TITLE: Calculating $\int_{- \infty}^{\infty} \frac{\sin x \,dx}{x+i} $
QUESTION [1 upvotes]: I'm having trouble calculating the integral
$$\int_{- \infty}^\infty \frac{\sin x}{x+i}\,dx $$
using residue calculus. I've previously encountered expressions of the form
$$\int_{- \infty}^\infty f(x) \sin x \,dx $$
where you would consider $f(z)e^{iz}$ on an appropriate contour (half circle), do away with the part of the contour that wasn't on the real axis by letting the radius go to infinity, then recover the imaginary part of the answer to get back the sine. However here, I can't replace the sine with $e^{iz}$ in my complex function because $$\operatorname{Im} \frac{e^{ix}}{x+i} \neq \frac{\sin x}{x+i}\cdots$$
How to remedy this? I'm not sure if substituting $\sin x = \frac{1}{2i}(e^{ix}-e^{-ix})$ and solving two integrals is how this problem is meant to be solved, although I'm 99% sure it would work

REPLY [2 votes]: Using the identity $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$ and separating into two integrals is a very reasonable way to solve this problem, and I would imagine it's the easiest way. For comparison, I want to describe another possibility, which involves separating $\frac{1}{x+i}$ into its real and imaginary parts.
We have
$$
\frac{1}{x+i}=\frac{x-i}{x^2+1}=\frac{x}{x^2+1}-i\frac{1}{x^2+1},
$$
so we can break up our integral as a sum in the following way:
\begin{eqnarray*}
\int_{-\infty}^{\infty}\frac{\sin(x)}{x+i}\,dx&=&\int_{-\infty}^\infty\frac{x\sin(x)}{x^2+1}\,dx-i\int_{-\infty}^\infty\frac{\sin(x)}{x^2+1}\,dx\\
&=&\mathrm{Im}\int_{-\infty}^\infty\frac{xe^{ix}}{x^2+1}\,dx-i\cdot\mathrm{Im}\int_{-\infty}^\infty \frac{e^{ix}}{x^2+1}\,dx.
\end{eqnarray*}
Both of these integrals can be evaluated using the method of contours.
I don't necessarily think this is easier than writing $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$.


text:
\begin{document}
\title{Borchers' Commutation Relations for Sectors with Braid Group
  Statistics in Low Dimensions}
\author{Jens Mund\thanks{Supported by FAPEMIG.}
\\
\scriptsize
Departamento de F\'{\i}sica, Universidade Federal de Juiz de Fora,\\

\scriptsize
36036-900 Juiz de Fora, MG, Brazil.\\

\scriptsize  E-mail: {\tt mund@fisica.ufjf.br}}

\date{\version
}
\maketitle
\begin{abstract}

Borchers has shown that in a translation covariant vacuum
representation of a theory of local observables with positive energy
the following holds: The (Tomita) modular
objects associated with the observable algebra of a fixed wedge
region give rise to a representation of the subgroup of the Poincar\'e
group generated by the boosts and the reflection associated to the
wedge, and the translations.

We prove here that Borchers' theorem also holds in charged sectors with
(possibly non-Abelian) braid group statistics in
low space-time dimensions. Our result is a crucial step towards
the Bisognano-Wichmann theorem for Plektons in $d=3$,
namely that the mentioned modular objects
generate a representation of the proper Poincar\'e group,
including a \CPT operator.
Our main assumptions are Haag duality of the observable algebra, and
translation covariance with positive energy as well as finite
statistics of the sector under consideration.
\end{abstract}

\section*{Introduction} \label{secIntro}

Borchers has shown~\cite{Borchers92} that in a theory of local
observables, which is translation covariant with positive energy,
the modular objects associated with the observable algebra of a
(Rindler) wedge region and the vacuum state have certain specific
commutation relations with the representers of the translations.
Namely, these commutation relations manifest that
the corresponding unitary modular group implements the
group of boosts which leave the wedge invariant, and that the corresponding
modular conjugation implements the reflection about the edge of the wedge.
Borchers' theorem has profound consequences. For example in two-dimensional
theories it means that the modular objects generate a representation
of the proper Poincar\'e group, under which the observables behave covariant,
and implies the \CPT theorem.
In higher dimensions, it is a crucial step towards the Bisognano-Wichmann
theorem in the general context of
local quantum physics~\cite{Borchers98,BY00,BGL93,GL00,Kuck00,SHW98,M01a}.
This theorem asserts that a certain class of Poincar\'e covariant
theories enjoys the property of modular covariance, namely that the
mentioned unitary modular group coincides with the
representers of the
boosts, and that the modular conjugation is a \CPT operator (where
`PT' means  the reflection about the edge of the wedge).

The hypothesis under which Borchers' theorem works is the double role
played by the vacuum vector within a theory of local algebras: The
vacuum is cyclic and separating for the local algebras, and it is
invariant under the positive energy representation
of the translation group under which these algebras are covariant.
In a charged sector, i.e., a non-vacuum representation of the
observables, this situation is not given. (This problem has been posed
by Borchers
in~\cite[Sect.\ VII.4]{Borchers2000}.) In the case of permutation group
statistics, one can use the field algebra instead of the observable
algebra to recover the result.
However, in low-dimensional space-time there may occur
superselection sectors with braid group statistics~\cite{F89,FM1}.
Then only in the Abelian case there is a field (C$^*$) algebra
for which the vacuum is cyclic and separating. In the case of
non-Abelian braid group statistics, there is no such field algebra.
Due to this complication, a general result corresponding to Borchers'
theorem has not been achieved yet.
In the present article, we prove an analogue of Borchers'
theorem for a superselection sector corresponding to a localizable
charge.
The implementers of the boosts and the reflection which we find are
the relative modular objects associated with the observable algebra of
the wedge, the vacuum state and some specific state in the conjugate sector.

We assume that the observable algebra satisfies Haag duality, see
Eq.~\eqref{eqHD}, and that the sector under consideration has
finite statistics and positive energy, and is irreducible.
We also need a slightly stronger irreducibility
property~\eqref{eqRhoIrred}, which may be
ensured by requiring for example Lorentz covariance or the split
property.

We consider charges which are localizable in space-like
cones\footnote{A space-like cone is, in $d\geq 3$,
a convex cone in Minkowski space generated by a double cone and a point
  in its causal complement, and in $d=2$ the causal completion
  thereof, which is a wedge region.},
and admit the case of non-Abelian braid group statistics which can
occur in low space-time dimensions, $d=2$ and $3$.

It must be noted that in two dimensions, our result is already
practically covered by the work of Guido and
Longo~\cite{GL92}. Namely, they show how a certain condition of
modular covariance in the vacuum sector allows, under the
same hypothesis as in the present article, for the construction of a
(ray) representation of the proper Poincar\'e group in charged
sectors. But in two dimensions, their modular covariance condition is
satisfied due to Borchers' theorem (in the vacuum sector), so their
analysis goes through, even in sectors with Braid group statistics.
However, it must be noted that in $d=2$ the assumption of Haag duality
excludes some massive models with braid group statistics as e.g.\ the
anyonic sectors of the CAR algebra~\cite{Adler}, and together with the
split property for wedges (expected to hold in massive models) excludes
localizable charges altogether~\cite{Mueger98}.

Our result shall be used to derive the \CPT and Bisognano-Wichmann theorems for
particles with braid group statistics in three-dimensional
space-time~\cite{Mu_BiWiAny}.
It would be gratifying to extend our analysis to soliton sectors in
2 dimensions, which would extend the range (and simplify the proof) of
Rehrens' \CPT theorem for solitons~\cite{Re98}.

\section{General Setting, Assumptions and Results} \label{Ass}
We consider a theory of local observables, given by a family of
von Neumann algebras $\AO(\calO)$ of operators acting in the
vacuum Hilbert space $\HO$, indexed by the double cones
$\calO$ in Minkowski space, and satisfying the conditions of isotony
and locality:
\begin{align*}
\AO(\calO_1) \subset \AO(\calO_2)\;\text{ if }\; \calO_1\subset
\calO_2\quad \text{ and } \quad
\AO(\calO_1) \subset \AO(\calO_2)' \;\text{ if }\; \calO_1\subset
\calO_2' ,
\end{align*}
where the prime denotes the commutant or the causal complement, respectively.
The vacuum Hilbert space $\calH_0$ carries a unitary
representation $\UO$ of the group of space-time translations $\RR^d$ with
positive energy, i.e.\ its spectrum\footnote{By spectrum of a
  representation of the translation group we
mean the energy-momentum spectrum, namely the joint
spectrum of the generators.} lies in the forward light cone.
It has a unique, up to a phase, invariant vector $\Omega\in\HO$,
corresponding to the vacuum state.
The representation $\UO$ implements
automorphisms
under which the net $\calO\to \AO(\calO)$ is covariant:
\begin{align}

\Ad \UO(x)\,\AO(\calO) &= \AO(x+\calO) \label{eqCovObs}
\end{align}
for all $ x\in\RR^d$. (By $\Ad U$ we denote the adjoint action of a
unitary $U$.)

Borchers' theorem, which we wish to generalize to charged sectors,
asserts that the representation $\UO$ has specific commutation
relations with certain algebraic objects, the so-called modular
group and conjugation, which suggest a geometric interpretation
of the latter.
Let us recall Borchers' commutation relations in this setting.
Let $W_1$ be the wedge defined as
\begin{align} \label{eqWe}
W_1 & := \{\,x\in\RR^d:\,|x^0|<x^1\;\}\,.
\end{align}
By the Reeh-Schlieder property, $\Omega$ is cyclic and
separating for the von Neumann algebra $\AO(W_1)$ generated by all
$\AO(\calO)$, $\calO\subset W_1$.
This allows for the definition of the Tomita operator~\cite{BraRob}, $\SO$,
associated to $\AO(W_1)$:
It is the  closed anti-linear involution satisfying
\begin{align}  \label{eqSTomO}
\SO\,A\Omega&=A^*\Omega, \quad A\in\AO(W_1)\,.
\end{align}

Its polar decomposition,
$\SO = \JO\,\DO^{1/2}$,
defines an  anti-unitary involution $\JO$, the so-called {modular conjugation},
and a positive operator $\DO$ giving rise to the so-called
{modular unitary group} $\DO^{it}$ associated to the wedge $W_1$.
By Tomita's Theorem, see e.g.~\cite{BraRob},
the adjoint action of $\DO^{it}$ leaves $\AO(W_1)$ invariant,
and the adjoint action of $\JO$ maps $\AO(W_1)$
onto its commutant $\AO(W_1)'$.

The mentioned theorem of Borchers now asserts that $\DO^{it}$ and
$\JO$, together with the representation $\UO$ of the translations,
induce a representation of the subgroup of
$\Poj$ generated by the boosts $\Boox{t}$ and the
reflection $j$ associated to the wedge, and the translations. More precisely,
let $\Boox{t}$ be the (rescaled) 1-boosts, leaving $W_1$ invariant and
acting on the coordinates $x^0,x^1$ as
\begin{equation} \label{eqBoox}
\left( \begin{array}{cc}
 \cosh(2\pi t) &  \sinh(-2\pi t)  \\
 \sinh(-2\pi t) &  \cosh(2\pi t)
 \end{array} \right),
\end{equation}
and let $j$ be the reflection about the edge of $W_1$,
acting on the coordinates $x^0,x^1$ as $-\unity$ and leaving the other
coordinates unchanged (if $d>2$).
Then Borchers' theorem asserts that
\begin{align}
\DO^{it}\,\UO(x)\,\DO^{-it}&= \UO(\Boox{t}x), \label{eqDUD0}\\
\JO\,\UO(x)\,\JO &= \UO(jx) \label{eqJUJ0}
\end{align}
for all $t\in\RR$ and $x\in\RR^d$.
These relations implement the group relations
of the translations with the boosts and reflections, respectively.
Modular theory further implies that $\JO$ is an involution and
commutes with the modular unitary group, implementing the
group relations $\J^2=1$ and $\J\Boox{t}\J^{-1}=\Boox{t}$.
Altogether, $U_0(x)$, $\DO^{it}$ and $\JO$ constitute a
representation of the subgroup of the Poincar\'e group generated
by the translations, the boosts $\Boox{t}$ and the reflection $\J$
(which is the direct product of the proper Poincar\'e group in the
time-like $x^0$-$x^1$ plane and the translation group in the
remaining $d-2$ dimensions).

Our aim is to find a similar result in a charged sector,
that is in a representation of the abstract $C^*$-algebra generated by
the local algebras $\AO(\calO)$, which is inequivalent from the defining vacuum
representation.

We shall consider an irreducible representation $\pi$, which is
localizable in space-like cones. That means that $\pi$ and
the vacuum representation are unitarily equivalent in restriction to
the  observable algebra associated with the causal complement of any space-like
cone.\footnote{It is known that every purely massive
representation is localizable in space-like cones~\cite{BuF}.}

We assume that the observable algebra satisfies Haag duality for
space-like cones and wedges, i.e., regions which arise by a proper
Poincar\'e transformation from $W_1$. Namely, denoting by $\Spc$ the class
of space-like cones, their causal complements, and wedges, we require
\begin{equation} \label{eqHD}
\AO(\spc')= \AO(\spc)',\quad \spc \in\Spc.
\end{equation}

A localizable representation can then be described by an endomorphism of
the so-called universal algebra $\Auni$ generated by isomorphic
images $\Auni(\spc)$ of the $\AO(\spc)$, $\spc\in\Spc$,
see~\cite{F90,GL92,FRSII}.
The family of isomorphisms $\Auni(\spc)\cong \AO(\spc)$ extends to a
representation $\pi_0$ of $\Auni$, the vacuum representation. We then have
\begin{equation} \label{eqAO}
\AO(\spc)= \pi_0\Auni(\spc),
\end{equation}
and the vacuum representation is faithful\footnote{However, $\pi_0$
  is in general not faithful on the global algebra $\Auni$
due to the  existence of global intertwiners~\cite{FRSII}.}
 and normal on the local\footnote{We call the algebras
  $\Auni(\spc)$ ``local'' although the regions $\spc$ extend to
  infinity in some direction, just in distinction
  from the ``global'' algebra $\Auni$.} algebras
$\Auni(\spc)$.
The adjoint action~\eqref{eqCovObs} of the translations on the local algebras
lifts to a representation by automorphisms $\alphaO_x$:
\begin{align} \label{eqU0impl}
\Ad \UO(x) \circ \pi_0 &= \pi_0\circ \alphaO_x,\\
\alphaO_x\,  \Auni(\spc)&=   \Auni(x+\spc). \label{eqCovObs'}
\end{align}
Our localizable representation $\pi$ is then
equivalent~\cite{DHRIII,F90} with a
representation of the form $\pi_0\circ \rho$ acting in $\HO$,
where $\rho$ is an endomorphism of $\Auni$ {localized} in some
specific space-time region $\spc_0\in\Spc$ in the sense that
\begin{equation} \label{eqRhoInC0}
\rho(A)=A\qquad \text{ if } \; A \in\Auni(C_0').
\end{equation}
We shall take the localization region of $\rho$ to be properly contained in
$W_1$, which implies by Haag duality~\eqref{eqHD} that $\rho$
restricts to an endomorphism of $\Auni(W_1)$.

We shall require that this endomorphism of $\calA(W_1)$ be
irreducible, namely that
\begin{equation} \label{eqRhoIrred}
\pi_0 \calA(W_1)\cap \big(\pi_0\rho\calA(W_1)\big)' = \CC \unity.
\end{equation}
This is a slightly stronger requirement than irreducibility of the
representation $\pi_0\rho$ of $\Auni$.
It has been shown by Guido and Longo that irreducibility of
$\pi_0\rho$, together with finite statistics, imply irreducibility in
the sense of Eq.~\eqref{eqRhoIrred} if $\rho$ is covariant under the
(proper orthochronous) Poincar\'e
group~\cite[Cor.~2.10]{GL96}~\footnote{Although not explicitly mentioned
in~\cite{GL96}, the proof does not depend on covariance of $\rho$
under the full Moebius group. See also~\cite[Thm.~2.2]{Longo96}.}
or if $\rho$ satisfies the split
property~~\cite[Prop.~6.3]{GL92}.

We further assume the representation $\pi\cong \pi_0\rho$ to be
translation covariant with positive energy. That means that there is a
unitary representation
$U_\rho$ of the translation group $\RR^d$ with spectrum
contained in the forward light cone such that
\begin{equation} \label{eqCovRho}
\Ad U_\rho(x) \circ \pi_0\rho =
\pi_0\rho \circ \alphaO_x, \quad x\in\RR^d.
\end{equation}

We finally assume that $\rho$ has finite statistics, i.e.\
that the so-called

statistics parameter $\lambda_\rho$~\cite{DHRIII} be non-zero.
This holds automatically if $\rho$ is massive~\cite{F81}, and
implies~\cite{DHRIV} the existence of a
{\em conjugate} morphism $\bar\rho$
characterized, up to equivalence, by the fact that the composite
sector $\pi_0\bar\rho\rho$
contains the vacuum representation $\pi_0$ precisely once.
Thus there is a unique, up to a factor, intertwiner $R_\rho\in\Auni(\spc_0)$
satisfying $\barrho\rho(A) R_\rho = R_\rho A$ for all $A\in\Auni$.
The conjugate $\barrho$ shares with $\rho$ the properties of
covariance~\eqref{eqCovRho}, finite statistics, and
localization~\eqref{eqRhoInC0} in
some space-like cone which we choose to be $\spc_0$.
Using the normalization convention of \cite[Eq.~(3.14)]{DHRIV},
namely $R_\rho^*R_\rho=|\lambda_\rho|^{-1}\unity$,
the positive linear endomorphism $\phi_\rho$ of $\Auni$  defined as
\begin{equation} \label{eqLeftInvR}
\phi_\rho(A)= |\lambda_\rho|\;
R_\rho^*\bar\rho(A) R_\rho
\end{equation}
is the unique left inverse~\cite{DHRIV,BuF} of $\rho$.
In the low-dimensional situation, $d=2,3$, the statistics parameter
$\lambda_\rho$ may be a complex non-real number,
corresponding to braid group statistics. We admit the case when its
modulus is different from one (namely when $\rho$ is not surjective),
corresponding to non-Abelian braid group statistics.

The modular objects for which we shall prove Borchers' commutation
relations are defined  as follows.
Let $S_\rho$ be the  closed anti-linear operator satisfying
\begin{align}  \label{eqSTomRho}
S_\rho \, \pi_0(A)\Om &:= \pi_0\barrho(A^*)R_\rho\Om, \quad
A\in\Auni(W_1),
\end{align}
and denote the polar decomposition of $S_\rho$ by $S_\rho=
J_\rho\Delta_\rho^{1/2}$.
$S_\rho$ is just the relative Tomita operator~\cite{Stratila}
with respect to a certain pair of (non-normalized) states. Namely,
consider the vacuum state $\omega_0:=\lsp \Om,\pi_0(\cdot)\Om\rsp$,
and the positive functional
$$
\varphi_\rho:= |\lambda_\rho|^{-1}\; \omega_0\circ \phi_\rho
= \big(R_\rho\Om,\pi_0\bar\rho(\cdot) R_\rho\Om\big).
$$
The restriction of $\varphi_\rho$ to $\Auni(W_1)$ is faithful and
normal, and has the
GNS-triple $(\HO,\pi_0\barrho,R_\rho\Omega)$.
Thus, $S_\rho$ is the relative Tomita operator associated with the
algebra $\Auni(W_1)$ and the pair of states $\omega_0$ and
$\varphi_\rho$, see Appendix~\ref{secRelMod}.

The motivation to consider these objects (instead of the modular
objects associated with $\Auni(W_1)$ and one
suitable state, e.g.\ $\varphi_\rho$) is that the so-defined relative
modular unitary
group $\Delta_\rho^{it}$ implements the modular automorphism group
associated with $\Auni(W_1)$ and $\omega_0$ in the
same way as the representation $U_\rho(x)$ implements the translations
$\alphaO_x$, see Eq.~\eqref{eqDeltaImplement} below.

This opens up the possibility to lift Borchers' commutation
relations~\eqref{eqDUD0} in the vacuum representation to the
representation $\pi_0\rho$.
In fact, pursuing this strategy, we shall find the following result.
Let $G$ be the subgroup of the proper Poincar\'e group generated by
the translations,
the boosts $\Boox{t}$ and the reflection $\J$.
Recalling that the representation $U_\rho$ may be shifted to a
representation $e^{ik\cdot x} U_\rho(x)$ whose spectrum has a Lorentz
invariant lower boundary~\cite{BoBu85},\footnote{This is automatically
  the case if $\rho$ is localizable in double cones and $d>2$ by a result of
  Borchers~\cite{Borch65}, which is applicable since in this case
  $\rho$ is implemented by local
  charged field operators~\cite{DR90}. It is also the case of
  course if $U_\rho$ extends to the Poincar\'e group.} we show
under the above-mentioned assumptions:
\begin{Thm}[Commutation Relations.]  \label{Borchers}
Assume that the lower boundary of the spectrum of $U_\rho$ is
Lorentz-invariant.
Then $U_\rho(x)$, $\Delta_\rho^{it}$, $J_\rho$ and the counterparts
for $\barrho$ constitute a continuous (anti-) unitary
representation\footnote{Strictly speaking, a ray representation since
  $J_\rho J_\barrho$ is only a multiple of unity.} of
$G$. More specifically, there hold the commutation relations
\begin{align}
\Delta_\rho^{it}\,U_\rho(x)\,\Delta_\rho^{-it}&= U_\rho(\Boox{t}x),
\label{eqDUD}\\
J_\rho\,U_\rho(x)\,J_\rho^{-1} &= U_\barrho(jx), \label{eqJUJ} \\
J_\rho\;\Delta_\rho^{it}\; J_\rho^{-1}&=\Delta_\barrho^{it}, \label{eqDJD} \\
J_\rho\,J_\barrho &=\chi_\rho \unity , \label{eqJJ}
\end{align}
for all $t\in\RR$ and $x\in\RR^d$.
The complex number  $\chi_\rho$ in Eq.~\eqref{eqJJ} has modulus one,
conjugate $\bar \chi_\rho=\chi_\barrho$ and is a root of unity
if $\barrho=\rho$.
\end{Thm}
(Note that Eq.~\eqref{eqDJD} corresponds to a standard property of
{\em modular} objects, but needs to be proved for our relative modular
objects.)

We also show that this representation of $G$ acts geometrically
correctly on the wedge algebras, namely for $W$ in the family $\calW_1$ of
translates of $W_1$ and $W_1'$,
$$
\calW_1:=\{ x+W_1,\, x\in\RR^d\}\cup\{ x+W_1',\, x\in\RR^d\},
$$
there holds
\begin{align} \label{eqDGeo}
\Ad \Delta_\rho^{it} :\,\pi_0\rho\Auni(W)&\to  \pi_0\rho \Auni(\Boox{t}W), \\
\Ad J_\rho           :\,\pi_0\rho\Auni(W) &\to
\pi_0\barrho\Auni(jW). \label{eqJGeo}
\end{align}
To this end, observe that modular theory~\cite{BraRob} and the relation
$\pi_0^{-1}(\AO(W_1)')=\Auni(W_1')$ imply that $\DO^{it}$ and $\JO$
implement an automorphism $\sigma_t$ of $\Auni(W_1)$ and
$\Auni(W_1')$, and an
anti-isomorphism from $\Auni(W_1)$ onto $\Auni(W_1')$ and vice versa,
respectively, defined by
\begin{align}  \label{eqDelta0Sig}
\Ad \DO^{it}\circ\pi_0&=\pi_0\circ\sigma_t \\%\quad\text{ on }\;\Auni(W_1),\\
\Ad J_0\circ \pi_0 &= \pi_0\circ \alphaO_j\label{eqJ0alphaj}
\end{align}
on $\Auni(W_1)\cup\Auni(W_1')$. By Borchers' commutation relations,
the same equations extend $\sigma_t$ and $\alphaO_j$ to the family $\Auni(W)$,
$W\in\calW_1$, acting in a geometrically correct way:
\begin{align} \label{eqD0Geo}
\sigma_t :\,\Auni(W)&\to  \Auni(\Boox{t}W), \\
\alphaO_j:\,\Auni(W)&\to \Auni(jW), \label{eqJ0Geo}
\end{align}
$W\in\calW_1$, see~\cite[Lem.~III.2]{Borchers92}.
But our representers $\Delta_\rho^{it}$ and $J_\rho$
implement these isomorphisms $\sigma_t$ and
$\alphaO_j$, respectively, in the direct product representation
$\pi_0\rho\oplus\pi_0\barrho$, namely:
\begin{Prop}[Implementation Properties.] \label{DeltaJImplement}
There holds
\begin{align}\label{eqDeltaImplement}
\Ad \Delta_\rho^{it} \circ \pi_0\rho &=  \pi_0\rho \circ \sigma_t \\
\Ad J_\rho \circ \pi_0\rho & =\pi_0\barrho\circ \alphaO_j \label{eqJImplement}
\end{align}
on the family of algebras $\Auni(W)$, $W\in\calW_1$.
\end{Prop}
Since $\sigma_t$ and $\alpha_j$ act geometrically correctly,
c.f.~Eq.s~\eqref{eqD0Geo} and \eqref{eqJ0Geo}, this implies  that
$\Delta_\rho^{it}$ and $J_\rho$ act geometrically correctly, as
claimed in Eq.s~\eqref{eqDGeo} and \eqref{eqJGeo}.

In two space-time dimensions, our group $G$
already coincides with
the proper Poincar\'e group $\Poj$, and our results therefore imply that
the translations and the relative modular objects constitute an
(anti-) unitary representation of the latter.
By our assumption of Haag duality~\eqref{eqHD} for wedges, the
so-called dual net
$$
\Auni^d(\calO) :=  \bigcap_{W\supset \calO} \Auni(W)
$$
is still local. (One needs to intersect in fact only the algebras of
one ``right wedge'' of the form $W_1+x$ and one
``left wedge'' of the form $W_1'+y$.)
The modular (anti-) automorphisms $\sigma_t$ and
$\alpha_j$ act on it in a geometrically correct way,
see~\cite[Prop.~III.3]{Borchers92}.
If the original net satisfies Haag duality also for double cones,
it coincides with the dual net.
Then the implementation properties~\eqref{eqDeltaImplement}
and \eqref{eqJImplement} hold, and therefore the
representation of $\Poj$ constructed in Theorem~\ref{Borchers} acts
geometrically correctly, namely there holds for any double cone $\calO$:
\begin{align}
\Ad U_\rho(g)&: \pi_0\rho\Auni(\calO) \to \pi_0\rho\Auni(g \,
\calO),\quad  g\in \Po, \nonumber\\
\Ad J_\rho&:\pi_0\rho\Auni(\calO)\to\pi_0\barrho\Auni(\J\, \calO).
\label{eqJRhoO}
\end{align}
Here we have written $U_\rho(a,\Boox{t}):=U_\rho(a)\Delta_\rho^{it}$.
In  particular, $J_\rho$ is a \CPT operator.\footnote{If the net does
  not satisfy Haag duality for double cones, it does not coincide
 with the dual net. Then our endomorphism $\rho$ has two generally
 distinct extensions $\rho_{R/L}$ to the dual net, according a choice
 of the right or left wedge~\cite{Roberts76}. (Each of them is
 localizable only in one type of wedges.) In this case, $J_\rho$
 intertwines $\pi_0\rho_R$ with $\pi_0\barrho_L\alphaO_j$, and
 in Eq.~\eqref{eqJRhoO} there appears $\rho_R$ on one side and $\bar\rho_L$ on
the other side.}
Again, it must be noted that these results (in $d=2$) are already
implicit in the work of Guido and Longo~\cite{GL92}, and also that the
split property
would exclude any charged sectors in our sense.

\section{Proofs}
\label{secProof}

We now prove Theorem~\ref{Borchers} and
Proposition~\ref{DeltaJImplement}.
Instead of proving Borchers' commutation relations directly
(e.g.\ paralleling Florig's nice
proof~\cite{Florig98}), we show how they lift from the
vacuum sector to our charged sector.

We shall use some well-known facts about relative modular objects,
which we recall in the Appendix for the convenience of the reader,
see also~\cite{Stratila} for a review.
Namely, the operator $\Delta_\rho^{it}\Delta_0^{-it}$ is in
$\pi_0\Auni(W_1)$ for $t\in\RR$, and we define
\begin{equation} \label{eqCoc}
\coc_\rho(t):= \pi_0^{-1}\big(\Delta_\rho^{it}\Delta_0^{-it}\big)
\quad \in \Auni(W_1).
\end{equation}
This family of observables coincides with the
Connes cocycle $(D\varphi_\rho:D\omega_0)_t$ with respect to the pair of
weights $\omega_0$ and $\varphi_\rho$, see Eq.~\eqref{eqConnesCoc}.

In the present context, it satisfies
\begin{equation} \label{eqCocInt}
\Ad \coc_\rho(t)\circ \sigma_t \circ \rho = \rho \circ \sigma_t
\qquad \text{ on } \calA(W_1),
\end{equation}
see  Proposition 1.1 in~\cite{Longo97}.
The definition~\eqref{eqCoc} and Eq.~\eqref{eqCocInt}
are analogous to well-known properties of the translation cocycles
which we shall use in the sequel. Observe that for $a\in W_1^\clo$, the closure
of $W_1$, we have $W_1+a\subset W_1$ and $W_1'-a \subset
W_1'$. Since $\rho$ acts trivially on $W_1'$, this implies that
the operator $U_\rho(a)\UO(-a)$ is in $\pi_0\Auni(W_1')'$ which
coincides with $\pi_0\Auni(W_1)$ by Haag duality.

This gives rise to the translation cocycle
\begin{equation} \label{eqCocY}
Y_\rho(a):= \pi_0^{-1}\big(U_\rho(a) \UO(-a)\big)\quad \in
\Auni(W_1),\; a\in W_1^\clo.
\end{equation}
By virtue of Eq.s~\eqref{eqU0impl} and \eqref{eqCovRho}, it satisfies
the intertwiner relation
\begin{equation} \label{eqCocYInt}
\Ad Y_\rho(x) \circ \alphaO_x \circ \rho = \rho \circ \alphaO_x,
\quad x\in \RR^d.
\end{equation}
The definitions of the cocycles $\coc_\rho(t)$ and $Y_\rho(x)$, the
intertwiner relations~\eqref{eqCocInt} and \eqref{eqCocYInt}, and
invariance of $\Om$ under $\DO^{it}$ and $\UO(x)$ imply the
identities
\begin{align} \label{eqZpi0AOm}
\Delta_\rho^{it}\,\pi_0(A)\Om &= \pi_0\big(\coc_\rho(t)\sigma_t(A)\big)\Om,
\;\quad A\in \Auni(W_1),\\
U_\rho(x)\,\pi_0(A)\Om &= \pi_0\big(Y_\rho(x)\alphaO_x(A)\big)\Om,
\quad A\in \Auni, \label{eqYpi0AOm}
\end{align}
which we shall frequently use in the sequel.
We shall also use the fact that Borchers' theorem applied to the
observable algebra implies that
\begin{equation}\label{eqSigmaAlphaSigma}
\sigma_t \alphaO_{\Boox{-t}x} \sigma_{-t} = \alphaO_{x}
\end{equation}
holds as an isomorphism from $\Auni(W)$ onto $\Auni(W+x)$,
$x\in\RR^d$, $W\in\calW_1$.
Finally, we make the interesting observation that
$S_\rho$ is the relative Tomita operator associated
not only with the pair of states $(\omega_0,\varphi_\rho)$, but
also with  pair of states $(\varphi_{\barrho},\omega_0)$:
\begin{Lem} \label{SRhoOm}
The span, $D$,  of vectors of the form
$\pi_0[\rho(A)R_\barrho]\Omega$, $A\in\Auni(W_1)$, is
a core for the relative Tomita operator $S_\rho$, and $S_\rho$ acts
on $D$ as
\begin{align}  \label{eqSTomRho'}
S_\rho\; \pi_0\rho(A)R_\barrho\Omega =  \chi_\rho \; \pi_0(A^*)\Omega, \quad
A\in\Auni(W_1),
\end{align}
where $\chi_\rho$ is a complex number of modulus one, with
$\bar\chi_\rho= \chi_\barrho$, and is a root of unity if $\barrho=\rho$.
\end{Lem}
\begin{Proof}
Eq.s~\eqref{eqCocInt}, \eqref{eqZpi0AOm} and \eqref{eqZZRR} imply that for
$A\in\Auni(W_1)$ there holds
$$
\Delta_\rho^{it}\;\pi_0[\rho(A)R_\barrho]\Omega =
\pi_0\big[ \rho\sigma_t(A\coc_\barrho(-t))

 R_\barrho\big]\Omega.
$$
Thus,  the domain $D$ is invariant under the unitary group
$\Delta_\rho^{it}$.
It is therefore a core for $\Delta_\rho^{1/2}$ and hence for $S_\rho$.
On this core, we have by definition
$$
S_\rho\; \pi_0[\rho(A)R_\barrho]\Omega=
\pi_0[\barrho(R_\barrho^*)R_\rho A^*]\Omega.
$$
But $\barrho(R_\barrho^*)R_\rho$ is a self-intertwiner of $\rho$, hence a
multiple of unity, $\chi_\rho\unity$. This proves Eq.~\eqref{eqSTomRho'}.
For the stated properties of $\chi_\rho$, see~\cite[Eq.~(3.2)]{FRSII}.

\end{Proof}

Since $(\HO,\pi_0\rho,R_\barrho\Omega)$ is the GNS triple for the
(non-normalized) state $\varphi_\barrho$ and $\chi_\rho\Omega$ is a
GNS vector for $\omega_0$, the Lemma implies that
$S_\rho$ is the relative Tomita operator associated

with the pair of states $(\varphi_{\barrho},\omega_0)$.

\paragraph{Proof of Theorem~\ref{Borchers}.}
To prove Eq.~\eqref{eqDUD} of Theorem~\ref{Borchers}, let
$A\in\Auni(W_1)$ and $a\in W_1^\clo$.
Using Eq.s~\eqref{eqZpi0AOm}, \eqref{eqYpi0AOm} and
\eqref{eqSigmaAlphaSigma}, we then have
\begin{align} \label{eqDUDOm}
\Delta_\rho^{it} U_\rho(\Boox{-t}a)\Delta_\rho^{-it} \,\pi_0(A)\Om
& \,=
\pi_0\big(\hat Y_\rho(a,t) \alphaO_{a}(A)\big)\Om,  \\
\hat Y_\rho(a,t)& :=
\coc_\rho(t)\sigma_t\big(Y_\rho(\Boox{-t}a)\alphaO_{\Boox{-t}a}
(\coc_\rho(-t)\big) . \nonumber
\end{align}
The intertwiner relations~\eqref{eqCocInt} and \eqref{eqCocYInt} imply
that
on $\Auni(W_1)$ there holds
\begin{align*}
\Ad \hat Y_\rho(a,t) \circ \alphaO_{a} \circ \rho
& \equiv
\Ad \coc_\rho(t) \circ \sigma_t \circ \Ad Y_\rho(\Boox{-t}a) \circ
\alphaO_{\Boox{-t}a} \circ \Ad \coc_\rho(-t)
\circ \sigma_{-t}  \circ \rho \\
&= \rho \circ \sigma_t \circ \alphaO_{\Boox{-t}a} \circ \sigma_{-t}
= \rho \circ \alphaO_a.
\end{align*}
That is, $\hat Y_\rho(a,t)$ satisfies the same intertwiner
relation~\eqref{eqCocYInt} on $\Auni(W_1)$ as $Y_\rho(a)$.
On the other hand, $\hat Y_\rho(a,t)$ is also contained in
$\Auni(W_1)$.
Therefore $\hat Y_\rho(a,t)Y_\rho(a)^*$ is in
  $(\rho\Auni(W_1))'\cap\Auni(W_1)$ which is trivial by our
assumption~\eqref{eqRhoIrred} of irreducibility.
Thus $\hat Y_\rho(a,t)$ coincides with
$Y_\rho(a)$ up to a
scalar function $c(a,t)$. Hence Eq.~\eqref{eqDUDOm} reads
\begin{align*}
\Delta_\rho^{it} U_\rho(\Boox{-t}a)\Delta_\rho^{-it} \,\pi_0(A)\Om
& = c(a,t)\; \pi_0\big(Y_\rho(a) \alphaO_{a}(A)\big)\Om
\\
&\equiv  c(a,t)\; U_\rho(a) \, \pi_0(A)\Om.
\end{align*}
Since the vacuum is cyclic for $\pi_0\Auni(W_1)$ by the Reeh-Schlieder
property, this shows that
\begin{align} \label{eqDUD'}
\Delta_\rho^{it} U_\rho(\Boox{-t}a)\Delta_\rho^{-it}=
c(a,t)\; U_\rho(a)
\end{align}
for $a\in W_1^\clo$. By adjoining, we get an analogous equation for
$-a\in W_1^\clo$. Since the closures of $W_1$ and $-W_1$ span the
whole Minkowski space, this shows that there is a function $c(a,t)$
such that Eq.~\eqref{eqDUD'} holds for all $a\in \RR^d$.

It remains to show that $c(a,t)\equiv 1$.
Eq.~\eqref{eqDUD'} gives us a ray representation of the group
$G$ generated by
the boosts $\Boox{t}$ and the translations in the $0,1$-plane, defined by
\begin{equation*}
 U(a,\Boox{t}) := U_\rho(a)\Delta_\rho^{it}.
\end{equation*}
(The group $G$ is a subgroup of $\Po$ in $d=3$ and coincides with
$\Po$ in $d=2$. The product in $G$ is
$
(a,\Boox{t})\cdot (a',\Boox{t'})= (a+\Boox{t}a',\Boox{t+t'})$.)
Now $G$ is simply connected, and its second cohomology group is known
to be trivial.
Therefore there exists a function $\nu$ from
$G$ into the unit circle such that

$\hat U(g):= \nu(g)\,  U(g)$ is a true representation of $G$.
Eq.~\eqref{eqDUD'} then implies that
\begin{equation} \label{eqcnua}
c(a,t) = \nu(a,\unity) \,
\nu(\Boox{-t}a,\unity)^{-1}.
\end{equation}
Since $U_\rho$ is a true representation of the translations, the
restriction of $\nu$ to the
translations is a one-dimensional representation, that is of the form
$\nu(a,\unity)=e^{ik\cdot a}$.
Therefore, the spectra of the representations $\hat U=\nu\otimes
 U$ and $ U$ differ by a translation about a vector $k$.
But the spectrum of $\hat U$ is invariant under the 1-boosts
since $\hat U$ extends to a true
representation of the (2-dimensional) Poincar\'e group $G$, and the
lower boundary of the spectrum of $ U$ is also Lorentz\ invariant
since it coincides with
the spectrum of $U_\rho$. This implies that $k=0$ and hence, by
Eq.~\eqref{eqcnua}, that $c(a,t)\equiv 1$.
This completes the proof of Eq.~\eqref{eqDUD} of the Theorem.

We now prove Eq.~\eqref{eqDJD} of the Theorem.
For $A\in\Auni(W_1)$, we have by Eq.~\eqref{eqZpi0AOm} and
the intertwiner relation~\eqref{eqCocInt}
\begin{align} \label{eqDSD'}
\Delta_\barrho^{it}\, S_\rho\,\Delta_\rho^{-it}\,\pi_0(A)\Om
&=
\pi_0\big(\barrho(A^*)\coc_\barrho(t)\sigma_t[\barrho(\coc_\rho(-t)^*)R_\rho]
\big)\,\Om.
\end{align}
We shall now use a result of Longo~\cite{Longo97}. Namely, we are in
the situation where Propositions 1.3 and 1.4 in~\cite{Longo97} apply, yielding
\begin{equation*}
R_\rho^* \barrho\big(\coc_\rho(-t)\big)\coc_\barrho(-t) = \sigma_{-t}
(R_\rho^*).
\end{equation*}
Applying $\sigma_t$, adjoining, and using the cocycle identity
$\coc_\barrho(t)\sigma_t(\coc_\barrho(-t))=1,
$
see Eq.~\eqref{eqCoc'}, yields
\begin{equation} \label{eqZZRR}
\coc_\barrho(t)\sigma_t[\barrho(\coc_\rho(-t)^*)R_\rho]=R_\rho.
\end{equation}
Hence Eq.~\eqref{eqDSD'} reads
$$
\Delta_\barrho^{it}\, S_\rho\,\Delta_\rho^{-it}\,\pi_0(A)\Om
=\pi_0\big(\barrho(A^*)R_\rho\big)\,\Om \equiv S_\rho \,\pi_0(A)\Om.
$$
Since $\Delta_\rho^{it}$ maps the core $\pi_0\Auni(W_1)\Om$ of
$S_\rho$ onto itself by Eq.~\eqref{eqZpi0AOm}, this shows that
\begin{align*}
\Delta_\barrho^{it}\, S_\rho\,\Delta_\rho^{-it} &=  S_\rho,
\end{align*}
which implies Eq.~\eqref{eqDJD} of the Theorem.

{}For the proof of Eq.~\eqref{eqJUJ} we need the following Lemma.
\begin{Lem} \label{DSDUSU}
For $a$ in the closure of $W_1$, there holds
\begin{align}
U_\barrho(a)^{-1}\, S_\rho\,U_\rho(a) &\subset   S_\rho. \label{eqUSU}
\end{align}
\end{Lem}
\begin{Proof}
First recall from~\cite{F81,BuF} that the representation
$\hat U_\barrho$ defined by
\begin{equation} \label{eqHatUBarrho}
\hat U_\barrho(x)\,\pi_0[\barrho(A)R_\rho]\Om :=
\pi_0[\barrho(\alpha_x(A)Y_\rho(x)^*)R_\rho]\Om
\end{equation}
implements $\alpha_x$ in the representation $\pi_0\barrho$, i.e.\
$\Ad\hat U_\barrho(x)\circ\pi_0\barrho=\pi_0\barrho\circ\alphaO_x$.
\footnote{We recall the argument in the present setting. The endomorphism
  $\alphaO_{-x}\circ\phi_\rho\circ \beta_x$, where
$\beta_x:=\Ad Y_\rho(x)\circ \alpha_x$, is a left inverse of $\rho$
  and therefore coincides with $\phi_\rho$ by uniqueness. This implies
  that the state $\varphi_\rho$ is invariant under the automorphism
  group $\beta_x$ and hence that
$$
U_{\barrho\rho}(x) \pi_0[\barrho(A)R_\rho]\Omega  :=
\pi_0[\barrho\beta_x(A)R_\rho]\Omega
$$
defines a unitary representation of the translations.
But $\hat U_\barrho(x)$ defined above coincides with
$\pi_0\barrho(Y_\rho(x)^*) U_{\barrho\rho}(x)$, hence is a
well-defined unitary operator. The implementation property is checked
directly from the definition~\eqref{eqHatUBarrho}, and implies in turn
the representation property.}

The representation $\hat U_\barrho$ therefore coincides with $U_\barrho$
up to a one-dimensional representation $c(\cdot)$.
We now have, for $a$ in the closure of $W_1$ and $A\in\Auni(W_1-a)$,
\begin{align*}
S_\rho\,U_\rho(a)\,\pi_0(A)\Om &=
\pi_0\big[\barrho(\alpha_a(A^*)Y_\rho(a)^*)R_\rho\big]\Om
= \hat U_\barrho(a)\,\pi_0[\barrho(A^*)R_\rho]\Om \\
& = \hat U_\barrho(a)\,S_\rho\,\pi_0(A)\Om.
\end{align*}
Since $\hat U_\barrho$ and $U_\barrho$ coincide up to the character $c$
as discussed above, we therefore have
$$
U_\barrho(-a)\,S_\rho \,U_\rho(a)= c(a)\, S_\rho
\quad \text{ on } \; D:=\calA_0(W_1-a)\Om.
$$
Applying $\Delta_\barrho^{it}\,\cdot\,\Delta_\rho^{-it}$ to this
equation and using the by now established Eq.s~\eqref{eqDUD} and
\eqref{eqDJD} of the Theorem, yields $c(\Boox{t}a)=c(a)$ or
$c((\unity-\Boox{t})a)=1$.
By the representation property of $c$, the same holds for $-a\in W_1^\clo$.
Since $W_1^\clo$ and $-W_1^\clo$ span the whole Minkowski space and
$1-\Boox{t}$ is invertible for $t\neq0$, this shows that $c$ is
trivial. Since $D$ is a core for the left hand side of relation~\eqref{eqUSU},
this completes the proof.
\end{Proof}
We are now ready to prove Eq.~\eqref{eqJUJ} of the Theorem.
To this end, let $a\in W_1^\clo$ and $\phi\in D:=\calA_0(W_1-a)\Om$.
By Eq~\eqref{eqDUD}, we have for all $t\in\RR$
\begin{equation} \label{eqJUJ'}

\Delta_\rho^{it}\,U_\rho(a) \, \phi
= U_\rho(\Boox{t}a)\,\Delta_\rho^{it} \phi.
\end{equation}
Now by Lemma~\ref{DSDUSU}, the vector
$U_\rho(a) \, \phi$ is in the domain of the
operator $\Delta_\rho^{1/2}$, hence the left hand side is bounded for
$t$ in the strip $\RR - i[0,1/2]$ and analytic in its
interior. The same holds for the vector valued function
$t\mapsto \Delta_\rho^{it} \,\phi$ on the right hand side.
Further, for $a\in W_1^\clo$ the
operator valued function $t\mapsto U_\rho(\Boox{t}a)$ is norm-bounded on
the strip $\RR - i[0,1/2]$ and analytic in its interior, and at
$t=-i/2$ has the value $U_\rho(ja)$, see e.g.~\cite[Section V.4.1]{H96}.
Therefore, Eq.~\eqref{eqJUJ'} implies that
$$
\Delta_\rho^{1/2}\,U_\rho(a) \,\phi
= U_\rho(ja)\,\Delta_\rho^{1/2} \,\phi.
$$
Multiplying with $J_\rho$ and using relation~\eqref{eqUSU} of
Lemma~\ref{DSDUSU} yields
\begin{equation} \label{eqJUJ''}
U_\barrho(a) \,S_\rho \,\phi
= J_\rho U_\rho(ja) J_\rho^{-1}\,S_\rho \,\phi.
\end{equation}
Since $S_\rho$ has dense range, this shows Eq.~\eqref{eqJUJ} for $x=ja$
in the closure of $jW_1$ and, by adjoining, also for
arbitrary $x$. This completes the proof of Eq.~\eqref{eqJUJ} of the Theorem.
To prove Eq.~\eqref{eqJJ}, note that Lemma~\ref{SRhoOm}
implies that $S_\rho=\chi_\rho\,S_\barrho^{-1}$. Using that
$\Delta_\barrho^{-1/2}J_\barrho^{-1}=J_\barrho^{-1}\,\Delta_\rho^{1/2}$
by Eq.~\eqref{eqDJD} and that $J_\barrho$ is anti-linear, one gets
Eq.~\eqref{eqJJ}. This completes the proof of the Theorem.

\paragraph{Proof of Proposition~\ref{DeltaJImplement}.}
We now turn to Eq.~\eqref{eqDeltaImplement} of
Proposition~\ref{DeltaJImplement}. On $\Auni(W_1)$, this equation follows from
Eq.~\eqref{eqCocInt} by applying $\pi_0$ to the latter. Further, the fact that
$\pi_0^{-1}(\Delta_\rho^{it}\DO^{-it})$ is in $\Auni(W_1)$ and hence
commutes with $\Auni(W_1')$ implies that
$\Ad\Delta_\rho^{it}\circ\pi_0=\pi_0\circ \sigma_t$ on $\Auni(W_1')$.
Since $\rho$ acts as the identity on $\Auni(W_1')$, this implies
Eq.~\eqref{eqDeltaImplement} on $\Auni(W_1')$. For translates of $W_1$
or $W_1'$, the equation follows from Borchers' commutation relations,
Eq.s~\eqref{eqDUD} and \eqref{eqSigmaAlphaSigma}.
Before proving Eq.~\eqref{eqJImplement} of the proposition, we
establish the following
intertwiner properties of the relative modular conjugation.
\begin{Lem}[Intertwiner Properties of $\boldsymbol{J_\rho}$.] \label{JInt}
The unitary operators $J_\rho J_0$ and $J_0J_\rho$
have the intertwiner properties
\begin{align}
 \pi_0\barrho(A) \; J_\rho J_0 & = J_\rho J_0 \; \pi_0(A),
\label{eqModConjInt} \\
\pi_0(A) \; J_0 J_\rho  &= J_0 J_\rho \; \pi_0\rho(A)  \label{eqModConjInt'}

\end{align}
for $A\in \Auni(W_1)$.
\end{Lem}
\begin{Proof}
These are consequences of a standard result~\cite{Stratila} which
relates the conjugations of relative Tomita operators, see
Eq.~\eqref{eqJ21J1} in the Appendix.
Here, in Eq.~\eqref{eqModConjInt} $S_\rho$ is being considered
as the relative Tomita operator associated with the pair of states
$(\omega_0,\varphi_\rho)$, characterized by Eq.~\eqref{eqSTomRho}, and
in Eq.~\eqref{eqModConjInt'} as the relative Tomita operator
associated with the pair $(\varphi_\barrho,\omega_0)$, characterized
by Eq.~\eqref{eqSTomRho'} of Lemma~\ref{SRhoOm}.
\end{Proof}
We are now ready to prove Eq.~\eqref{eqJImplement} of
Proposition~\ref{DeltaJImplement}.

By Eq.s~\eqref{eqModConjInt'} and Eq.~\eqref{eqJ0alphaj}
we have on $\Auni(W_1)$
$$
\Ad J_\rho \circ \pi_0\rho
\equiv \Ad J_0 \circ\Ad(J_0J_\rho)\circ \pi_0\rho
=\pi_0\circ \alphaO_j =\pi_0\barrho\circ \alphaO_j,
$$
since $\barrho$ acts as the identity on
$\alphaO_j\Auni(W_1)\equiv\Auni(W_1')$, while by
Eq.~\eqref{eqModConjInt} and Eq.~\eqref{eqJ0alphaj} we have on $\Auni(W_1')$
$$
\Ad J_\rho \circ \pi_0\rho =\Ad J_\rho \circ \pi_0
\equiv \Ad (J_\rho J_0)\circ\Ad J_0\circ\pi_0

=\pi_0\barrho\circ \alphaO_j.
$$
This shows that Eq.~\eqref{eqJImplement} holds on
$\Auni(W_1)\cup\Auni(W_1')$. Borchers' commutation relations then
imply that it holds on $\Auni(W)$, $W\in\calW_1$, completing the proof
of Proposition~\ref{DeltaJImplement}.

\appendix
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\section{Relative Tomita Operators} \label{secRelMod}
We recall the relevant notions from relative Tomita theory,
following~\cite{Stratila}. (For the standard Tomita theory,
see e.g.~\cite{BraRob} and Eq.~\eqref{eqSTomO} above.)
Let $\calM$ be a von Neumann
algebra and $\varphi_1$, $\varphi_2$ two faithful normal positive
functionals on $\calM$, and denote by $\sigma_t^1$ and $\sigma_t^2$ the
respective modular automorphism groups.
Then there exists a family of unitaries
$Z_{21}(t)\in \calM$ satisfying the intertwiner and cocycle properties
\begin{align} \nonumber
\sigma_t^2(A)\, Z_{21}(t)&= Z_{21}(t)\, \sigma_t^1(A) ,\\
Z_{21}(t+s)&=Z_{21}(t)\sigma_t^1(Z_{21}(s)), \label{eqCoc'}
\end{align}
respectively, and characterized by a certain KMS property. These facts
have been shown by Connes~\cite{Connes73} and are reviewed
in~\cite[Sect.~I.3.1]{Stratila}.
The family $Z_{21}(t)$ is called the
Connes-cocycle associated with  the pair $\varphi_1$ and $\varphi_2$
and usually denoted by $(D\varphi_1:D\varphi_2)_t$.
This cocycle may be expressed in terms of the corresponding GNS
representations as follows~\cite[Sect.~I.3.11]{Stratila}.
Let $(\calH_i,\pi_i,\xi_i)$ be the GNS triples of $\varphi_i$,
$i=1,2$. Then the operator $S_{21}$ from $\calH_1$ to $\calH_2$ defined by
\begin{equation*}
S_{21}\,\pi_1(A)\xi_1 := \pi_2(A^*)\xi_2, \quad A\in \calM,
\end{equation*}
is closable. We denote its closure by the same symbol, and its polar
decomposition by
$$
S_{21}= J_{21} \, \Delta_{21}^{1/2}.
$$
These operators are called the relative Tomita modular objects
associated with the pair $\varphi_1$ and $\varphi_2$.
Let now $\Delta_1^{it}$ denote the unitary modular group of
$\pi_1(\calM)$ and $\xi_1$. Then $\Delta_{21}^{it}\Delta_1^{-it}$ is
in $\pi_1(\calM)$ and coincides with $\pi_1(Z_{21}(t))$, i.e.\ there
holds~\cite[Sect.~I.3.11]{Stratila}
\begin{equation} \label{eqConnesCoc}
Z_{21}(t) = \pi_1^{-1}\big(\Delta_{21}^{it}\Delta_1^{-it}\big).
\end{equation}
Finally, as shown in~\cite[Sect.~I.3.16]{Stratila}, the unitary operator
$$
V_{21} := J_{21}\,J_1 \equiv J_2\,J_{21},
$$
where $J_i$ is the modular conjugation of $\pi_i(\calM)$ and $\xi_i$,
$i=1,2$, is an intertwiner from $\pi_1$ to $\pi_2$, that means it satisfies
\begin{equation} \label{eqJ21J1}
\pi_2(A)\, V_{21} = V_{21}\, \pi_1(A),\quad A\in \calM.
\end{equation}

\paragraph{Acknowledgements.}
I gratefully acknowledge financial support by
FAPEMIG.
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config: arxiv


TITLE: Calculating a derivative from a Maclaurin series
QUESTION [0 upvotes]: I'm attempting to find $f^{(100)}(x)$ for $f(x)=\frac{1}{1+x^2}$. The Maclaurin series is $\sum\limits_{n=0}^{\infty}(-1)^nx^{2n}$.
I figure that $(-1)^nx^{2n}=\frac{f^{(n)}(0)\;x^n}{n!}$. Setting $n=100$,
$x^{200}=\frac{f^{(100)}(0)\;x^{100}}{100!}$
$f^{(100)}(0)=x^{100}100!$
Clearly I'm doing something wrong, because that should be a numerical answer (in fact, it should be $100!$).

REPLY [2 votes]: Your setting was wrong: it should be $$(-1)^nx^{2n}=\frac{f^{(2n)}(0)x^{2n}}{(2n)!}.$$
So $f^{(100)}=(-1)^{50}(100)!=(100)!$.
Tell me if I miss something. Thanks in advance.
P.S. In case it is still ambiguous, let explain more:
$\sum\limits_0^{\infty}(-1)^nx^{2n}$ in fact is equal to $\sum\limits_0^{\infty}\text{Re}(i^n)x^n$, where $\text{Re}$ means to take the real part of a comple number. Now you can see what the terms of the series are, and find out why the $2n$-th term is $(-1)^nx^{2n}$.


TITLE: Probability that the man is the first to die
QUESTION [0 upvotes]: If p is the probability that a man aged x will die in a year then the
  probability that out of n men $A_1, A_2 ,A_3 ...A_n$, each aged x,
  $A_1$ will die in an year and be the first to die is: ?
a) $1- (1- p)^{n-1}$
b)$(1-p)^n$
c) $\dfrac 1n [1- (1-p)^n]$
d) $\dfrac 1n (1-p)^n$

Attempt:
Order of dying matters.
Choose $r$  out of $n-1$ and arrange them (i.e. their deaths) in $r!$ ways.
So $P(E)  = p\times{^{n-1}C_0} (1- p)^{n-1}+ p \times p\times {^{n-1}C_1} (1-p)^{n-1} + p\times p^2 \times 2!\times {^{n-1}C_2} (1- p)^{n-2}+...$
$\implies P(E) = p \sum_{r=0}^{n-1} {^{n-1}}C_r (1-p)^{n-1-r}p^r \times r!$
But this doesn't match any of the options.
Could someone please tell me where I have gone wrong?
Edit: My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate

REPLY [2 votes]: Your expression is almost correct. Let $N$ be the number of deaths out of $A_2, \ldots, A_n$. We know that $A_1$ must die. After that, we decompose the event with respect to $N$ (notice we are assuming all deaths are independent) and use that given $N = r$, the probability that the first to die is $A_1$ is $\frac{1}{r+1}$ (here assuming that any order of deaths is equally likely). Thus:
\begin{align*}
\mathbb{P}[\text{given event}]
  &= \mathbb{P}[A_1\ \text{dies}] \cdot \sum_{r=0}^{n-1} \mathbb{P}[N=r] \cdot \mathbb{P}[A_1\ \text{dies first} \mid N=r]\\
  &= p \cdot \sum_{r=0}^{n-1} \binom{n-1}{r} p^r (1-p)^{n-1-r} \cdot \frac{1}{r+1}
\end{align*}
To simplify the expression above, we notice that
$$
\binom{n-1}{r}\frac{1}{r+1} = \frac1n \binom{n}{r+1}.
$$
From the above, we can write
\begin{align*}
\mathbb{P}[\text{given event}]
  &= \frac{1}{n} \cdot \sum_{r=0}^{n-1} \binom{n}{r+1} p^{r+1} (1-p)^{n-1-r} \\
  &= \frac1n [(p + (1-p))^n - (1-p)^n] = \frac1n [1 - (1-p)^n].
\end{align*}

REPLY [1 votes]: They probability for a person aged $x$ to not die is $1-p$. Hence, the probability that none of them is $(1-p)^n$. Therefore, the probability that at least one of them will die is $1-(1-p)^n$. SInce every one of them has the same probability to be that first person (as all are identical), it being $A_1$ has probability ${1 \over n}[1-(1-p)^n]$.
Hence, option $(c)$ is correct.


TITLE: What may be the number of connected components of this randomly generated topological space?
QUESTION [1 upvotes]: I hope there exists some answer to this but I am not very good in probability so i do not know how to tackle this problem. It's contemporary-art related so it might not have a good answer. Furthermore, it is clear that this question
What can be said about the number of connected components of $G(n,p)$ random graphs?
is related, however my situation will be quite different.
Begin with the square $X:=[0,1]^2$ and take $n\geq 1$. It is clear that I can divide $X$ into $n^2$ squares, the $(i,j)$-th square being given by :
$$[\frac{i-1}{n},\frac{i}{n}]\times [\frac{j-1}{n},\frac{j}{n}]$$
Let us say we denote $S_{i,j}$ this square. Now take $0\leq p\leq 1$ and take $X_{i,j}$ a random variable which go to $0$ with probability $p$ and to $1$ with probability $1-p$. Of course all the $X_{i,j}$ are taken mutually independent. Then you can construct :
$$Y:=\bigcup_{(i,j)\mid X_{i,j}=1}S_{i,j} $$
Now my question is the following what can be said about the number of connected components of $Y$ ? Even with $p=\frac{1}{2}$, I have trouble finding the law of $|\pi_0(Y)|$...
I have though about the following simplifications one might want to make :

1 The square $S_{i,j}$ is changed to $T_{i,j}$, the square $S_{i,j}$ with its four vertices removed.

Using this a square can be adjacent to at most four squares.

2 Consider the toric $\mathbb{T}$ version of $X$, because in $X$ we get side effects (the square $S_{1,1}$ is in a corner and hence has a very particular role whereas in $\mathbb{T}$ everyone is at the center).

My attempt  starts with those two simplifications hence we start from the graph $G_n$ with $n^2$ vertices (each vertex is a square) and each vertices is linked to four vertices (we draw the incidence graph of the CW-complex induced by the squares).
Now the random construction appears, for each vertex there is a probability of $p$ that we delete it with all its contiguous edges. And then I am stuck. I think what I am trying to understand is percolation but it may not be true (do not hesitate to delete the tag if it is not)... Any help will be appreciated.

REPLY [1 votes]: This is a highly nontrivial question. You should look at the book "Percolation" from Grimmett for more Details, but here are some Facts that are known (which I took from the book). If $K_n$ is the number of conn. comp. (called Clusters) then $K_n/n^2 \to \kappa(p)$ where $\kappa(p)$ is a constant depending on $p$. There is also a central Limit Theorem for $K_n$.


TITLE: Proving $\phi (N)$ is a normal subgroup of $\phi (G)$
QUESTION [0 upvotes]: Let $\phi \colon  G → K$ be a group homomorphism and let $N$ be a normal subgroup of $G$. Show that $\phi (N)$ is a normal subgroup of $\phi (G$).
Wondering what a good way to go about this is... I know a lot of different small things that can be derived from this information but not a way I can utilize this information to arrive at the intended conclusion.

REPLY [0 votes]: $\phi(N)$ is invariant under any inner automorphism $i_{\phi(g)}$ defined by an element of $\phi(G)$, since $i_{\phi(g)}(\phi(N))=\phi(g)\phi(N)\phi(g)^{-1}=\phi(gNg^{-1})=\phi(N)$, since $gNg^{-1}=N$ by normality of $N$.


TITLE: Qutrits and spin operators
QUESTION [0 upvotes]: Please I have a problem:
When dealing with qutrits as three level systems, is it not appropriate to use spin $S=1$ operators to describe the qutrit system?
I have read somewhere that it is not appropriate to describe a qutrit using these operators, and I don't understand why. I am used to find this representation in the literature and I don't understand the above issue.
Can you please help me clarifying this issue?

REPLY [0 votes]: In a qutrit you should be able to access transitions between any two of the states.  You cannot do this with spin operators: you can verify for yourself that no combination of $S_x,S_y$ or $S_z$, or combination of $S_+,S_-,S_z$ can connect - say - the states $\vert 1\rangle$ and $\vert 2\rangle$ only, without including $\vert 3\rangle$.  This follows immediately from
\begin{align}
S_-=\sqrt{2}\left(\begin{array}{ccc}
0&0&0\\
1&0&0\\
0&1&0
\end{array}\right)\, ,\qquad S_+=(S_-)^\dagger
\end{align}
so you cannot construct the operator
\begin{align}
\left(\begin{array}{ccc}
0&1&0\\
1&0&0\\
0&0&0
\end{array}\right)
\end{align}
for instance, which would only connect states $\vert 1\rangle$ and $\vert 2\rangle$.
Clearly there are $6$ possible pairs of states in a $3$-level system, so you need $6$ off-diagonal operators.  In addition, you also need $3$ diagonal operators for a total of $9$.
These $9$ operators are spanned by several sets of matrices.  One example is the Gell-Mann matrices plus the identity matrix.  Another very useful set is obtained using the generalized Pauli matrices and closely related to the clock and shift matrices.


TITLE: Is this a Karusch-Kuhn-Tucker method or something else?
QUESTION [0 upvotes]: This is a question that originated with World of Warcraft.  I have a solution, but I don't know where to look up other problems of the same kind for a better explanation.
There is a plane with p axis and q axis.
There is a function f(x,y) = 2px+qy^2 such that f(x,y) is only defined on the curve x^3+y^3=1.
Claim: There are regions of the plane for which f has exactly one critical point and exactly three critical points.
The proposed solution was:
F(x,y,k)=2px+qy^2+k(x^3+y^3-1)
This was not explained, but the last term multiplies k by zero.
I think this is some kind of Lagrange multiplier method, but it's not like any use of that method that I've seen before.
The solution continued:
Fx=2p+3kx^2=0
Fy=2qy++3ky^2=0
I assume that Fx means "derivative of F with respect to x" rather than a simple product.
This part of the solution looked like a Karusch-Kuhn-Tucker method, but I have only seen simple examples of that.
The most applicable book I have found for this kind of problem is Optimization Concepts and Applications in Engineering by Belegundu and Chandrupatla.
But perhaps this is not a KKT method specifically - perhaps it's just advanced calculus.
I have one book that touches on this kind of problem but doesn't really explain much.  I want to find a book (or better yet, a free online wiki) that explains this kind of problem thoroughly.
Thanks

REPLY [1 votes]: It is using KKT. For details, I'd recommend the Convex Optimization book by Boyd and Vandenberghe.
As a side note, this is not research level mathematics, and I would like to vote to close, but since I am still a new user, I don't know how to do it and I just leave this comment as an answer (can't comment yet either).


TITLE: Fourier transforms of characteristic functions
QUESTION [8 upvotes]: I am wondering how badly summable the Fourier transform of the characteristic function of a measurable subset of $S^1$ can be.

Question: Let $\alpha \colon \mathbb N \to [1,\infty)$ be a monotone increasing function with $\lim_{n \to \infty} \alpha(n) = \infty$. Is there a measurable subset $E \subset S^1$, such that
  $$\sum_{n \in \mathbb Z} | \widehat \chi_E(n)|^2 \cdot \alpha(|n|) = \infty \ ?$$
  Here, $\widehat \chi(n)$ are the usual moments $$\widehat \chi_E(n):= \int_E z^n \ dz.$$

The only example I know is the Fourier transform of the characteristic function of an interval, which grows like $1/n$. On the other hand, one can easily see that the growth cannot be better than $1/n$ (something like $1/n^{1 + \varepsilon}$), since $\ell^1 \mathbb Z \subset C(S^1)$.
More concretely:

Question: Can anyone compute the growth of the Fourier transform of the characteristic function of something like a Cantor set of non-zero measure?

Again, more abstractly:

Question: What can be said about the growth of the Fourier transform of the characteristic function of a generic subset of $S^1$?

REPLY [2 votes]: It is a difficut problem to compute the Fourier transform of the characteristic function of a union of open intervals (in the general case) and it is known that such Fourier transform can converge to 0 with a very slow growth rate. I am currently working on that problem. Probably the growth rate may depend on some arithmetical properties of the boundary of the set as it is the case in studying the Fourier dimension of sets.


TITLE: Write an integral giving the amount of work required to empty the full tank of fuel to a height 2 feet above the top of the tank
QUESTION [0 upvotes]: The tank shown below is in the shape of a right triangular prism where cross sections
parallel to the ground are rectangles (2 ft by 4 ft) and cross sections parallel to the front
face are right triangles (6ft tall and 2 ft wide at the top). The tank is initially filled to
the top with a fuel whose weight-density is 80 lb/ft3. The fuel is to be pumped out of the
tank to a height 2 feet above the top of the tank. Write an integral giving the amount of
work required to empty the full tank of fuel to a height 2 feet above the top of the tank.

REPLY [0 votes]: Well, first we must know the work formula, which is F times d.
In order to find the work done here, we are trying to integrate F with respect with d.
I'm not sure what you mean by the cross sections being 2 by 4 feet, since the cross sections will vary in length in width as you go up in height. Same idea goes for the right triangle's cross sections.
Please either clarify your answer or provide a photo of some sort to aid understanding.


TITLE: a countable and connected metric space $X$ has only one point
QUESTION [0 upvotes]: Prove that a countable and connected metric space $X$ has only one point.
Could you check my attempt? Thanks!
Suppose not. Consider two distinct point $x$ and $y$. Define $d:=d(x,y)>0$. Let $a<d$. By the connectedness of $X$, for each $\epsilon\in (0,a)$ there exists $z_{\epsilon}\in\partial B(x,\epsilon)$ where $\partial B(x,\epsilon)$ is the boundary of the open ball $B(x,\epsilon)=\{z\in X: d(z,x)<\epsilon\}$. This defines a map $\epsilon\mapsto z_{\epsilon}$ from $(0,a)$ to $X$. But this map is injective because for any distinct $\epsilon, \epsilon_1\in (0,a)$ we have $$d(z_{\epsilon},z_{\epsilon_1})\geq \big|d(z_{\epsilon},x)-d(x,z_{\epsilon_1})\big|=\big|\epsilon-\epsilon_1\big|>0$$So $z_{\epsilon}\neq z_{\epsilon_1}$.

REPLY [1 votes]: The idea is OK, but some slight simplification suggestions:
It's probably easiest to note that $B(x,\varepsilon)$ and $\{z: d(x,z) > \varepsilon\}$ are both open and non-empty, so there must be some point $z_\varepsilon$ with $d(x,z_\varepsilon) = \varepsilon$ to avoid having a disconnection for $X$. (Otherwise you need a small argument why for $z \in \partial B(x,\varepsilon)$ we have $d(x,z) = \varepsilon$).
The injectivity is then also clear: if two points have different distance to $x$, they are different points (no need for the reverse triangle inequality).


TITLE: Isomorphism of modules implies isomorphism of tensor products
QUESTION [2 upvotes]: Let $R$ be a Noetherian ring, $M$ a finitely generated left $R$-module and $I,J$ two sided ideals in $R$ such that $I \subset J$. Suppose that $IM=JM$. What are some conditions that will allow us to conclude that $$ I \otimes_R M \cong J \otimes_R M?$$
I know that $M$ being a flat $R$-module is enough, but I am looking for some weaker conditions. Many thanks.

REPLY [1 votes]: A more precise statement is: supposing that $I \subseteq J \subseteq R$ are ideals with $IM=JM$, the canonical map $I \otimes M \to J \otimes M$ is an isomorphism if the following two Tor groups are $0$,
$$\mathrm{Tor}_R^1(J/I,M)=0 \quad \text{and} \quad \mathrm{Tor}_R^1(R/J,M)=0.$$ In particular, this holds if $M$ is acyclic for $\otimes_R$ (that is, flat as an $R$-module).
I don't think the Noetherian hypotheses are necessary.
Proof: There is an exact sequence $0 \to I \to J \to J/I \to 0$ of $R$-modules. Tensoring this with $M$ gives an exact sequence
$$\mathrm{Tor}_R^1(J/I,M) \to I \otimes M \to J \otimes M \to J/I \otimes M \to 0.$$ Thus the canonical map is surjective if and only if $J/I \otimes M=0$, and is injective if and only if the map $\mathrm{Tor}_R^1(J/I,M) \to I \otimes M$ is the zero map. On the other hand, tensoring the exact sequence
$$0 \to J/I \to R/I \to R/J \to 0$$ by $M$ produces
$$\mathrm{Tor}_R^1(R/J,M) \to J/I \otimes M \to M/IM \to M/JM \to 0,$$ where we have used the canonical isomorphisms $R/I \otimes M \cong M/IM$ and $R/J \otimes M \cong M/JM$. By hypothesis the map $M/IM \to M/JM$ is an isomorphism, so by exactness
$$\mathrm{Tor}^1_R(R/J,M) \to J/I \otimes M$$ is surjective, finishing the proof.


TITLE: Mathematical Description for Steam Rising from a Cup
QUESTION [3 upvotes]: I was staring at a cup of coffee I have on the desk just now. The light shines through the water vapor as they rise from the cup. The shape of the steam is not completely random, as it drift from hot area to cold area according to the second law of thermodynamics. If you whiff your hand over, the steam swirls with a distinct center of rotation.
How would one go about describing the shape and characteristics of the steam as it rises from a hot cup of water? What branches of math would be involved in creating such a model?

REPLY [0 votes]: Computer graphics has been simulating this using fluids dynamics for some time now. See for instance Vortex Fluid for Gaseous Phenomena and Cuppa Coffee and The Lonely Cigarette.


TITLE: Finding the kernel of an action on conjugate subgroups
QUESTION [4 upvotes]: I'm trying to solve the following problem:

Let $G$ be a group of order 12. Assume the 3-Sylow subgroups of $G$ are not normal.  Prove that $G\cong A_4$.

Here's my attempt: let $\mathscr S$ be the set of 3-Sylow subgroups of $G$. Since the elements of $\mathscr S$ are not normal, by Sylow's theorem, $\# \mathscr S > 1$. Again by Sylow's theorem, $\#\mathscr S = 4$ and the elements of $\mathscr S$ are conjugate to each other.  Hence, one can define a group action of $G$ on $\mathscr S$ by conjugation, and this defines a homomorphism $\phi : G\rightarrow \mathrm{Sym}(\mathscr S)\cong S_4$.  Thus, it suffices to show that $\phi$ is injective and its image is $A_4$.
But I'm stuck at this last step.  I tried to find the kernel of $\phi$ and found that $a\in\mathrm{Ker}\phi\Leftrightarrow \forall H\in\mathscr S\ aHa^{-1} = H$, but I do not understand what this leads to.
I would be most grateful if you could provide a clue (not necessarily a complete solution).

REPLY [1 votes]: If you are still asking why the image of $G$ is (contained in) $A_4$, consider the homomorphism $G \to S_4 \to \mathbb{Z}/2$, where the latter map is the parity morphism. If it were surjective, there would be a normal subgroup of order 6 in $G$. This leads to a contradiction.


TITLE: Does there exist a linear transformation from $\mathbb{R}^3 \to\mathbb{R}^3$ such that $T((1,1,1)) = (1,2,3), T((1,2,1) = (1,1,1)$?
QUESTION [2 upvotes]: I tried making the column vectors $(1,1,1)$ and $(1,2,1)$ into a matrix, but because it's $3\times2$, it's noninvertible. Does this mean that I cannot recover the linear transformation without a third vector? Do the vectors that define the transformation $(1,1,1)$ and $(1,2,1)$ have to form a basis for the subspace they're being transformed from for the transformation to be defined? Sorry if I'm mixing things up or wording things in a confusing manner.

REPLY [2 votes]: You're looking for a matrix  $ \begin{pmatrix}
a&b&c\\
e&f&g\\
h&i&j
\end{pmatrix}$ such that $$\begin{pmatrix}
a&b&c\\
e&f&g\\
h&i&j
\end{pmatrix}
\begin{pmatrix}
1\\
1\\
1
\end{pmatrix}
= \begin{pmatrix}
1\\
2\\
3
\end{pmatrix}
$$
and
$$\begin{pmatrix}
a&b&c\\
e&f&g\\
h&i&j
\end{pmatrix}
\begin{pmatrix}
1\\
2\\
1
\end{pmatrix}
= \begin{pmatrix}
1\\
1\\
1
\end{pmatrix}
.$$ If you do the multiplications you get six equations
$$\begin{cases}
a+b+c=1 \\
e+f+g=2 \\
h+i+j=3 \\
a+2b+c=1 \\
e+2f+g= 1 \\
h+2i+j=1
\end{cases}.$$ You can easily find (by subtraction [e.g. $a+b+c=1$ and $a+2b+c=1$]) that $b=0$, $f=-1$, $i=-2$. Then for example choose $a=1, c=0, e=3,g=0, h=5,j=0$ and you can conclude that $$  \begin{pmatrix}
1&0&0\\
3&-1&0\\
5&-2&0
\end{pmatrix}$$ is a solution to your problem.


TITLE: Frequency of wave related with the speed of wave
QUESTION [0 upvotes]: Today I am studying about waves. So, we learn the formula "speed = frequency * wavelength". There is a question come out in my book

If the frequency of water waves is decreased, what will happened to its speed?

So, this is what did,

If speed = frequency * wavelength
That means speed is directly proportional to frequency
When frequency increases, speed increases and vice versa.

Eventually, I write the speed of the water waves will decrease. But the answer from the book is the speed of water waves will stay constant.
So, my question is why the speed of water waves will stay constant if  the frequency of water waves is decreased?

REPLY [2 votes]: I will delay the use of water waves, rather use waves on a string, as in case of water waves, nonlinear effects can be there which can have different dispersion relations like in the case of shallow water/ deep water waves and thereby a different relation for group velocity than you mentioned. So I will derive the case for waves on a string and then sketch the analogous case for water waves, taking into account appropriate approximations. So for waves on a string we have
$$c = \nu \lambda$$
Now we know that the speed of the waves is dependent on the tension and the linear mass density.
$$c = \sqrt{T/\mu}$$ Since neither of them are changing in your example, we know that the speed of the waves is constant. Therefore from the first relation the only thing that changes with change with frequency is the wavelength and not speed.
For water waves, similarly, using appropriate limits if the dispersion relation yields a similar group velocity relation with frequency and wavelength, we can say that since the pressure and the density are not changed, and as a result, the only thing that changes with change with frequency is the wavelength and not speed.

REPLY [0 votes]: The relation $c = \lambda f$ is derived from the propagation equation :
$$
\frac{1}{c^2} \frac{\partial^2u}{\partial t^2} = \frac{\partial^2u}{\partial x^2}
$$
Where $c$ is in a fact a constant, by mathematical nature (it is defined from physical constants, not changing during propagation). So when stating that $c = \lambda f$, it is implicit that $c$ is a constant, which is not very clear in the lesson/book/whatever you are reading.
Therefore, your question's answer is : the wavelength $\lambda$ will increase to compensate for $f$ decrease.
And that's why in mechanical propagation as well as radiowaves, lower frequencies mean higher wavelengths.


TITLE: Circle in the plane becomes cylinder in space?
QUESTION [0 upvotes]: I understand how the equation for example, $x^2 + (y-1)^2 = 5$, describes a circle on a 2d plane; but I was told that in space (3d) the same equation describes a cylinder, and I don't understand that.
Is the idea that in 3d space there is always an implicit $z$, here of any value since not assigned one, and therefore you get a cylinder of infinite height/depth?

REPLY [3 votes]: A not-so-technical explanation:
View the cylinder from bird's eye view, at an arbitrary height $z=h$ (say e.g. $z=3$, or $z=9$). No matter what value of $h$ you prescribe, the cylinder--when viewed orthogonally--will still look like a circle of radius $r$, and it takes the circle equation $x^2+y^2=r^2$, like below:

And this circle is indeed a cylinder that's pointing up or down from the page, at some arbitrary height $z=h$.


TITLE: Expected value of squared random variable if expected value of random variable is $0$
QUESTION [2 upvotes]: Let
$E[X]=0$
them How is Expected value of $E[X^2]$ ?  in my opinion is $0$ But I need confirmation and explanation
In general, if $ (\Omega,\Sigma,P) $ is a probability space and $ X: (\Omega,\Sigma) \to (\mathbb{R},\mathcal{B}(\mathbb{R})) $ is a real-valued random variable, then
$$
\text{E}[X^{2}] = \int_{\Omega} X^{2} ~ d{P}.
$$
see for more about this in this quetion Computing the Expectation of the Square of a Random Variable: $E[X^2]$.

REPLY [0 votes]: No way. Since $X^2\geq 0$, $\mathbb{E}[X^2]=0$ iff $X$ is zero almost surely.
Otherwise, if $\mathbb{E}[X^2]$ is well defined, $\mathbb{E}[X^2]\color{red}{>}0.$


text:
\section{Distances -- Lower bounds}
\label{lowBound}

To prove the lower bounds in Theorem \ref{distThm}, we relate the graph distance $H_N$ to the coalescence of two walkers. By symmetry, we may assume one of the randomly chosen nodes in the initial layer to be at position 0. Next, let $\Xln_h \in \{0, \dots, N - 1\}$ denote the position after $h \ge 0$ steps of a walker starting from $0$ and following the arrows. More precisely, we put recursively $\Xln_0 = 1$, and then $\Xln_{h + 1} \in \{0, \dots, N - 1\}$ such that
$$\Fln_{h + 1} := F_{\Xln_{h + 1}} = \max_{i:\, d_N(i,\Xln_h) \le \lc\Fln_h\rc}F_{i, h}.$$
Additionally,
$$\Gln_h := \lm(2\Fln_h)$$
denotes the log-fitness. Similarly to $\{\Xln_h\}_{h \ge 0}$, we let $\{\Xrn_h\}_{h \ge 0}$ denote the walker started from a {uniformly chosen } random position in layer 0.

To prove the lower bound of Theorem \ref{distThm}, note that for every $h \ge 0$ and $\eta > 0$,
\begin{align}
	\P(H_N \le h) &\le \P\bi(d_N\bi(0, \Xln_h\bi) \ge \eta N\bi) + \P\bi(d_N\bi(\Xrn_0, \Xrn_h\bi) \ge \eta N\bi) + \P\bi(d_N\bi(0, \Xrn_0\bi) \le 2\eta N\bi) \nonumber \\
	&\le  2\P\bi(d_N\bi(0, \Xln_h\bi) \ge \eta N) + 2\eta. \label{eqHnBd}
\end{align}
For $\de \ge1$, we establish highly accurate upper bounds on the growth of the fitnesses $\Fln_h$ as a function of the layer $h$. Consequently, we also obtain bounds on the location
\begin{align}
	\label{locBoundEq}
	d_N(0, \Xln_h) \le \Fln_0 + \cd + \Fln_{h - 1} + h
\end{align}
after $h$ steps, where the addition of $h$ on the right hand side arises from rounding.

Using the same fitnesses attached to the nodes yields a natural coupling between the model on a finite torus and the limit model on the integers $\Z$. We write $\{\Fl_h\}_{h \ge 0}$ for the fitnesses in this limit model. As long as $\vp(\Fln_i)\le N$, the wrapping around the torus is not observable so that, for $\eta\in(0,1)$,
$$\Fln_0 + \cd + \Fln_{h - 1} + h \ge \eta N$$
if and only if in the coupled limit model
$$\Fl_0 + \cd + \Fl_{h - 1} + h \ge \eta N.$$
Hence, it suffices to study the limit model.

\subsection{Proof for $\de = 1$.}
\label{lowCritSec}
To bound the right-hand side in \eqref{locBoundEq}, we show that the log-fitnesses concentrate sharply around the current layer $h$.

\begin{lemma}[Fluctuations of log-fitnesses]
	\label{critTightLem}
	For $\delta=1$ and $\Gl_h :=\log_\mu(2\Fl_h)$,
	$$\P\bi(\limsup_{h \to \i} h^{-2/3}\bi|\Gl_h - h\bi| = 0\bi) = 1.$$
\end{lemma}
Before proving Lemma \ref{critTightLem}, we explain how to derive from it the lower bound on the distances.

\begin{proof}[Proof of Theorem \ref{distThm}; lower bound; $\de = 1$]
	Let $I \ge 0$ be an almost surely finite random variable such that, by Lemma \ref{critTightLem}, $\bi|\Gl_h - h\bi|\le h^{2/3}$ holds for all $h \ge I$.
	Then, inserting into \eqref{locBoundEq},
	\begin{align*}
		\P\Big(h + \sum_{i < h}\Fl_i \ge \eta N\Big) &\le \P\Bi(\sum_{i \le I} \Fl_i > \eta N /2\Bi) + \one\Bi\{h + \sum_{i \le h}\mu^{i + i^{2/3}} > \eta N/2\Bi\}\\
		&\le \P\Bi(\sum_{i \le I} \Fl_i > \eta N /2\Bi) + \one\bi\{h + h\mu^{h + h^{2/3}} > \eta N/2\bi\}.
	\end{align*}
The probability on the right hand side vanishes as $N\to\infty$. Further, we fix $\e' > 0$ and let $h = \lm(N) (1- \e')$, then the indicator vanishes as well in the limit $N\to\infty$ for arbitrary $\eta>0$.
Inserting this into \eqref{eqHnBd} proves the claim.
\end{proof}

The key towards obtaining the bounds on the scopes in Lemma \ref{critTightLem} is the max-stability of the Fr\'echet distribution:
 if $F_1, \dots, F_m$ are i.i.d.\ Fr\'echet random variables with tail index $\de = 1$, then $\max\{F_1, \dots, F_m\}$ has the same distribution as $mF$, where $F$ is again a Fr\'echet random variable with tail index 1. Moreover, $G_i = \lm(2F_i)$ follows a Gumbel distribution.

In particular, writing $m = \vp(\Fl_h)$ (recall \eqref{eqPhiDef} for the definition of $\vp$) and $F = F_{h + 1}$, we represent $\Fl_{h+1}$ recursively as
	\begin{align}
		\label{cDiffEq}
		\Fl_{h + 1} = \vp(\Fl_h)F_{h + 1},
	\end{align}
	so that
	\begin{align}
		\label{cDiffEq2}
		0 \le \Gl_{h + 1} - \Gl_h - G_{h + 1}
		\le \lm\!\left(\f{1 + 2\lceil \Fl_h\rceil}{2\Fl_h}\right)
		\le \lm(\r_h),
	\end{align}
	where $\r_h := 1 + \f3{2\Fl_h}.$
	Starting from this observation, we now prove Lemma \ref{critTightLem}.

	\begin{proof}[Proof of Lemma \ref{critTightLem}]
		In order to develop an intuition for the proof, we first establish the asserted lower bound on the growth. That is,
		$$\P\bi(\li_{h \to \i} h^{-2/3}(\Gl_h - h) \ge 0\bi) = 1.$$
		Indeed, applying the bound \eqref{cDiffEq2},
		$$\Gl_h - h\ge \sum_{i \le h} (G_i - 1),$$
		where $\{G_h - 1\}_{h \ge 0}$ is an i.i.d.\ sequence of centered random variables with finite exponential moments. Hence, by moderate deviations \cite[Theorem 11.2]{mdp}, almost surely,
		$$\lim_{h \to \i}h^{-2/3}\sum_{j \le h}(G_j - 1) = 0.$$

		Moreover, now the lower bound on the growth of $\Gl_h$ implies that the error terms of the form $h^{-2/3}\sum_{j \le h} \lm(\r_h)$ in \eqref{cDiffEq2} tend to 0 as $h \to \i$, thereby concluding the proof.
	\end{proof}

\subsection{Proof for $\de < 1$.}
\label{lowHeavSec}

In the heavy-tailed setting, we need a strong tightness property for the log-fitnesses.

\begin{lemma}[Tightness for heavy tails]
	\label{heavTightLem}
	For $\delta<1$,
	$$\P\bi(\lim_{h \to \i}\de^h\Gl_h \in (0, \i) \bi) = 1.$$
\end{lemma}

First, we explain how Lemma \ref{heavTightLem} enters the proof of the lower bound.
\begin{proof}[Proof of Theorem \ref{distThm}; lower bound; $\de < 1$]
	Let $I$ be a strictly positive random variable such that $\de^h\Gl_h \le I$ holds almost surely for all $h \ge 0$.
	Hence, writing $h = \ll(N) - K$ and inserting the bound from Lemma \ref{heavTightLem} into the representation from \eqref{locBoundEq} gives that
	\begin{align*}
\P\big(h + \sum_{j < h}\Fl_i \hspace{-.05cm}>\hspace{-.05cm} \eta N\big)
		 \le \P\bi(h + \sum_{j < h}\mu^{\de^{-j}I} \hspace{-.05cm}>\hspace{-.05cm} \eta N\bi)
		 \le \P\bi(h + h\mu^{\de^{-h}I} \hspace{-.05cm}>\hspace{-.05cm} \eta N\bi)
		 = \P\bi(h(1 + N^{\de^K I}) \hspace{-.05cm}>\hspace{-.05cm} \eta N \bi).
	\end{align*}
	We conclude the proof by noting that the right-hand side tends to 0 if we first take $N$ sufficiently large and then let $K$ tend to $\i$.
\end{proof}

It remains to show Lemma \ref{heavTightLem}. The proof mimics the arguments presented in Lemma \ref{critTightLem}. Therefore, we present in detail only those arguments that are substantially different. The key identity now reads
	\begin{align}
		\label{hDiffEq}
		\Fl_{h + 1} = \bi(\vp(\Fl_h)F_{h + 1}\bi)^{1/\de},
	\end{align}
	so that, as in \eqref{cDiffEq2},
	\begin{align}\label{hDiffEq2}
		0 \le \de\Gl_{h + 1} - \Gl_h - G_{h + 1} \le \lm(\r_h),
	\end{align}
	where $\r_h := 1 + \f3{2\Fl_h}.$
\begin{proof}[Proof of Lemma \ref{heavTightLem}]
 First, we iterate \eqref{hDiffEq2} to get that for every $h_2 \ge h_1 \ge 1$,
	\begin{equation}
		\label{heavRecEq}
		0 \le 		\de^{h_2}\Gl_{h_2} - \de^{h_1}\Gl_{h_1} - \sum_{h_1 \le j < h_2}\de^jG_{j + 1}  \le   \sum_{h_1 \le j < h_2 }\de^j\lm(\r_j).
	\end{equation}
Now,	the key step is to show that
		\begin{align}
			\label{heavLimEq}
			\P\bi(\li_{h \to \i}\de^h\Gl_h > 0 \bi) = 1.
		\end{align}
		Then, almost surely,
		$$\lim_{h \to \i}\sup_{h_2 \ge h_1 \ge h}\sum_{h_1 \le j < h_2} \de^j \lm(\r_j) = 0.$$
		Moreover, by the Borel-Cantelli lemma also
		$$\sup_{h_2 \ge h_1 \ge h}\sum_{h_1 \le j < h_2} \de^j G_{j + 1}$$
		tends to 0 almost surely as $h \to \i$. Hence, $\de^h\Gl_h$ converges to an almost surely finite limit.

		It remains to show \eqref{heavLimEq}. To achieve this goal, we assert that there exists an almost surely finite random variable $I$ such that
		\begin{align}
			\label{kol_eq}
			\min\Bi\{\Gl_I, \sum_{j \ge I}\de^j G_{j + 1}\Bi\}> 0.
		\end{align}
		Once \eqref{kol_eq} is established, we obtain that
	\begin{align*}
		\de^h\Gl_{h + 1}  \ge  \de^I\Gl_I + \sum_{I \le j < h + 1} \de^jG_{j + 1}\ge \sum_{I \le j < h + 1}\de^jG_{j + 1},
		\end{align*}
		so that taking the limit as $h \to \i$ concludes the proof.

		To prove \eqref{kol_eq}, we may first apply the Borel-Cantelli lemma to see that the sum $\sum_{j \ge 0}\de^j G_{j + 1}$ converges almost surely. Hence,
	$$\inf_{i \ge 1}\P\Bi(\min\Bi\{G_i, \sum_{j \ge i}\de^j G_{j + 1}\Bi\}> 0\Bi) = \P\Bi(\min\Bi\{G_1, \sum_{j \ge 1}\de^j G_{j + 1}\Bi\}> 0\Bi) > 0.$$
	In particular, the Kolmogorov 0-1-law yields almost surely finite random variable $I$ such that
			$$\min\Bi\{G_I, \sum_{j \ge I}\de^j G_{j + 1}\Bi\}> 0.$$
			Since $\Gl_i \ge G_i$ for every $i \ge 1$, this observation concludes the proof of \eqref{kol_eq}.
\end{proof}

\ssec{Proof for $\de > 2$.}
\label{lowLightSec}
In the light-tailed setting, we show that the suitably rescaled walker $\{\Xl_h\}_{h \ge 0}$ satisfies the invariance principle.

\begin{lemma}[Invariance principle]
	\label{invPrinLem}
	Let $\de > 2$.
	{Then, $\big\{h^{-1/2}\Xl_{ht} \big\}_{t \le 1}$ converges in distribution as $h\to \ff$ to some Brownian motion $\{B_t\}_{t \le 1}$.}
\end{lemma}

{Throughout we write $ht$ for $\lfloor ht\rfloor$.} Before establishing Lemma \ref{invPrinLem}, we explain how to conclude the proof of the lower bound. {Mind that, for the lower bound, a central limit theorem suffices. However, for the proof of the upper bound in the next section, we need a full functional CLT. }
\begin{proof}[Proof of Theorem \ref{distThm}; lower bound; $\de > 2$]
	The invariance principle in the form of Lemma \ref{invPrinLem} {for $t=1$} gives
	$$\lim_{\e \to 0}\lim_{N \to \ff}\P(\Xl_{\e N^2} \ge \e^{1/4} N) = \lim_{\e \to 0}\lim_{N \to \ff}\P(\Xl_{\e N^2}(\e N^2)^{-1/2} \ge \e^{-1/4} )= 0,$$
	as asserted.
\end{proof}

In order to prove Lemma \ref{invPrinLem}, we rely on the general martingale functional CLT from \cite[Theorem D.6.4]{meyn}. To cast this problem in the setting of the present context, we let
$$\FF_h = \s(\{\Fl_j, \Xl_j\}_{j \le h}).$$
denote the information provided by the positions of the walker and the corresponding fitnesses up to layer $h$. Then, $M_h := \Xl_h$ form a square-integrable martingale with respect to the filtration $\{\FF_h\}_{h \ge 0}$. In order to apply \cite[Theorem D.6.4]{meyn}, we need to verify two conditions.

\begin{enumerate}
	\item[]{\bf (M1).} Almost surely,
		$$\lim_{h \to \ff} \f1h \sum_{j \le h} \E[(M_j - M_{j - 1})^2\, |\, \FF_{j - 1 }] = \g^2,$$
		for some constant $0 < \g^2 < \ff$.
	\item[]{\bf (M2).} Almost surely, for every $\e > 0$,
		$$\lim_{h \to \ff} \f1h \sum_{j \le h} \E[(M_j - M_{j - 1})^2\one\{(M_j - M_{j - 1})^2 \ge \e n\}\, |\, \FF_{j - 1}] = 0.$$
\end{enumerate}
Note that, when fixing any $\e_0 > 0$, condition {\bf (M2)} follows from the following Lyapunov-type condition.
\begin{enumerate}
	\item[]{\bf (M2').} Almost surely,
		$$\lim_{h \to \ff} \f1{h^{1 + \e_0}} \sum_{j \le h} \E[(M_j - M_{j - 1})^{2 + \e_0}\, |\, \FF_{j - 1}] = 0.$$
\end{enumerate}

Before establishing the invariance principle for the random walk $\{\Xl_h\}_{h \ge 0}$, we first show that the underlying Markov chain of fitnesses $\{\Fl_h\}_{h \ge 0}$ satisfies a Foster-Lyapunov drift condition, thereby forming the basis for a Markov-chain LLN.
\bel[Drift condition]
Let $\de > 1$, $\a < \de$ and set $V(f) = 1 + f^\a$. Then, there exists $K = K(\a) > 0$ such that for all $f > 0$,
	\begin{align}
		\label{driftEq}
		\E\bi[V(\Fl_1)\,|\, \Fl_0 = f\bi] \le \f12 V(f) + K \one\{f \le K\}.
	\end{align}
\enl
\bep
	To bound
	$ \E\bi[V(\Fl_1)\,|\, \Fl_0 = f\bi],$
	we leverage recursion \eqref{hDiffEq} to deduce that for every $f > 3$,
	$$\E\bi[(\Fl_1)^\a\,|\, \Fl_0 = f\bi] = \vp(f)^{\a/\de} \E\bi[F_1^{\a/\de}\bi] \le  \E\bi[F_1^{\a/\de}\bi] 3^{\a/\de} f^{\a/\de}.$$
	Hence, since $\de > 1$, there exists $K > 3$ such that
	$$\E\bi[(\Fl_1)^\a\,|\, \Fl_0 = f\bi] \le \f12 f^\a$$
	holds for all $f > K$, thereby verifying the drift equation \eqref{driftEq}.
\enp

Now, we have collected all ingredients to prove the invariance principle, i.e., Lemma \ref{invPrinLem}.
\begin{proof}[Proof of Lemma \ref{invPrinLem}]\phantom{a}~\\
	{\bf Condition (M1).}
First, $\E[(M_j - M_{j - 1})^2\, |\, \FF_{j - 1}] = w(\Fl_j)$, where
	$$w(r) = \f{\vp(r)^2 - 1}{12}$$
	denotes the variance of the uniform distribution on $\vp(r)$ consecutive integers.
	Then, condition {\bf (M1)} becomes
	\begin{align}
		\label{m1Eq}
		\lim_{h \to \ff} \f1h \sum_{j \le h} w(\Fl_j)  = \g^2.
	\end{align}
	This is a prototypical Markov-chain LLN that follows from the drift condition \eqref{driftEq} through \cite[Theorem 17.0.1]{meyn}. More precisely, we deduce from \eqref{driftEq} and \cite[Theorem 9.1.8]{meyn} that the chain $\{\Fl_h\}_{h \ge 0}$ is Harris recurrent. Next, by \cite[Theorem 14.0.1]{meyn}, it is also positive recurrent with an invariant measure $\pi$ satisfying  $\int_0^\ff x^2 \pi(\d x) < \ff$.  Hence, the asserted LLN in \eqref{m1Eq} follows  from \cite[Theorem 17.1.7]{meyn}.

	{\bf Condition (M2').} Similarly, let now $w_{\e_0}(r)$ denote the centered $(2 + \e_0)$-th moment of a uniform random variable on $\vp(r)$ consecutive integers. Then, {\bf (M2')} becomes
	$$\lim_{h \to \ff} \f1{h^{1 + \e_0}} \sum_{j \le h} w_{\e_0}(\Fl_j) = 0,$$
	which again follows from the Markov LLN \cite[Theorem 17.1.7]{meyn}.
\end{proof}

\ssec{Proof for $\de \in (1, 2)$}
\label{stab_sec}
Finally, we deal with the stable case, i.e.,  $1 < \de < 2$. In the light-tailed setting, a key ingredient was the invariance principle in the form of Lemma \ref{invPrinLem}, which stated that the rescaled walker $\{h^{-1/2}\Xl_{ht}\}_{ t \in [0, 1]}$ converges to Brownian motion as $h \to \ff$.
Now, we need a stable analog of this result. More precisely, we establish convergence to a symmetric $\de$-stable process with L\'evy measure
\begin{align}
	\label{levEq}
	\f{\nu(\d x)}{\d x} =  \f{c(\de)}{|x|^{\de + 1}} \one\{ x \ne 0\},
\end{align}
for some $c(\de) > 0$.

\begin{lemma}[Stable limit]
	        \label{stabPrinLem}
			Let $\de \in (1, 2)$. Then, $\big\{h^{-1/\de}\Xl_{ht} \big\}_{t \le 1}$ converges in distribution to symmetric $\de$-stable processes.
\end{lemma}

Before establishing Lemma \ref{stabPrinLem}, we elucidate how it gives the tightness of $(H_N / N^\de)$ away from 0. Essentially, this relies on the same line of arguments that we have seen in Section \ref{lowLightSec}.

\begin{proof}[Proof of Theorem \ref{distThm}; lower bound; $\de \in (1, 2)$]
Invoking Lemma \ref{stabPrinLem} gives that
			$$\lim_{\e \to 0}\lim_{N \to \ff}\P(\Xl_{\e N^\de} \ge \e^{1/(2\de)} N) = \lim_{\e \to 0}\lim_{N \to \ff}\P(\Xl_{\e N^\de}(\e N^\de)^{-1/\de} \ge \e^{-1/(2\de)} )= 0,$$
			        as asserted.
\end{proof}

In order to prove Lemma \ref{stabPrinLem}, we proceed as in \cite{rem} and apply the versatile functional limit theorem \cite[Theorem 4.1]{resnick}.

We now formulate a version of \cite[Theorem 4.1]{resnick}, where we adapted (actually simplified) the conditions to our needs:
Let $\{Z_{j, h}\}_{1 \le j \le h}$ be a triangular array of centered random variables and let $\{\FF_{j, h}\}_{1 \le j \le h}$ be a triangular array of $\s$-algebras such that $Z_{j, h}$ is $\FF_{j, h}$ measurable.
Now, assume the following conditions:
\begin{enumerate}
	\item[]{\bf (D1)} There exists a symmetric measure $\nu$ such that for all $x > 0$ and $t \le 1$,
		$$\sum_{j \le ht}\P(Z_{j, h} > x\,|\, \FF_{j-1,h}) \to t \nu([x, \ff))
		\qquad\text{ in probability as $h \to \ff$.}$$
        \item[]{\bf (D2)}
                For $\e > 0$,
		$$\sum_{j \le h}\P(|Z_{j, h}| > \e\,|\, \FF_{j-1,h})^2 \to 0
		\qquad\text{ in probability as $h \to \ff$.}$$
        \item[]{\bf (D3)}
                For $\eta, \e > 0$,
		$$\lim_{\eta \to 0}\ls_{h \to \ff}\P\Big(\sum_{j \le h}\E\big[Z_{j, h}^2\one\{|Z_{j, h}| \le \eta\}\,| \FF_{j-1,h}\big] > \e\Big) = 0.$$
\end{enumerate}
Then, $\big\{\sum_{j \le ht}Z_{j, h}\big\}_{t \le 1}$ converges in distribution as $h\to\infty$ to a symmetric stable process with L\'evy measure $\nu$.

We now use this criterion to prove convergence with the $\delta$-stable L\'evy measure $\nu$ as in \eqref{levEq}.
To this end, we set
$$Z_{j, h} := h^{-1/\de}(\Xl_j - \Xl_{j - 1})$$
and let
$$\FF_{j, h}:= \s(Z_{1, h}, \dots, Z_{j, h}, \Fl_{1, h}, \dots, \Fl_{j - 1, h})$$
be the $\s$-algebra generated by the increments up to layer $h$ and the fitnesses up to layer $h - 1$.
To verify conditions {\bf (D1)}--{\bf (D3)}, we rely on explicit computations with Fr\'echet random variables that we present as a separate auxiliary result.
\bel[Fr\'echet computations]
\label{fre_comp_lem}
Let  $F$ be a standard Fr\'echet random variable with tail index 1. Then,
\been
\im for every $\de > 1$,
$$	\lim_{a \to \ff} a\E[(1 - (a/F)^{1/\de})_+] = \f1{\de + 1},$$
\im for every $\de \in (1, 2)$ and $\eta > 0$,
	$$\lim_{a \to \ff}a\E\big[\eta^2  \wedge (F/a)^{2/\de}\big] = \f{2\eta^{2 - \de}}{2 - \de}.$$
\enen
\enl

\medskip
We postpone the proof of the lemma and first show how it implies the proof of Lemma \ref{stabPrinLem}.

\subsubsection*{Verification of condition {\bf (D1)}}
\label{cd1_sec}
To verify condition {\bf (D1)}, we need to compute the conditional expectation $\P(Z_h > x\,|\,\FF_{j-1,h})$. Now, similar to the proof for $\de>2$, the key insight is that $\Xl_j$ is distributed uniformly in the scope of size $\vp(\Fl_{j - 1})$. As an initial observation, we note that $\max_{j\le h}\Fl_j / h\in o_h(1)$ with high probability. Indeed, by the Markov inequality, for any $\a \in (1, \de)$,
$$\P(\max_{j \le h} \Fl_j > h) \le \sum_{j \le h}\P((\Fl_j)^\a > h^\a) \le \f1{h^\a}\sum_{j \le h}\E[(\Fl_j)^\a],$$
so that similarly to the arguments in Section \ref{lowLightSec}, we may invoke the Markov ergodic theorem, \cite[Theorem 14.0.1]{meyn}.

We also recall from \eqref{hDiffEq} that
$\Fl_{j - 1} = \vp(\Fl_{j - 1})^{1/\de}F_{j - 1}^{1/\de}$.
Hence, by part (1) of Lemma \ref{fre_comp_lem},
\begin{align}
	\P(Z_{j, h} > x\,|\, \FF_{j-1,h}) &= \E\Big[\Big(1 - \f{h^{1/\de}x}{2\vp(\Fl_{j - 2})^{1/\de}F_{j - 1}^{1/\de}}\Big)_+\,\big|\, \FF_{j-1,h} \Big]\,(1 + o_h(1)) \nonumber \\
	&= \f{\Fl_{j - 2}}{2^\de hx^\de(\de + 1)}(1 + o_h(1)).\label{eqPZkBd}
\end{align}
Finally, as in the computations in Section \ref{lowLightSec}, we deduce that there is an LLN, so that
$\f1h\sum_{j \le h}\Fl_{j - 2}$
converges weakly.

\subsubsection*{Verification of condition {\bf (D2)}}
We use \eqref{eqPZkBd} to get
$$\P(Z_{j, h} > x\,|\, \FF_{j-1,h}) = \f{\Fl_{j - 2}}{2^\de hx^\de(\de + 1)} (1 + o_h(1)).$$
We observed already before that $\max_{j\le h}\Fl_j / h\in o_h(1)$ with high probability. Hence,
$$\max_{j \le h}\P(Z_{j, h} > x\,|\, \FF_{j-1,h}) \to 0,$$
which, together with \textbf{(D1)}, implies the desired claim.

\subsubsection*{Verification of condition {\bf (D3)}}
Finally, we show that

	$$\lim_{\eta \to 0}\ls_{h \to \ff}\P\Big(\sum_{j \le h}\E\big[Z_{j, h}^2\one\{|Z_{j, h}| \le \eta\}\,| \FF_{j-1,h}\big] > \e\Big) = 0.$$
		First, conditioned on the event $\{|Z_{j, h}| \le \eta\}$, the increment $Z_{j, h}$ is uniformly distributed in an interval of length $\eta \wedge \Fl_{j - 1}h^{-1/\de}$. Hence, it suffices to show that
	\begin{align}
		\label{d3_eq}
		\ls_{h \to \ff}\P\Big(\sum_{j \le h}\E\big[\eta^2  \wedge(\Fl_{j - 1}h^{-1/\de})^2\,| \FF_{j-1,h}\big] > \e\Big)
	\end{align}
	tends to 0 as $\eta \to \ff$.
By part (2) of Lemma \ref{fre_comp_lem}, the conditional expectation inside the probability becomes
$$\f{2\eta^{2 - \de}\Fl_{j - 2}}{(2 - \de)hx^\de} $$
We may once more cite the Markov LLN for the weak convergence of
$\f1h\sum_{j \le h}{\Fl_{j-2}}$
to deduce that \eqref{d3_eq} tends to 0 as $\eta \to \ff$.
\qed

\medskip

It remains to establish the limits in Lemma \ref{fre_comp_lem}.
\bep[Proof of Lemma \ref{fre_comp_lem}]\phantom{a}~\\

{\bf Part (1).} Integration with respect to the Fr\'echet density yields that
\begin{align*}
	\E[(1 - (a/F)^{1/\de})_+] &= \int_a^\ff (1 - (a/x)^{1/\de})x^{-2}\exp(-x^{-1}) \d x.
\end{align*}
For large $x$, the exponential factor approaches 1 and
$$\int_a^\ff (1 - (a/x)^{1/\de})x^{-2} \d x= a^{-1} - \f\de{\de + 1}a^{-1} = \f1{\de + 1} a^{-1}.$$

{\bf Part (2).}
	We split the  expectation depending on which of the two contributions in the minimum becomes relevant.
First, as in part (1),
	\begin{align}
		\label{fre_comp_eq}
		\lim_{a \to \ff} a\eta^2  \P(F \ge a \eta^\de) = \eta^{2 - \de}.
	\end{align}
	Hence, it remains to determine
	$$\lim_{a \to \ff}a^{1 - 2/\de}\E\big[F^{2/\de} \one\{F \le a \eta^\de\}\big].$$
	By l'H\^opital's rule, we see that
	$$
	\lim_{a \to \ff}(a\eta^\de)^{1 - 2/\de}\int_0^{a \eta^\de} x^{2/\de}x^{-2}\exp(-x^{-1}) \d x = \f\de{2 - \de}.
	$$
	Combining the latter with the result in \eqref{fre_comp_eq} concludes the proof.
\enp