi\left(\theta_{k}+\frac{2\pi
i}{3}\right)\eta\left(\theta_{k}-\frac{\pi i}{3}\right)$
$\displaystyle\qquad\hphantom{=2\pi|R|\int}{}\times\overline{\Phi_{n}\left(\theta_{1},\dots,\theta_{n}\right)}\Psi_{n}\left(\theta_{1},\dots,\theta_{n}\right).$

By recalling (S6) which implies $R=i|R|$, this cancels the first contribution
from the commutator $\langle\Phi,[\phi(f),\phi^{\prime}(g)]\Psi\rangle$.

The remaining terms in $\langle\chi(\xi)\Phi,\chi^{\prime}(\eta)\Psi\rangle$
are, by an analogous argument as above,

$\displaystyle\langle
D_{n}(\rho_{k})(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Phi,D_{n}(\rho^{\prime}_{n-k+1})({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))\Psi\rangle$
$\displaystyle\qquad{}=-2\pi iR\int d\theta_{1}\cdots d\theta_{n}\prod_{l\neq
k}S\left(\theta_{k}-\theta_{l}+\frac{\pi
i}{3}\right)\eta\left(\theta_{k}-\frac{2\pi
i}{3}\right)\xi\left(\theta_{k}+\frac{\pi i}{3}\right)$
$\displaystyle\qquad\hphantom{=-2\pi
iR\int}{}\times\overline{\Phi_{n}\left(\theta_{1},\dots,\theta_{n}\right)}\Psi_{n}\left(\theta_{1},\dots,\theta_{n}\right)$

which is equal to the second contribution from the commutator
$\langle\Phi,[\phi(f),\phi^{\prime}(g)]\Psi\rangle$ up to the sign, therefore,
they cancel each other. ∎

###  3.3 Estimates

Let us assume that $\chi_{n}(\xi)+\chi^{\prime}_{n}(\eta)$ is already
essentially self-adjoint on its natural domain
${\mathscr{D}}_{n}(\xi,\eta):=\operatorname{Dom}(\chi_{n}(\xi))\cap\operatorname{Dom}(\chi^{\prime}_{n}(\eta))$.
We denote the algebraic direct sum by
${\mathscr{D}}(\xi,\eta):=\underline{\bigoplus}_{n}{\mathscr{D}}_{n}(\xi,\eta)$.
We are going to show that Proposition 3.1 can be applied to show that
${\widetilde{\phi}}(\xi)$ and ${\widetilde{\phi}}^{\prime}(\eta)$ strongly
commute.

######  Proposition 3.3.

There is $c(\xi,\eta)>0$ such that, for $\Psi\in{\mathscr{D}}_{n}(\xi,\eta)$,
it holds that

$\displaystyle\operatorname{Re}\langle\chi(\xi)\Psi,\chi^{\prime}(\eta)\Psi\rangle\geq-
nc(\xi,\eta)\langle(\chi(\xi)+\chi^{\prime}(\eta)+2)\Psi,\Psi\rangle,$

and therefore,

$\displaystyle\|\chi(\xi)\Psi\|\leq\|(\chi(\xi)+\chi^{\prime}(\eta)+c(\xi,\eta)(N+{\mathbbm{1}})\Psi\|,$
$\displaystyle\|\chi^{\prime}(\eta)\Psi\|\leq\|(\chi(\xi)+\chi^{\prime}(\eta)+c(\xi,\eta)(N+{\mathbbm{1}})\Psi\|$

for $\Psi\in{\mathscr{D}}(\xi,\eta)$.

###### Proof.

We may assume that $\Psi\in{\mathscr{D}}_{n}(\xi,\eta)$. We use the following
expressions:

$\displaystyle\chi_{n}(\xi)=\sum_{1\leq k\leq
n}D_{n}(\rho_{k})(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})P_{n},$
$\displaystyle\chi^{\prime}_{n}(\eta)=\sum_{1\leq k\leq
n}D_{n}(\rho^{\prime}_{k})({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))P_{n}.$

By proceeding as in Section 3.2, we see that
$\langle\chi(\xi)\Psi,\chi^{\prime}(\eta)\Psi\rangle$ consists of $n(n-1)$
times

$\displaystyle\langle(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi,({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))\Psi\rangle$

and the following $n$ terms ($1\leq k\leq n$):

$\displaystyle\langle
D_{n}(\rho_{k})(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi,D_{n}(\rho^{\prime}_{n-k+1})({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))\Psi\rangle.$
(3.1)

We write $D_{n}(\rho_{k})=M_{S_{k}}F_{k}$, where

$\displaystyle
S_{k}\left(\boldsymbol{\theta}\right):=\prod_{j<k}S\left(\theta_{k}-\theta_{j}\right)$

and $F_{k}$ is the corresponding cyclic permutation of the variables. We have

$\displaystyle
D_{n}(\rho_{k})(\chi_{1}(\xi)\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi$
$\displaystyle\qquad{}=M_{S_{k}}F_{k}\big{(}\chi_{1}(\xi)^{\frac{1}{2}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}F_{k}^{*}F_{k}(\chi_{1}(\xi)^{\frac{1}{2}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi$
$\displaystyle\qquad{}=\big{(}{\mathbbm{1}}\otimes\cdots{\mathbbm{1}}\otimes\underset{k\textrm{-th}}{\chi_{1}(\xi)^{\frac{1}{2}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}M_{S_{k}(\boldsymbol{\theta}+\frac{\pi
i}{6})}F_{k}\big{(}\chi_{1}(\xi)^{\frac{1}{2}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}\Psi,$

where

$\displaystyle S_{k}\left(\boldsymbol{\theta}+\frac{\pi
i}{6}\right)=\prod_{j<k}S\left(\theta_{k}-\theta_{j}+\frac{\pi i}{6}\right),$

which is bounded by $1$ by (S8). Similarly,

$\displaystyle
D_{n}(\rho^{\prime}_{n-k+1})({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))\Psi$
$\displaystyle\qquad{}=M_{S^{\prime}_{n-k+1}}F^{\prime}_{n-k+1}\big{(}{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta)^{\frac{1}{2}}\big{)}F_{n-k+1}^{\prime*}F^{\prime}_{n-k+1}\big{(}{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta)^{\frac{1}{2}}\big{)}\Psi$
$\displaystyle\qquad{}=\big{(}{\mathbbm{1}}\otimes\cdots{\mathbbm{1}}\otimes\underset{k\textrm{-th}}{\chi^{\prime}_{1}(\eta)^{\frac{1}{2}}}\otimes{\mathbbm{1}}\cdots\otimes{\mathbbm{1}}\big{)}M_{S^{\prime}_{n-k+1}(\boldsymbol{\theta}+\frac{\pi
i}{6})}F_{n-k+1}\big{(}{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta)^{\frac{1}{2}}\big{)}\Psi,$

where $S^{\prime}_{n-k+1}(\boldsymbol{\theta}+\frac{\pi
i}{6})=\prod\limits_{j>k}S(\theta_{k}-\theta_{j}+\frac{\pi i}{6})$, which is
again bounded by $1$ and $F^{\prime}_{n-k+1}$ is the corresponding cyclic
permutation. We define
$c(\xi,\eta)=\|\chi_{1}(\xi)^{\frac{1}{2}}\cdot\chi^{\prime}_{1}(\eta)^{\frac{1}{2}}\|+1$,
which is finite. Now we have

$\displaystyle|\eqref{eq:cross}|\leq
c(\xi,\eta)\big{\|}(\chi_{1}(\xi)^{\frac{1}{2}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi\big{\|}^{\frac{1}{2}}\cdot\big{\|}({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta)^{\frac{1}{2}})\Psi\big{\|}^{\frac{1}{2}}$
$\displaystyle\hphantom{|\eqref{eq:cross}|}{}\leq
c(\xi,\eta)\big{(}\big{\|}(\chi_{1}(\xi)^{\frac{1}{2}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi\big{\|}+\big{\|}({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta)^{\frac{1}{2}})\Psi\big{\|}\big{)}$
$\displaystyle\hphantom{|\eqref{eq:cross}|}{}\leq
c(\xi,\eta)\big{(}\big{\|}(\chi_{1}(\xi)^{\frac{1}{2}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi\big{\|}^{2}+\big{\|}({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta)^{\frac{1}{2}})\Psi\big{\|}^{2}+2\big{)}$
$\displaystyle\hphantom{|\eqref{eq:cross}|}{}=c(\xi,\eta)\big{(}\langle\chi_{n}(\xi)\Psi,\Psi\rangle+\langle\chi^{\prime}_{n}(\eta)\Psi,\Psi\rangle+2\big{)},$

from which the first statement follows directly.

The second statement is easy because

$\displaystyle\|(\chi(\xi)+\chi^{\prime}(\eta)+(n+1)c(\xi,\eta)\Psi\|^{2}$
$\displaystyle\qquad{}=\|\chi(\xi)\Psi\|^{2}+2\operatorname{Re}\langle\chi(\xi)\Psi,\chi^{\prime}(\eta)\Psi\rangle+\|\chi^{\prime}(\eta)\Psi\|^{2}$
$\displaystyle\qquad\quad{}+2(n+1)c(\xi,\eta)\langle(\chi(\xi)+\chi^{\prime}(\eta)\Psi,\Psi\rangle+(n+1)^{2}c(\xi,\eta)^{2}\|\Psi\|^{2},$

which extends to $\Psi=(\Psi_{n})\in{\mathscr{D}}(\xi,\eta)$ by replacing $n$
by the number operator $N$, because the operators appearing here preserve
${\mathcal{H}}_{n}$. Therefore, by dropping the positive terms at each $n$ in
the right-hand side, we conclude:

$\displaystyle\|\chi(\xi)\Psi\|^{2}\leq\|(\chi(\xi)+\chi^{\prime}(\eta)+c(\xi,\eta)(N+{\mathbbm{1}})\Psi\|^{2},$
$\displaystyle\|\chi^{\prime}(\eta)\Psi\|^{2}\leq\|(\chi(\xi)+\chi^{\prime}(\eta)+c(\xi,\eta)(N+{\mathbbm{1}})\Psi\|^{2}.$
∎

It is important that the estimate does not grow too fast with $n$, thanks to
(S8).

######  Corollary 3.4.

Assume that $\chi_{n}(\xi)+\chi^{\prime}_{n}(\eta)$ is essentially self-
adjoint. Then ${\widetilde{\phi}}(\xi)$ and
${\widetilde{\phi}}^{\prime}(\eta)$ commute strongly.

###### Proof.

From the assumption, it is immediate that
$\chi(\xi)+\chi^{\prime}(\eta)+c(N+{\mathbbm{1}})$ is essentially self-
adjoint, where $N$ is the number operator and
$c=\|\xi\|+\|\eta\|+c(\xi,\eta)$, where $c(\xi,\eta)$ comes from Proposition
3.3. For simplicity we do not indicate the dependence of $c$ on $\xi$ and
$\eta$. As $\chi(\xi)$ and $\chi^{\prime}(\eta)$ are defined at each
$n$-particle space, they strongly commute with $N$.

Next, let us observe that
${\widetilde{\phi}}(\xi)+{\widetilde{\phi}}^{\prime}(\eta)+c(N+{\mathbbm{1}})$
is essentially self-adjoint. Indeed, this is equal to the sum of
$\chi(\xi)+\chi^{\prime}(\eta)+c(N+{\mathbbm{1}})$ and
$\phi(\xi)+\phi^{\prime}(\eta)$. As $\chi(\xi)+\chi^{\prime}(\eta)$ and
$N+{\mathbbm{1}}$ are both positive and strongly commute, we have for any
$\Psi\in{\mathscr{D}}(\xi,\eta)$ that
$\|c(N+{\mathbbm{1}})\Psi\|\leq\|(\chi(\xi)+\chi^{\prime}(\eta)+c(N+{\mathbbm{1}}))\Psi\|$.
On the other hand, as $\phi(\xi)+\phi^{\prime}(\eta)$ are the sum of creation
and annihilation operators, it holds by the above choice of $c$ that
$\|(\phi(\xi)+\phi^{\prime}(\eta))\Psi\|\leq\|(\|\xi\|+\|\eta\|)(N+{\mathbbm{1}})\Psi\|<\|c(N+{\mathbbm{1}})\Psi\|$.
Now by Kato–Rellich perturbation theorem [26, Theorem X.12]222Here the larger
operator is only essentially self-adjoint, but the estimate extends to the
whole domain, hence the theorem applies. Accordingly, the conclusion of the
theorem is essential self-adjointness.,
${\widetilde{\phi}}(\xi)+{\widetilde{\phi}}^{\prime}(\eta)+c(N+{\mathbbm{1}})$
is essentially self-adjoint on ${\mathscr{D}}(\xi,\eta)$.

In order to show the hypothesis (1) of Proposition 3.1 with
$A={\widetilde{\phi}}(\xi)$, $B={\widetilde{\phi}}^{\prime}(\eta)$,
$T_{0}=c(N+{\mathbbm{1}})$, note that, by Proposition 3.3 and the definition
of $c$, we have

$\displaystyle\big{\|}{\widetilde{\phi}}(\xi)\Psi\big{\|}=\|(\phi(\xi)+\chi(\xi))\Psi\|\leq\|c(N+{\mathbbm{1}})\Psi\|+\|(\chi(\xi)+\chi^{\prime}(\eta))+c(\xi,\eta)(N+{\mathbbm{1}})\Psi\|$
$\displaystyle\hphantom{\big{\|}{\widetilde{\phi}}(\xi)\Psi\big{\|}=\|(\phi(\xi)+\chi(\xi))\Psi\|}{}\leq
2\|(\chi(\xi)+\chi^{\prime}(\eta)+c(N+{\mathbbm{1}}))\Psi\|$

and a parallel estimate shows that
$\big{\|}{\widetilde{\phi}}^{\prime}(\eta)\Psi\big{\|}\leq
2\|(\chi(\xi)+\chi^{\prime}(\eta)+c(N+{\mathbbm{1}}))\Psi\|$.

As for (2) and (3), recall that $N$ commutes with $\chi(\xi)$ and
$\chi^{\prime}(\eta)$, therefore, the (weak) commutator between
$(N+{\mathbbm{1}})$ and ${\widetilde{\phi}}(\xi)$ is reduced to
$[N,\phi(\xi)]$, which is the sum of $[N,z^{\dagger}(\xi)]=z^{\dagger}(\xi)$
and $[N,z(\xi)]=-z(\xi)$. Namely, we have

$\displaystyle\big{|}\langle
N\Psi,{\widetilde{\phi}}(\xi)\varphi\rangle-\langle{\widetilde{\phi}}(\xi)\Psi,N\Phi\rangle\big{|}=|\langle\Psi,(z^{\dagger}(\xi)-z(\xi))\Phi\rangle|$

and it is easy to see that this can be bounded by both
$\big{\|}(c(N+{\mathbbm{1}}))^{\frac{1}{2}}\Psi\big{\|}\cdot\big{\|}(c(N+{\mathbbm{1}}))^{\frac{1}{2}}\Phi\big{\|}$
and $\|\Psi\|\cdot\|c(N+{\mathbbm{1}})\Phi\|$.

As we saw that ${\widetilde{\phi}}(\xi)$ and
${\widetilde{\phi}}^{\prime}(\eta)$ weakly commute on
${\mathscr{D}}(\xi,\eta)$ in Section 3.2, we can apply Proposition 3.1 and
they strongly commute. ∎

###  3.4 Construction of Borchers triple

We now proceed to von Neumann algebras, always assuming that
$\chi_{n}(\xi)+\chi^{\prime}_{n}(\eta)$ is essentially self-adjoint on
${\mathscr{D}}_{n}(\xi,\eta)$. Among the properties of Borchers triples,
important are the separating property of the vacuum $\Omega$ and the
endomorphic action of the Poincaré group. The former is an immediate
consequence of Corollary 3.4 and the latter follows from the canonical
correspondence from $\xi$ to $\chi(\xi)$.

For the next lemma, we need to recall the Beurling factorization [27, Theorems
17.15 and 17.17]. Any bounded analytic function $f(\zeta)$ on the unit circle
can be decomposed in to the product
$f(\zeta)=cf_{\mathrm{Bl}}(\zeta)f_{\mathrm{out}}(\zeta)f_{\mathrm{in}}(\zeta)$,
where the Blaschke product $f_{\mathrm{Bl}}$ is a possibly infinite product of
functions $\frac{|\alpha|}{\alpha}\frac{\alpha-\zeta}{1-\bar{\alpha}\zeta}$,
where $|\alpha|<1$, $f_{\mathrm{out}}$ is the Poisson integral of the boundary
value of $|f(\zeta)|$ on $S^{1}$ and $f_{\mathrm{in}}$ is the Poisson integral
of a certain singular measure. A corresponding decomposition holds for any
region in ${\mathbb{C}}$ which is conformally equivalent to the unit disk, in
particular, the strip $\SS_{-\frac{2\pi}{3},-\frac{\pi}{3}}$. In [35], we
exploited this decomposition in order to construct the analytic function
$\eta_{0}$ (see Section 2.4).

From here on, we switch to the objects with ′, because they are by convention
those which generate the von Neumann algebra.

######  Lemma 3.5.

$\chi^{\prime}(\eta)$ and ${\widetilde{\phi}}^{\prime}(\eta)$ are covariant in
the following sense: if $a\in W_{\mathrm{R}}$, then we have
$\chi^{\prime}(U_{1}(a,\lambda)\eta)=\operatorname{Ad}U(a,\lambda)(\chi^{\prime}(\eta))$
and
${\widetilde{\phi}}^{\prime}(U_{1}(a,\lambda)\eta)=\operatorname{Ad}U(a,\lambda)({\widetilde{\phi}}^{\prime}(\eta))$
as operators $($including the domains$)$.

###### Proof.

First note that, if
$\eta\in{H^{2}(\SS_{-\pi,0})}\cap{H^{\infty}(\SS_{-\pi,0})}$, then it follows
from $(U_{1}(a,\lambda)\eta)(\theta)=e^{ip(\theta)\cdot
a}\eta(\theta-\lambda)$ that
$U_{1}(a,\lambda)\eta\in{H^{2}(\SS_{-\pi,0})}\cap{H^{\infty}(\SS_{-\pi,0})}$,
as $a\in W_{\mathrm{R}}$ implies that333Our convention of the Minkowski metric
is $a\cdot b=a_{0}b_{0}-a_{1}b_{1}$. $|e^{ip(\zeta)\cdot a}|\leq 1$ for
$\zeta\in\SS_{-\pi,0}$. Furthermore, if $\eta=\underline{\eta}^{2}$, then
$(U_{1}(a,\lambda)\eta)(\theta)=\left(e^{ip(\theta)\cdot
a/2}\underline{\eta}(\theta-\lambda)\right)^{2}$, hence
$\chi^{\prime}(U_{1}(a,\lambda)\eta)$ can be defined.

Next, let us remember how
$\chi_{1}^{\prime}(\eta)=M_{\eta_{0}(\cdot\,-\frac{2\pi
i}{3})}^{\prime*}{\Delta_{1}^{-\frac{1}{6}}}M_{\eta_{0}(\cdot\,-\frac{2\pi
i}{3})}^{\prime}$ is constructed: $M_{\eta_{0}(\cdot\,-\frac{2\pi
i}{3})}=M_{\eta_{\mathrm{Bl}}}M_{\eta_{\mathrm{in}}}M_{\eta_{-}}$ (again up to
the scaling), where $M_{\eta_{\mathrm{Bl}}}$ is the Blaschke product
corresponding to the zeros of $\underline{\eta}$ in
$\SS_{-\frac{2\pi}{3},-\frac{\pi}{3}}$, $M_{\eta_{\mathrm{in}}}$ is the inner
singular part of $\underline{\eta}$ again in
$H^{\infty}\big{(}\SS_{-\frac{2\pi}{3},-\frac{\pi}{3}}\big{)}$ and $\eta_{-}$
is an outer function on $\SS_{-\frac{2\pi}{3},-\frac{\pi}{3}}$ such that
$\eta_{-}\big{(}\theta-\frac{\pi
i}{3}\big{)}\overline{\eta_{-}\big{(}\theta-\frac{2\pi
i}{3}\big{)}}=\eta_{\mathrm{out}}\big{(}\theta-\frac{\pi i}{3}\big{)}$, or
equivalently,

$\displaystyle\eta_{-}\left(\theta-\frac{2\pi
i}{3}\right)\overline{\eta_{-}\left(\theta-\frac{\pi
i}{3}\right)}=\eta_{\mathrm{out}}\left(\theta-\frac{2\pi i}{3}\right),$ (3.2)

where $\eta_{\mathrm{out}}$ is the outer part of $\underline{\eta}$.

First we consider a pure boost $U_{1}(0,\lambda)$. As the decomposition
$M_{\eta_{0}(\cdot\,-\frac{2\pi
i}{3})}=M_{\eta_{\mathrm{Bl}}}M_{\eta_{\mathrm{in}}}M_{\eta_{-}}$ is canonical
where ${\eta_{\mathrm{Bl}}}$ is a Blaschke product, $\eta_{\mathrm{in}}$ is
inner and $\eta_{-}$ is outer with the above equality, we also have the
corresponding decomposition for $U_{1}(0,\lambda)\eta$ and we obtain
$\chi^{\prime}_{1}(U_{1}(0,\lambda)\eta)=U_{1}(0,\lambda)\chi^{\prime}(\eta)U_{1}(0,\lambda)^{*}$.

Next (and more importantly), take a pure translation $U_{1}(a,0)$. We have
$(U_{1}(a,0)\eta)(\theta)=e^{ia\cdot p(\theta)}\eta(\theta)$, hence
$(U_{1}(a,0)\eta)\big{(}\theta-\frac{2\pi i}{3}\big{)}=e^{ia\cdot
p(\theta-\frac{2\pi i}{3})}\eta\big{(}\theta-\frac{2\pi i}{3}\big{)}$. The
function $e^{ia\cdot p(\theta)}$ is purely outer444$e^{ia\cdot p(\theta)}$ is
singular as a function on $\SS_{-\pi,0}$, but is outer when restricted to
$\SS_{-\frac{2\pi}{3},-\frac{\pi}{3}}$, as it gets scaled. on
$\SS_{-\frac{2\pi}{3},-\frac{\pi}{3}}$. It holds that
$-p(\theta)+p\big{(}\theta-\frac{\pi i}{3}\big{)}=p\big{(}\theta-\frac{2\pi
i}{3}\big{)}$, hence

$\displaystyle e^{-ia\cdot p(\theta)}\cdot\overline{e^{-ia\cdot
p(\theta+\frac{\pi i}{3})}}=e^{ia\cdot p(\theta-\frac{2\pi i}{3})},$

which, by comparison with (3.2), implies that $(U_{1}(a,0)\eta)_{-}$ is the
product $\eta_{-}$ and $e^{-ia\cdot p(\theta)}$, i.e.,

$\displaystyle(U_{1}(a,0)\eta)_{-}\left(\theta-\frac{\pi
i}{3}\right)=e^{-ia\cdot p(\theta)}\eta_{-}\left(\theta-\frac{\pi
i}{3}\right)$

and $(U(a,0)\eta)_{\mathrm{Bl}}=\eta_{\mathrm{Bl}}$,
$(U(a,0)\eta)_{\mathrm{in}}=\eta_{\mathrm{in}}$. In other words,

$\displaystyle M_{(U_{1}(a,0)\eta)_{-}(\cdot\,-\frac{\pi
i}{3})}=U_{1}(-a,0)M_{\eta_{-}(\cdot\,-\frac{\pi
i}{3})}=U_{1}(a,0)^{*}M_{\eta_{-}(\cdot\,-\frac{\pi i}{3})}.$

Now we have the decomposition

$\displaystyle M_{(U_{1}(a,0)\eta)_{0}(\cdot\,-\frac{2\pi
i}{3})}=M_{\eta_{\mathrm{Bl}}}M_{\eta_{\mathrm{in}}}M_{\eta_{-}}U_{1}(a,0)^{*},$

therefore,

$\displaystyle\chi^{\prime}_{1}(U_{1}(a,0)\eta)=U_{1}(a,0)\chi^{\prime}(\eta)U_{1}(a,0)^{*}.$

The covariance of the full operators
$\chi^{\prime}(\eta)=\bigoplus_{n}P_{n}({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))P_{n}$
and
${\widetilde{\phi}}^{\prime}(\eta)=\phi^{\prime}(\eta)+\chi^{\prime}(\eta)$
follows from the covariance of $\chi^{\prime}_{1}(\eta)$. ∎

######  Theorem 3.6.

Assume that ${\mathcal{V}}$ is a subset of
${H^{2}(\SS_{0,\pi})}\cap{H^{\infty}(\SS_{0,\pi})}$, whose elements $\xi$ have
the form $\xi=\underline{\xi}^{2}$, satisfy
$\underline{\xi}(\theta+i\pi)=\overline{\underline{\xi}(\theta)}$, and
${\mathcal{V}}$ is invariant under
$U_{1}(a,\lambda),(a,\lambda)\in{{\mathcal{P}}^{\uparrow}_{+}}$ such that
$a\in W_{\mathrm{L}}$ and is total in ${\mathcal{H}}_{1}$. Assume further that
$\chi(\xi)+\chi^{\prime}(\eta)$ is essentially self-adjoint on
${\mathscr{D}}(\xi,\eta)$ for any pair of $\xi\in{\mathcal{V}}$, $\eta\in
J_{1}{\mathcal{V}}$.

We define

$\displaystyle{\mathcal{M}}=\big{\\{}e^{i{\widetilde{\phi}}^{\prime}(\eta)}\colon\eta\in
J_{1}{\mathcal{V}}\big{\\}}^{\prime\prime}.$

Then, $({\mathcal{M}},U,\Omega)$ is a Borchers triple.

###### Proof.

$\Omega$ is cyclic for ${\mathcal{M}}$ by the argument of [6, Section 3,
Reeh–Schlieder property] and the density of the linear span of ${\mathcal{V}}$
in ${\mathcal{H}}_{1}$. It is separating because $\Omega$ is cyclic for
$\big{\\{}e^{i{\widetilde{\phi}}(\xi)}\colon\xi\in{\mathcal{V}}\big{\\}}^{\prime\prime}$,
which commutes with ${\mathcal{M}}$ by Corollary 3.4. Finally, for $a\in
W_{\mathrm{R}}$,
$\operatorname{Ad}U(a,\lambda){\mathcal{M}}\subset{\mathcal{M}}$ follows
immediately from Lemma 3.5 and the assumption that ${\mathcal{V}}$ is
invariant under $U_{1}(a,\lambda)$ for $a\in W_{\mathrm{L}}$, hence
$J_{1}{\mathcal{V}}$ is invariant under $U_{1}(a,\lambda)$, $a\in
W_{\mathrm{R}}$. ∎

Note that, if the assumption of strong commutativity is dropped, the totality
of ${\mathcal{V}}$ is easy, as (cf. [6]). We only have to find sufficiently
many $\xi$’s and $\eta$’s which satisfy strong commutativity.

##  4 Self-adjointness of bound state operators

Here we consider $\chi(\xi)+\chi^{\prime}(\eta)$. Although our fields is
${\widetilde{\phi}}(\xi)=\phi(\xi)+\chi(\xi)$ and not $\chi(\xi)$ or
$\chi^{\prime}(\eta)$, the question of the domain of ${\widetilde{\phi}}(\xi)$
can be solved by Corollary 3.4, once that of $\chi(\xi)+\chi^{\prime}(\eta)$
is settled. We show that $\chi_{n}(\xi)+\chi^{\prime}_{n}(\eta)$ is
essentially self-adjoint for $n=1,2$ if $\xi=\underline{\xi}^{2}$ and
$\eta=\underline{\eta}^{2}$ as in Section 2.4, and give some observations for
$n\geq 3$.

#### Observations on closure

######  Lemma 4.1.

Let $X$, $Y$ be closed operators and assume that there is $c\in{\mathbb{R}}$
such that $c\|\Psi\|^{2}\leq\operatorname{Re}\langle X\Psi,Y\Psi\rangle$ for
$\Psi\in\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$. Then $X+Y$ is closed.

###### Proof.

Let $\Psi_{n}\in\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$ such that
$\Psi_{n}\to\Psi$ and $(X+Y)\Psi_{n}\to\Phi$. Then

$\displaystyle\|(X+Y)(\Psi_{m}-\Psi_{n})\|^{2}$
$\displaystyle\qquad{}=\|X(\Psi_{m}-\Psi_{n})\|^{2}+\|Y(\Psi_{m}-\Psi_{n})\|^{2}+2\operatorname{Re}\langle
X(\Psi_{m}-\Psi_{n}),Y(\Psi_{m}-\Psi_{n})\rangle$
$\displaystyle\qquad{}\geq\|X(\Psi_{m}-\Psi_{n})\|^{2}+\|Y(\Psi_{m}-\Psi_{n})\|^{2}+2c\|\Psi_{m}-\Psi_{n}\|^{2},$

from which it follows that $\\{X\Psi_{m}\\}$ and $\\{Y\Psi_{m}\\}$ are both
convergent, and therefore,
$\Psi\in\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$, namely, $X+Y$ is
closed. ∎

######  Lemma 4.2.

Let $X$, $Y$ be closable operators and assume that there is $c\in{\mathbb{R}}$
such that $c\|\Psi\|^{2}\leq\operatorname{Re}\langle X\Psi,Y\Psi\rangle$ for
$\Psi\in\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$. Then $X+Y$ is
closable and $\overline{X+Y}\subset\overline{X}+\overline{Y}$.

###### Proof.

For $\Psi\in\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$, we have
$\|(X+Y)\Psi\|^{2}=\|X\Psi\|^{2}+\|Y\Psi\|^{2}+2\operatorname{Re}\langle
X\Psi,Y\Psi\rangle$ where the last term is bounded below by $c\|\Psi\|^{2}$ by
assumption. If $\Psi_{n}\to 0$ and $(X+Y)\Psi_{n}\to\Phi$, then
$\|(X+Y)(\Psi_{n}-\Psi_{m})\|\to 0$ and

$\displaystyle\lim_{m,n}\,\inf\left(2\operatorname{Re}\langle
X(\Psi_{n}-\Psi_{m}),Y(\Psi_{n}-\Psi_{m})\rangle\right)\geq 0.$

From this it follows that $\|X(\Psi_{n}-\Psi_{m})\|\to 0$ and
$\|Y(\Psi_{n}-\Psi_{m})\|\to 0$, namely $X\Psi_{n}$ and $Y\Psi_{n}$ are
separately convergent. As $X$ and $Y$ are closable, we obtain $X\Psi_{n}\to 0$
and $Y\Psi_{n}\to 0$ and $\Phi=0$, namely $X+Y$ is closable.

To show $\overline{X+Y}\subset\overline{X}+\overline{Y}$, we only have to
repeat the argument with $\Psi_{n}\to\Psi$ to obtain
$\Psi\in\operatorname{Dom}(\overline{X})\cap\operatorname{Dom}(\overline{Y})=\operatorname{Dom}(\overline{X}+\overline{Y})$.
∎

######  Lemma 4.3.

Let $X$, $Y$ be closed operators and assume that there is $c>0$ such that
$0\leq
c^{2}\|\Psi\|^{2}+2c(\langle\Psi,X\Psi\rangle+\langle\Psi,Y\Psi\rangle)+2\operatorname{Re}\langle
X\Psi,Y\Psi\rangle$ for
$\Psi\in\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$. Then $X+Y$ is closed.

###### Proof.

Let $\Psi_{n}\in\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$ such that
$\Psi_{n}\to\Psi$ and $(X+Y+c)\Psi_{n}\to\Phi$. Hence we have

$\displaystyle\|(X+Y+c)(\Psi_{m}-\Psi_{n})\|^{2}$
$\displaystyle\qquad{}=\|X(\Psi_{m}-\Psi_{n})\|^{2}+\|Y(\Psi_{m}-\Psi_{n})\|^{2}+2\operatorname{Re}\langle
X(\Psi_{m}-\Psi_{n}),Y(\Psi_{m}-\Psi_{n})\rangle$
$\displaystyle\qquad\quad{}+2c(\langle(\Psi_{m}-\Psi_{n}),X(\Psi_{m}-\Psi_{n})\rangle+\langle(\Psi_{m}-\Psi_{n}),Y(\Psi_{m}-\Psi_{n})\rangle)+c^{2}\|(\Psi_{m}-\Psi_{n})\|^{2}$
$\displaystyle\qquad{}\geq\|X(\Psi_{m}-\Psi_{n})\|^{2}+\|Y(\Psi_{m}-\Psi_{n})\|^{2},$

from which it follows that $\\{X\Psi_{m}\\}$ and $\\{Y\Psi_{m}\\}$ are both
convergent, and therefore,
$\Psi\in\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$, namely, $X+Y+c$ is
closed. This is equivalent to that $X+Y$ is closed. ∎

###  4.1 One-particle components

Now we show that $\chi_{1}(\xi)+\chi^{\prime}_{1}(\eta)$ is self-adjoint on
the intersection
$\operatorname{Dom}(\chi_{1}(\xi))\cap\operatorname{Dom}(\chi^{\prime}_{1}(\eta))$.
A key observation is that $\chi_{1}(\xi)$ can be written in the form $X^{*}X$
where $X={\Delta_{1}^{\frac{1}{12}}}M_{\xi_{0}(\cdot\,+\frac{2\pi i}{3})}$. It
is clear that $X$ is densely defined, as $\xi_{0}\in
H^{\infty}(\SS_{\frac{\pi}{3},\frac{2\pi}{3}})$, and closed as
$M_{\xi_{0}(\cdot\,+\frac{2\pi i}{3})}$ is bounded. Furthermore, it holds that
$X^{*}=M_{\xi_{0}(\cdot\,+\frac{2\pi i}{3})}^{*}{\Delta_{1}^{\frac{1}{12}}}$
without closure, since $M_{\xi_{0}(\cdot\,+\frac{2\pi i}{3})}^{*}$ is unitary.
Similarly, one can write $\chi^{\prime}_{1}(g)=Y^{*}Y$, where
$Y={\Delta_{1}^{-\frac{1}{12}}}M_{\eta_{0}(\cdot\,-\frac{2\pi i}{3})}$ and
$Y^{*}=M_{\eta_{0}(\cdot\,-\frac{2\pi
i}{3})}^{*}{\Delta_{1}^{-\frac{1}{12}}}$.

Some ideas in the following lemma are due to Henning Bostelmann.

######  Lemma 4.4.

Let $X$, $Y$ be closed, densely defined operators such that

  * •

$X\operatorname{Dom}(X)\subset\operatorname{Dom}(Y^{*})$ and
$Y\operatorname{Dom}(Y)\subset\operatorname{Dom}(X^{*})$,

  * •

$Y^{*}X$ and $X^{*}Y$ are bounded,

  * •

$X+Y$ and $X^{*}+Y^{*}$ are densely defined,

  * •

there is a decomposition of
${\mathcal{H}}={\mathcal{H}}_{X}\oplus{\mathcal{H}}_{Y}$ such that

    * –

any $\Psi\in\operatorname{Dom}({X})$ is the sum $\Psi=\Psi_{X}\oplus\Psi_{Y}$,
$\Psi_{X}\in{\mathcal{H}}_{X}\cap\operatorname{Dom}({X})$ and
$\Psi_{Y}\in{\mathcal{H}}_{Y}\cap\operatorname{Dom}({X})$,

    * –

any $\Phi\in\operatorname{Dom}({Y})$ is the sum $\Phi=\Phi_{X}\oplus\Phi_{Y}$,
$\Phi_{X}\in{\mathcal{H}}_{X}\cap\operatorname{Dom}({Y})$ and
$\Phi_{Y}\in{\mathcal{H}}_{Y}\cap\operatorname{Dom}({Y})$,

    * –

${X}$ is bounded on ${\mathcal{H}}_{X}$ and ${Y}$ is bounded on
${\mathcal{H}}_{Y}$.

Then the sum $X^{*}X+Y^{*}Y$ is self-adjoint on
$\operatorname{Dom}(X^{*}X)\cap\operatorname{Dom}(Y^{*}Y)$.

###### Proof.

By definition,
$\operatorname{Dom}(X+Y)=\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$.
Under the assumption, $X+Y$ is closed. Indeed, we have $\langle
X\Psi,Y\Psi\rangle=\langle\Psi,X^{*}Y\Psi\rangle$ which is bounded by
$\|X^{*}Y\|\cdot\|\Psi\|^{2}$ and we can apply Lemma 4.1.

Next, note that
$\operatorname{Dom}({X}+{Y})=\operatorname{Dom}({X})\cap\operatorname{Dom}({Y})$
is a core of ${X}$. Indeed, any $\Psi\in\operatorname{Dom}({X})$ can be
decomposed as $\Psi_{X}\oplus\Psi_{Y}$ by assumption. On the other hand,
$\operatorname{Dom}({Y})$ is dense and admits the decomposition
$\operatorname{Dom}({Y})=(\operatorname{Dom}({Y})\cap{\mathcal{H}}_{X})\oplus(\operatorname{Dom}({Y})\cap{\mathcal{H}}_{Y})$.
As $\operatorname{Dom}({X}+{Y})$ is dense by assumption,
$\operatorname{Dom}({X}+{Y})\cap{\mathcal{H}}_{X}$ is dense in
${\mathcal{H}}_{X}$. But ${X}$ is bounded on ${\mathcal{H}}_{X}$, thus
$\operatorname{Dom}({X})$ includes ${\mathcal{H}}_{X}$. Namely,
$\operatorname{Dom}({X})={\mathcal{H}}_{X}\oplus(\operatorname{Dom}({X})\cap{\mathcal{H}}_{Y})$.
Similarly, we have
$\operatorname{Dom}({Y})=(\operatorname{Dom}({Y})\cap{\mathcal{H}}_{X})\oplus{\mathcal{H}}_{Y}$.
Therefore,
$\operatorname{Dom}({X}+{Y})=(\operatorname{Dom}({Y})\cap{\mathcal{H}}_{X})\oplus(\operatorname{Dom}({X})\cap{\mathcal{H}}_{Y})$.
We saw that
$\operatorname{Dom}({X}+{Y})\cap{\mathcal{H}}_{X}=\operatorname{Dom}({Y})\cap{\mathcal{H}}_{X}$
is dense in ${\mathcal{H}}_{X}$ and ${X}$ is bounded here, hence, the closure
of ${X}$ restricted on $\operatorname{Dom}({X}+{Y})$ is
${\mathcal{H}}_{X}\oplus\operatorname{Dom}({X})\cap{\mathcal{H}}_{Y}=\operatorname{Dom}({X})$.
Similarly, it holds that $\operatorname{Dom}({X}+{Y})$ is a core of ${Y}$.

Next we see that
$\operatorname{Dom}((X+Y)^{*})=\operatorname{Dom}(X^{*})\cap\operatorname{Dom}(Y^{*})$.
For each vector
$\Psi\in\operatorname{Dom}({X}+{Y})=\operatorname{Dom}({X})\cap\operatorname{Dom}({Y})$,
we take the decomposition $\Psi=\Psi_{X}\oplus\Psi_{Y}$ which is given by the
assumption. By the definition of the adjoint operator, a vector $\Phi$ is in
$\operatorname{Dom}(({X}+{Y})^{*})$ if and only if the map
$\Psi\mapsto\langle\Phi,({X}+{Y})\Psi\rangle$ is continuous for
$\Psi\in\operatorname{Dom}({X}+{Y})$ (namely, there is $c$ such that
$|\langle\Phi,({X}+{Y})\Psi\rangle|\leq c\|\Psi\|$). Then we have the
following chain of equivalences:

$\displaystyle\Phi\in\operatorname{Dom}(({X}+{Y})^{*})$
$\displaystyle\qquad\Longleftrightarrow\
\Psi\mapsto\langle\Phi,({X}+{Y})\Psi\rangle\textrm{ is continuous for
}\Psi\in\operatorname{Dom}({X}+{Y})$ $\displaystyle\qquad\Longleftrightarrow\
\Psi_{X}\mapsto\langle\Phi,({X}+{Y})\Psi_{X}\rangle\textrm{ and
}\Psi_{Y}\mapsto\langle\Phi,({X}+{Y})\Psi_{Y}\rangle$
$\displaystyle\qquad{}\hphantom{\Longleftrightarrow}{}\ \ \textrm{are
continuous for
}\Psi_{X}\in\operatorname{Dom}({X}+{Y})\cap{\mathcal{H}}_{X},\Psi_{Y}\in\operatorname{Dom}({X}+{Y})\cap{\mathcal{H}}_{Y}.$
$\displaystyle\qquad\Longleftrightarrow\
\Psi_{X}\mapsto\langle\Phi,{Y}\Psi_{X}\rangle\textrm{ and
}\Psi_{Y}\mapsto\langle\Phi,{X}\Psi_{Y}\rangle$
$\displaystyle\qquad{}\hphantom{\Longleftrightarrow}{}\ \ \textrm{are
continuous for
}\Psi_{X}\in\operatorname{Dom}({X}+{Y})\cap{\mathcal{H}}_{X},\Psi_{Y}\in\operatorname{Dom}({X}+{Y})\cap{\mathcal{H}}_{Y}.$
$\displaystyle\qquad\Longleftrightarrow\
\Psi_{X}\oplus\Psi_{Y}\mapsto\langle\Phi,{Y}(\Psi_{X}\oplus\Psi_{Y})\rangle\textrm{
and
}\Psi_{X}\oplus\Psi_{Y}\mapsto\langle\Phi,{X}(\Psi_{X}\oplus\Psi_{Y})\rangle$
$\displaystyle\qquad{}\hphantom{\Longleftrightarrow}{}\ \ \textrm{are
continuous for }\Psi=\Psi_{X}\oplus\Psi_{Y}\in\operatorname{Dom}({X}+{Y}).$

We saw that $\operatorname{Dom}({X}+{Y})$ is a core both for ${X}$ and ${Y}$,
thus, the continuity of $\Phi\mapsto\langle\Phi,{X}\Psi\rangle$ for
$\Psi\in\operatorname{Dom}({X}+{Y})$ extends to
$\Psi\in\operatorname{Dom}({X})$, which implies that
$\Phi\in\operatorname{Dom}(X^{*})$. Analogously, we obtain
$\Phi\in\operatorname{Dom}(Y^{*})$. This implies that
$({X}+{Y})^{*}=X^{*}+Y^{*}$.

It is well-known [26, Theorem X.25] that, because $X+Y$ is closed, on the
domain
$\operatorname{Dom}((X+Y)^{*}(X+Y))=\\{\Psi\in\operatorname{Dom}(X+Y):(X+Y)\Psi\in\operatorname{Dom}((X+Y)^{*}))\\}$
the operator $(X+Y)^{*}(X+Y)$ is self-adjoint. From the above observations,
this domain is actually

$\displaystyle\operatorname{Dom}((X+Y)^{*}(X+Y))=\\{\Psi\in\operatorname{Dom}(X+Y)\colon(X+Y)\Psi\in\operatorname{Dom}((X+Y)^{*}))\\}$
$\displaystyle\hphantom{\operatorname{Dom}((X+Y)^{*}(X+Y))}{}=\\{\Psi\in\operatorname{Dom}(X+Y)\colon(X+Y)\Psi\in\operatorname{Dom}(X^{*})\cap\operatorname{Dom}(Y^{*})\\}$
$\displaystyle\hphantom{\operatorname{Dom}((X+Y)^{*}(X+Y))}{}=\\{\Psi\in\operatorname{Dom}(X+Y)\colon
X\Psi\in\operatorname{Dom}(X^{*}),Y\Psi\in\operatorname{Dom}(Y^{*})\\},$

where in the last line we used the assumptions that
$X\operatorname{Dom}(X)\subset\operatorname{Dom}(Y^{*})$ and
$Y\operatorname{Dom}(Y)\subset\operatorname{Dom}(X^{*})$.

Now we have $(X+Y)^{*}(X+Y)=X^{*}X+X^{*}Y+Y^{*}X+Y^{*}Y$ and this is different
from $X^{*}X+Y^{*}Y$ only by a bounded operator. ∎

######  Proposition 4.5.

The sum of one-particle components $\chi_{1}(\xi)+\chi_{1}^{\prime}(\eta)$ is
self-adjoint on the domain
$\operatorname{Dom}(\chi_{1}(\xi))\cap\operatorname{Dom}(\chi_{1}^{\prime}(\eta))$.

###### Proof.

This is a straightforward consequence of this Lemma 4.4.

Indeed, we only check the existence of the decomposition
${\mathcal{H}}_{1}={\mathcal{H}}_{X}\oplus{\mathcal{H}}_{Y}$. We consider
$\log\Delta$ and its spectral decomposition: we take ${\mathcal{H}}_{X}$ as
the spectral subspace of $\log\Delta$ corresponding to $(-\infty,0]$, while
${\mathcal{H}}_{Y}$ corresponds to $(0,\infty)$. Now the required properties
are shown as follows.

Let us denote the spectral projections onto these subspaces by $P_{X}$ and
$P_{Y}$. Let us take an element
$\Psi\in\operatorname{Dom}(X)=\operatorname{Dom}\big{(}{\Delta^{\frac{1}{12}}}M_{\xi_{0}(\cdot\,+\frac{2\pi
i}{3})}\big{)}$. This means, by definition, $\xi_{0}(\theta+\frac{2\pi
i}{3})\Psi(\theta)$ is in ${\mathcal{H}}^{2}\left(-\frac{\pi}{6},0\right)$. In
addition, $\Psi(\theta)$ alone is $L^{2}$. While it holds that
$\Psi=P_{X}\Psi\oplus P_{Y}\Psi$, $P_{X}\Psi$ is obviously in the domain of
${\Delta^{\frac{1}{12}}}$, hence also in the domain of
$X={\Delta^{\frac{1}{12}}}M_{\xi_{0}(\cdot\,+\frac{2\pi i}{3})}$. This means
that $P_{Y}\Psi=\Psi-P_{X}\Psi$ is also in the domain of $X$. The restriction
of $X$ on ${\mathcal{H}}_{X}$ is bounded, since there it holds that
$X\Psi_{X}=M_{\xi_{0}(\cdot\,+\frac{\pi i}{2})}{\Delta^{\frac{1}{12}}}$, where
$\xi_{0}(\theta+\frac{\pi i}{2})$ is bounded.

Similarly, the decomposition ${\mathcal{H}}_{X}\oplus{\mathcal{H}}_{Y}$ is
compatible with $\operatorname{Dom}(Y)$ and $Y$ is bounded on
${\mathcal{H}}_{Y}$. ∎

###  4.2 Two-particle components

In this Section, under (S7), we show that
$\chi_{2}(\xi)+\chi^{\prime}_{2}(\eta)=P_{2}(\chi_{1}(\xi)\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))P_{2}$
is essentially self-adjoint. It should be stressed that this expression does
not involve Friedrichs extension. Simply the restriction to
$P_{2}{\mathcal{H}}_{1}^{\otimes 2}$ is already essentially self-adjoint. This
is important, because we have to be able to compute the weak commutator. In
the course we also show that $\chi_{2}(\xi)$ is essentially self-adjoint. It
is also closed under (S7).

We provide different proofs. The first one is longer and tricky, but one may
hope that with some more ideas the cases $n\geq 3$ could be treated. The other
two works only for $\chi_{2}(\xi)$, but are relatively simple, and give the
hints of why the general case requires a better understanding of the poles and
zeros of $S$. It should be noted that operators of the form
${\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}+M_{f}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{6}}}\big{)}M_{f}^{*}$
may in general fail to be self-adjoint (see Appendix D). We have to properly
use the properties of $S$.

####  4.2.1 Fubini’s theorem

We first show that $\chi_{2}(\xi)$ is self-adjoint. For this purpose, let us
note that the unitary operator $M_{\xi_{0}}\otimes M_{\xi_{0}}$ obviously
commutes with $M_{S}^{*}$, hence also with $D_{2}(\tau_{1})$ and $P_{2}$. As
we have

$\displaystyle\chi_{2}(\xi)=P_{2}(M_{\xi_{0}}^{*}\otimes
M_{\xi_{0}}^{*})\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}(M_{\xi_{0}}\otimes
M_{\xi_{0}})P_{2}$
$\displaystyle\hphantom{\chi_{2}(\xi)}{}=(M_{\xi_{0}}^{*}\otimes
M_{\xi_{0}}^{*})P_{2}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}(M_{\xi_{0}}\otimes
M_{\xi_{0}}),$

the question is now reduced to the self-adjointness of
$P_{2}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}$.

We start with some observations on $L^{2}$-functions.

######  Lemma 4.6.

If $\Psi(\theta_{1},\theta_{2})$ and $\Phi(\theta_{1},\theta_{2})$ belong to
$L^{2}({\mathbb{R}}^{2})$, and
$e^{\theta_{1}}\Psi(\theta_{1},\theta_{2})+e^{\theta_{2}}\Phi(\theta_{1},\theta_{2})$
is $L^{2}$, then on any region bounded with respect to $\theta_{2}$,
$e^{\theta_{1}}\Psi(\theta_{1},\theta_{2})$ is $L^{2}$. Similarly, on any
region bounded with respect to $\theta_{1}$,
$e^{\theta_{2}}\Phi(\theta_{1},\theta_{2})$ is $L^{2}$. On the region
$\theta_{1}+\theta_{2}<N$, both $e^{\theta_{1}}\Psi(\theta_{1},\theta_{2})$
and $e^{\theta_{2}}\Phi(\theta_{1},\theta_{2})$ are $L^{2}$.

###### Proof.

Consider the region $\theta_{2}<N$. Here,
$e^{\theta_{2}}\Phi(\theta_{1},\theta_{2})$ is obviously $L^{2}$, and
$e^{\theta_{1}}\Psi(\theta_{1},\theta_{2})+e^{\theta_{2}}\Phi(\theta_{1},\theta_{2})$
is by assumption $L^{2}$, therefore, the difference

$\displaystyle
e^{\theta_{1}}\Psi(\theta_{1},\theta_{2})=\big{(}e^{\theta_{1}}\Psi(\theta_{1},\theta_{2})+e^{\theta_{2}}\Phi(\theta_{1},\theta_{2})\big{)}-e^{\theta_{2}}\Phi(\theta_{1},\theta_{2})$

is again $L^{2}$. The other case is similar.

The region $\theta_{1}+\theta_{2}<N$ is a subset of the union of
$\theta_{1}<\frac{N}{2}$ and $\theta_{2}<\frac{N}{2}$. The function
$e^{\theta_{1}}\Psi(\theta_{1},\theta_{2})$ is obviously $L^{2}$ on the former
and also on the latter by the first paragraph. The other case is similar. ∎

Coming back to our operators, let us summarize the situation.

  * •

$M_{S}$: multiplication operator on
${\mathcal{H}}^{1}\otimes{\mathcal{H}}^{1}=L^{2}({\mathbb{R}}^{2})$ by
$S(\theta_{2}-\theta_{1})$. Similarly, we denote by $M_{S(\cdot\,+i\lambda)}$
the multiplication operator by $S(\theta_{2}-\theta_{1}+i\lambda)$.

  * •

$\big{(}{\Delta_{1}^{\frac{1}{6}}}\Psi\big{)}(\theta)=\Psi\big{(}\theta-\frac{\pi
i}{3}\big{)}$ on the domain $H^{2}\big{(}{-}\frac{\pi}{3},0\big{)}$.

  * •

$\Delta_{1}\otimes\Delta_{1}$ and $M_{S}$ strongly commute. It holds that
$M_{S}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}\subset\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}M_{S(\cdot\,+\frac{\pi
i}{6})}$, since $S(\zeta)$ is bounded on ${\mathbb{R}}+i\big{(}0,\frac{\pi
i}{6}\big{)}$.

Let us denote the flip operator by
$F\colon\Psi\otimes\Phi\mapsto\Phi\otimes\Psi$ where
$\Psi,\Phi\in{\mathcal{H}}_{1}$. Then we can write the $S$-symmetrization as
follows: $P_{2}=\frac{1}{2}({\mathbbm{1}}+M_{S}F)$.

######  Proposition 4.7.

$P_{2}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}$ is
essentially self-adjoint. Under (S7), it is even self-adjoint.

###### Proof.

It is well-known that $A^{*}A$ is self-adjoint for a closed operator $A$ [26,
Theorem X.25]. The operator
$\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}$ is closed
(because ${\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}$ is closed (self-
adjoint) and $P_{2}$ is bounded) and densely defined, hence
$P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}$ is
closable and
$\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}=\big{(}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}\big{)}^{*}$
[28, Theorem 13.2]. Therefore, the product
$\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}\cdot\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}$
is self-adjoint. The proof will be complete once we show that its domain
coincides with that of
$P_{2}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}$. The
domain of
$\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}\cdot\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}$
consists of vectors
$\underline{\Psi}\in\operatorname{Dom}\big{(}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}\big{)}$
such that
$\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}\Psi\in\operatorname{Dom}\Big{(}\,\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}\,\Big{)}$.
We put $\Psi=\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes
I\big{)}P_{2}\underline{\Psi}$. Now, what we have to prove is that
$\Psi\in\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes I\big{)}$.

If $\Psi$ belongs to the domain of
$\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}$,
then there is a convergent sequence
$\Psi_{n}\in\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}$
such that
$P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\Psi_{n}$
is again convergent. By the remark before this Proposition, the latter
sequence is equal to
$\frac{1}{2}\big{(}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\Psi_{n}+\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S(\cdot\,+\frac{\pi
i}{6})}F\Psi_{n}\big{)}$.

Let us consider the joint spectral decomposition with respect to
$\log\Delta_{1}\otimes{\mathbbm{1}}$ and ${\mathbbm{1}}\otimes\log\Delta_{1}$.
Any vector in ${\mathcal{H}}_{1}\otimes{\mathcal{H}}_{1}$ can be considered as
a two-variable $L^{2}$-function on the spectral space of $\log\Delta_{1}$.
Now, note that $M_{S(\cdot\,+\frac{\pi i}{6})}F$ is a bounded operator. If we
put $\Phi_{n}=M_{S(\cdot\,+\frac{\pi i}{6})}F\Psi_{n}$, then $\Psi_{n}$ and
$\Phi_{n}$ are convergent to $\Psi$ and $\Phi$ in the $L^{2}$-sense,
respectively.

Now, by considering ${\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}$ as a
multiplication operator on the function $\Psi$ in the spectral space
considered above,
$\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\Psi$ is a
priori not a vector in ${\mathcal{H}}^{1}\otimes{\mathcal{H}}^{1}$: this would
mean that
$\Psi\in\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}$,
which we have to prove. Yet, as $\Psi$ is a function and
$\Delta_{1}\otimes{\mathbbm{1}}$ is a multiplication operator on the above-
mentioned spectral space, it is well-defined as a function (almost
everywhere), a priori not $L^{2}$. Similarly, we can treat
$\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\Phi$ as a
function. In this interpretation, the defining property of $\Psi$ says that
$\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\Psi_{n}+\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\Phi_{n}$
is $L^{2}$-convergent, in particular, by taking a subsequence, we may assume
that the sequence is almost everywhere convergent and the pointwise limit is
again $L^{2}$. Namely,
$\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\Psi+\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\Phi$
is $L^{2}$.

Let $Q_{N}$ the spectral projection of $\Delta_{1}\otimes\Delta_{1}$
corresponding to $(e^{N-1},e^{N}]$. Note that on the joint spectral space of
$\log\Delta_{1}\otimes{\mathbbm{1}}$ and ${\mathbbm{1}}\otimes\log\Delta_{1}$,
this corresponds to the region $N-1<t_{1}+t_{2}\leq N$. It commutes strongly
with
$\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}\cdot\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}$.
We first prove that, if
$\underline{\Psi}\in\operatorname{Dom}\Big{(}\,\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}\cdot\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}\Big{)}$
and $\underline{\Psi}=P_{2}Q_{n}\underline{\Psi},$ then
$\underline{\Psi}\in\operatorname{Dom}\big{(}P_{2}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}\big{)}$.
Under this situation,
$\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\Psi+\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\Phi$
is $L^{2}$, hence Lemma 4.6 tells that
$\Psi\in\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}$
or equivalently,
$\underline{\Psi}\in\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}$.

In other words, any vector $\underline{\Psi}=Q_{N}\underline{\Psi}$ in the
domain of
$\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}\cdot\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}$
belongs to the domain of ${\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}$,
which means that we do not need the closure on the operator above for such
$\underline{\Psi}$. As $Q_{N}$ commutes with
$\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}\cdot\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}$,
any vector in its domain is the direct sum of such $Q_{N}\underline{\Psi}$’s,
hence we obtain
$\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}}\cdot\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}P_{2}=\overline{P_{2}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}}$
and $P_{n}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}$
is essentially self-adjoint.

We now prove that
$P_{2}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}$ is
closed under (S7). Indeed, For
$\underline{\Psi}\in\operatorname{Dom}\big{(}P_{2}({\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}})P_{2}\big{)}$,
we may assume that $P_{2}\underline{\Psi}=\underline{\Psi}$. We claim that
$\big{\langle}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}\underline{\Psi},M_{S}F\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}\underline{\Psi}\big{\rangle}\geq
0$. Note that we are not claiming that $M_{S}F$ is a positive operator (it is
not), but the positivity holds for
$\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}\underline{\Psi}$,
where $\underline{\Psi}=P_{2}\underline{\Psi}$. We have

$\displaystyle
M_{S}F\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}\underline{\Psi}=M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{6}}}\big{)}F\underline{\Psi}=M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{6}}}\big{)}M_{S}^{*}\underline{\Psi},$

since $\underline{\Psi}$ is $S$-symmetric:
$\underline{\Psi}=M_{S}F\underline{\Psi}$. Now, the inner product in question
is

$\displaystyle\big{\langle}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}\underline{\Psi},M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\underline{\Psi}\big{\rangle}$
$\displaystyle\qquad{}=\big{\langle}M_{S}^{*}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}\underline{\Psi},\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\underline{\Psi}\big{\rangle}$
$\displaystyle\qquad{}=\big{\langle}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}M_{S(\cdot\,+\frac{\pi}{6})}^{*}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\underline{\Psi},\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\underline{\Psi}\big{\rangle}.$
(4.1)

By Lemma E.6,
$M_{S(\cdot\,+\frac{\pi}{6})}^{*}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\underline{\Psi}\in\operatorname{Dom}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}$
and
$\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\underline{\Psi}\in\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}$.
Then by [6, Lemma B.1], we obtain

$\displaystyle\eqref{eq:cross1}=\big{\langle}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S(\cdot\,+\frac{\pi}{6})}^{*}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\underline{\Psi},\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\underline{\Psi}\big{\rangle}$
$\displaystyle\hphantom{\eqref{eq:cross1}}{}=\big{\langle}M_{S(\cdot\,+\frac{\pi}{3})}^{*}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\underline{\Psi},M_{S}^{*}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\underline{\Psi}\big{\rangle}$
$\displaystyle\hphantom{\eqref{eq:cross1}}{}=\big{\langle}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\underline{\Psi},M_{S(\cdot\,+\frac{\pi}{3})}M_{S}^{*}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\underline{\Psi}\big{\rangle},$

where in the second step we used Lemma E.6. Now, we assume (S7) that the real
part of $S\big{(}\theta+\frac{\pi
i}{3}\big{)}S(-\theta)=S\big{(}\theta+\frac{2\pi i}{3}\big{)}$ is positive.
Therefore, this scalar product has the positive real part. Now the closedness
of
$P_{2}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}=\frac{1}{2}({\mathbbm{1}}+M_{S}F)\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\big{)}P_{2}$
follows from Lemma 4.1. ∎

######  Proposition 4.8.

$\chi^{\prime}_{2}(\eta)=P_{2}({\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))P_{2}$
is essentially self-adjoint, and under (S7) it is self-adjoint.

###### Proof.

Parallel to the case of $\chi_{2}(\xi)$. One can also prove it by considering
the operator $J_{2}P_{2}(\chi_{1}(\xi)\otimes{\mathbbm{1}})P_{2}J_{2}$. ∎

######  Proposition 4.9.

Under (S7),
$\chi_{2}(\xi)+\chi^{\prime}_{2}(\eta)=P_{2}(\chi_{1}(\xi)\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes\chi^{\prime}_{1}(\eta))P_{2}$
is essentially self-adjoint.

###### Proof.

For simplicity of notation we write $A=\chi_{1}(\xi)^{\frac{1}{2}}$,
$B=\chi^{\prime}_{1}(\eta)^{\frac{1}{2}}$. Recall that
$A=M_{\xi_{0}(\cdot\,+\frac{2\pi
i}{3})}^{*}{\Delta_{1}^{\frac{1}{12}}}M_{\xi_{0}(\cdot\,+\frac{2\pi i}{3})}$
and $B=M_{\eta_{0}(\cdot\,-\frac{2\pi
i}{3})}^{*}{\Delta_{1}^{-\frac{1}{12}}}M_{\eta_{0}(\cdot\,-\frac{2\pi
i}{3})}$, where we have $\xi_{0}\in{\mathcal{H}}^{\infty}(\SS_{\frac{\pi
i}{3},\frac{2\pi i}{3}})$, $\eta_{0}\in{\mathcal{H}}^{\infty}(\SS_{-\frac{2\pi
i}{3},-\frac{\pi i}{3}})$. It is straightforward to observe that $AB$ and $BA$
are bounded. Our goal is the essential self-adjointness of
$P_{2}(A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B^{2})P_{2}$.

First we claim that
$\overline{P_{2}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)}\subset\overline{P_{2}(A\otimes{\mathbbm{1}})}+\overline{P_{2}({\mathbbm{1}}\otimes
B)}$. In order to apply Lemma 4.2, it is enough to check that
$c\|\Psi\|^{2}\leq\operatorname{Re}\langle
P_{2}(A\otimes{\mathbbm{1}})\Psi,P_{2}({\mathbbm{1}}\otimes B)\Psi\rangle$.
This holds indeed:

$\displaystyle\langle
P_{2}(A\otimes{\mathbbm{1}})\Psi,P_{2}({\mathbbm{1}}\otimes
B)\Psi\rangle=\frac{1}{4}\langle({\mathbbm{1}}+D_{2}(\tau_{1}))(A\otimes{\mathbbm{1}})\Psi,({\mathbbm{1}}+D_{2}(\tau_{1}))({\mathbbm{1}}\otimes
B)\Psi\rangle$ $\displaystyle\hphantom{\langle
P_{2}(A\otimes{\mathbbm{1}})\Psi,P_{2}({\mathbbm{1}}\otimes
B)\Psi\rangle}{}=\frac{1}{2}(\langle(A\otimes
B)\Psi,\Psi\rangle+\langle(A\otimes{\mathbbm{1}})\Psi,M_{S}(B\otimes{\mathbbm{1}})M_{S}^{*}\Psi\rangle),$

and by noting that $A\otimes B$ is positive, while
$(A\otimes{\mathbbm{1}})M_{S}(B\otimes{\mathbbm{1}})=M_{S(\cdot\,+\frac{\pi
i}{6})}(AB\otimes{\mathbbm{1}})$ is bounded.

We show that, if $(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)\Psi\in\operatorname{Dom}(\overline{P_{2}(A\otimes{\mathbbm{1}})})$, then it
is in $\operatorname{Dom}(A\otimes{\mathbbm{1}})$. This follows the idea of
Proposition 4.7 (see the third and fourth paragraphs). Namely, we consider the
following formal expression with
$\tilde{\Psi}=(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B)\Psi$,

$\displaystyle\big{\langle}\big{(}A\otimes{\mathbbm{1}}+({\mathbbm{1}}\otimes
A)M_{S(\cdot\,+\frac{\pi
i}{6})}F\big{)}\tilde{\Psi},\big{(}A\otimes{\mathbbm{1}}+({\mathbbm{1}}\otimes
A)M_{S(\cdot\,+\frac{\pi i}{6})}F\big{)}\tilde{\Psi}\big{\rangle}.$

This is formal, yet the whole expression has the meaning as an $L^{1}$
integral over the spectral space with respect to $\log A\otimes{\mathbbm{1}}$
and ${\mathbbm{1}}\otimes\log A$: it is $L^{1}$ because

$\displaystyle\big{(}A\otimes{\mathbbm{1}}+({\mathbbm{1}}\otimes
A)M_{S(\cdot\,+\frac{\pi
i}{6})}F\big{)}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B)\Psi$

has the meaning as an $L^{2}$-function. As
$\tilde{\Psi}=(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B)\Psi$, this inner
product can be formally decomposed as

$\displaystyle\big{\|}(A\otimes{\mathbbm{1}})\tilde{\Psi}\big{\|}^{2}+\big{\|}({\mathbbm{1}}\otimes
A)M_{S(\cdot\,+\frac{\pi
i}{6})}F\tilde{\Psi}\big{\|}^{2}+2\operatorname{Re}\big{\langle}(A\otimes{\mathbbm{1}})\tilde{\Psi},({\mathbbm{1}}\otimes
A)M_{S(\cdot\,+\frac{\pi i}{6})}F\tilde{\Psi}\big{\rangle},$

where each of these terms can be understood as an integral over the above-
mentioned spectral space. If the first term is finite, it means that
$\tilde{\Psi}$ is in the domain of $A\otimes{\mathbbm{1}}$ and we are done. If
not, namely the first term is not finite, it must be positive infinite in the
sense of Lebesgue integral. Yet the whole expression is $L^{1}$ by the
defining property of $\Psi$ (that $(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)\Psi\in\operatorname{Dom}(\overline{P_{2}(A\otimes{\mathbbm{1}})})$) and the
second term is positive (actually in this case it must be positive infinite as
well), thus the last term must be negative infinite again in the sense of
Lebesgue integral: a mixed infinity is impossible, because then the whole
integral would have a measurable set on which it is positive infinite, which
contradicts the assumed $L^{1}$-property.

On the spectral space of $\log A$, we can write $A$ as the multiplication
operator by $e^{t}$. The crossing term in question
$\langle(A\otimes{\mathbbm{1}})\tilde{\Psi},({\mathbbm{1}}\otimes
A)M_{S(\cdot\,+\frac{\pi i}{6})}F\tilde{\Psi}\rangle$ is the following form:

$\displaystyle\int
dt_{1}\,dt_{2}\,e^{t_{1}+t_{2}}\tilde{\Phi}(t_{1},t_{2})\overline{\tilde{\Psi}(t_{1},t_{2})},$

where $\tilde{\Phi}=M_{S(\cdot\,+\frac{\pi i}{6})}F\tilde{\Psi}$.

When the real part of an integral is negative infinite, one can apply the
Fubini theorem. We insert the spectral projection $Q_{N}$ of $(A\otimes A)$
corresponding to $\big{(}e^{N-1},e^{N}\big{]}$, which amounts to consider a
strip $N-1<t_{1}+t_{2}\leq N$. This is equivalent to the change of variable
and integrating first with respect to $t_{2}-t_{1}$, which is legitimate by
Fubini’s theorem. On this strip, the integral is $L^{1}$, because
$\tilde{\Psi}$ and $\tilde{\Phi}$ are $L^{2}$.

Now recall that $\tilde{\Psi}=(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)\Psi$, therefore, the above integral consists of four contributions. Even
though the expression above is formal, note that $Q_{N}$ commutes with
$A\otimes{\mathbbm{1}}$ and ${\mathbbm{1}}\otimes A$, while $AB$, $BA$ are
bounded. Therefore, if we take two of the contributions

$\displaystyle\big{\langle}Q_{N}(A\otimes A)({\mathbbm{1}}\otimes
B)\Psi,M_{S(\cdot\,+\frac{\pi
i}{6})}F(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B)\Psi\big{\rangle}$
$\displaystyle\qquad{}=\big{\langle}Q_{N}({\mathbbm{1}}\otimes
AB)(A\otimes{\mathbbm{1}})\Psi,M_{S(\cdot\,+\frac{\pi
i}{6})}F(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B)\Psi\big{\rangle},$

where $(A\otimes{\mathbbm{1}})\Psi$ and
$(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B)\Psi$ are vectors in
${\mathcal{H}}_{1}\otimes{\mathcal{H}}_{1}$ and the operators appearing there
are all bounded, hence it is finite in the sense of Lebesgue integral. The
same argument applies to the term

$\displaystyle\big{\langle}Q_{N}(A\otimes
A)(A\otimes{\mathbbm{1}})\Psi,M_{S(\cdot\,+\frac{\pi
i}{6})}F({\mathbbm{1}}\otimes B)\Psi\big{\rangle}$
$\displaystyle\qquad{}=\big{\langle}(A\otimes{\mathbbm{1}})\Psi,M_{S(\cdot\,+\frac{\pi
i}{6})}FQ_{N}({\mathbbm{1}}\otimes
AB)(A\otimes{\mathbbm{1}})\Psi\big{\rangle},$

which implies that the only remaining term

$\displaystyle\big{\langle}Q_{N}(A\otimes
A)(A\otimes{\mathbbm{1}})\Psi,M_{S(\cdot\,+\frac{\pi
i}{6})}F(A\otimes{\mathbbm{1}})\Psi\big{\rangle}$

must have the real part which is negative infinite in the sense of Lebesgue
integral. However, we have computed this and it is positive under (S7). Hence
the whole integral is bounded below, even after removing $Q_{N}$. This
contradicts the assumption that the whole integral was negative infinite,
hence we obtain that each term of the formal expression is finite, namely,
$(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)\Psi\in\operatorname{Dom}(A\otimes{\mathbbm{1}})$.

Furthermore, we note that $A\otimes{\mathbbm{1}}$ and
$(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B)$ commute strongly. By
considering the joint spectral decomposition, it is immediate to see that

$\displaystyle\Psi\in\operatorname{Dom}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)\textrm{ and }(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)\Psi\in\operatorname{Dom}(A\otimes{\mathbbm{1}})$
$\displaystyle\qquad\Longleftrightarrow\
\Psi\in\operatorname{Dom}(A\otimes{\mathbbm{1}})\textrm{ and
}(A\otimes{\mathbbm{1}})\Psi\in\operatorname{Dom}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B).$

To summarize, we proved

$\displaystyle\overline{P_{2}(A\otimes{\mathbbm{1}})}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)P_{2}=P_{2}(A\otimes{\mathbbm{1}})(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)P_{2}$
$\displaystyle\hphantom{\overline{P_{2}(A\otimes{\mathbbm{1}})}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)P_{2}}{}=P_{2}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)(A\otimes{\mathbbm{1}})P_{2}.$

Similarly, we obtain $\overline{P_{2}({\mathbbm{1}}\otimes
B)}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)P_{2}=P_{2}({\mathbbm{1}}\otimes
B)(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)P_{2}=P_{2}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)({\mathbbm{1}}\otimes B)P_{2}$.

By the above consideration, we obtain the equality

$\displaystyle\overline{P_{2}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)P_{2}\subset\big{(}\overline{P_{2}(A\otimes{\mathbbm{1}})}+\overline{P_{2}({\mathbbm{1}}\otimes
B)}\big{)}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B)P_{2}$
$\displaystyle\hphantom{\overline{P_{2}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)P_{2}}{}=\left(P_{2}(A\otimes{\mathbbm{1}})+P_{2}({\mathbbm{1}}\otimes
B)\right)(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes B)P_{2}$
$\displaystyle\hphantom{\overline{P_{2}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)}(A\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B)P_{2}}{}=P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+2(A\otimes
B)+{\mathbbm{1}}\otimes B^{2}\big{)}P_{2},$

which is self-adjoint, because of the first expression.

We show that $P_{2}(A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2})P_{2}$ is essentially self-adjoint, using the commutator Theorem C.1, by
taking $P_{2}\left(A^{2}\otimes{\mathbbm{1}}+2(A\otimes
B)+{\mathbbm{1}}\otimes B^{2}+c{\mathbbm{1}}\right)P_{2}$ as the reference
operator with some $c>0$. Let us check the required estimates

$\displaystyle\big{\|}P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+2(A\otimes
B)+{\mathbbm{1}}\otimes B^{2}+c{\mathbbm{1}}\big{)}\Psi\big{\|}^{2}$
$\displaystyle\qquad{}\geq\big{\|}P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2}\big{)}\Psi\big{\|}^{2}+4\|P_{2}(A\otimes B)\Psi\|^{2}$
$\displaystyle\quad\qquad{}+4\operatorname{Re}\big{\langle}P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2}\big{)}\Psi,P_{2}(A\otimes B)\Psi\big{\rangle}$
$\displaystyle\quad\qquad{}+2c\big{\langle}\big{(}A^{2}\otimes{\mathbbm{1}}+2(A\otimes
B)+{\mathbbm{1}}\otimes B^{2}\big{)}\Psi,\Psi\big{\rangle}.$

In order to show $\|P_{2}(A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2})\Psi\|^{2}\leq c_{1}\|P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+2(A\otimes
B)+{\mathbbm{1}}\otimes B^{2}+c{\mathbbm{1}}\big{)}\Psi\|^{2}$, it is enough
to have a lower bound on the following term:

$\displaystyle
4\big{\langle}P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2}\big{)}\Psi,P_{2}(A\otimes B)\Psi\big{\rangle}$
$\displaystyle\qquad{}=2\big{\langle}\big{(}A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2}\big{)}\Psi,(A\otimes B)\Psi\big{\rangle}$
$\displaystyle\quad\qquad{}+2\big{\langle}\big{(}A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2}\big{)}\Psi,M_{S}(B\otimes{\mathbbm{1}})M_{S}^{*}D_{2}(\tau_{1})(A\otimes{\mathbbm{1}})\Psi\big{\rangle},$

where the first term is positive, while we have
$\big{\langle}\big{(}A^{2}\otimes{\mathbbm{1}}\big{)}\Psi,M_{S}(B\otimes{\mathbbm{1}})M_{S}^{*}D_{2}(\tau_{1})(A\otimes{\mathbbm{1}})\Psi\big{\rangle}=\langle
x(A\otimes{\mathbbm{1}})\Psi,(A\otimes{\mathbbm{1}})\Psi\rangle$, where $x$ is
a bounded operator. This is bounded by
$\|x\|\cdot\|(A\otimes{\mathbbm{1}})\Psi\|^{2}=\|x\|\cdot\langle(A^{2}\otimes{\mathbbm{1}})\Psi,\Psi\rangle$,
therefore, we take $c>2\|x\|$ and $c_{1}=1$. We can estimate analogously the
term containing ${\mathbbm{1}}\otimes B^{2}$.

A bound of the weak commutator

$\displaystyle\big{[}P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+2(A\otimes
B)+{\mathbbm{1}}\otimes
B^{2}\big{)}P_{2},P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2}\big{)}P_{2}\big{]}$

reduces to the weak commutator $[P_{2}(A\otimes
B)P_{2},P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2}\big{)}P_{2}]$, and its bound by

$\displaystyle\big{\|}\big{(}P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+2(A\otimes
B)+{\mathbbm{1}}\otimes
B^{2}+c{\mathbbm{1}}\big{)}P_{2}\big{)}^{\frac{1}{2}}\Psi\big{\|}^{2}$
$\displaystyle\qquad{}=\big{\langle}\big{(}A^{2}\otimes{\mathbbm{1}}+2(A\otimes
B)+{\mathbbm{1}}\otimes B^{2}+c{\mathbbm{1}}\big{)}\Psi,\Psi\big{\rangle}$

follows from the estimate above of the term
$\big{\langle}P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2}\big{)}\Psi,P_{2}(A\otimes B)\Psi\big{\rangle}$, up to the constant $c$
above, by noting that $A\otimes B$ commutes with $A^{2}\otimes{\mathbbm{1}}$
and ${\mathbbm{1}}\otimes B^{2}$. ∎

In Proposition 4.14 we show that
$\chi_{2}(\xi)+\chi^{\prime}_{2}(\eta)+4\sqrt{c(\xi,\eta)}$ is closed, hence
self-adjoint.

####  4.2.2 A shorter proof for $\boldsymbol{\chi_{2}(\xi)}$

We here present a proof of the essential self-adjointness of $\chi_{2}(\xi)$.
Although it is not of our direct interest as we need the self-adjointness of
$\chi_{2}(\xi)+\chi^{\prime}_{2}(\eta)$, we believe it clarifies the reason
why our method does not work for $n\geq 3$.

We observe that it is enough to show that
$\chi_{1}(\xi)\otimes{\mathbbm{1}}+M_{S}({\mathbbm{1}}\otimes\chi_{1}(\xi))M_{S}^{*}$
is self-adjoint, because this operator and
$P_{2}=\frac{1}{2}({\mathbbm{1}}+D_{2}(\tau_{1}))$ (strongly) commute. We
denote again $\chi_{1}(\xi)=A^{2}$.

######  Proposition 4.10.

$A^{2}\otimes{\mathbbm{1}}+M_{S}\big{(}{\mathbbm{1}}\otimes
A^{2}\big{)}M_{S}^{*}$ is essentially self-adjoint.

###### Proof.

We consider the operator $A\otimes A^{-1}+M_{S}\big{(}A^{-1}\otimes
A\big{)}M_{S}^{*}$. Let $X=A^{\frac{1}{2}}\otimes A^{-\frac{1}{2}}$,
$Y=A^{-\frac{1}{2}}\otimes A^{\frac{1}{2}}M_{S}^{*}$. Then
$Y^{*}=M_{S}\big{(}A^{-\frac{1}{2}}\otimes A^{\frac{1}{2}}\big{)}$ and
$\operatorname{Dom}(Y^{*})=\operatorname{Dom}\big{(}A^{-\frac{1}{2}}\otimes
A^{\frac{1}{2}}\big{)}$. One can easily check the assumptions of Lemma 4.4: as
for the decomposition of the Hilbert space, we change the variables
$t_{+}:=\theta_{2}+\theta_{1}$, $t_{-}:=\theta_{2}-\theta_{1}$. Then the
problem reduces to one variable $t_{-}$ and one can take the same
decomposition as Proposition 4.5, by noting that $S$ is bounded in
${\mathbb{R}}+i\big{(}0,\frac{\pi}{6}\big{)}$.

Finally, both $A\otimes A^{-1}$ and $M_{S}\big{(}A^{-1}\otimes
A\big{)}M_{S}^{*}$ commute with $A\otimes A$ strongly. Therefore, the bounded
operator $Q_{N}(A\otimes A)$ commute with both of them, where $Q_{N}$ is the
spectral projection of $A\otimes A$ onto $\big{(}e^{N-1},e^{N}\big{]}$ and the
product is

$\displaystyle Q_{N}(A\otimes A)\big{(}A\otimes
A^{-1}+M_{S}\big{(}A^{-1}\otimes
A\big{)}M_{S}^{*}\big{)}=Q_{N}\big{(}A^{2}\otimes{\mathbbm{1}}+M_{S}\big{(}{\mathbbm{1}}\otimes
A^{2}\big{)}M_{S}^{*}\big{)},$

which is still self-adjoint. Our operator $A\otimes
A^{-1}+M_{S}\big{(}A^{-1}\otimes A\big{)}M_{S}^{*}$ is an extension of the
direct sum of these components, therefore, it is essentially self-adjoint. ∎

It appears difficult to apply the same idea to
$P_{2}\big{(}A^{2}\otimes{\mathbbm{1}}+{\mathbbm{1}}\otimes
B^{2}\big{)}P_{2}$, as $A$ and $B$ do not commute.

If $n\geq 3$, $A\otimes A\otimes A$ cannot reduce the problem to one variable.
We know no other convenient operator.

####  4.2.3 Konrady’s trick

We give a less simpler proof which however gives an insight of why the problem
is complicated, cf. Section 4.3. For it, we need an additional assumption on
$S$.

$\displaystyle\operatorname{Re}S\left(\theta+\frac{\pi
i}{6}\right)S(-\theta)\geq 0.$ (4.2)

We indicate some supporting evidence in Appendix A.3 that there should exist
examples of $S$ which satisfy this condition. We do not pursue this condition
further, as our main result (Proposition 4.9) has been proved without it.

The general idea used here is called Konrady’s trick [17], [26, Section X.2].

######  Lemma 4.11.

Let $X$ and $Y$ be symmetric operators on $\operatorname{Dom}(X)$ and
$\operatorname{Dom}(Y)$, respectively. Assume that $X+Y$ is self-adjoint on
$\operatorname{Dom}(X+Y)=\operatorname{Dom}(X)\cap\operatorname{Dom}(Y)$ and
it holds that $\operatorname{Re}\langle X\Psi,Y\Psi\rangle\geq 0$ for
$\Psi\in\operatorname{Dom}(X+Y)$. Then $X$ and $Y$ are essentially self-
adjoint on $\operatorname{Dom}(X+Y)$.

###### Proof.

Compute:

$\displaystyle\|(X+Y)\Psi\|^{2}=\|X\Psi\|^{2}+\|Y\Psi\|^{2}+2\operatorname{Re}\langle
X\Psi,Y\Psi\rangle\geq\max\big{\\{}\|X\Psi\|^{2},\|Y\Psi\|^{2}\big{\\}},$

now one can use Wüst’s theorem [26, Theorem X.14] to see that $X=(X+Y)-Y$ and
$Y=(X+Y)-X$ are essentially self-adjoint on $\operatorname{Dom}(X+Y)$. ∎

######  Proposition 4.12.

For $S$ satisfying (4.2),
${\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{6}}}\big{)}M_{S}^{*}$
is essentially self-adjoint.

###### Proof.

First,
${\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}+c\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}$
is self-adjoint by the spectral theorem for a constant $c>0$. We choose
$c>|S(\theta+\frac{\pi i}{6})|$, $\theta\in{\mathbb{R}}$.

Then, we claim that
${\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}+c\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}$
is self-adjoint. Indeed, for
$\Psi\in\operatorname{Dom}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}$,
we have
$\big{\|}M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\Psi\big{\|}=\big{\|}M_{S(\cdot\,+\frac{\pi
i}{6})}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\Psi\big{\|}$
which is bounded by
$c\big{\|}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\Psi\big{\|}$.
Therefore, one can apply Kato–Rellich theorem [26, Theorem X.12].

Next, we show that
${\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}$
is essentially self-adjoint on the domain
$\operatorname{Dom}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}+c\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\big{)}$.
In order to apply Lemma 4.11, we only have to check that
$\big{\langle}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\Psi,\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\Psi\big{\rangle}\geq
0$ (trivial) and
$\big{\langle}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\Psi,M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\Psi\big{\rangle}\geq
0$. As
$\Psi\in\operatorname{Dom}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}$,
it holds that
$M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\Psi=M_{S}M_{\overline{S(\cdot\,+\frac{\pi
i}{6})}}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\Psi$.
The inequality above follows from (4.2) and hermitian analyticity (S2).
Actually, by the same assumption (4.2), it follows from Lemma 4.1 and the
argument of Proposition 4.7,
${\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}$
is closed on its natural domain, hence it is self-adjoint there.

Note that
${\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}$
commutes strongly with $\Delta_{1}\otimes\Delta_{1}$. Consider its spectral
projection $Q_{[0,N]}$ corresponding to the interval $[0,N]$.

As $S(\theta_{2}-\theta_{1}+i\lambda)$ is uniformly bounded for
$\lambda\in[0,\frac{\pi}{6}]$, we have
$M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\subset\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S(\cdot\,+\frac{\pi
i}{6})}$ and

$\displaystyle\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}\cdot
M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}Q_{[0,N]}\subset\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}\cdot
M_{S(\cdot\,+\frac{\pi i}{6})}M_{S}^{*}Q_{[0,N]}$

is bounded. Similarly,

$\displaystyle
M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\cdot\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}\big{)}Q_{[0,N]}\subset
M_{S}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S(\cdot\,+\frac{\pi
i}{6})}^{*}Q_{[0,N]}.$

Now
$\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*})^{2}Q_{[0,N]}$
is self-adjoint. From the boundedness above, we can expand the square:

$\displaystyle\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\mathbbm{1}}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S}^{*}\big{)}^{2}Q_{[0,N]}$
$\displaystyle\qquad{}\subset\Big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}+\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S(\cdot\,+\frac{\pi
i}{6})}M_{S}^{*}$
$\displaystyle\qquad\hphantom{\subset\Big{(}}{}+M_{S}\big{(}{\Delta_{1}^{\frac{1}{12}}}\otimes{\Delta_{1}^{\frac{1}{12}}}\big{)}M_{S(\cdot\,+\frac{\pi
i}{6})}^{*}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{6}}}\big{)}M_{S}^{*}\Big{)}Q_{[0,N]}.$

Here, the left-hand side is self-adjoint, and the right-hand side is
symmetric, hence it must be an equality and the right-hand side is self-
adjoint. The second and the third terms are bounded, therefore, they have no
effect on domains and
$\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}+M_{S}\big{(}{\mathbbm{1}}\otimes{\Delta_{1}^{\frac{1}{6}}}\big{)}M_{S}^{*}\big{)}Q_{[0,N]}$
is self-adjoint. As $N$ is arbitrary, we obtain the claim. ∎

###  4.3 Many-particle components

######  Proposition 4.13.

Under (S7), $\chi_{n}(\xi)$ is closed.

###### Proof.

We show the closedness of $\sum
D_{n}(\rho_{k})\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}D_{n}(\rho_{k})^{*}$
on $P_{n}{\mathcal{H}}_{n}$.

To prove it by induction, we may assume that $\chi_{n-1}(\xi)$, hence
$\chi_{n-1}(\xi)\otimes{\mathbbm{1}}$ is closed. To apply Lemma 4.1, we only
have to check that the following is positive:

$\displaystyle\operatorname{Re}\langle
D_{n}(\rho_{k})\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}D_{n}(\rho_{k})^{*}\Psi_{n},D_{n}(\rho_{n})\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}D_{n}(\rho_{n})^{*}\Psi_{n}\rangle$
$\displaystyle\qquad{}=\operatorname{Re}\langle
D_{n}(\rho_{k})\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}\Psi_{n},D_{n}(\rho_{n})\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}\Psi_{n}\rangle$
$\displaystyle\qquad{}=\operatorname{Re}\langle({\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi_{n},D_{n}(\tau_{1,n})({\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})\Psi_{n}\rangle$
$\displaystyle\qquad{}=\operatorname{Re}\langle\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}\Psi_{n},D_{n}(\tau_{1,2})\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}\Psi_{n}\rangle,$

where $\tau_{k,l}$ is the transposition between $k$ and $l$, we used the fact
that $\rho_{k}$ and $\rho_{n}$ may be replaced by $\tau_{1,k}$, $\tau_{1,n}$,
respectively (see [6, proof of Theorem 3.4]) and we can further rewrite
$\tau_{1,2}=\tau_{2,3}\cdots\tau_{n-1,n}\tau_{1,n}\tau_{n-1,n}\cdots\tau_{2,3}$,
and
$D_{n}(\tau_{k,k+1})\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}\Psi_{n}=\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\big{)}\Psi_{n}$
for $k\geq 2$. From here, we can use Lemma E.6 and the same computation as at
the end of Proposition 4.7 under (S7) to see that the real part of the above
inner product is positive. ∎

######  Proposition 4.14.

Under (S7), $\chi_{n}(\xi)+\chi^{\prime}_{n}(\eta)$ is closed.

###### Proof.

This can be done by the estimate of Proposition 3.3

$\displaystyle\operatorname{Re}\langle\chi_{n}(\xi)\Psi,\chi^{\prime}_{n}(\eta)\Psi\rangle\geq-
nc(\xi,\eta)\langle(\chi(\xi)+\chi^{\prime}(\eta)+2)\Psi,\Psi\rangle$

and by Lemma 4.3 applied to $X=\chi_{n}(\xi)$, $Y=\chi^{\prime}_{n}(\eta)$
(here $X$ and $Y$ are positive, hence we can take the constant $c$ in Lemma
4.3 larger if necessary). ∎

######  Proposition 4.15.

We assume (S9). For a fixed $n$, there is $\epsilon$ such that
$P_{n}(\Delta_{1}^{\epsilon}\otimes{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}})P_{n}$
is self-adjoint.

###### Proof.

We follow the idea of Proposition 4.12. We set
$S_{k}(\boldsymbol{\theta}):=\prod\limits_{j<k}S(\theta_{k}-\theta_{j})$ as
before. There is $\epsilon>0$ such that

$\displaystyle\operatorname{Re}\overline{S_{k}(\theta_{1}+2\pi\epsilon
i,\dots,\theta_{k}+2\pi\epsilon i)}S_{k}(\theta_{1},\dots,\theta_{k})\geq 0,$

because $S_{k}$ is uniformly continuous in a small neighborhood of
${\mathbb{R}}^{n}$ as we assume (S9) and
$\overline{S_{k}(\boldsymbol{\theta})}S_{k}(\boldsymbol{\theta})=1$.

First consider
$\Delta_{1}^{\epsilon}\otimes\cdots\otimes{\mathbbm{1}}+\cdots+c{\mathbbm{1}}\otimes\cdots\otimes\Delta_{1}^{\epsilon}\otimes\cdots\otimes{\mathbbm{1}}+\cdots+c{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\Delta_{1}^{\epsilon}$,
which is self-adjoint, where $c\geq 0$.

Next,

$\displaystyle\Delta_{1}^{\epsilon}\otimes\cdots\otimes{\mathbbm{1}}+\cdots+c{\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\Delta_{1}^{\epsilon}+\sum_{k}M_{S_{k}}({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\underset{k\mathrm{-th}}{\Delta_{1}^{\epsilon}}\otimes{\mathbbm{1}}\cdots\otimes{\mathbbm{1}})M_{S_{k}}^{*}$

is self-adjoint if $c$ is sufficiently large by Kato–Rellich perturbation, as
in Proposition 4.12.

Now in order to prove that
$\sum_{k}M_{S_{k}}({\mathbbm{1}}\otimes\cdots\otimes{\mathbbm{1}}\otimes\underset{k\mathrm{-th}}{\Delta_{1}^{\epsilon}}\otimes{\mathbbm{1}}\cdots\otimes{\mathbbm{1}})M_{S_{k}}^{*}$
is essentially self-adjoint and to apply Lemma 4.11, it is enough to check
that the real part of the following is positive:

$\displaystyle\Big{\langle}({\mathbbm{1}}\otimes\cdots\otimes\underset{j\mathrm{-th}}{\Delta_{1}^{\epsilon}}\otimes\cdots\otimes{\mathbbm{1}})\Psi,M_{S_{k}}({\mathbbm{1}}\otimes\cdots\otimes\underset{k\mathrm{-th}}{\Delta_{1}^{\epsilon}}\otimes\cdots\otimes{\mathbbm{1}})M_{S_{k}}^{*}\Psi\Big{\rangle}.$

If $j>k$, the operators in the product commute and it is positive. If $j=k$,
it reduces to

$\displaystyle\Big{\langle}({\mathbbm{1}}\otimes\cdots\otimes\underset{k\mathrm{-th}}{\Delta_{1}^{\epsilon}}\otimes\cdots\otimes{\mathbbm{1}})\Psi,M_{S_{k}\overline{S_{k}(\cdot\,+2\pi\epsilon
i)}}({\mathbbm{1}}\otimes\cdots\otimes\underset{k\mathrm{-th}}{\Delta_{1}^{\epsilon}}\otimes\cdots\otimes{\mathbbm{1}})\Psi\Big{\rangle},$

and by the choice of $\epsilon$, this is positive. If $j<k$, we get

$\displaystyle\Big{\langle}({\mathbbm{1}}\otimes\cdots\otimes\underset{j\mathrm{-th}}{\Delta_{1}^{\epsilon}}\otimes\cdots\otimes{\mathbbm{1}})M_{\check{S}_{k,j}}^{*}\Psi,\;M_{S_{k,j}\overline{S_{k,j}(\cdot\,+2\pi\epsilon
i)}}({\mathbbm{1}}\otimes\cdots\otimes\underset{k\mathrm{-th}}{\Delta_{1}^{\epsilon}}\otimes\cdots\otimes{\mathbbm{1}})M_{\check{S}_{k,j}}^{*}\Psi\Big{\rangle}$
$\displaystyle\qquad{}=\Big{\langle}\Delta_{1,k,j}^{\frac{\epsilon}{2}}M_{\check{S}_{k,j}}^{*}\Psi,\;M_{S_{k,j}(\cdot\,+2\pi\epsilon
i)\overline{S_{k,j}(\,\cdot)}}\Delta_{1,k,j}^{\frac{\epsilon}{2}}M_{\check{S}_{k,j}}^{*}\Psi\Big{\rangle},$

where
$\Delta_{1,k,j}={\mathbbm{1}}\otimes\cdots\otimes\underset{j\mathrm{-th}}{\Delta_{1}}\otimes\cdots\otimes\underset{k\mathrm{-th}}{\Delta_{1}}\cdots\otimes{\mathbbm{1}}$,
which is positive again by the choice of $\epsilon$, where we introduced
$S_{k,j}(\underline{\theta})=S(\theta_{k}-\theta_{j})$ and
$\check{S}_{k,j}(\underline{\theta})=S_{k}(\underline{\theta})S(\theta_{k}-\theta_{j})^{-1}$.

The closedness for sufficiently small $\epsilon$ follows by arguing as in
Proposition 4.13. ∎

The method of this proof is invalid for $\epsilon=\frac{1}{6}$, or any fixed
$\epsilon$. Indeed, as $k$ increases,
$\operatorname{Re}S_{k}(\boldsymbol{\theta}+\frac{\pi}{6})S_{k}(-\boldsymbol{\theta})$
may take negative values and the Konrady’s trick fails.

In order to have the right domain of self-adjointness of
$P_{n}\big{(}{\Delta_{1}^{\frac{1}{6}}}\otimes{\mathbbm{1}}\cdots\otimes{\mathbbm{1}}\big{)}P_{n}$,
we may have to correctly incorporate the zeros and poles of $S$ to the domain
(cf. [35]).

##  5 Outlook

In order to complete the construction of Borchers triples, we have to prove
that $\chi_{n}(\xi)+\chi^{\prime}_{n}(\eta)$ is (essentially) self-adjoint. We
also hope to clarify the role played by the assumption (S7), and to drop it if
possible. With the same assumption, one should be able to prove the
Bisognano–Wichmann property and modular nuclearity [4, 19] for an inclusion
with a minimal distance [1], hence it should be possible to construct
Haag–Kastler nets with minimal size, which will be published elsewhere [8].

On the other hand, if the strong commutativity of our candidates fails, then
the construction of the Haag–Kastler nets for a given $S$-matrix with poles
will be really a hard problem: if the net exists at all, the polarization-free
generators are canonically constructed [3], while we checked that our
${\widetilde{\phi}}(f)$’s are formally compatible with the form factor program
[6, Section 4.1]. If our candidates are not the right polarization-free
generators, the formal computations must fail, which means that the right
polarization-free generators should have even subtler domains. A related
problem is whether the $S$-matrix is a complete invariant for asymptotically
complete nets. This is in general open even for the simplest case where $S=1$,
if the temperateness of the polarization-free generators is not assumed [24].

##  Appendix A Properties of the $\boldsymbol{S}$-matrix

Here we show the existence of two-particle $S$-matrix which satisfies some
additional properties. Recall that the most general $S$-matrix satisfying
(S1)–(S6) and (S9) takes the form555In [6, Appendix A], we used a parameter
$B=\frac{1}{2}+\frac{3\varepsilon}{\pi i}$, $0<B<1$, and had the equivalent
expression $S(\theta)=\frac{\tanh\frac{1}{2}\left(\theta+\frac{2\pi
i}{3}\right)}{\tanh\frac{1}{2}\left(\theta-\frac{2\pi
i}{3}\right)}\cdot\frac{\tanh\frac{1}{2}\left(\theta+\frac{(B-2)\pi
i}{3}\right)}{\tanh\frac{1}{2}\left(\theta-\frac{(B-2)\pi
i}{3}\right)}\frac{\tanh\frac{1}{2}\left(\theta-\frac{B\pi
i}{3}\right)}{\tanh\frac{1}{2}\left(\theta+\frac{B\pi i}{3}\right)}\cdot
S_{\mathrm{Blaschke}}(\theta)$.

$\displaystyle S(\theta)=\frac{\tanh\frac{1}{2}\left(\theta+\frac{2\pi
i}{3}\right)}{\tanh\frac{1}{2}\left(\theta-\frac{2\pi
i}{3}\right)}\cdot\frac{\tanh\frac{1}{2}\left(\theta-\frac{\pi
i}{2}-i\varepsilon\right)}{\tanh\frac{1}{2}\left(\theta-\frac{\pi