TITLE: Functional weakly lower-semicontinuous
QUESTION [3 upvotes]: If $X$ is a topological space, then a functional $\varphi:X\to\mathbb{R}$ is lower-semicontinuous (l.s.c) if $\varphi^{-1}(a,\infty)$ is open in $X$ for any $a\in\mathbb{R}$.
If $X$ is a Hilbert space, then $\varphi$ is weakly l.s.c if it is l.s.c on $X$ with its weak topology.
My question: If $X$ is a Hilbert space and $\varphi:X\to\mathbb{R}$, then
$\varphi(x)\leq \liminf{x_n}$ whenever $x_n$ converges weakly to $x$ $\Rightarrow$ $\varphi$ weakly l.s.c?
I read this in "A invitation to Variational Methods in Differential Equations". It isn't a exercise, maybe a definition.

REPLY [4 votes]: I think the author is being a little sloppy in this paragraph, which raised your question:

Lower semicontinuity was earlier defined by the openness of upper level sets. The weak topology being non-metrizable, one should not casually insert "in other words" followed by the definition of sequential weak lower semicontinuity.
For example, Ciarlet (Mathematical Elasticity: Three-dimensional elasticity, Volume 1) is rather clear on this point.

I have not seen anyone actually using  weak lower semicontinuity. It's sequential weak lower semicontinuity that's being used all the time, and because of this, many authors  drop "sequential" altogether. Unsurprisingly, confusion ensues.
Unfortunately I do not have an example of a sequentially weakly lsc functional on a Hilbert space that is not weakly lsc.


TITLE: Need a second help in understanding of theorem7.1 in chapter2 in Israel Gohberg.
QUESTION [0 upvotes]: I do not understand why $I - (I - K)^{-I}$ is of finite rank, could anyone explain this for me please?

REPLY [1 votes]: Let $A(y)$ be the matrix on the RHS of $(7.1)$.
$A$ is of finite rank !
We have, by $(7.1)$:
$(I-K)^{-1}y=y- \frac{1}{\det(a_{ij})}A(y)$,
hence
$I-(I-K)^{-1}=\frac{1}{\det(a_{ij})}A$.


TITLE: Use projection to create $A\in M_{2,2}(\mathbb{Q})$ with $A\cdot A=A$ and $A\cdot\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}$
QUESTION [0 upvotes]: How to use

Vector space $V = U \oplus W$
Linear map $f: V \to V, x \mapsto \left \{ \begin{matrix} x ~~, & x \in U \\ 0 ~~, & x \in W \\ \end{matrix} \right. $
$f = f \circ f$

to create

Matrix $A \in M_{2,2}(\mathbb{Q})$
$A \cdot A = A$
$A \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

?

REPLY [0 votes]: We know that the dot product with an unit vector gives back the projection in the direction of that unit vector. This means that the mapping
$$x \mapsto e(e \cdot x)$$
Is the linear projection to $e$. If we write it out component-wise, we get that
$$x_i =\sum_j e_i (e_j x_j)=\sum_j e_i e_j x_j$$
We can notice that the $(i,j)$-th component of the projection matrix is $e_i e_j$.
In your case: $e=\frac{1}{\sqrt{2}} [1,1]^t$, which means that $e_i e_j=\frac{1}{2}$.


TITLE: Deriviative of natural log help finding
QUESTION [1 upvotes]: $$y=7\ln\frac{11}x$$
I need to use the product rule please
$$\frac{d}{dx} (7) \cdot(\ln(11/x) + \frac{d}{dx}\left(\ln\frac{11}x\right) \cdot 7$$
then
$$0+\frac{d}{dx}\frac{11}x \cdot7$$
what do i do

REPLY [2 votes]: $$
\frac{d}{dx} \ln \frac{11}x = \frac{d}{dx}(\ln 11 - \ln x) = 0 - \frac1x.
$$
The derivative of the first term is $0$ since $\ln 11$ is a constant.
You can also use the chain rule:
$$
\frac{d}{dx}\ln\frac{11}x = \frac{1}{11/x}\cdot\frac{d}{dx}\frac{11}x
$$
and then you have to find that last derivative.  When you're done simplifying you'll get the same answer.  But this method is more complicate than what is necessary, since you can use the first method above.
You don't really need the full-fledged product rule since $7$ is a constant, so you can say
$$
\frac{d}{dx}\left(7\ln\frac{11}x\right) = 7\frac{d}{dx}\ln\frac{11}x.
$$


TITLE: Integral $\int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx$
QUESTION [2 upvotes]: I would like to ask how to integrate the following integration.
\begin{gather*}
    \int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx
\end{gather*}
I try to use integration by parts and I stuck with the following
\begin{gather*}
   \frac{\ln(x^2+1)}{x} - \int_{0}^{1} d\frac{\ln(x^2+1)}{x^2} \
\end{gather*}
I am confused with the continue step.

REPLY [0 votes]: Good idea using the integration by part.
Let's recall that integration by Parts is a special method of integration that is often useful when two functions are multiplied together.
At first let us see the rule: $$\int f(x)g(x) dx=u(x)\int v(x) dx-\int u'(x) \Big(\int v(x) dx\Big) dx$$
Now for your problem apply this rule with $v(x)=\frac{1}{x^2}$ and $u(x)=\log{(1+x^2)}$.  So:
$$\int \frac{\log{(1+x^2)}}{x^2}=\log{(1+x^2)}\int \frac{1}{x^2} dx-\int (\log{(1+x^2)})' \Big(\int \frac{1}{x^2} dx\Big) dx=$$ $$=-\frac{1}{x}\log{(1+x^2)}+\int \frac{1}{x}\frac{2x}{1+x^2}=-\frac{1}{x}\ln{(1+x^2)}+2\arctan{(x)}+c$$ Obviously I have shown the indefinite integral but in your case you have to evaluate the integral with the extremes ($0$ and $1$) imposed, in fact your problem I think was the integration itself.


text:
\begin{document}

\title{Quantitative propagation of chaos for generalized Kac particle systems}

\author{Roberto Cortez\footnote{Department of Mathematical Engineering and Center for Mathematical Modeling,  UMI(2807) UCHILE-CNRS, Universidad de Chile. Casilla 170-3, Correo 3, Santiago-Chile.} \footnote{Supported by Proyecto Mecesup UCH0607 Doctoral Fellowship. Email: \texttt{rcortez@dim.uchile.cl}}  \, and Joaquin Fontbona\footnotemark[1] \footnote{Supported by Fondecyt Grant 1110923,  Basal-CONICYT Center for Mathematical Modeling (CMM), and  Millenium Nucleus  NC120062. Email: \texttt{fontbona@dim.uchile.cl}}
}

\maketitle

\begin{abstract}
We study a class of one dimensional  particle systems with true (Bird type) binary interactions, which includes Kac's  model of the Boltzmann equation and  nonlinear equations for the evolution of wealth distribution  arising in kinetic economic models. We obtain explicit rates of convergence for the Wasserstein distance between the law of the particles and their limiting law, which are linear  in time and depend in a mild polynomial manner  on the number of particles. The proof is based on a novel coupling between the particle system and a suitable system of non-independent nonlinear processes, as well as  on recent sharp estimates for empirical measures.
\end{abstract}

\noindent {\bf Keywords:} propagation of chaos, Kac equation, wealth distribution equations, stochastic particle systems, Wasserstein distance, optimal coupling.

\medskip
\noindent {\bf AMS classification (2010):} 60K35, 82C22, 82C40.

\section{Introduction and main result}

\subsection{The kinetic equation}
We consider the collection $(P_t)_{t\geq 0}$ of probability measures on $\RR$, solution of the following nonlinear kinetic-type equation:
\begin{equation} \label{eq:Pt}
 \partial_t P_t = - P_t + \Q^+ (P_t).
\end{equation}
Here $\Q^+$ is a generalized Wild convolution, which associates with every measure $\mu$ on $\RR$ a new measure $\Q^+(\mu)$ given by
\begin{equation} \label{eq:Q+}
  \int \phi(u) \Q^+(\mu)(du) = \iint \frac{1}{2}\ELR (\phi(\L u + \R v) + \phi(\tL v + \tR u)) \mu(dv) \mu(du),
\end{equation}
for all bounded  measurable functions $\phi$, where $(\L,\R,\tL,\tR)$  is a given  random vector in $\RR^4$ (with known distribution) and $\ELR$ denotes  the expectation with respect to it.

Equations \eqref{eq:Pt}-\eqref{eq:Q+} describe the behaviour of an infinite  number of  objects or ``particles'' subjected to binary interactions. The state of each particle is characterized by a scalar $u\in\RR$, and $P_t(du)$ represents the proportion of particles  in state $u$
at time $t\geq 0$. The microscopic binary interactions, which occur randomly at constant rate, are heuristically described as follows: when a particle  at state $u$ interacts with
a particle at state  $v$,  their states  change according to the rule
\begin{equation}
\label{eq:inter_rule}
(u,v) \mapsto (\L u + \R v, \tL v + \tR u).
\end{equation}
This model is a generalization of Kac's  one dimensional  simplification of the (more realistic) Boltzmann equation for a spatially homogeneous dilute gas in $\RR^3$, in which the interacting objects represent actual physical particles.
Specifically,  in Kac's model introduced in \cite{kac1956}, the state of a particle is its one dimensional velocity, and the  interactions correspond to random exchanges of velocities that occur  at binary collisions that preserve kinetic energy, so that $\L = \cos \theta= \tL$, $\R = -\sin \theta=-\tR$, with $\theta$ randomly chosen in $[0,2\pi)$.
We refer the reader to \cite{mischler-mouhot2013} and the references therein for historical background on Kac and Boltzmann's equations.

A further source of models of the type described by equations \eqref{eq:Pt}-\eqref{eq:Q+} is the  kinetic  description  of the evolution of the wealth distribution in a simplified economy, studied for instance in  \cite{matthes-toscani2008} (see also the references therein). In that setting the state of a particle represents the wealth of an economic agent, and the binary interactions correspond to trades or  economic exchanges between them. Early versions of that model assumed
\begin{equation}
\label{eq:LR}
|\L|^p + |\tR|^p = 1 \quad \text{a.s.}, \qquad |\tL|^p + |\R|^p  = 1 \quad \text{a.s.},
\end{equation}
for some $p\geq 1$ (notice that in  Kac's model  \eqref{eq:LR}  is satisfied with $p=2$). In the case $p=1$,  for non-negative $\L$, $\R$, $\tL$ and $\tR$, condition \eqref{eq:LR} can be seen as \emph{exact} conservation of total wealth in each interaction. The weaker condition
\begin{equation}
\label{eq:ELR}
\ELR(|\L|^p + |\tR|^p) = 1,  \qquad \ELR(|\tL|^p + |\R|^p)  = 1,
\end{equation}
interpreted as conservation of wealth only \emph{in the mean}  (so that risky trades with  possible gain or loss of total wealth in each interaction are allowed), has also been considered  in order to obtain wider classes of  equilibrium distributions for the nonlinear dynamics (see \cite{matthes-toscani2008}, \cite{bassetti-ladelli-matthes2011}).

\subsection{Particle system and propagation of chaos}
In order to rigorously justify the interpretation of the model \eqref{eq:Pt}-\eqref{eq:Q+} as representing the evolution of an infinite number of interacting particles or agents, one
considers a finite system of $N$ of such particles, which we denote $\mathbf{X}_t = (X_t^1, \ldots, X_t^N)$, starting independently with common law $P_0$ and such that, at each binary interaction, the states of both involved   particles are modified according to the rule \eqref{eq:inter_rule}.  In the terminology of particle approximations of the Boltzmann equation,  a particle system  with such (true) binary interactions is  called of \emph{Bird type}, as opposed to particle systems of \emph{Nanbu type}, in which only one particle changes its state  after interaction with some other.

Specifically, the particle system $\mathbf{X}$ has infinitesimal generator
\begin{equation} \label{eq:generator}
\mathcal{A}^N \phi (\mathbf{x})
= \frac{1}{2(N-1)} \sum_{i\neq j} \int_{\RR^4} [\phi(\mathbf{x}+a_{ij}(\eta,x^i,x^j))
                           - \phi(\mathbf{x})] \lr(d\eta)
\end{equation}
for all $\mathbf{x} = (x^1,\ldots,x^N) \in\RR^N$ and every test function $\phi$ on $\RR^N$,
where $\a = (\l,\r,\tl,\tr)$ denotes a generic point in $\RR^4$,
$\lr$ is the joint law of  $(\L,\R,\tL,\tR)$ and
$a_{ij}(\a,u,v)$ is the vector of $\RR^N$ whose $i$-th and $j$-th components are $(\l-1)u + \r v$
and $(\tl-1)v + \tr u$, respectively, and which is equal to  $0$ in the other components.

Convergence of such a particle system,  more precisely of its empirical measures  $\frac{1}{N}\sum_{i=1}^N \delta_{X^i_t}$  towards the unique solution  $(P_t)_{t\geq 0}$ of the nonlinear  evolution \eqref{eq:Pt}  as $N$ goes to infinity, has been studied in more general frameworks and  from several points of view, see for instance \cite{graham-meleard1997},  \cite{mischler-mouhot2013} and the references therein (in particular, well posedness of \eqref{eq:Pt}  is by now standard).  Since the particles are exchangeable, the convergence  of the  empirical measure to $P_t$ for large $N$,  as a random variable in the space of probability measures in $\RR$ endowed with the weak topology, is equivalent to the property of \emph{propagation of chaos} of $\mathbf{X}_t$ with respect to $P_t$ (see \cite{sznitman1989} for background): for every fixed $k\in\NN$, the joint law of $X^1_t,\ldots,X^k_t$ converges
weakly to $P_t^{\otimes k}$ as $N$ goes to $\infty$. That is, when $N$ is large, any fixed number of particles of the system behaves at time $t$ approximately like independent random variables of law $P_t$.   This property was introduced and first established by  Kac himself in \cite{kac1956} for the particle system bearing his name, and is nowadays known to hold  for  a large class of particle models, under  general mild assumptions.

\subsection{Main result}
Typically, weak convergence results are not sufficiently informative, and one looks for more quantitative statements.
In this article we will study the  Bird-type $N$-particle system $\mathbf{X}=(X^1,\ldots,X^N)$
and its propagation
of chaos property,
in the cases $p=1$ and $p=2$.
Our main goal is to obtain rates of convergence, as $N\to\infty$,
for the Wasserstein distance between the empirical measure of the particle system
at time $t$ and its limiting law $P_t$,
with explicit estimates on $N$ and $t$ that grow reasonably fast as functions of $t$.

Let $p\in\{1,2\}$ be fixed. In the case $p=2$, we will assume the additional condition
$
\ELR(\L\R+\tL\tR) = 0,
$
which is certainly satisfied in Kac's model.
As a generalization of \eqref{eq:ELR}, we will work under the assumption
\begin{equation}
\label{eq:elast_and_inelast}
\frac{1}{2} \ELR( |\L|^p + |\R|^p + |\tL|^p + |\tR|^p ) \leq 1.
\end{equation}
With some abuse of language, for each value of $p\in \{1,2\}$  we will say that the model is \emph{inelastic} if the latter inequality is strict. In that case, the interacction between particles produce an average loss of energy when $p=2$ (see for instance the inelastic Kac model in \cite{pulvirenti-toscani2004}) or of  ``wealth'' (in the context of \cite{matthes-toscani2008}) when $p=1$.

Also, to avoid trivial situations, in all what follows
we will assume that the model is non-degenerate, that is,
$\ELR(|\R|+|\tR|) > 0$; this means
that the system produces at least some effective interactions.

Let us fix some notation.
$\P(E)$ denotes the space of probability measures on the metric space $E$.
For $\mathbf{x}\in\RR^N$ and any $i=1,\ldots,N$ we define the empirical
measures $\EM{\mathbf{x}} = \frac{1}{N} \sum_j \delta_{x^j}$
and $\EMI{\mathbf{x}}{i} = \frac{1}{N-1} \sum_{j\neq i} \delta_{x^j}$, both being elements
of $\P(\RR)$.
Define $\mom_q(\mu) = \int |u|^q \mu(du)$ the absolute $q$-moment  of $\mu\in \P(\RR)$.
Given a random vector $\mathbf{Z}$ on $\RR^N$ we denote its law by $\law(\mathbf{Z}) \in \P(\RR^N)$,
and the joint law of its first $k$ components by $\law^k(\mathbf{Z}) \in \P(\RR^k)$.

Recall that for $\mu,\nu\in\P(\RR^k)$
their $p$-Wasserstein distance $\W_p(\mu,\nu)$ is defined to be the cost of the optimal transfer plan between $\mu$ and $\nu$, that is,
\[
\W_p(\mu,\nu)
= \left(\inf_{\pi} \int_{\RR^k \times \RR^k} d_{k,p}(\mathbf{x},\mathbf{y})^p
                    \pi(d\mathbf{x},d\mathbf{y}) \right)^{1/p}
= \left(\inf_{\theta,\vartheta} \EE d_{k,p}(\theta,\vartheta)^p \right)^{1/p},
\]
where the first infimum is taken over all measures $\pi$ on $\RR^k \times \RR^k$
with marginals  $\mu$ and $\nu$, and the second infimum is taken over all pairs of
random vectors $\theta$ and $\vartheta$ such that $\law(\theta) = \mu$ and $\law(\vartheta) = \nu$ (see for instance \cite{villani2009} for  background on Wasserstein distances).
We will use the \emph{normalized} distance $d_{k,p}$ on $\RR^k$
 given by
\begin{equation}
\label{eq:dkp}
d_{k,p}(\mathbf{x},\mathbf{y}) = \left(\frac{1}{k} \sum_{i=1}^k |x^i - y^i|^p \right)^{1/p},
\end{equation}
which is natural when one cares about the dependence on the dimension.

In order to obtain good rates of convergence in $N$ which moreover are well behaved with respect to $t$, it is
convenient to introduce the concave function
\[
\cq{q} = 1 - \frac{1}{2} \ELR(|\L|^q + |\R|^q + |\tL|^q + |\tR|^q), \qquad \forall q\geq 0.
\]
We also define
\[
q^* = \sup \{q: \mom_q(P_0)<\infty, \cq{q}>0\}.
\]

These objects play an important role in \cite{matthes-toscani2008},
since, when $p=1$ and $q^*$ is non-trivial (that is, $1<q^*<\infty$),
$q^*$ corresponds to the Pareto index of the stationary distribution of $P_t$.
More importantly in the present context,
the moments of order $q<q^*$ of $P_t$ can be controlled uniformly in time (see Lemma 5 below).
Assuming \eqref{eq:elast_and_inelast} and $\mom_p(P_0)<\infty$, the concavity of $\cq{q}$ implies that either $q^* \in [p,\infty]$
or $q^* = -\infty$. Also, define for all $q \in \{p\} \cup (p,q^*)$
\[
\barcq{p,q} = \inf_{p\leq r \leq q} \cq{r} = \min(\cq{p},\cq{q}).
\]
Note that if $\cq{p} = 0$, then $\barcq{p,q} = 0$ for all such $q$,
so this function is meaningful only in the case $\cq{p} > 0$, in which case it will be useful   to obtain uniform  (in time) estimates.

We are now ready to state our main theorem (see also Corollary \ref{cor:trajectorial}
for a trajectorial result):

\begin{thm}
\label{thm:main2}
Let $(P_t)_{t\geq 0}$ be the unique solution of \eqref{eq:Pt}
and let $\mathbf{X}$ be the particle system starting with law $P_0^{\otimes N}$ and with generator \eqref{eq:generator}.
For $p=1$ or $p=2$,
assume $\cq{p}\geq 0$ and $\mom_p(P_0) < \infty$.
If $p=2$, assume also that $\ELR(\L\R + \tL\tR) = 0$ and $q^*>2$.
Then
\begin{itemize}
\item for any $q\in \{1\}\cup (1,q^*)$ and
any $\gamma < (2+1/q)^{-1}$ in the case $p=1$, or
\item for any $q\in(2,q^*)$, $q\neq 4$ and for $\gamma = \min(1/3,\frac{q-2}{2q-2})$
in the case $p=2$,
\end{itemize}
there exists a constant $C$,
depending on $p$, $q$, $\gamma$ and some moments of $P_0$ and $(\L,\R,\tL,\tR)$
of order at most $q$, such that:
\begin{enumerate}
\item \label{item:thm_main2_i} for all $k \leq N$ and for all $t\geq 0$,
\begin{align*}
\W_p^p(\law^k(\mathbf{X}_t),P_t^{\otimes k} )
\leq C  \left( \frac{t(1+t)^{p-1}e^{-\frac{p}{q}\barcq{p,q} t}}{N^\gamma} + \ind_{k\neq 1}\frac{k \min(1,t) e^{-\cq{p} t}}{N}\right),
\end{align*}
\item \label{item:thm_main2_ii} for all $t\geq 0$,
\[
\EE \W_p^p(\EM{\mathbf{X}}_t,P_t) \leq \frac{C (1+t)^p e^{-\frac{p}{q}\barcq{p,q} t} }{N^\gamma}.
\]
\end{enumerate}
\end{thm}

\begin{rmk}
\begin{itemize}
\item The power $\gamma$ in  Theorem \ref{thm:main2} is a consequence of using recently established   sharp quantitative  estimates in Wasserstein distance for  the empirical measures of  exchangeable or i.i.d.\ collections of random variables (which improve or extend a classical result  in \cite{rachev-ruschendorf1998}).
More specifically,
the rate $N^{-\gamma}$ with $\gamma < (2+1/q)^{-1}$ in the case $p=1$ comes from
Theorem 1.2 of \cite{hauray-mischler2014}, whereas the value $\gamma = \min(1/3,\frac{q-2}{2q-2})$
in the case $p=2$ comes from Theorem 1 of \cite{fournier-guillin2013}.
On the other hand, the dependence on $t$ results from our estimates,
which   rely on Gronwall's lemma.

\item The restriction $q\neq 4$ in the case $p=2$ comes from Theorem 1 of \cite{fournier-guillin2013}.
As those authors mention, the case $q=4$ would produce additional logarithmic terms, which in
our case translate  into a rate of order $N^{-1/3}$ times a logarithmic function of $N$.

\item In the elastic case (that is, $\cq{p} = 0 = \barcq{p,q}$),
\ref{item:thm_main2_i} and \ref{item:thm_main2_ii}
give estimates that grow linearly with time (in the case $p=2$ both sides
are squared).
In the inelastic case, which corresponds to  $\cq{p}, \barcq{p,q} >0$, all  estimates are uniform in time.

\item From a physical point of view, it is interesting to consider
models where infinitely many particles interact over finite time intervals,
such as the Kac equation without cutoff.
The techniques used in the proof of Theorem \ref{thm:main2} can also be applied to cutoffed approximations of that equation and, in the case that (in the notation of \cite{desvillettes-graham-meleard1999}) the classical condition  $\int_0^\pi \theta^2 \beta(\theta)d\theta <\infty$ on the cross-section function $\beta:[-\pi,\pi]\to\RR_+$ is satisfied, they yield a constant that does not
depend on the cutoff parameter, see Remark \ref{rmk:non-cutoff}.

\end{itemize}
\end{rmk}

\subsection{Particular cases and comparison with known results}

\subsubsection{The Kac equation}
Note that if the stronger condition \eqref{eq:LR} is satisfied (or holds with $\leq$ instead of equality),
then $|\L|$, $|\R|$, $|\tL|$ and $|\tR|$ are all $\leq 1$ a.s., which
implies that $\cq{q}$ is strictly increasing with $q$. Thus, $\barcq{p,q} = \cq{p}$ for $q\geq p$
and the value of $q^*$ will depend only on the finiteness of the moments of $P_0$.
In Kac's model, since \eqref{eq:LR} is satisfied for $p=2$, if $P_0$ has finite moment of order $4+\eps$, then $q^*>4$ and  Theorem \ref{thm:main2} gives
\[
\EE \W_2^2(\EM{\mathbf{X}}_t,P_t) \leq \frac{C (1+t)^2}{ N^{1/3}}.
\]
Several similar results can be found in the literature.
The closest one corresponds to quantitative rates for the Nanbu system associated with Kac's model, which are found for instance in  the proof of Proposition 6.2 of \cite{fournier-godinho2014}. The authors state there a  $\W_2^2$ convergence rate   that also depends quadratically on $t$ and is  optimal on $N$, in the sense that it is equal to the   $\W_2^2$ rate of convergence of the empirical measure of an i.i.d.\ sample towards their common law. The latter is of order $N^{-1/2}$, according to Theorem 1 of \cite{fournier-guillin2013}. Thus, the Bird-type particle system seems to produce a slower rate of convergence than  the corresponding Nanbu-type system.
An interesting question is whether
this difference is a mere consequence of the techniques used in our proof
 (more specifically, some order is lost when one uses Lemma \ref{lem:W2Ymu})
or is intrinsically related  to the type of binary
interactions (Bird or Nanbu) in the system.

A similar result as the one of \cite{fournier-godinho2014} can be found in \cite{fournier-mischler2013} where, motivated by the numerical approximation of the Boltzmann equation for  hard spheres, hard potentials and Maxwellian gases, a pathwise coupling argument was developed  for Nanbu particle systems,  which extends a coupling construction based on optimal transport developed in  \cite{fontbona-guerin-meleard2009}. That pathwise approach however does not readily extend to  the particle systems of Bird type we are interested in, which in turn  provide a physically more transparent description of the relevant interaction phenomena.

As for the Bird particle system, in \cite{graham-meleard1997} the authors obtain an explicit rate in
total variation distance on the path space, between the law of one particle
and the law of the nonlinear process (to be introduced later). However, due to the generality of their hypotheses and the strong pathwise distance they use,  the convergence rate  depends exponentially  on the length of the time interval  that is considered.
Similarly, in Theorem 4.3 of \cite{desvillettes-graham-meleard1999}
the authors state a propagation of chaos result in $\W_2$ for the law at time $t$ of one particle in the system with
cutoff, towards the law $P_t$ of the nonlinear dynamics without cutoff. Since some relations between
 $N$  and the cutoff parameter must be satisfied when removing the latter, that result gives estimates that are logarithmic in $N$ and grow
exponentially with $t$.

On the other hand,
the general theory  developed in \cite{mischler-mouhot2013} provides a framework and  a methodology to establish quantitative (in $t$ and $N$) propagation of chaos estimates which can be applied in the present framework. For instance, in their Theorem 5.2,  a $\W_1$ estimate for the Boltzmann equation in the Maxwell molecules case is obtained, which is uniform in time and decays    with $N$ in a polynomial way (see also Step 3 of the proof of Theorem 8 in \cite{carrapatoso2014B} for results in $\W_2$ distance); we expect that similar bounds can be  obtained with their techniques for the Kac model.
The actual dependence on  $N$  they give   seems however hard  to trace in general,  and we have not  been able to deduce with their techniques an estimate in Wasserstein distance  as sharp as ours in terms of  $N$.
Also, their
approach  does not provide  any information on the way in which trajectories  of  particles  get closer to  those of the limiting processes. On the other side,  unfortunately our techniques (ultimately relying on Gronwall's lemma)   do not seem to yield uniform in time estimates for  the elastic Kac equation,
 even if  $P_0$ were compactly supported.

Finally, we observe that for the inelastic Kac model,  $\barcq{2,q} = \cq{2} >0$ for all $q$,  hence Theorem \ref{thm:main2}  does  give a uniform-in-time rate in that case:
\[
\EE \W_2^2(\EM{\mathbf{X}}_t,P_t) \leq \frac{C e^{-\frac{\cq{2}}{2+\eps}t}}{ N^{1/3}},
\]
if  $P_0$ is  again assumed to have finite moment of order $q>4$.
This exponential decay is not surprising: when $P_0$ has finite moment of
order 2 it is known that $M_2(P_t)$ decays exponentially fast, see \cite{pulvirenti-toscani2004}.
Nevertheless, to our knowledge, our quantitative (in $N$) propagation of chaos result is new for the inelastic Kac model.

\subsubsection{Models for economic exchanges and wealth distribution}
Working  with $p=1$ as in \cite{matthes-toscani2008} and assuming only that the first moment  of $P_0$ is finite, Theorem \ref{thm:main2} gives  in the elastic case
  a propagation
of chaos result for $\W_1$ of order almost $N^{-1/3}$,  with estimates growing linearly with time.
Any additional finite $q$-moment
of $P_0$ (with $q<q^*$) can be used to improve the rate in $N$, up to almost $N^{-1/(2+1/q^*)}$.
In the case of exact conservation of wealth (condition \eqref{eq:LR})
we have $q^*=\infty$
and we obtain a rate of $N^{-(1/2 -\eps)}$, which is almost optimal
according to Theorem 1 of \cite{fournier-guillin2013}.
To our knowledge, this is the first quantitative propagation of chaos result
for kinetic equations modeling the evolution of wealth distribution.

\subsection{The nonlinear process and idea of the proof}

Following ideas pioneered by Tanaka in the case of the Boltzmann equation (see \cite{tanaka1978} and \cite{tanaka1979}), it is also possible to establish the convergence of the \emph{pathwise} law  of a particle, to the law
of some process obtained by the following construction:
consider a Poisson point measure $\M$ on
$\RR_+ \times \RR^2 \times \RR$ with intensity $dt \barlr(d\l,d\r) P_t(dv)$,
where $\barlr = \frac{1}{2}(\law(\L,\R) + \law(\tL,\tR))$,
and let $(V_t)_{t\geq 0}$ be the jump process on $\RR$ defined as the unique
solution starting with law $P_0$ of the stochastic equation
\begin{equation} \label{eq:nlpr}
dV_t = \int_{\RR^2} \int_{\RR} [(\l -1) V_{t^-} + \r v] \M(dt,d\l,d\r,dv).
\end{equation}
It is not hard to see that such a  jump process  $V$  exists, it is uniquely defined
and it satisfies $\law(V_t) = P_t$ for all $t\geq 0$.
We call $P$ the pathwise law of $V$, and any process with law $P$ is
called a \emph{nonlinear process};
it represents the trajectory of any fixed particle
in the (infinite) population subjected to the random interactions described above in \eqref{eq:inter_rule}.

To prove our results, we will couple the Bird particle system
$\mathbf{X}_t$ with a system $\mathbf{U}_t = (U_t^1,\ldots,U_t^N)$ where each $U^i$ is
a copy of the nonlinear process  $V$, constructed in such a way that it remains close to $X^i$.
To achieve this,  we will  use techniques of optimal coupling  inspired by those used
in \cite{fontbona-guerin-meleard2009} and \cite{fournier-mischler2013}, in order to carefully choose the jumps of the nonlinear process $U^i$ as similar as possible to those of  the particle $X^i$. However,  contrary to those papers which   deal with Nanbu-type particle systems (in which each  randomness source  acts on the trajectory of only one of the particles),  ensuring closeness of $X^i$ and $U^i$
 simultaneously for all $i=1,\ldots,N$ will imply
that the processes $U^1,\ldots,U^N$ are \emph{not}  independent. Therefore, to obtain the desired estimates we will need, in a second step, to ``decouple''
the system $\mathbf{U}_t$ as $N$ goes to infinity, which we will be able to do with estimates that
are uniform in time, see Lemma \ref{lem:decoupling} below.

Let us point out that the coupling construction  we will introduce can in principle be replicated in
higher dimensions, and with more general interaction rules, which is why  we preferred to avoid the use of  specific one-dimensional features in its construction, see for instance Remark \ref{rmk:post_thm_coupling}. We thus expect  these techniques
to be applicable in physically  more relevant  situations, hopefully including  (at least some instances of) the Boltzmann equation. Also, we think it should be possible to adapt  this coupling construction  in order to quantitatively  study  ``Bird-type'' Brownian particle approximations of a  certain Gaussian white-noise   driven nonlinear process,  associated  with  the  Landau equation arising  in the grazing collisions limit of the Boltzmann equation. Such a process was   studied  in \cite{funaki1984} and \cite{guerin2004}, and a particle approximation result  with  a   ``Nanbu type''    Brownian particle system  was proved in   \cite{fontbona-guerin-meleard2009}, by means of a coupling construction   based  on optimal transport. The corresponding   particle system   of  Bird-type    is studied in  \cite{carrapatoso2014} using the functional tools  developed in \cite{mischler-mouhot2013}, but  there seems to be  so far no suitable coupling argument available in order to  deal  with such class of particle systems.\footnote{When the present work was just finished, the authors learnt from Nicolas Fournier that the latter  question was currently  being studied by him, Fran\c cois Bolley and Arnaud Guillin.}

\subsection{Plan of the paper}
In Section \ref{sec:prel} we give the explicit construction of the
particle system $\mathbf{X}_t$, and more importantly, we couple  it with the system
$\mathbf{U}_t = (U_t^1,\ldots,U_t^N)$ of  dependent nonlinear processes that we will use
throughout the rest of this article. In Section \ref{sec:scheme} we

prove Theorem  \ref{thm:main2}.

The proof of some intermediate lemmas, including  statements of  Section \ref{sec:prel}  and the  ``decoupling'' of the process $\mathbf{U}_t$,
is left for the final Section \ref{sec:proofs}.

\section{Coupling of the particle system and the nonlinear processes}
\label{sec:prel}

\subsection{The particle system}
Let us fix the number of particles $N\in\NN$.
Although most of the subsequent objects will depend on $N$, for notational simplicity we will not
make this dependence explicit.
We will define both the particle system $\mathbf{X}$ and the nonlinear processes $\mathbf{U}$
by means of integral equations driven by the same
Poisson point measure. To this end, let us first
introduce the function $\i:[0,N) \to \{1,\ldots,N\}$ given by
$\i(\rho) = \lfloor \rho \rfloor + 1$, and the set $\C \subseteq [0,N)^2$
\[
 \C = \{ (\rho,\sigma) \in  [0,N)^2: \i(\rho) \neq \i(\sigma) \}.
\]
Note that $|\C| = N(N-1)$.

As in \eqref{eq:generator}, denote $\a = (\l,\r,\tl,\tr)$ a generic point in $\RR^4$
and  $\lr = \law(\L,\R,\tL,\tR)$.
Now, let $\N(dt,d\a,d\rho,d\sigma)$ be a Poisson point measure on $[0,\infty)\times \RR^4 \times [0,N)^2$
with intensity
\[
 \frac{N}{2} dt \lr(d\a) d\rho d\sigma \frac{1}{|\C|} \ind_{\C}(\rho,\sigma)
 = \frac{1}{2(N-1)} dt \lr(d\a) d\rho d\sigma \ind_{\C}(\rho,\sigma).
\]
In words, $\N$ picks atoms in $[0,\infty)$ at constant rate of $N/2$, and
for each such atom it also independently samples a tuple $(\l,\r,\tl,\tr)$ from $\lr$
and a pair $(\rho,\sigma)$ uniformly on $\C$.
We will use $(\rho,\sigma)$
to choose the indices of the particles that interact at each jump.
Consider also $N$ independent random variables $(X_0^1, \ldots, X_0^N) =: \mathbf{X}_0$, independent
from $\N$, each having distribution $P_0$.
Finally, set $\F = (\F_t)_{t\geq 0}$ to be the complete right continuous filtration generated by $\mathbf{X}_0$ and $\N$. We denote $\PP$ and $\EE$ the probability and expectation in the corresponding probability space.

The particle system $\mathbf{X} = (X^1,\ldots,X^N)$ is defined as the
solution, starting from $\mathbf{X}_0$, of the following integral equation:
\begin{equation} \label{eq:Xt}
d\mathbf{X}_t = \int_{\RR^4} \int_{[0,N)^2} \sum_{i,j=1}^N \ind_{\{\i(\rho) = i, \i(\sigma) = j\}}
         a_{ij} (\a, X_{t^-}^i, X_{t^-}^j) \N(dt,d\a,d\rho,d\sigma).
\end{equation}
(Recall that $a_{ij}(\a,u,v)$ is the vector of $\RR^N$ whose $i$-th and $j$-th components are $(\l-1)u + \r v$
and $(\tl-1)v + \tr u$, respectively, and is equal to  $0$ in the other components).
Given the timely ordered  atoms $(t_n,\a_n,\rho_n,\sigma_n)_{n\geq 0}$ of $\N$ (that is, $t_n\leq t_{n+1}$ for all $n\geq 0$),
a solution of this equation can be constructed as follows:
recursively define $\mathbf{X}_{t_n}$ as
\begin{equation}
\label{eq:Xtn}
X_{t_n}^\ell =
\begin{cases}
\l_n X_{t_{n-1}}^{i} + \r_n X_{t_{n-1}}^{j}, & \ell=i, \\
\tl_n X_{t_{n-1}}^{j} + \tr_n X_{t_{n-1}}^{i}, & \ell=j, \\
X_{t_{n-1}}^\ell, & \ell \neq i,j,
\end{cases}
\end{equation}
where $(i,j) = (\i(\rho_n),\i(\sigma_n))$, and set $\mathbf{X}_t = \mathbf{X}_{t_n}$ for all $t\in(t_n,t_{n+1})$.
Uniqueness for \eqref{eq:Xt} also holds, since there is
no choice to make in this construction. It is straightforward to verify that
$\mathbf{X}$ has generator \eqref{eq:generator}.

Thus, the system $\mathbf{X}$ is what we want it to be: at rate $N/2$ we choose two distinct indices $i=\i(\rho)$
and $j=\i(\sigma)$, and then we update the particles $X^i$ and $X^j$ according to the rule
described in \eqref{eq:inter_rule}. The fact that we use continuous variables $(\rho,\sigma)$ to
choose the indices $(i,j)$ (instead of a discrete pair chosen uniformly from
the set $\{1,\ldots,N\}^2\setminus \{i=j\}$) will be crucial to define our
system $\mathbf{U}$ of $N$ nonlinear processes.

\subsection{Coupling with the nonlinear processes}
From \eqref{eq:Xt} it follows that for any $i = 1,\ldots,N$, the process $X^i$ satisfies
\begin{equation} \label{eq:Xtk}
dX_t^i
= \int_{\RR^2} \int_{[0,N)} [(\l-1)X_{t^-}^i + \r X_{t^-}^{\i(\tau)}] \N^i(dt,d\l,d\r,d\tau),
\end{equation}
where $\N^i$ is defined as
\begin{align} \label{eq:Nk}
\begin{split}
\N^i(dt,d\l,d\r,d\tau)
&= \N(dt,(d\l \times d\r \times \RR^2),[i-1,i),d\tau) \\
&\quad + \N(dt,(\RR^2 \times d\l \times d\r),d\tau,[i-1,i)).
\end{split}
\end{align}
Clearly, $\N^i$ is a Poisson point measure on $[0,\infty)\times\RR^2\times [0,N)$ with intensity
\[
dt \barlr(d\l , d\r) \frac{ d\tau}{N-1} \ind_{A^i}(\tau),
\]
where $\barlr = \frac{1}{2}(\law(\L,\R) + \law(\tL,\tR))$,
and $A^i = [0,N)\setminus [i-1,i)$. In other words, $\N^i$ selects only the atoms
of $\N$ that produce a jump of $X^i$, that is, the atoms in which $\i(\rho) = i$ or $\i(\sigma) =i$.

Let us examine the expression \eqref{eq:Xtk} in more detail. First note that since $\tau$ is chosen uniformly
in $A^i$, the variable $X_{t^-}^{\i(\tau)}$ corresponds to a sample from the (random)
probability measure
$\EMI{\mathbf{X}}{i}_{t^-} = \frac{1}{N-1} \sum_{j\neq i} \delta_{X_{t^-}^j}$.
Thus, from the point of view of the process $X^i$, the dynamics
is as follows: at rate $1$, a number $v = X_{t^-}^{\i(\tau)}$ is sampled from the measure $\EMI{\mathbf{X}}{i}_{t^-}$,
and then the value of the process is updated  according to the rule $X_{t^-}^i \mapsto \l X_{t^-}^i + \r v$,
where $(\l,\r)$ is chosen with law $\barlr$.

Comparing \eqref{eq:nlpr} and \eqref{eq:Xtk},
the key observation is the following: if for each jump time $t$ one replaces $X_{t^-}^{\i(\tau)}$ in \eqref{eq:Xtk}
with a realization $v$ of the law $P_t(dv)$, the resulting process has law $P$.
In view of this, we would like to define the system of nonlinear
processes $\mathbf{U}=(U^1,\ldots,U^N)$ based on this idea,
but using a realization of $P_t$ that is optimally coupled to
the realization
$X_{t^-}^{\i(\tau)}$ of the measure $\EMI{\mathbf{X}}{i}_{t^-}$.
In doing this, some measurability issues need to be taken into account.

\begin{lem}[coupling] \label{lem:coupling}
 For every $p\geq 1$ and $i\in\{1,\ldots,N\}$
 there exist a  measurable  mapping $\G^i :\RR_+ \times \RR^N \times A^i \to \RR$,
 $(t,\mathbf{x},\tau)\mapsto \G_t^i(\mathbf{x},\tau)$,
 with the following property: for every $t\geq 0$ and $\mathbf{x}\in\RR^N$,
 if $\tau$ is uniformly chosen from $A^i$, then the pair $(\G_t^i(\mathbf{x},\tau),x^{\i(\tau)})$
 is an optimal coupling between $P_t$ and $\EMI{\mathbf{x}}{i} = \frac{1}{N-1} \sum_{j\neq i} \delta_{x^j}$
 with respect to the cost function $c(u,v) = |u-v|^p$.
 Moreover, if $\mathbf{Y}$ is any exchangeable random vector in $\RR^N$, then
 $\EE \int_{j-1}^j \phi(\G_t^i(\mathbf{Y},\tau)) d\tau  = \langle P_t, \phi \rangle$
 for any $j\in\{1,\ldots,N\}$,
 $j\neq i$, and any  bounded measurable function $\phi$.
\end{lem}

For simplicity, in our notation we have not made explicit the dependence of $\G_t^i$ on $p$ (however, see Remark \ref{rmk:post_thm_coupling}).
Now, we can define $U^i$ as the solution of
\begin{equation} \label{eq:Utk}
dU_t^i
= \int_{\RR^2} \int_{[0,N)} [(\l-1)U_{t^-}^i + \r\G_t^i(\mathbf{X}_{t^-},\tau)] \N^i(dt,d\l,d\r,d\tau),
\end{equation}
where $\N^i$ is the same Poisson point measure as in \eqref{eq:Xtk}.  The proof of Lemma  \ref{lem:coupling}  will imply that the  mapping $((t,\omega),\l,\r ,\tau)\mapsto  (\l-1)U_{t^-}^i (\omega)+ \r \G_t^i(\mathbf{X}_{t^-}(\omega),\tau)$    above  is  measurable with respect to the product of the predictable sigma field  (in $(t,\omega)$)  and the Borel sigma field of $\RR^2\times [0,N)$. This ensures that the  integral in \eqref{eq:Utk}  has the usual properties of  integrals with respect to Poisson point processes.

We summarize our construction in the following

\begin{lem} \label{lem:XU}
Let $p\geq 1$ be fixed. For each $i=1,\ldots,N$ there is a unique solution $U^i$ of \eqref{eq:Utk},
and it is a nonlinear process.
Moreover, the collection $(X^1,U^1),\ldots,(X^N,U^N)$ is exchangeable.
\end{lem}

Thus, the system $\mathbf{U} = (U^1,\ldots,U^N)$
is indeed a tuple of $N$ nonlinear processes. However, as we already mentioned, they are \emph{not
independent}, since $\N^i$ and $\N^j$ share a portion of $\N$,
namely, the atoms of $\N$ whose coordinates $(\rho,\sigma)$ lie in
$[i-1,i)\times [j-1,j)$ or $[j-1,j)\times [i-1,i)$.
In particular, whenever such an atom occurs the processes
$U^i$ and $U^j$ jump simultaneously, using a single realization of $(\L,\R,\tL,\tR)$,
and  samples of $P_t$ that  also are correlated.

\section{Proof of the main result}
\label{sec:scheme}

Before proving our results, let us first state two lemmas
that constitute our basic tools; they will be proven in Section \ref{sec:proofs}.
The first one provides uniform bounds for
the moments of $P_t$; it can be seen as a version
of Theorem 3.2 in \cite{matthes-toscani2008}:

\begin{lem}[moment bounds]
\label{lem:moment_bounds}
For $p=1$ or $p=2$, assume $\cq{p} \geq 0$ and $\mom_p(P_0) < \infty$.
If $p=2$, assume also that $\ELR(\L\R + \tL\tR) = 0$.
Then for any $q\in \{p\}\cup (p,q^*)$ there exists
a constant $C$, depending on $q$ and some moments of $P_0$ and $(\L,\R,\tL,\tR)$
of order at most $q$, such that
\[
\mom_q(P_t) \leq C e^{- \barcq{p,q} t } \quad \forall t \geq 0.
\]
\end{lem}

The second lemma is fundamental in our developments since it decouples
the non-independent nonlinear processes uniformly in time, even in
the case $\cq{p} = 0$:

\begin{lem}[decoupling] \label{lem:decoupling}
For $p=1$ or $p=2$,
assume $\cq{p}\geq 0$ and $\mom_p(P_0) < \infty$.
If $p=2$, assume also that $\ELR(\L\R + \tL\tR) = 0$.
Then there exists a constant $C$, depending only on the $p$-moment
of $P_0$ and $(\L,\R,\tL,\tR)$, such that for all $k=2,\ldots,N$ and $t\geq 0$,
\[
\W_p^p( \law^k(\mathbf{U}_t), P_t^{\otimes k})
\leq \frac{C (k-1) \min(1,t) e^{-\cq{p} t}}{N-1}.
\]
\end{lem}

\begin{proof}[Proof of Theorem \ref{thm:main2}]
Define the constants $\cql{p} = \frac{1}{2}\ELR(|\L|^p + |\tL|^p)$
and $\cqr{p} = \frac{1}{2} \ELR(|\R|^p + |\tR|^p)$, so $\cq{p} = 1-\cql{p}-\cqr{p}$.
We first treat the case $p=1$.
Thus, we work with the processes $U^i$ solution of \eqref{eq:Utk} using
the functions $\G_t^i$ of Lemma \ref{lem:coupling} with $p=1$.
Let us prove \ref{item:thm_main2_i} first. We  estimate the quantity $f_t = \EE |X_t^1 - U_t^1|$
which provides an upper bound for $\W_1(\law^1(\mathbf{X}_t),P_t)$.
Using \eqref{eq:Xtk} and \eqref{eq:Utk}, for all $0\leq s \leq t$ we have
\begin{align}
\label{eq:XtUt}
\begin{split}
& |X_t^1 - U_t^1| - |X_s^1 - U_s^1| \\
& =  \int_{(s,t]} \int_{\RR^2}\int_{[0,N)}
    \left( |\l(X_{r^-}^1 - U_{r^-}^1) + \r (X_{r^-}^{\i(\tau)} - \G_r^1(\mathbf{X}_{r^-},\tau))|  \right. \\
& \qquad \qquad \qquad \qquad {}  \left. - |X_{r^-}^1 - U_{r^-}^1| \right) \N^1(dr,d\l,d\r,d\tau).
\end{split}
\end{align}
Recall that the intensity of $\N^1$ is $(N-1)^{-1} dt\barlr(d\l,d\r) d\tau \ind_{A^1}(\tau)$,
where $\barlr = (\law(\L,\R) + \law(\tL,\tR))/2$.
By the compensation formula,  $t \mapsto f_t$ is absolutely continuous and we obtain
\begin{align}
f_t - f_s
& \leq \EE \int_s^t \int_{\RR^2} \int_{A^1} \left( (|\l|-1)|X_r^1 - U_r^1| \right. \notag \\
& \qquad \qquad \qquad {}  \left. + |\r| |X_r^{\i(\tau)} - \G_r^1(\mathbf{X}_r,\tau)| \right)
                                     \frac{d\tau}{N-1} \barlr(d\l,d\r) dr \notag \\
& = \EE \int_s^t \left( (\cql{1} - 1)|X_r^1 - U_r^1| + \cqr{1} \W_1(\EMI{\mathbf{X}}{1}_r,P_r) \right) dr, \label{eq:EXtUt}
\end{align}
where in the last step we have used the fact that when $\tau$ is uniform in $A^1$,
$(\G_s^1(\mathbf{x},\tau), x^{\i(\tau)})$ is an optimal coupling between $P_s$ and $\EMI{\mathbf{x}}{1}$.
We deduce  that for almost all $t\geq 0$
\begin{equation}
\label{eq:dft1}
\partial_t f_t \leq -(1-\cql{1}) f_t + \cqr{1} \EE \W_1(\EMI{\mathbf{X}}{1}_t,P_t).
\end{equation}
Recall that $\EMI{\mathbf{U}}{i}_t = \frac{1}{N-1} \sum_{j\neq i} \delta_{U_t^j}$
for $i=1,\ldots,N$. The triangle inequality for $\W_1$ gives us
\begin{align}
\EE \W_1(\EMI{\mathbf{X}}{1}_t,P_t)
&\leq \EE \W_1(\EMI{\mathbf{X}}{1}_t,\EMI{\mathbf{U}}{1}_t) + \EE \W_1(\EMI{\mathbf{U}}{1}_t,P_t) \notag \\
&\leq \EE |X_t^1 - U_t^1| + \EE \W_1(\EMI{\mathbf{U}}{1}_t,P_t), \label{eq:W1XPt}
\end{align}
where the last inequality comes from the fact that $(X_t^{\i(\tau)},U_t^{\i(\tau)})$
is a coupling between $\EMI{\mathbf{X}}{1}_t$ and $\EMI{\mathbf{U}}{1}_t$ when $\tau$ is
uniformly chosen in $A^1$, and from the exchangeability
of $(X^i,U^i)_{i=1,\ldots,N}$. Putting this together with \eqref{eq:dft1},
we obtain
\begin{equation} \label{eq:EXU}
\partial_t f_t
\leq - \cq{1} f_t
      + \cqr{1} \EE \W_1(\EMI{\mathbf{U}}{1}_t,P_t).
\end{equation}
Next, we need an estimate for $\EE\W_1(\EMI{\mathbf{U}}{1}_t,P_t)$. Since the system
$(U^2,\ldots,U^N)$ is exchangeable, using a recent result
(Theorem 1.2 of \cite{hauray-mischler2014}), we obtain the
following: for each $q>0$ and each $\gamma < (2+1/q)^{-1}$, there exists a constant $C_{q,\gamma}$ such that
\begin{equation} \label{eq:H&M}
\EE \W_1(\EMI{\mathbf{U}}{1}_t,P_t)
\leq C_{q,\gamma} \mom_q(P_t)^{1/q} \left( \W_1( \law(U_t^2,U_t^3), P_t^{\otimes 2}) + \frac{1}{N-1} \right)^\gamma.
\end{equation}
Now, Lemma \ref{lem:decoupling} in the case $p=1$ and $k=2$ implies
$\W_1( \law(U_t^1,U_t^2), P_t^{\otimes 2}) \leq C/N$,
where $C$ is some constant, which can change from line to line in what follows.
From this, Lemma \ref{lem:moment_bounds}, and \eqref{eq:EXU}-\eqref{eq:H&M} we have
$\partial_t f_t \leq - \cq{1} f_t  +  C N^{-\gamma} e^{-\frac{1}{q}\barcq{1,q} t}$,
and then Gronwall's lemma yields
\[
f_t
\leq \frac{C}{N^\gamma} \int_0^t e^{-\cq{1}(t-s)} e^{-\frac{1}{q}\barcq{1,q} s}ds,
\]
since $f_0 = 0$.
Bounding $e^{-\cq{1} (t-s)} \leq e^{-\frac{1}{q}\barcq{1,q} (t-s)}$
gives \ref{item:thm_main2_i} in the case $p=1$ and $k=1$.
From this and Lemma \ref{lem:decoupling}, the case $k\geq 2$ follows:
\begin{align}
\W_1(\law^k(\mathbf{X}_t),P_t^{\otimes k} ) \notag
&\leq \W_1(\law^k(\mathbf{X}_t),\law^k(\mathbf{U}_t) )
 + \W_1(\law^k(\mathbf{U}_t),P_t^{\otimes k} ) \notag \\
&\leq \EE|X_t^1 - U_t^1| + \frac{C k \min(1,t) e^{-\cq{1} t}}{N}. \label{eq:case_k_geq_2}
\end{align}

We now prove \ref{item:thm_main2_ii}: as in \eqref{eq:W1XPt} we have
\begin{align*}
\EE \W_1(\EM{\mathbf{X}}_t,P_t)
&\leq \EE|X_t^1 - U_t^1| + \EE \W_1(\EM{\mathbf{U}}_t,P_t) \\
&\leq \frac{Ct e^{-\frac{1}{q}\barcq{1,q} t }}{N^\gamma} + \frac{C e^{-\frac{1}{q} \barcq{1,q} t}}{N^\gamma},
\end{align*}
where the last inequality comes from \ref{item:thm_main2_i} in the case $k=1$, and
from \eqref{eq:H&M} (with $\EM{\mathbf{U}}_t$ and $N$ in place of $\EMI{\mathbf{U}}{1}_t$ and $N-1$)
together with Lemma \ref{lem:decoupling} in the case $k=2$.
From the previous inequality, \ref{item:thm_main2_ii} follows;
moreover, the same estimate is also valid for $\EE \W_1(\EMI{\mathbf{X}}{1}_t,P_t)$.

Now we treat the case $p=2$.
The proof is similar to the previous case,
with adaptations where required.
We work with the processes $U^i$ solution of \eqref{eq:Utk} using
the functions $\G_t^i$ of Lemma \ref{lem:coupling} with $p=2$.
As before, to prove the case $k=1$ we want to estimate $f_t = \EE (X^1_t-U^1_t)^2$.
We proceed as in \eqref{eq:XtUt}:
from \eqref{eq:Xtk} and \eqref{eq:Utk}, we have for all $0\leq s \leq t$
\begin{align}
& (X_t^1 - U_t^1)^2 - (X_s^1 - U_s^1)^2 \label{eq:XU2} \\
& =  \int_{(s,t]} \int_{\RR^2}\int_{[0,N)}
    \left( [\l(X_{r^-}^1 - U_{r^-}^1) + \r (X_{r^-}^{\i(\tau)} - \G_r^1(\mathbf{X}_{r^-},\tau)) ]^2  \right. \notag \\
& \qquad \qquad {}  \left. - [X_{r^-}^1 - U_{r^-}^1]^2 \right) \N^1(dr,d\l,d\r,d\tau) \notag \\
& =  \int_{(s,t]} \int_{\RR^2}\int_{[0,N)}
    \left( [\l^2 - 1] (X_{r^-}^1 - U_{r^-}^1)^2 + \r^2 (X_{r^-}^{\i(\tau)} - \G_r^1(\mathbf{X}_{r^-},\tau))^2  \right. \notag \\
& \qquad \qquad {}  \left. + 2 \l\r (X_{r^-}^1 - U_{r^-}^1)(X_{r^-}^{\i(\tau)} - \G_r^1(\mathbf{X}_{r^-},\tau)) \right) \N^1(dr,d\l,d\r,d\tau). \notag
\end{align}
Taking expectations, the last term in the integral vanishes thanks to
condition $\ELR(\L\R+\tL\tR) = 0$. As in \eqref{eq:EXtUt}-\eqref{eq:dft1}, this yields
\begin{equation}
\label{eq:dft2}
\partial_t f_t \leq -(1-\cql{2}) f_t  + \cqr{2} \EE \W_2^2(\EMI{\mathbf{X}}{1}_t,P_t)
\end{equation}
Defining $g_t = \EE\W_2^2(\EMI{\mathbf{U}}{1}_t,P_t)$ and using the triangle inequality of $\W_2$ we have
\begin{align}
\EE \W_2^2(\EMI{\mathbf{X}}{1}_t,P_t)
 & \leq    \EE \W_2^2(\EMI{\mathbf{X}}{1}_t,\EMI{\mathbf{U}}{1}_t) +  2 \EE \W_2(\EMI{\mathbf{X}}{1}_t,\EMI{\mathbf{U}}{1}_t)\W_2(\EMI{\mathbf{U}}{1}_t,P_t)
  + \EE \W_2^2(\EMI{\mathbf{U}}{1}_t,P_t) \notag \\
 & \leq f_t + 2 f_t^{1/2} g_t^{1/2} + g_t, \label{eq:EW2XPt}
\end{align}
where in the last inequality the term $f_t$ is obtained with the same argument
as in \eqref{eq:W1XPt}, and the term $f_t^{1/2}g_t^{1/2}$ comes from the Cauchy-Schwarz inequality.
From this and \eqref{eq:dft2} we obtain
\[
\partial_t f_t \leq -\cq{2} f_t + 2 \cqr{2}f_t^{1/2}g_t^{1/2} + \cqr{2}g_t.
\]
Using a version of Gronwall's lemma (see e.g.\  Lemma 4.1.8 of \cite{ambrosio-gigli-savare2008})
together with Jensen's inequality, we  obtain
\begin{equation}
\label{eq:EXU2}
f_t \leq \cqr{2} e^{-\cq{2} t} (2+8\cqr{2} t) \int_0^t e^{\cq{2} s}  g_s ds.
\end{equation}
Now, we need an estimate for $g_t = \EE \W_2^2(\EMI{\mathbf{U}}{1}_t,P_t)$.
Unfortunately, we do not have at our disposal a result similar to \eqref{eq:H&M},
which is valid only for $\W_1$.
To bypass this, we will make use of the following lemma (proved in Section  \ref{sec:proofs}); it has the spirit of \eqref{eq:H&M}
in the sense that it will allow us to work with $\W_2^2(\law^n(\mathbf{U}_t) , P_t^{\otimes n})$
instead of $\EE \W_2^2(\EMI{\mathbf{U}}{1}_t,P_t)$, but at the price of the extra term $\eps_{n,2}(P_t)$.

\begin{lem}
\label{lem:W2Ymu}
Let $\mathbf{Y} = (Y^1,\ldots,Y^m)$ be an exchangeable random vector,
and let $\mu$ be a probability measure on $\RR$.

Then, for any $p\geq 1$ and $n \leq m$, $n\in\NN$, we have
\begin{align*}
\frac{1}{2^{p-1}} \EE \W_p^p( \EM{\mathbf{Y}}, \mu)
\leq& \frac{kn}{m} \left( \W_p^p(\law^n(\mathbf{Y}),\mu^{\otimes n}) + \eps_{n,p}(\mu) \right) \\
    &     + \frac{\ell }{m} \left( \W_p^p(\law^\ell (\mathbf{Y}),\mu^{\otimes \ell }) + \eps_{\ell ,p}(\mu) \right)
\end{align*}
where $k$ and $\ell $ are the unique non-negative integers satisfying
$m = kn + \ell $, with $\ell \leq n-1$.
Here $\eps_{n,p}(\mu) := \EE \W_p^p(\EM{\mathbf{Z}}, \mu )$,
where $\mathbf{Z}=(Z^1,\ldots,Z^n)$ are i.i.d.\ and $\mu$ distributed.
\end{lem}

Note that $\W_2^2(\law^\ell (\mathbf{U}_t),P_t^{\otimes \ell }) + \eps_{\ell ,2}(P_t) \leq 8 M_2(P_t)$.
Using this Lemma with $p=2$, $m=N-1$, $\mathbf{Y} = (U^2_t,\ldots,U^{N-1}_t)$ and $\mu = P_t$, we obtain
that for every $n\leq N-1$
\begin{align*}
\EE \W_2^2(\EMI{\mathbf{U}}{1}_t,P_t)
\leq& \W_2^2(\law^n(\mathbf{U}_t),P_t^{\otimes n}) + \eps_{n,p}(P_t)
     + \frac{n-1}{N-1} 8 M_2(P_t) \\
\leq& C \left( \frac{ne^{-\cq{2} t}}{N} + \eps_{n,2}(P_t) \right),
\end{align*}
where in the last inequality we have used Lemma \ref{lem:moment_bounds} and
Lemma \ref{lem:decoupling} with $p=2$ and $k=n$;
again $C$ is some constant that can change from line to line.
Putting this into \eqref{eq:EXU2} gives

\[
f_t \leq C (1+t) \left(\frac{nte^{-\cq{2} t}}{N} + \int_0^t e^{-\cq{2}(t-s)} \eps_{n,2}(P_s) ds \right).
\]

Given $q \in (2,q^*)$, $q\neq 4$, from Theorem 1 of \cite{fournier-guillin2013}
we know that $\eps_{n,2}(P_t) \leq C M_q^{2/q}(P_t) n^{-\eta}$, where $\eta = \min(1/2,\frac{q-2}{q})$.
Choosing $n = \lfloor N^{1/(1+\eta)} \rfloor$ and using Lemma \ref{lem:moment_bounds}
with $p=2$ yields

\[
f_t \leq C (1+t) \left(\frac{te^{-\cq{2} t}}{N^\gamma}
+ \frac{1}{N^\gamma}\int_0^t e^{-\cq{2}(t-s)} e^{-\frac{2}{q} \barcq{2,q} s} ds \right).
\]

where $\gamma = \eta /(1+\eta) = \min(1/3, \frac{q-2}{2q-2})$.
Bounding $e^{-\cq{2}(t-s)} \leq e^{-\frac{2}{q} \barcq{2,q} (t-s)}$
gives \ref{item:thm_main2_i} in the case $p=2$ and $k=1$.
The case $k\geq 2$ follows as in \eqref{eq:case_k_geq_2}.

Finally, \ref{item:thm_main2_ii} in the case $p=2$ follows from \eqref{eq:EW2XPt}
with a similar argument as in the case $p=1$. This concludes the proof.
\end{proof}

\begin{cor} \label{cor:trajectorial}
Under the same hypotheses and notation of Theorem \ref{thm:main2},
we have for all $T\geq 0$,
\[

\EE \sup_{t\in[0,T]}
|X_t^1 - U_t^1|^p
\leq \frac{C}{N^\gamma} \int_0^T (1+t)^p e^{-\frac{p}{q}\barcq{p,q}t } dt.
\]
\end{cor}

\begin{proof}
From \eqref{eq:XtUt}, discarding
the negative term in the integral, we have
\begin{align*}
\sup_{t\in[0,T]} |X_t^1 - U_t^1|
& \leq \int_{(0,T]} \int_{\RR^2}\int_{[0,N)}
    \left( |\l| |X_{t^-}^1 - U_{t^-}^1| \right. \\
& \qquad \qquad {} \left. + |\r| |X_{t^-}^{\i(\tau)} - \G_t^1(\mathbf{X}_{t^-},\tau)| \right) \N^1(dt,d\l,d\r,d\tau).
\end{align*}
With the same argument that produced the term $\W_1(\EMI{\mathbf{X}}{1}_r,P_r)$ in \eqref{eq:EXtUt},
the conclusion follows taking expectations and using
the previous estimates for $\EE|X_t^1 - U_t^1|$ and $\EE \W_1(\EMI{\mathbf{X}}{1}_t,P_t)$.
This proves the case $p=1$,
and the case $p=2$ follows from \eqref{eq:XU2} with a similar argument.
\end{proof}

\begin{rmk} \label{rmk:non-cutoff}
To illustrate how our methods can indeed be used in non-cutoff
contexts, consider Kac's model: $\L = \cos \theta = \tL$ and $\R = -\sin \theta = -\tR$,
where $\theta$ is chosen according to an even cross-section function $\beta:[-\pi,\pi] \to \RR_+$
that possibly is singular at $0$, but
satisfies the classical condition $\int_0^\pi \theta^2 \beta(\theta)d\theta < \infty$,
see \cite{desvillettes-graham-meleard1999} for details.
Define $\beta_\eps(\theta) = \ind_{|\theta|>\eps} \beta(\theta)$ for a given cutoff level  $\eps>0$,
and associate with it the collection $(P_t^\eps)_{t\geq 0}$ solving

$\partial_t P_t^\eps = \kappa_\eps (-P_t^\eps + \Q^+_\eps(P_t^\epsilon))$,
where $\kappa_\eps = \int_{-\pi}^\pi \beta_\eps(\theta) d\theta$ and
$\Q_\eps^+$ is defined as
\[
  \int \phi(u) \Q_\eps^+(\mu)(du) = \int_\RR \int_\RR \int_{-\pi}^\pi \phi(u \cos\theta - v \sin\theta ) \frac{\beta_\eps(\theta)d\theta}{\kappa_\eps}\mu(dv) \mu(du).
\]
The particle system $\mathbf{X}^\eps$ and
nonlinear processes $\mathbf{U}^\eps$
are constructed
in a way similar as in \eqref{eq:Xtk} and \eqref{eq:Utk}
but now using a Poisson measure $\N^{\eps,i}(dt,d\theta,d\tau)$ with intensity
$(N-1)^{-1}dt \beta_\eps(\theta)d\theta d\tau \ind_{A^i}(\tau)$
and functions $\G_t^{\eps,i}$ that couple optimally with $P_t^\eps$ instead of $P_t$.
Note that:
\begin{itemize}
\item The even moments of $P_t^\eps$ are controlled uniformly
in time and independently of $\eps$ (see for instance Lemma A.5 in \cite{fournier-godinho2014}
in the case of the non-cutoff nonlinear process; also, an induction similar
to the one used in the proof of Lemma \ref{lem:moment_bounds} yields the
desired uniform bounds for $M_q(P_t^\eps)$ when $q$ is even).

\item The decoupling property of Lemma \ref{lem:decoupling} is also valid for the system $\mathbf{U}_t^\eps$,
with constants independent of $\eps$:
in \eqref{eq:EUV2}, all the terms involve either $1-L$ or $R^2$, which
correspond to $1-\cos\theta$ and $\sin^2 \theta$, respectively, both of order $\theta^2$.

\item In \eqref{eq:EXU2} the constant $\cqr{2}$ corresponds to $\int_{-\pi}^\pi \sin^2 \theta \beta_\eps(\theta)$.
\end{itemize}

Thus, the argument can be replicated and the final constant will depend
on $\int_0^\pi \theta^2 \beta_\eps(\theta) d\theta$, which remains bounded
as we let the cutoff $\eps \to 0$.
Assuming for instance that $M_6(P_0) < \infty$,
this yields a constant $C$ independent of $\eps>0$ such that
\[
\EE\W_2^2(\EM{\mathbf{X}}_t^\eps, P_t^\eps ) \leq \frac{C(1+t)^2 }{ N^{1/3}}.
\]

However, we have not been able to obtain a trajectorial
result in the non-cutoff case: discarding the negative term in the integral of \eqref{eq:XU2}
produces the term $\int_0^\pi \cos^2\theta \beta_\eps(\theta) d\theta$ which no
longer stays bounded when $\eps \to 0$.
\end{rmk}

\section{Proof of intermediate lemmas}
\label{sec:proofs}

\begin{proof}[Proof of Lemma \ref{lem:coupling}]
For fixed   $n\in \NN$, given $\mathbf{y} = (y^1,\ldots,y^n) \in\RR^n$ recall that we write $\EM{\mathbf{y}} = \frac{1}{n} \sum_j \delta_{y^j}$.
 The mapping $(t,\mathbf{y}) \mapsto (P_t,\EM{\mathbf{y}})$
 from $\RR_+ \times \RR^n$ to $\P(\RR)\times\P(\RR)$ is continuous  when $\P(\RR)$ is endowed with the weak topology (weak continuity of  $t\mapsto P_t$ is clear from the pathwise properties of the nonlinear process).  Thus, thanks to a measurable selection
 result   (see for instance Corollary 5.22 of \cite{villani2009}),
 there exists a   measurable  mapping $(t,\mathbf{y}) \mapsto \pi_{t,\EM{\mathbf{y}}}$  such that
 $\pi_{t,\EM{\mathbf{y}}} \in \P(\RR\times\RR)$ is an optimal transference plan
 between $P_t$ and $\EM{\mathbf{y}}$.
 We now define
 \[
 G(t,\mathbf{y},B)
 = \frac{\pi_{t,\EM{\mathbf{y}}}(B\times\{y^1\})}{\pi_{t,\EM{\mathbf{y}}}(\RR\times\{y^1\})}
 = \pi_{t,\EM{\mathbf{y}}}(B\times\{y^1\} \mid \RR\times\{y^1\} ),
 \]
 for $t\geq 0$, $\mathbf{y}\in\RR^n$ and any Borel set $B \subseteq \RR$.
 We claim that $G$ is a probability kernel from $\RR_+ \times \RR^n$ into $\RR$.
 Indeed, it suffices to show that for every such $B$
 the mapping $(t,\mathbf{y}) \mapsto \pi_{t,\EM{\mathbf{y}}}(B\times \{y^1\})$ is measurable,
 which in turn follows from the measurability of $(t,\mathbf{y}) \mapsto (\pi_{t,\EM{\mathbf{y}}},\mathbf{y})$ and the identity
 \[
  \pi_{t,\EM{\mathbf{y}}}(B\times\{y^1\})
  = \lim_{\eps \to 0} \sum_{\ell\in\NN} \pi_{t,\EM{\mathbf{y}}}(B \times D_\ell^{\eps}) \ind_{D_\ell^{\eps}}(y^1),
 \]
 where $(D_\ell^{\eps})_{\ell\in\NN}$ is a measurable partition of $\RR$ with $\textrm{diam}(D_\ell^{\eps}) \leq \eps$.

 Now, given  $N\geq 1$, with the kernel $G$ defined above for $n=N-1$ we can associate  a measurable mapping $g:\RR_+ \times\RR^{N-1}\times[0,1]\to\RR$ or randomization of $G$
 such that $g(t,\mathbf{y},\theta)$ has distribution $G(t,\mathbf{y},\cdot)$
 whenever $\theta$ is a uniform random variable in $[0,1]$
 (see for instance Lemma 3.22 of \cite{kallenberg2002}). For $\mathbf{x}\in \RR^N$ we  now put
 \begin{equation}
 \label{eq:def_coup_funct}
  \G_t^i(\mathbf{x},\tau) = \sum_{j\neq i}^N \ind_{\{\i(\tau)=j\}} g(t,\mathbf{x}^{(ij)}, \tau - \lfloor \tau \rfloor), \quad \tau \in A^i,
 \end{equation}
 where $\mathbf{x}^{(ij)} \in \RR^{N-1}$ denotes the vector $\mathbf{x}$ with its $i$ coordinate removed,
  the $j$ coordinate in the first position, and the remaining coordinates in positions $2,\ldots,N-1$
 in increasing order.
 We now  show that when $\tau$ is uniform in $A^i$,  $\G_t^i(\mathbf{x},\tau)$ and $x^{\i(\tau)}$ have joint distribution
 $\pi_{t,\EMI{\mathbf{x}}{i}}$.

 Denoting $\IP^i$ the law of this
 random variable $\tau$ and using the fact that $g(t,\mathbf{x}^{(ij)},\theta)$ has law
 $\pi_{t,\EMI{\mathbf{x}}{i}}(du \times \{ x^j \} \mid \RR \times \{x^j\})$ when
 $\theta$ is uniform in $[0,1]$,
 we have for every fixed measurable set $B\subseteq\RR$ and every $j\neq i$:
 \begin{align*}
   \IP^i(\G_t^i(\mathbf{x},\tau) \in B, x^{\i(\tau)} = x^j)
  &= \sum_{\ell: x^\ell = x^j, \ell \neq i} \int_{\ell-1}^\ell
     \ind_B(g(t,\mathbf{x}^{(i\ell)},\tau - \lfloor \tau \rfloor)) \frac{d\tau}{N-1} \\
  &= \frac{1}{N-1} \sum_{\ell: x^\ell = x^j, \ell\neq i}
     \frac{\pi_{t,\EMI{\mathbf{x}}{i}}(B\times\{x^\ell\})}{\pi_{t,\EMI{\mathbf{x}}{i}}(\RR\times\{x^\ell\})} \\
  &= \frac{|\{\ell: x^\ell = x^j, \ell\neq i \}|}{(N-1)\pi_{t,\EMI{\mathbf{x}}{i}}(\RR\times\{x^j\})} \pi_{t,\EMI{\mathbf{x}}{i}}(B\times\{x^j\}),
 \end{align*}
 where the quotient in the last line equals 1. This shows that $(\G_t^i(\mathbf{x},\tau),x^{\i(\tau)})$ has distribution $\pi_{t,\EMI{\mathbf{x}}{i}}$
 and concludes the proof of the existence of $\G^i$.

 It remains to show that $\EE \int_{j-1}^j \phi(\G_t^i(\mathbf{Y},\tau)) d\tau  = \langle P_t, \phi \rangle$
 when $\mathbf{Y}$ is exchangeable, $j\neq i$ and $\phi$ is bounded and measurable.

 We get from \eqref{eq:def_coup_funct} that
 \begin{align*}
\int_{j-1}^j \phi(\G_t^i(\mathbf{Y},\tau)) d\tau
 &= \int_0^1 \phi( g(t,\mathbf{Y}^{(ij)}, \tau )) d\tau \\
 &= \int_\RR \phi(u) \pi_{t,\EMI{\mathbf{Y}}{i}}(du \times \{Y^j\} \mid \RR \times \{Y^j\}),
 \end{align*}
 where we have again used that $g(t,\mathbf{Y}^{(ij)},\theta)$ has distribution
 $\pi_{t,\EMI{\mathbf{Y}}{i}}(du \times \{Y^j\} \mid \RR \times \{Y^j\})$ when $\theta$ is uniform in $[0,1]$.
 From the exchangeability of $\mathbf{Y}$, it is clear that the last expression has
 the same distribution, for all $j\neq i$. Thus, its expected value must be the
 same for all $j\neq i$, and since
 \[
 \langle P_t, \phi \rangle
 = \int_{A^i} \phi(\G_t^i(\mathbf{Y},\tau)) \frac{d\tau}{N-1}
 = \sum_{j\neq i} \int_{j-1}^j \phi(\G_t^i(\mathbf{Y},\tau)) \frac{d\tau}{N-1},
 \]
 the conclusion follows.
 \end{proof}

\begin{rmk}\label{rmk:post_thm_coupling}
 Since we are working on $\RR$, the increasing coupling between $P_t$ and $\EMI{\mathbf{x}}{i}$
 is in fact an optimal coupling (see for instance Theorem 6.0.2 in \cite{ambrosio-gigli-savare2008}),
 which allows for a simpler proof of Lemma \ref{lem:coupling}. However, we opted to give a proof
 that remains valid on $\RR^d$ with the hope that this coupling  can be used in a more general setting.
\end{rmk}

\begin{proof}[Proof of Lemma \ref{lem:XU}]
Existence and uniqueness for \eqref{eq:Utk} are obtained with a construction similar to \eqref{eq:Xtn}.
To show that $U^i$ is a nonlinear process,
define $\tilde{\N}^i(dt,d\l,d\r,dv)$ to be the point measure on $\RR_+ \times \RR^2 \times \RR$
with atoms $(t,\l,\r,\G^i_t(\mathbf{X}_{t^-},\tau))$ for every atom $(t,\l,\r,\tau)$ of $\N^i$;
since the dependence on $\mathbf{X}$ is predictable, one can use the
compensation formula to compute the Laplace functional of $\tilde{\N}^i$
and conclude that $\tilde{\N}^i$ is a Poisson point measure with intensity
$dt \barlr(d\l,d\r) P_t(dv)$.
Then, \eqref{eq:nlpr} is satisfied for $V=U^i$ with
$\M = \tilde{\N}^i$, implying that $\law(U^i) = P$. The collection
$(X^1,U^1),\ldots,(X^N,U^N)$ is obviously exchangeable.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:moment_bounds}]
Call $h_t^q = \int |u|^q P_t(du)$.
We first prove the statement for the case $p=2$.
Using \eqref{eq:Pt}-\eqref{eq:Q+} with $\phi = |\cdot|^2$
yields $\partial_t h_t^2 = - \cq{2} h_t^2 + \ELR(\L\R + \tL\tR) (h_t^1)^2$,
and since $\ELR(\L\R + \tL\tR) = 0$ this implies $h_t^2 = h_0^2 e^{-\cq{2}t}$.
Assume now that $q \in (2,q^*)$ is an integer.
Using \eqref{eq:Pt}-\eqref{eq:Q+} with $\phi = |\cdot|^q$, we have
\begin{align}
\partial_t h_t^q
&=    - h_t^q + \frac{1}{2} \iint \ELR( |\L u + \R v|^q + |\tL v + \tR u|^q ) P_t(du) P_t(dv) \label{eq:htq} \\
&\leq - \cq{q} h_t^q
 + \frac{1}{2} \sum_{i=1}^{q-1} \binom{q}{i} h_t^i h_t^{q-i} \ELR( |\L|^i|\R|^{q-i} + |\tL|^i|\tR|^{q-i} ). \notag
\end{align}
Using loose bounds for $\binom{q}{i}$ we obtain
\[
h_t^q
\leq h_0^q e^{-\cq{q}t}
 + C \sum_{i=1}^{q-1} \int_0^t e^{ -\cq{q}(t-s)} h_s^i h_s^{q-i} ds,
\]
where $C$ is a constant that does not depend on $t$, and may change
from line to line.
We now apply induction: the case $q=2$ was already proven, and for $q\in(2,q^*)$ integer,
assuming the desired property for all integer in $\{2,\ldots,q-1\}$ and
using the bound $h_t^1 \leq (h_t^2)^{1/2} \leq C e^{-\cq{2}t/2}$, we obtain
\begin{align*}
h_t^q
&\leq h_0^q e^{-\cq{q}t}
 + C \int_0^t e^{ -\cq{q}(t-s)} e^{-\cq{2}s/2} e^{-\barcq{2,q-1} s} ds \\
& \qquad {} + C \sum_{i=2}^{q-2}  \int_0^t e^{ -\cq{q}(t-s)} e^{-\barcq{2,i} s} e^{-\barcq{2,q-i} s} ds.
\end{align*}
Note that  $\cq{q}>0$, since $2<q<q^*$, and recall that
$\barcq{2,q} := \inf_{2\leq r \leq q} \cq{r} = \min(\cq{2},\cq{q})$. Thus,
if $\cq{2} = 0$ then $\barcq{2,i} = \barcq{2,q-i} = \barcq{2,q} = 0$ and
the last inequality yields
$h_t^q \leq h_0^q + C \int_0^t e^{-\cq{q}(t-s)}ds \leq C$, as desired.
On the other hand, if $\cq{2} > 0$, we bound $\cq{2}$, $\cq{q}$, $\barcq{2,i}$ and $\barcq{2,q-i}$
from below by $\barcq{2,q}>0$ and obtain
$h_t^q \leq h_0^q e^{-\barcq{2,q}t} + C \int_0^t e^{-\barcq{2,q}(t+s/2)}ds \leq C e^{-\barcq{2,q}t}$,
which concludes the induction and the proof in the case $p=2$ and integer $q\in\{2\} \cup (2,q^*)$.

Asume now that $2<q=m+\eps<q^*$ with $m\in\{2,\ldots\}$ and $\eps\in(0,1)$.
Bounding $|x+y|^{q} \leq (|x|+|y|)^m(|x|^\eps+|y|^\eps)$ in \eqref{eq:htq}
and using the binomial theorem as before, we obtain
\begin{align*}
\partial_t h_t^q
&=    - h_t^q + \frac{1}{2} \sum_{i=0}^m \binom{m}{i} \iint
      \ELR\left( (|\L u|^\eps + |\R v|^\eps) |\L u|^i |\R v|^{m-i} \right. \\
& \qquad \qquad \qquad {} \left. + (|\tL v|^\eps + |\tR u|^\eps) |\tL v|^i |\tR u|^{m-i}
      \right) P_t(du) P_t(dv) \\
&= -\cq{q} h_t^q + \sum_{i=0}^{m-1} \binom{m}{i}
    \ELR\left(|\L|^{i+\eps}|\R|^{m-i} + |\L|^{m-i}|\R|^{i+\eps} \right. \\
& \qquad\qquad\qquad {} \left.
                      + |\tL|^{i+\eps}|\tR|^{m-i} + |\tL|^{m-i}|\tR|^{i+\eps}\right) h_t^{i+\eps}h_t^{m-i},
\end{align*}
which yields
\[
h_t^q
\leq h_0^q e^{-\cq{q}t}
 + C \sum_{i=0}^{m-1} \int_0^t e^{ -\cq{q}(t-s)} h_s^{i+\eps}h_s^{m-i} ds.
\]
Note that $h_t^r \leq (h_t^2)^{r/2} \leq C e^{-r \cq{2} t /2}$ for $r \in (0,2)$.
This and the fact that the property is true for the integers,
allow us to use induction on $m$ in a way similar as before, and conclude the proof in the case $p=2$.

A similar argument, with the induction starting at $q=1$, proves the case $p=1$.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:decoupling}]
Let us first prove the case $p=1$.
Given $k\in\{2,\ldots,N\}$ fixed, we want to construct $k$ independent nonlinear processes
$V^1,\ldots,V^k$ such that $\EE|U_t^i - V_t^i|$ is small. To achieve this, we will decouple
$U^1,\ldots,U^k$ by replacing the shared atoms of $\N^1,\ldots,\N^k$
with new, independent atoms. To this end, let $\M$ be an independent copy of
$\N$ (also independent from $\mathbf{X}_0$),
and define for each $i\in\{1,\ldots,k\}$
\begin{align} \label{eq:Mkl}
\begin{split}
\M^{i}(dt,d\l,d\r,d\tau)
&= \N(dt,(d\l\times d\r \times \RR^2),[i-1,i),d\tau) \\
&\quad + \N(dt,(\RR^2 \times d\l \times d\r),d\tau,[i-1,i)) \ind_{[k,N)}(\tau) \\
&\quad + \M(dt,(\RR^2 \times d\l \times d\r),d\tau,[i-1,i)) \ind_{[0,k)}(\tau).
\end{split}
\end{align}
Note that $\M^{i}$ is, like $\N^i$, a Poisson point
measure on $\RR_+ \times \RR^2 \times [0,N)$ with intensity $(N-1)^{-1}dt \barlr(\l,\r) d\tau \ind_{A^i}(\tau)$,
and that $\M^{1},\ldots,\M^{k}$ are independent.
Following \eqref{eq:Utk}, we define $V^{i}$ as the solution of
\begin{equation} \label{eq:Vtkl}
dV_t^{i}
= \int_{\RR^2} \int_{[0,N)} [(\l-1)V_{t^-}^{i} + \r\G_t^i(\mathbf{X}_{t^-},\tau)] \M^{i}(dt,d\l,d\r,d\tau),
\end{equation}
with $V_0^{i} = U_0^i$.
If we define $\tilde{\M}^i$ to be the point process in $\RR_+ \times \RR^2 \times \RR$
with atoms $(t,\l,\r,\G_t^i (\mathbf{X}_{t^-},\tau))$ for every atom $(t,\l,\r,\tau)$ of $\M^i$,
it is clear that $V^i$ depends only on $\tilde{\M}^i$ and $X_0^i$.
Since: (i) the dependence on $\mathbf{X}$ is predictable,
(ii) the Poisson measures $\M^1,\ldots,\M^k$ are independent and
(iii) the $\tau$-law of $\G_t^i(\mathbf{x},\tau)$ is $P_t$ for every $\mathbf{x}\in\RR^N$,
one can use the compensation formula to compute the joint
Laplace functional of $\tilde{\M}^1,\ldots,\tilde{\M}^k$
and conclude that they are independent Poisson point measures,
all with intensity $dt \barlr(d\l,d\r) P_t(dv)$.
This shows that each $V^i$ is a nonlinear process and that
they are independent.

Consequently, we have
\[
\W_1( \law^k(\mathbf{U}_t), P_t^{\otimes k})
\leq \EE \left( \frac{1}{k} \sum_{i=1}^k |U_t^i - V_t^{i}| \right)
= \EE |U_t^1 - V_t^{1}|,
\]
where in the last step we used the fact that all the $(U^i,V^{i})$'s
have the same law. To estimate the last term
$h_t = \EE |U_t^1 - V_t^{1}|$, we proceed as in \eqref{eq:XtUt}:
from \eqref{eq:Nk}, \eqref{eq:Utk}, \eqref{eq:Mkl} and \eqref{eq:Vtkl}, we have
for all $0\leq s\leq t$:
\begin{equation}
h_t
= h_s + \EE \int_{(s,t]} \int_{\RR^2}\int_{[0,N)} ( J_r^1 + J_r^2 + J_r^3 ), \label{eq:UtkVtkl}
\end{equation}
where $J_r^1$ is the term associated with the simultaneous jumps of $U^1$ and $V^1$,
$J_r^2$ corresponds to the jumps of $U^1$ alone, and $J_r^3$ gives the jumps of $V^1$ alone.
Specifically,
\begin{align*}
J_r^1
&= (|\l|-1) |U_{r^-}^1 - V_{r^-}^{1}| \left( \vphantom{\ind_{[k,N)}(\tau)}
   \N(dr,(d\l \times d\r \times \RR^2),[0,1),d\tau) \right. \\
& \qquad \qquad \qquad \qquad \qquad {} \left. + \N(dr,(\RR^2 \times d\l \times d\r),d\tau,[0,1)) \ind_{[k,N)}(\tau) \right), \\
J_r^2
&= \left(|\l(U_{r^-}^1 - V_{r^-}^{1}) + \r \G_r^1(\mathbf{X}_{r^-},\tau) + (\l-1)V_{r^-}^{1}| - |U_{r^-}^1 - V_{r^-}^{1}|\right) \\
& \qquad \qquad \qquad \N(dr,(\RR^2 \times d\l \times d\r),d\tau,[0,1)) \ind_{[0,k)}(\tau), \\
J_r^3
&= \left( |\l(U_{r^-}^1 - V_{r^-}^{1}) - \r \G_r^1(\mathbf{X}_{r^-},\tau) - (\l-1)U_{r^-}^1| - |U_{r^-}^1 - V_{r^-}^{1}|\right) \\
& \qquad \qquad \qquad \M(dr,(\RR^2 \times d\l \times d\r),d\tau,[0,1)) \ind_{[0,k)}(\tau).
\end{align*}
Then:
\[
\EE \int_{(s,t]} \int_{\RR^2} \int_{[0,N)} J_r^1
=  \left( \frac{1}{2}(\ELR|\L|-1)  + \frac{1}{2}(\ELR|\tL|-1) \frac{N-k}{N-1} \right) \int_s^t h_r dr.
\]
Using the triangle inequality in the term $J_r^2$:
\begin{align*}
\EE \int_{(s,t]} \int_{\RR^2} \int_{[0,N)} J_r^2
&\leq \EE \int_s^t \int_1^k \left( (\ELR|\tL|-1) |U_{r}^1 - V_{r}^{1}| \right. \\
&    \qquad {} \left. + \ELR|\tR| |\G_r^1(\mathbf{X}_r,\tau)| + \ELR|\tL-1| |V_r^{1}| \right)
      \frac{dr  d\tau }{2(N-1)}.
\end{align*}
From Lemma \ref{lem:coupling}
we know that $\EE \int_{i-1}^i |\G_r^1(\mathbf{X}_r,\tau)| d\tau = \mom_1(P_r)$ for all $i=2,\ldots,k$.
Using that $V_r^{1}$ has law $P_r$, we obtain
\begin{align*}
&\EE \int_{(s,t]} \int_{\RR^2} \int_{[0,N)} J_r^2  \\
&\leq \frac{k-1}{2(N-1)} \left( (\ELR|\tL|-1) \int_s^t h_r  dr
+ (\ELR|\tL-1| + \ELR|\tR|) \int_s^t \mom_1(P_r) dr \right).
\end{align*}
With a similar argument, the last inequality is also valid with $J_r^3$ in the left hand side.
Putting all this into \eqref{eq:UtkVtkl}, we have
\begin{align*}
h_t
 &\leq h_s - \left(1-\frac{1}{2}\ELR(|\L|+|\tL|) + \frac{1}{2}(1-\ELR|\tL|)\frac{k-1}{N-1} \right)
  \int_s^t h_r  dr \\
 & \qquad {} + \frac{(\ELR|\tL-1| + \ELR|\tR|) (k-1)}{N-1}\int_s^t \mom_1(P_r)dr.
\end{align*}
Recall the constants $\cql{1} = \frac{1}{2}\ELR(|\L|+|\tL|)$,
$\cqr{1} = \frac{1}{2}\ELR(|\R|+|\tR|)$ and $\cq{1} = 1- \cql{1} - \cqr{1}$.
Also, put $\bi = \frac{1}{2}(1-\ELR|\tL|)$, which can be assumed
non-negative without loss of generality
(if not, exchange the roles of $(\L,\R)$ and $(\tL,\tR)$).
From the previous inequality and from Lemma \ref{lem:moment_bounds} in the case $q=1$,
it follows that for almost all $t\geq 0$,
\[
\partial_t h_t \leq - \left( \cq{1} + \cqr{1} + \bi \frac{k-1}{N-1} \right) h_t
+ \frac{C (k-1) e^{-\cq{1} t} }{N-1},
\]
and now Gronwall's lemma gives
\[
h_t
\leq \frac{C(k-1)e^{-\cq{1} t}}{(N-1)(\cqr{1} + \bi \frac{k-1}{N-1})}
     \left[ 1 - e^{-(\cqr{1} + \bi \frac{k-1}{N-1})t} \right].
\]
Using the inequality $1-e^{-x} \leq x$, the desired result follows for the case $p=1$.

In the case $p=2$, we construct the system $V^1,\ldots,V^k$ exactly as before,
but using the functions $\G_t^i$ provided by Lemma \ref{lem:coupling} with cost $|x-y|^2$.
To obtain the desired inequality for
$\W_2^2(\law^k(\mathbf{U}_t),P_t^{\otimes k})$ it suffices to work with $h_t = \EE(U_t^1-V_t^1)^2$.
We also have \eqref{eq:UtkVtkl}, where $J_r^1$, $J_r^2$ and $J_r^3$ now are given by
\begin{align*}
J_r^1
&= (\l^2-1) (U_{r^-}^1 - V_{r^-}^{1})^2 \left( \vphantom{\ind_{[k,N)}(\tau)}
   \N(dr,(d\l \times d\r \times \RR^2),[0,1),d\tau) \right. \\
& \qquad \qquad \qquad \qquad \qquad {} \left. + \N(dr,(\RR^2 \times d\l \times d\r),d\tau,[0,1)) \ind_{[k,N)}(\tau) \right), \\
J_r^2
&= \left((\l(U_{r^-}^1 - V_{r^-}^{1}) + \r \G_r^1(\mathbf{X}_{r^-},\tau) + (\l-1)V_{r^-}^{1})^2 - (U_{r^-}^1 - V_{r^-}^{1})^2\right) \\
& \qquad \qquad \qquad \N(dr,(\RR^2 \times d\l \times d\r),d\tau,[0,1)) \ind_{[0,k)}(\tau), \\
J_r^3
&= \left( (\l(U_{r^-}^1 - V_{r^-}^{1}) - \r \G_r^1(\mathbf{X}_{r^-},\tau) - (\l-1)U_{r^-}^1)^2 - (U_{r^-}^1 - V_{r^-}^{1})^2\right) \\
& \qquad \qquad \qquad \M(dr,(\RR^2 \times d\l \times d\r),d\tau,[0,1)) \ind_{[0,k)}(\tau).
\end{align*}
Using that $\EE \int_{i-1}^i \G_t^1(\mathbf{X}_t,\tau)^2 d\tau = \mom_2(P_t)$ for all $i=2,\ldots,k$,
we obtain:
\begin{align*}
\EE \int_{(s,t]} \int_{\RR^2} \int_{[0,N)} J_r^1
&=  \left( \frac{1}{2}(\ELR \L^2-1)  + \frac{1}{2}(\ELR \tL^2-1) \frac{N-k}{N-1} \right) \int_s^t h_r dr, \\
\EE \int_{(s,t]} \int_{\RR^2} \int_{[0,N)} J_r^2
&= \int_s^t \left( \vphantom{\int_1^k} (\ELR \tL^2 -1) h_r + \ELR \tR^2 \mom_2(P_r) + \ELR(\tL-1)^2 \mom_2(P_r) \right. \\
&    \quad {} + 2 \ELR (\tL\tR) \EE(U_r^1-V_r^1)\int_1^k \G_r^1(\mathbf{X}_r,\tau) \frac{d\tau}{k-1} \\
&    \quad {} + 2\ELR(\tL(\tL-1)) \EE (U_r^1-V_r^1) V_r^{1} \\
&    \quad {} + \left. 2\ELR ((\tL-1)\tR) \EE V_r^1 \int_1^k \G_r^1(\mathbf{X}_r,\tau)\frac{d\tau}{k-1} \right) \frac{(k-1) dr}{2(N-1)}, \\
\EE \int_{(s,t]} \int_{\RR^2} \int_{[0,N)} J_r^3
&= \int_s^t \left( \vphantom{\int_1^k} (\ELR \tL^2 -1) h_r + \ELR \tR^2 \mom_2(P_r) + \ELR(\tL-1)^2 \mom_2(P_r) \right. \\
&    \quad {} - 2 \ELR (\tL\tR) \EE(U_r^1-V_r^1)\int_1^k \G_r^1(\mathbf{X}_r,\tau) \frac{d\tau}{k-1} \\
&    \quad {} - 2\ELR(\tL(\tL-1)) \EE (U_r^1-V_r^1) U_r^{1} \\
&    \quad {} + \left. 2\ELR ((\tL-1)\tR) \EE U_r^1 \int_1^k \G_r^1(\mathbf{X}_r,\tau)\frac{d\tau}{k-1} \right) \frac{(k-1) dr}{2(N-1)}.
\end{align*}
From this and \eqref{eq:UtkVtkl}, we have for almost all $t\geq 0$
\begin{align}
\label{eq:EUV2}
\begin{split}
\partial_t h_t
&= -h_t \left( \left(1-\frac{1}{2}\ELR(\L^2 + \tL^2) \right) + \frac{k-1}{2(N-1)}\ELR(\tL-1)^2 \right) \\
& \qquad {} + \mom_2(P_t) \frac{k-1}{N-1} (\ELR(\tL-1)^2 + \ELR\tR^2) \\
& \qquad {} + \ELR((\tL-1)\tR) \EE (U_t^1 + V_t^1) \int_1^k \G_t^1(\mathbf{X}_t,\tau)\frac{d\tau}{N-1}.
\end{split}
\end{align}
We also have $\int_{i-1}^i  \EE (U_t^1 + V_t^1) \G_t^1(\mathbf{X}_t,\tau) d\tau \leq 2  \mom_2(P_t)$
for all $i=2,\ldots,k$, thanks to the Cauchy-Schwarz and Jensen's inequalities.
Recall the constants
$\cql{2} = \frac{1}{2}\ELR(\L^2 + \tL^2)$, $\cqr{2} = \frac{1}{2}\ELR(\R^2 + \tR^2)$,
$\cq{2} = 1-\cql{2} - \cqr{2}$,
and put $\bii = \frac{1}{2}\ELR(\tL-1)^2$.
Using Lemma \ref{lem:moment_bounds} in the case $p=q=2$,
we thus obtain
\[
\partial_t h_t \leq -\left(\cq{2} + \cqr{2} + b\frac{k-1}{N-1} \right)h_t
 + \frac{C (k-1) \mom_2(P_0) e^{-\cq{2} t}}{N-1},
\]
and the conclusion follows from Gronwall's lemma as before.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:W2Ymu}]
For simplicity, we will prove only the case $\ell =0$,
that is,  when $n$ divides $m$. Let us arrange a vector $\mathbf{y}\in\RR^m$ as a matrix
with $k$ rows and $n$ columns, that is, $\mathbf{y} = (y^{ij})$, with $i=1,\ldots,k$,
$j=1,\ldots,n$, and write $\mathbf{y}_i = (y^{i1},\ldots,y^{in})$
and $\EM{\mathbf{y}}_i = \frac{1}{n}\sum_{j=1}^n \delta_{y^{ij}}$.

Let us couple $\mathbf{Y}$ with a random vector $\mathbf{Z} \in (\RR^n)^k$
in such a way that each $(\mathbf{Y}_i,\mathbf{Z}_i)$ is an optimal coupling between

$\law^n(\mathbf{Y})$
and $\mu^{\otimes n}$
(with respect to the cost function $d_{n,p}^p(\cdot,\cdot)$ of \eqref{eq:dkp}, as usual).
Using the latter, we have:
\begin{equation}
\label{eq:EWpYZ}
\EE \W_p^p(\EM{\mathbf{Y}},\EM{\mathbf{Z}})
\leq \frac{1}{k} \sum_{i=1}^k \EE \frac{1}{n} \sum_{j=1}^n |Y^{ij}-Z^{ij}|^p
= \W_p^p(\nu^n,\mu^{\otimes n}).
\end{equation}
On the other hand, for each $i=1,\ldots,k$ there is a function $q^i : \RR^m \times [0,1] \to \RR$
such that for all $\mathbf{z} \in \RR^m$,  the pair $(z^{i\i^n(\theta)},q^i(\mathbf{z},\theta))$ with $\i^n(\theta) = \i(n\theta) = \lfloor n \theta \rfloor + 1$,  is an optimal coupling
between $\EM{\mathbf{z}}_i$ and $\mu$ when $\theta$ is uniformly chosen in $[0,1]$.
Now we randomize the choice of $i$ with a uniform variable $\vartheta \in [0,1]$
independent of $\theta$, so $z^{\i^k(\vartheta) \i^n(\theta)}$ and $q^{\i^k(\vartheta)}(\mathbf{z},\theta)$
are $(\theta,\vartheta)$-realizations of $\EM{\mathbf{z}}$ and $\mu$, respectively.
Putting $\mathbf{Z}$ in place of $\mathbf{z}$, this construction gives:
\begin{align*}
\EE \W_p^p(\EM{\mathbf{Z}},\mu)
&\leq \EE \iint_{[0,1]^2} |Z^{\i^k(\vartheta) \i^n(\theta)} - q^{\i^k(\vartheta)}(\mathbf{Z},\theta))|^p d\vartheta d\theta \\
&= \EE \frac{1}{k} \sum_{i=1}^k \W_p^p(\EM{\mathbf{Z}}_i,\mu)
= \eps_{n,p}(\mu).
\end{align*}
(Recall that $\eps_{n,p}(\mu) = \EE\W_p^p(\frac{1}{n}\sum_i \delta_{\zeta^i},\mu)$,
with $\zeta^1,\ldots,\zeta^n$ independent and $\mu$-distributed).
With this and \eqref{eq:EWpYZ} we conclude in the case $\ell =0$:
\begin{align*}
\EE \W_p^p(\EM{\mathbf{Y}},\mu)
&\leq \EE\left(\W_p(\EM{\mathbf{Y}},\EM{\mathbf{Z}}) + \W_p(\EM{\mathbf{Z}},\mu) \right)^p \\
&\leq 2^{p-1}\left(\EE \W_p^p(\EM{\mathbf{Y}},\EM{\mathbf{Z}}) + \EE \W_p^p(\EM{\mathbf{Z}},\mu) \right).
\end{align*}
In the case $\ell >0$ the construction is similar, but now $(\mathbf{Y},\mathbf{Z})$
must include an additional optimal coupling between

$\law^\ell (\mathbf{Y})$
and $\mu^{\otimes \ell }$,
which gives the extra term.
\end{proof}

\section*{Acknowledgements}

We  thank two anonymous referees for carefully reading a former version of this  work and for their questions and remarks that allowed us to improve its  presentation.

\bibliographystyle{acm}
\bibliography{probability.bib}{}

\end{document}

config: arxiv


TITLE: specific magma examples
QUESTION [5 upvotes]: Give an example of a magma $S$ such that $S$ has a zero and $S$ has a left zero divisor that is not a right zero divisor
an example of a magma with an identity such that there is an element with exactly $2$ left inverses but only one right inverse
For the first, I was thinking $\{0,a,b\}$ where $ab=0$, $ba=a$, $b^2=b$, $a^2=a$.
Thanks

REPLY [1 votes]: Your answer to the first question may be right (thanks @celtschk for commenting):
$$
\begin{array}{c|ccc}
\cdot & 0 & a & b\\
\hline
0 & 0 & 0 & 0\\
a & 0 & a & 0\\
b & 0 & a & b
\end{array}
$$
Here $0$ denotes an absorbing element, $a$ is only a left zero divisor, while $b$ is only a right zero divisor.

Second question: here they ask for an identity, let  us call it $1$. From your comment

@drhab a^2=a, ab=b, ba=1, b^2=a?

I get
$$
\begin{array}{c|ccc}
\bullet & 1 & a & b\\
\hline
1 & 1 & a & b\\
a & a & a & b\\
b & b & 1 & a
\end{array}
$$
but to obtain

an element with exactly 2 left inverses but only one right inverse

we must have, in the Cayley table, $1$ appearing exactly twice in the column labeled by the element and exactly once in the line labeled by the same element, for example:
$$
\begin{array}{c|ccc}
\bullet & 1 & a & b\\
\hline
1 & 1 & a & b\\
a & a & 1 & b\\
b & b & 1 & a
\end{array}
$$
In this last table (I got this from the previous changing only one cell, $a\bullet a$), $1$ is the identity, $a$ and $b$ are both left inverses of $a$, while only $a$ is a right inverse of $a$.
With four or more elements in the underlying set we could get more various examples, like:
$$
\begin{array}{c|cccccc}
* & 1 & 2 & 3 & 4 & 5 & 6\\
\hline
1 & \color{blue}{1} & \color{blue}{2} & \color{blue}{3} & \color{blue}{4} & \color{blue}{5} & \color{blue}{6}\\
2 & \color{blue}{2} & 3 & 2 & 3 & 2 & 3\\
3 & \color{blue}{3} & 3 & 2 & 3 & 2 & \color{red}{1}\\
4 & \color{blue}{4} & 3 & 2 & 3 & 2 & \color{red}{1}\\
5 & \color{blue}{5} & 3 & 2 & 3 & 2 & 3\\
6 & \color{blue}{6} & 3 & 2 & 3 & \color{red}{1} & 3\\
\end{array}
$$
Here, $1$ is the identity (in blue), $3$ and $4$ are left inverses of $6$ and $5$ is right inverse of $6$ (in red).


TITLE: $\phi$ invariant Borel probability measure
QUESTION [2 upvotes]: Given $X$ and  $\phi:[0,1]\rightarrow[0,1]$
$\phi(x)= \begin{cases} \frac{1}{2}x & x\neq0\\ 1&x=0\end{cases}$
I've to show that there can't be a Borel-probability measure $\mu$ which is $\phi$ invariant (i.e. $\mu(\phi^{-1}(A))=\mu(A)$ $\forall A\in \mathcal{B}(X)$)

My first idea was proving it by contradiction. But without success.

REPLY [2 votes]: Suppose such probability measure $P$ on $[0, 1]$ exists. Suppose $F(x)$ is its CDF. By our supposition:
$F(x) = \begin{cases}
    1 - P(\{0\} )  & \quad x > \frac{1}{2}\\
    F(2x) - P(\{0\})  & \quad x \leq \frac{1}{2}
  \end{cases}$
If $P(\{0\}) > 0$, then $\exists x > 0 \text{ such that } F(x) < 0$, which is impossible.
If $P(\{0\}) = 0$, then $\forall x > 0 \text{ } F(x) = 1$, which is also impossible.
Thus we receive a contradiction.


TITLE: How can i rewrite my specific $F_{n,2d}^a$ polynomial to be a sum of $(3n-4)$ squares?
QUESTION [2 upvotes]: So, i've been messing around with the following $n$-variate polynomials of degree $2d$:
$$F_{n,2d}^a = \sum_{i=1}^n a_ix_i^{2d} + 2d\prod_{i=1}^n x_i^{a_i} $$
Where $\sum_{i=1}^n a_i=2d$
Now, i want to show that this $F_{n,2d}^a$ can be written at most as the sum of $3n-4$ squares.
I thought of doing it per induction, but there were two problems:

$F_{2,2d}^a$ can be written as 2 sums of squares (which works) , but i couldn't prove it
even if i could, it doesnt work because if you presume $\sum_{i=1}^n a_i=2d$, you cant say this about $\sum_{i=1}^{n+1} a_i$ anymore.

What can I do? (I would like to show the first one anyway, just to understand whats going on. so if you have any ideas on that, let me know)
thanks!!

REPLY [0 votes]: (1) First, we prove $F_{2,2d}^a$ is the sum of two squares:
So $F_{2,2d}^a=a_1x_1^{2d} +a_2x_2^{2d} -2dx_1^{a_1}x_2^{a_2} $
We can take the homonization $P(x) = a_1x^{2d} + a_2 -2dx^{a_1}$, which is univariate. If we now show that $P(x)\geq 0$, then it follows from excercise 11.3(b) in the lecture notes that $P$ is a sum of at most two squares. And if we then rewrite this sum of square polynomial it to its homogeniousation, $F_{2,2d}^a$, we attain that this can also only be a sum of at most two squares as well.
Now, to prove $P\geq0$, there are 3 cases we can distinguish between:

$a_1=0$, In this case we'd have: $P(x) =  0 + 2d - 2dx^0 = 0$, which is clearly a sum of zero squares.

$a_1=2d$, In this case we'd have: $P(x) =  2dx^{2d} - 2dx^{2d} = 0$, which is clearly also a sum of zero squares.

$0<a_1<2d$, In that case we see that \begin{align*}
        \lim\limits_{x\to \infty} P(x) &=  \lim\limits_{x\to \infty} a_1x^{2d} + a_2 -2dx^{a_1}\\
        &=  \lim\limits_{x\to \infty} a_1x^{2d} \\
        &=  \lim\limits_{x\to \infty} a_1(-x)^{2d} \\
        &=  \lim\limits_{x\to -\infty} a_1x^{2d} \\
        &=  \lim\limits_{x\to -\infty} a_1x^{2d} + a_2 -2dx^{a_1}\\
        &= \lim\limits_{x\to -\infty} P(x)
    \end{align*}
Then, $\frac{\partial}{\partial x} P(x) = 2da_1x^{2d-1} - 2da_1x^{a_1-1} = 2da_1x^{a_1-1}(x^{2d-a_1} - 1)$, so our extreme points are attained at 0 and 1 (and possibly -1, in the case that $a_1$ is even). Which are $P(0)=a_2$ and $P(1) = a_1 + a_2 - 2d=0$. And as we know $a_2>0$, we know that $P(x)\geq 0$.

\begin{pmatrix}
0 & \frac{a_1}{2} &  & -d \\
\frac{a_1}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{a_1}{2} \\
-d & 0 & \frac{a_1}{2} & 0 \\
\end{pmatrix}
and take the vector \begin{pmatrix}
x_1^{a_1} \\
x_1^{a_2} \\
x_2^{a_1} \\
x_2^{a_2} \\
\end{pmatrix}
But this is not a PSD matrix because the vector (1,-1,0,0) makes things go off.
(2) Now we show that if $a\in\mathbb{N}^n$ with $|a|= 2d$ that $a$ can be decomposed as $a=b+c$, where $b,c\in\mathbb{N}^n$, $|b|=|c|= d$ and there is at most one index $i \in[n]$ such that $b_i,c_i>̨0$.
There are two options: there exists an $i$ such that $a_i\geq d$ or $\forall i$ $a_i<d$

In the case there exists an $ i$ such that $a_i\geq d$ set \begin{equation*}
    b_j = \begin{cases} 0 \quad & j\neq i\\ d & j=i \end{cases} \qquad \text{    and    } \qquad
    c_j = \begin{cases} a_j & j\neq i\\ a_i-d & j=i \end{cases}
\end{equation*}
Then $|b| = 0+\dots+0+d =d$ and $|c| = \smash{\sum_{j\neq i} a_j + a_i -d = \smash{\sum_{i=1}^n} a_i} -d = 2d-d=d$. And it is also trivial that $(b+c)_j = \begin{cases} 0 +a_j \quad & j\neq i\\ d + a_i-d& j=i \end{cases} = a_j \quad \Rightarrow b+c=a$

In the case $\forall i$ $a_i<d$, set
$$m = \max\Bigg\{i\in[n] \mid \smash{\sum_{j=1} ^i} a_i \leq d\Bigg\}$$
Then set
\begin{equation*}
    b_j = \begin{cases}
    a_i \quad & i\leq m\\
    d-\smash{\sum_{j=1}^m} a_j  & i=m+1 \\
    0 & i>m+1\end{cases}
    \qquad \text{    and    } \qquad
    c_j = \begin{cases}
    0 \quad & i\leq m\\
    d-\smash{\sum_{j=m+2}^n} a_j  & i=m+1 \\
    a_i & i>m+1\end{cases}
\end{equation*}
Then  $|b| =\smash{\sum_{j=1}^m} a_j +d-\smash{\sum_{j=1}^m} a_j  =d$ and $|c| = \smash{\sum_{j=m+2}^n} a_j +d-\smash{\sum_{j=m+2}^n} a_j  =d$.
And lastly, \begin{align*}
    (b+c)_i & = \begin{cases}
    a_i+0 \quad & i\leq m\\
    d-\smash{\sum_{j=1}^m} a_j +d-\smash{\sum_{j=m+2}^n} a_j   & i=m+1 \\
    0+a_i & i>m+1\end{cases}\\
    &= \begin{cases}
    a_i \quad & i\leq m\\
    2d-\smash{\sum_{j\neq m+1}} a_j  & i=m+1 \\
    a_i & i>m+1\end{cases}\\
    &= \begin{cases}
    a_i \quad & i\leq m\\
    \smash{\sum_{j=1}^n} a_j-\smash{\sum_{j\neq m+1}} a_j  & i=m+1 \\
    a_i & i>m+1\end{cases}\\
    &= \begin{cases}
    a_i \quad & i\leq m\\
    a_{m+1}  & i=m+1 \\
    a_i & i>m+1\end{cases}\\
    &= a_i \end{align*}
$\Rightarrow b+c=a$
Thus, for both cases there is at most one index $i \in[n]$ such that $b_i,c_i>̨0$

(3) Now we show that with $a, b, c$ as in (2), show that $$F_{n,2d}^a(x) =\frac{1}{2}\Big(F_{n,2d}^{2b}(x)+F_{n,2d}^{2c}(x)\Big) + d(x^b-x^c)^2$$
\begin{align*}
  \frac{1}{2}\Big(F_{n,2d}^{2b}(x)&+F_{n,2d}^{2c}(x)\Big) + d(x^b-x^c)^2\\
  &=  \frac{1}{2}\Bigg(\smash{\sum_{i=1}^n} 2b_ix_i^{2d} - 2dx^{2b} + \smash{\sum_{i=1}^n} 2c_ix_i^{2d} - 2dx^{2c} \Bigg) + d(x^b-x^c)^2\\
  &= \smash{\sum_{i=1}^n} (b_i+c_i)x_i^{2d} - d(x^{2b}+ x^{2c})  + d(x^b-x^c)^2\\ &\\
  (*)&= \smash{\sum_{i=1}^n} a_ix_i^{2d} - d(\smash{\prod_{i=1}^n} x_i^{2b_i} + \smash{\prod_{i=1}^n} x_i^{2c_i} )  + d(\smash{\prod_{i=1}^n} x_i^{b_i} -\smash{\prod_{i=1}^n} x_i^{c_i})^2\\
\end{align*}
Then we have \begin{align*}
\Big(\smash{\prod_{i=1}^n} x_i^{b_i} -\smash{\prod_{i=1}^n} x_i^{c_i} \Big)^2& = \smash{\prod_{i=1}^n} x_i^{2b_i} + \smash{\prod_{i=1}^n} x_i^{2c_i} - 2\smash{\prod_{i=1}^n} x_i^{b_i}\smash{\prod_{i=1}^n} x_i^{c_i} \\
&\\
&= \smash{\prod_{i=1}^n} x_i^{2b_i} + \smash{\prod_{i=1}^n} x_i^{2c_i} - 2\smash{\prod_{i=1}^n} x_i^{b_i+c_i}\\
&\\
&= \smash{\prod_{i=1}^n} x_i^{2b_i} + \smash{\prod_{i=1}^n} x_i^{2c_i} - 2\smash{\prod_{i=1}^n} x_i^{a_i}\\
\end{align*}
$\iff$ $\smash{\prod_{i=1}^n} x_i^{2b_i} + \smash{\prod_{i=1}^n} x_i^{2c_i} = \Big(\smash{\prod_{i=1}^n} x_i^{b_i} -\smash{\prod_{i=1}^n} x_i^{c_i} \Big)^2 + 2\smash{\prod_{i=1}^n} x_i^{a_i}$
So we can continue \begin{align*}
\smash{\sum_{i=1}^n} a_ix_i^{2d} - &d(\smash{\prod_{i=1}^n} x_i^{2b_i} + \smash{\prod_{i=1}^n} x_i^{2c_i} )  + d(\smash{\prod_{i=1}^n} x_i^{b_i} -\smash{\prod_{i=1}^n} x_i^{c_i})^2\\ \\
&= \smash{\sum_{i=1}^n} a_ix_i^{2d} - d\Bigg(\Big(\smash{\prod_{i=1}^n} x_i^{b_i} -\smash{\prod_{i=1}^n} x_i^{c_i} \Big)^2 + 2\smash{\prod_{i=1}^n} x_i^{a_i} \Bigg)  + d(\smash{\prod_{i=1}^n} x_i^{b_i} -\smash{\prod_{i=1}^n} x_i^{c_i})^2\\ \\
&= \smash{\sum_{i=1}^n} a_ix_i^{2d} - 2d \smash{\prod_{i=1}^n} x_i^{a_i} \\ \\
&= F_{n,2d}^a(x)
\end{align*}
(4) And now we show that, for any $a\in\mathbb{N}^n$ with $|a|=2d$, the polynomial $F_{n,2d}^a$ can be written as the sum of at most $3n - 4$ squares.
We prove this by induction.
We know it holds for $n=2$, as can be assumed from (1) (because $F_{2,2d}^a$ is sum of 2 squares, and $3\times 2 -4 = 2$.)
Now, we assume it holds for all $k\leq n$, and we show it holds for $n+1$.
From (3) we know $F_{n+1,2d}^a(x) =\frac{1}{2}\Big(F_{n+1,2d}^{2b}(x)+F_{n+1,2d}^{2c}(x)\Big) + d(x^b-x^c)^2$ with the assumtions on $b$ and $c$ as in (2). Define $I_b = \{i\in[n+1] \mid b_i\neq 0\}$ and $I_c = \{i\in[n+1] \mid c_i\neq 0\}$ and set $n_b=|I_b|$ and $n_c=|I_c|$. Then, from (2) we know that $n_b+n_c \leq n+2$ as they have at most one component double.
So there are two cases: $n_b$, $n_c<n+1$ or $n_b=n=1$ and $n_c=1$ or the other way around.
We'll first show the case of $n_b$, $n_c<n+1$.
Define the vectors $b'=b_{I_b}\in \mathbb{N}^{n_b}$ and $c'=c_{I_c}\in \mathbb{N}^{n_c}$ as the vectors $b$ and $c$ with the zero components deleted.
Then the polynomials $F_{n+1,2d}^{2b}$ and $F_{n+1,2d}^{2c}$ are equivalent to the polynomials $F_{n_b,2d}^{2b'}$ and $F_{n_c,2d}^{2c'}$, because $b_k=0 \quad \forall k\notin I_b$. So for these $k\notin I_b$ we have $b_kx_k =0$ and $x_k^{b_k}=1$. Thus $F_{n+1,2d}^{2b}=F_{n_b,2d}^{2b'}$. Same thus holds for $c'$. Because $n_b$, $n_c<n+1$, we can apply the induction hypotheses. So, for these $F_{n+1,2d}^{2b}$ and $F_{n+1,2d}^{2c}$ we know from the induction hypothesis that these are the sums of at most $3n_b-4$ and $3n_c-4$ squares resp.
Then, as $F_{n,2d}^a(x) =\frac{1}{2}\Big(F_{n,2d}^{2b}(x)+F_{n,2d}^{2c}(x)\Big) + d(x^b-x^c)^2$, this is the sum of at most $3n_b-4+3n_c-4 + 1$ squares. Which is the same as $3(n_b+n_c)-7 = 3(n+2)-7=3(n+1)-4$. Thus $F_{n+1,2d}^a(x)$ is the sum of at most $3(n+1)-4$ squares.
Now, in the case where $n_b=n=1$ and $n_c=1$ or the other way around, assume wlog that $n_b=n=1$ and $n_c=1$. Then we can keep splitting $2b$ until we do have that it splits into two vectors $b_1$ and $b_2$ such that $2b=b_1+b_2$ and $|b_1|=d$ and $|b_2|=d$. And then we can just apply the induction hypothesis as above. For $c$ an possible other offsplittings of $b$ with $n_c=1$, we know $c=de_i$ (with $e_i$ being the $i^{th}$ canonical basis vector), because it only has one non zero component and $|c|=d$. But in this case $F_{n+1,2d}^{2c} = 2dx_i^{2d} - 2dx_i^{2d} =0$. So this is just the zero polynomial and has zero sums of squares. Therefore we know that in the end $F_{n+1,2d}^{2b}$ is at most $3n_b-4$ sums of squares.


TITLE: Number of ways to color a sequence of squares so that no two black squares are adjacent
QUESTION [0 upvotes]: A sequence of squares may be colored so that each square is black or white. Let $S_n$ be the number of ways of coloring the sequence so that no two black squares are adjacent. Find a recursive relation for $S_n$.

Progress
$S_1= \{b,w\}$, $S_2=\{bw,wb,ww\}$, ... $S_3=\{bww,bwb,wbw,www,,wwb,\}$... This is what I have so far. Not even sure if it's correct.. this is new to me.

REPLY [0 votes]: S1: (b),(w) =2 .....S2:(b,w),(w,w),(w,b)=3.....S3: (w,w,w),(w,b,w)(w,w,b),(b,w,w), (b,w,b)=5
case 1: A valid sequence of n−1 squares followed by a white square;
case 2: valid sequence of n−2 squares followed by a white and then a black square.
Sn=Sn-1 + Sn-2
initial conditions : 2, when n=1..3 when n=2 when n>=3


TITLE: Diffeomorphism invariance in GR
QUESTION [2 upvotes]: I apologize for asking the duplicate question, but I still get confused after reading the related post In general relativity, are two pseudo-Riemannian manifolds physically equivalent if they are isometric, or just diffeomorphic?. In particular, for the statement

if universe is represented by manifold $M$ with matter fields $\psi$ and metric $g$, and $\phi: M \rightarrow M$ is a diffeomorphism, then sets $(M, g, \psi)$ and $(M, \phi^*g, \phi^* \psi)$ represents the same physical situation.

what does "same physical situation" really mean? Does it have some mathematical formulation? Is a diffeomorphism equivalent to a general coordinate transformation?
Also, in https://physics.stackexchange.com/a/596779, it said that we can actually prove the diffeomorphism invariance of the Hilbert action, can anyone gives a derivation?

REPLY [3 votes]: I want to add one additional point that is usually glossed over, that being the behavior of the diffeomorphisms at infinity, in the case in which we can define a good notion of infinity in the spacetime under study. This will be somewhat long, but I hope instructive.
In my opinion the most precise way to state this is via the Covariant Phase Space formalism. This formalism allows you to, starting from a Lagrangian density ${\cal L}$ construct one phase space $(\Gamma,{\pmb \Omega})$. Here $\Gamma$ is a symplectic manifold and ${\pmb \Omega}$ is the associated symplectic form, which is just a closed and non-degenerate two-form on $\Gamma$. Symplectic manifolds are the underlying structure behind standard Hamiltonian mechanics.
In particular they give a clear description of what are true symmetries and what are transformations merely relating redundant descriptions of the same physics. This starts as follows: we define a canonical transformation to be a phase space diffeomorphism $T: \Gamma\to \Gamma$ preserving the symplectic form $T^\ast{\pmb\Omega}={\pmb\Omega}$. As you probably known, a vector field ${\bf X}:\Gamma\to T\Gamma$ generates a one-parameter family of diffeomorphisms. These diffeomorphisms will be canonical transformations if and only if $L_{\bf X}{\pmb \Omega}=0$ where $L_{\bf X}$ is the Lie derivative.
Now if you use Cartan's magic formula we have $$L_{\bf X}{\pmb \Omega}={\bf d}\iota_{\bf X} {\pmb \Omega}+\iota_{\bf X} {\bf d}{\pmb \Omega}$$
where $\iota_{\bf X}{\pmb \Omega}({\bf Y})={\pmb \Omega}({\bf X},{\bf Y })$ is the interior product. Now since ${\pmb \Omega}$ is closed we find that ${\bf X}$ generates canonical transformations if and only if ${\iota_{\bf X}}{\pmb \Omega}$ is a closed one-form. Assuming trivial topology of the phase space this means that there should be some phase space function $Q$ with the property that $$\iota_{\bf X}{\pmb \Omega}=-{\bf d}Q.$$
This establishes a correspondence between symmetries and phase space functions. It is the correspondence between symmetries and conserved quantities. We call $X$ a Hamiltonian vector field and $Q$ its associated Hamiltonian charge. Now we also see that if the right-hand side vanishes ${\iota_{\bf X}}{\pmb \Omega}=0$. This means that ${\bf X}$ is a degenerate vector of ${\pmb \Omega}$ and all configurations along its flow are physically equivalent.
So that is the clear way to state the physical symmetry vs redundancy issue. You should evaluate ${\iota}_{\bf X}{\pmb \Omega}$. If it vanishes identically you have a redundancy, if it gives you some nonzero $-{\bf d}Q$ it gives you a symmetry and you have found its charge.
Now this is very general and in fact all of this structure was there when you studied Hamiltonian mechanics even if you haven't seem it! The new stuff really comes with the Covariant Phase Space formalism. The key aspect is that you start with the space of all field configurations ${\scr F}$ and inside it you define the space of solutions to the equations of motion $\scr S$ and you'll construct the phase space as an appropriate subspace of $\scr S$.
In particular variations of fields $\delta \phi$ define tangent vectors to $\scr S$. Given the Lagrangian density ${\cal L}$ if you define the Lagrangian form $L={\cal L}\epsilon$ where $\epsilon$ is the spacetime volume form and if you take its variation it can always be put in the form $$\delta L = E[\phi]\delta \phi + d{\pmb\theta[\phi,\delta\phi]}.$$
The object ${\pmb \theta}[\phi,\delta\phi]$ depends on $\phi$ and is linear in $\delta \phi$ so you identify it as a one-form in field space. It is also a $(D-1)$-form on spacetime, where $D$ is the spacetime dimension. We thus refer to it as $(D-1,1)$-form. Taking a one phase space exterior derivative we obtain a $(D-1,2)$-form ${\pmb \omega}={\bf d}{\pmb \theta}$. Finally if the spacetime has one Cauchy slice $\Sigma$ we define the pre-sympletic form
\begin{eqnarray}
    {\pmb \Omega}[\phi;\delta_1 \phi,\delta_2\phi]=\int_{\Sigma}{\pmb \omega}[\phi,\delta_1\phi,\delta_2\phi].
\end{eqnarray}
The key aspect now is that in general, if you have local transformations $\delta_\varepsilon$, like diffeomorphisms in the case of gravity, or gauge transformations in the cause of gauge theories, then ${\pmb \omega}[\phi;\delta_\varepsilon\phi,\delta\phi]$ will turn out to be a total derivative and ${\pmb \Omega}[\phi;\delta_\varepsilon\phi,\delta\phi]$ will turn out to be a boundary term at $\partial \Sigma$. Intuitively you can think of a Cauchy slice as "all of space at an instant of time" and then $\partial \Sigma$ is at infinity.
Now you must study this integral over $\partial \Sigma$. Can it be written as $\delta Q$ for some $Q$? If the answer is no, then the transformation is not canonical and hence not a symmetry. If the answer is yes then we study the charge $Q$. Transformations with parameter $\varepsilon|_{\partial \Sigma}$ such that $Q$ vanishes will give rise to redundancies. These are the trivial gauge transformations. Transformations with parameter $\varepsilon|_{\partial\Sigma}$ such that $Q$ doesn't vanish will give rise to true physical symmetries. These are large gauge transformations.
So the answer is: by "same physical situation" we mean points in phase space related by the flow of degenerate vectors of the symplectic form. Diffeomorphisms which act trivially at $\partial \Sigma$ would give rise to such degenerate vectors and hence relate redundant descriptions of the same physical situation. On the other hand, diffeomorphisms with non-trivial action at $\partial\Sigma$ will be true physical symmetries.
One central example are the BMS transformations in asymptotically flat spacetimes. They are diffeomorphisms, but they act non-trivially at the celestial sphere at null infinity and therefore they are true physical symmetries. Finally for more details on that I recommend arXiv:1703.05448 and arXiv:2009.14334.


TITLE: To establish the truth of a logical predicate expression by the specification of variables
QUESTION [0 upvotes]: Please, help me to establish the truth of a logical expression by specifying method.

∀x A(x) ∨ ∀x B(x) = ∀x∀y(A(x) ∨ B(y)))

As i know, i should get equal expressians on both sides of "=". So, i used some logical rules to solve it, but i can not get it equal.
Specifying:
x={a,b}
y={c,d}
∀x A(x) = A(a) ∧ A(b)
∀x B(x) = B(a) ∧ B(b)
Left side:
 A(a) ∧ A(b) ∨ B(a) ∧ B(b)
Right side: ∀x[∀y(A(x) ∨ B(y))]= ∀x [(A(x) ∨ B(c)) ∧ (A(x) ∨ B(d))] = ...  = A(a) ∨ B(d), B(d) ∨ B(c) ∨ A(d), A(b) ∨ B(d), A(a) ∨ B(d) ∨ B(c)
As you can see, this two parts are not equal to each other. What is more, there are some "extra" variables: B(d) and B(c).  They do not even exist on the left side..
Probably i miss some key points. I asked my teacher about it, but i did not get appropriate answer.
So, why it happens and how to prove it?
Please help me. Your help will be greatly appreciated!

REPLY [0 votes]: Whenever you have multiple quantifiers in the same statement, the quantifiers have to range over the exact same domain. So, you can't do what you did, and say that $x$ ranges over the objects $a$ and $b$, but that $y$ ranges over the objects $c$ and $d$.
Here is how you can make sense of the equivalence. In general, we could say that we have objects $a,b,c,...$ in our domain. So:
$$\forall x A(x) \lor \forall x B(x) =$$
$$(A(a) \land A(b) \land ...) \lor (B(a) \land B(b) \land ...)=$$
$$(A(a) \lor B(a)) \land (A(a) \lor B(b)) \land ... \land (A(b) \lor B(a)) \land (A(b) \lor B(b)) \land ... =$$
$$\forall y (A(a) \lor B(y)) \land \forall y (A(b) \lor B(y)) \land ... =$$
$$\forall x \forall y (A(x) \lor B(y))$$


TITLE: In a Circle, given segments AC = 13, BE = 6, BE=DE, find AE (E point where segments AC DB cross)
QUESTION [0 upvotes]: In a Circle, given segments $\overline{AC}$ = 13, $\overline{BE}$ = 6, $\overline{BE}$=$\overline{DE}$, find $\overline{AE}$ (E point where segments $\overline{AC}$ and $\overline{DB}$ cross)
It is known that when two chords intersect each other inside a circle, the products of their segments are equal. So:
$\overline{DE}$ $\overline{EB}$ = $\overline{AE}$ $\overline{EC}$
Tried Pitagorian and other relations but could not find a way to get the result

REPLY [0 votes]: $$\overline { DE } \overline { EB } =\overline { AE } \overline { EC } \\ { 6 }^{ 2 }=\left( 13-x \right) x\\ { x }^{ 2 }-13x+36=0\\ \left( x-4 \right) \left( x-9 \right) =0\\ { x }_{ 1 }=4,{ x }_{ 2 }=9\\ \overline { AE } =9,\overline { EC } =4\\ $$