Measure-valued urn sequences (MVPS) are a generalization of the observation processes generated by $k$-color urn models, where the space of colors $\mathbb{X}$ is a complete separable metric space and the urn composition is a finite measure on $\mathbb{X}$, in which case reinforcement reduces to a summation of measures. In this paper, we prove a representation theorem for the reinforcement measures $R$ of all exchangeable MVPSs, which leads to a characterization result for their directing random measures $\tilde{P}$. In particular, when $\mathbb{X}$ is countable or $R$ is dominated by the initial distribution $\nu$, then any exchangeable MVPS is a Dirichlet process mixture model over a family of probability distributions with disjoint supports. Furthermore, for all exchangeable MVPSs, the predictive distributions converge on a set of probability one in total variation to $\tilde{P}$. A final result shows that $\tilde{P}$ can be decomposed into an absolutely continuous and a mutually singular measure with respect to $\nu$, whose support is universal and does not depend on the particular instance of $\tilde{P}$.

Exchangeability, Reinforced processes, sequences, Directing random measures, Urn models

MSC2020 Classification: 60G09, 60G25, 60G57, 62G99

§ INTRODUCTION

The classical two-color urn model, which describes the evolution of an urn reinforced with one ball of the observed color, has a fundamental role in the predictive construction of prior distributions for Bayesian inference [Fortini and Petrone, 2012]. Suppose we have an urn that initially contains $w_0$ and $w_1$ balls with colors $0$ and $1$, respectively. Let us denote by $X_n\in\{0,1\}$ the color of the sampled ball at step $n$. Then $X_1=1$ with probability $\frac{w_1}{w_0+w_1}$ and, for all $n\geq1$,

\begin{equation}\label{intro:predictive:polya_urn}
\mathbb{P}(X_{n+1}=1|X_1,\ldots,X_n)=\frac{w_1+\sumN X_i}{w_0+w_1+n},
\end{equation}

which is the proportion of $1$-balls in the urn at time $n$. It is well-known that the process $(X_n)_{n\geq1}$ is exchangeable and satisfies, as $n\rightarrow\infty$,

\[\frac{1}{n}\sumN X_i\overset{a.s.}{\longrightarrow}\tilde{p},\]

where $\tilde{p}$ has a Beta distribution with parameters $(w_0,w_1)$, see, e.g., <cit.>; therefore, by de Finetti representation theorem for exchangeable sequences <cit.>, given $\tilde{p}$, the $X_n$ are conditionally independent and identically distributed with probability of "success" $\tilde{p}$. The converse is also true – any exchangeable sequence of Bernoulli random variables with a Beta prior distribution has the predictive structure (<ref>).

There has been much interest in generalizing the construction (<ref>) to obtain greater model flexibility and, at the same time, retain tractability of the process dynamics, see [Fortini and Petrone, 2012, Mahmoud, 2008, Pemantle, 2007] and references therein. Various extensions of the urn scheme consider time-dependent, randomized, or generalized reinforcement mechanisms where colors other than those observed are added to the urn. More recently, [Bandyopadhyay and Thacker, 2022, Janson, 2019, Mailler and Marckert, 2017] have proposed an extended class of measure-valued urn processes that are very general in nature, yet retain characteristic features of urn processes and contain, as a special case, most $k$-color urn models. The idea is to consider the urn composition as a finite measure $\mu$ on the space of colors $\mathbb{X}$, in the sense that, for any measurable $B\subseteq\mathbb{X}$, the quantity $\mu(B)$ records the total mass of balls in the urn whose colors lie in the set $B$. Reinforcement is then reduced to a summation of measures, so that the updated urn composition, given that a ball with color $x$ has been observed, becomes $\mu+R_x$, where $R$ is a transition kernel on $\mathbb{X}$ that maps colors $x\mapsto R_x$ to finite (random) measures. Assuming $\mu+R_x$ as the new urn composition, the next draw proceeds in the same way as before, independent of all previous draws. Thus, a sequence of finite measures $(\mu_n)_{n\geq0}$ is a measure-valued urn processes (MVPP) if it is a Markov process with the aforementioned additive structure for a given reinforcement kernel $R$. Then Theorem 1 in [Fortini et al., 2021] implies the existence of a companion observation process $(X_n)_{n\geq 1}$ such that $X_1\sim\frac{\mu_0}{\mu_0(\mathbb{X})}=:\mu_0'$ and, for all $n\geq1$,

\[\mathbb{P}(X_{n+1}\in\cdot\mid \mu_0,X_1,\mu_1,\ldots,X_n,\mu_n)=\frac{\mu_n(\cdot)}{\mu_n(\mathbb{X})}=:\mu_n'(\cdot).\]

In particular, when the reinforcement $R$ is non-random, the above becomes

\begin{equation}\label{intro:predictive:mvps}
\mathbb{P}(X_{n+1}\in\cdot\mid X_1,\ldots,X_n)=\frac{\mu_0(\cdot)+\sumN R_{X_i}(\cdot)}{\mu_0(\mathbb{X})+\sumN R_{X_i}(\mathbb{X})},
\end{equation}

which corresponds to the normalized urn composition at time $n$ in the urn analogy. We call such a sequence $(X_n)_{n\geq1}$ a measure-valued urn sequence (MVPS) to distinguish it from the process $(\mu_n)_{n\geq0}$.

Analysis of MVPPs has been centered around two very distinct cases. In [Fortini et al., 2021, Sariev et al., 2023], the authors study the so-called "diagonal" model, where only the observed color is reinforced, i.e. $R_x=w(x)\cdot\delta_x$, with $w(x)>0$, for $x\in\mathbb{X}$, and $\delta_x$ is the unit mass at $x$. It follows under certain conditions that there exists a random probability measure $\tilde{P}$ on $\mathbb{X}$ such that, as $n\rightarrow\infty$, the normalized urn composition $\mu_n'$ converges almost surely (a.s.) in total variation to $\tilde{P}$. On the other hand, [Bandyopadhyay and Thacker, 2022, Mailler and Marckert, 2017, Mailler and Villemonais, 2020] consider measure-valued urn processes, for which there exists an underlying Markov chain with kernel $R$, which satisfies some "irreducibility"-type conditions (see <cit.> for more details). Then, under additional assumptions, they prove that $\mu_n'$ converges a.s. weakly to a (deterministic) probability measure on $\mathbb{X}$.

In this work, we consider MVPPs that generate an exchangeable observation sequence $(X_n)_{n\geq1}$ via (<ref>). Standard results in exchangeability theory then guarantee that the predictive distributions of $(X_n)_{n\geq1}$ converge a.s. weakly to a random probability measure $\tilde{P}$ on $\mathbb{X}$, whose distribution determines the law of the exchangeable process. Therefore, the only exchangeable "irreducible" MVPSs are sequences of i.i.d. random variables. Our goal is to characterize all $R$, for which $(X_n)_{n\geq1}$ is exchangeable, and in the process prove some new facts about $\tilde{P}$ and hence about the distribution of $(X_n)_{n\geq1}$. This will help us assess the limitations of the whole class of exchangeable MVPSs when used as models for Bayesian analysis. Key to our understanding are the kernel based Dirichlet sequences studied by Berti et al. [Berti et al., 2023], which are exchangeable MVPSs whose reinforcement is a regular version of the conditional distribution for $\mu_0'$ given some sub-$\sigma$-algebra $\mathcal{G}$,

\begin{equation}\label{intro:reinforcement:kds}
R_x(\cdot)=\mu_0'(\cdot\mid\mathcal{G})(x)\qquad\mbox{for }x\in\mathbb{X}.
\end{equation}

In fact, our main theorem states that a normalized version of condition (<ref>) is true for every member of the class of exchangeable MVPSs. This allows us to use results from [Berti et al., 2023], which were developed under the assumption (<ref>), to characterize the directing random measure $\tilde{P}$ of any exchangeable MVPS. In the process, we prove that $R_x(\mathbb{X})$ must be constant for almost every $x$, which implies that a "diagonal" MVPS will be exchangeable only if $w(x)$ is constant, see also Example <ref>. On the other hand, using a variant of the representation (<ref>), we provide a complete description of all possible exchangeable $k$-color urn models with positive time-homogeneous reinforcement, which to our knowledge has not been done before.

The rest of the paper is structured as follows. In Section <ref>, we formally define our model and state some general facts about exchangeable sequences. Results and examples are given in Section <ref>, with proofs postponed to Section <ref>. We state our representation theorem for general $R_x$ in Section <ref>, while in Section <ref> we study the case when $R_x$ is dominated by $\mu_0'$, which includes $k$-color urns, and in Section <ref> when $R_x$ and $\mu_0'$ are mutually singular. It should be noted that the proofs rely heavily on the abstraction of the classical urn models to measure-valued objects. A final section concludes the paper.

§ THE MODEL

Let $(\Omega,\mathcal{H},\mathbb{P})$ be a probability space, and $\mathbb{X}$ a complete separable metric space, endowed with its Borel $\sigma$-algebra $\mathcal{X}$. We will call any sequence $(X_n)_{n\geq1}$ of $\mathbb{X}$-valued random variables on $(\Omega,\mathcal{H},\mathbb{P})$ a measure-valued urn sequence with parameters $\theta$, $\nu$ and $R$, denoted MVPS$(\theta,\nu,R)$, if $X_1\sim\nu$ and, for all $n\geq1$,

\begin{equation}\label{model:predictive}
\mathbb{P}(X_{n+1}\in\cdot\mid X_1,\ldots,X_n)=\frac{\theta\nu(\cdot)+\sumN R_{X_i}(\cdot)}{\theta+\sumN R_{X_i}(\mathbb{X})}\qquad\mbox{a.s.},
\end{equation}

where $R$ is a finite transition kernel on $\mathbb{X}$, called the replacement/reinforcement kernel of the process, $\nu$ is a probability measure on $\mathbb{X}$, known as the base measure, and $\theta>0$ is a positive constant. We will say that the MVPS is balanced if there is, in addition, some $m>0$ such that $R_x(\mathbb{X})=m$ for $\nu$-a.e. $x$, which in the urn analogy means that we add the same total number of balls each time. Moreover, unlike (<ref>), we decompose the initial urn composition into two parameters, $\theta$ and $\nu$, each one having an independent effect on the process (see, e.g., the construction in (<ref>)).

When $(X_n)_{n\geq1}$ is exchangeable, then there exists a random probability measure $\tilde{P}$ on $\mathbb{X}$, called the directing random measure of the sequence, such that, given $\tilde{P}$, the random variables $X_1,X_2,\ldots$ are conditionally independent and identically distributed (i.i.d.) with marginal distribution $\tilde{P}$,

\begin{eqnarray}
X_n\mid\tilde{P}& \overset{i.i.d.}{\sim} & \tilde{P} \nonumber\\
\tilde{P}& \sim & Q \nonumber
\end{eqnarray}

where $Q$ is the (nonparametric) prior distribution of $\tilde{P}$, see <cit.>. In addition, the predictive distributions of $(X_n)_{n\geq1}$ converge weakly on a set of probability one to $\tilde{P}$,

\begin{equation}\label{model:exchangeable:convergence}
\mathbb{P}(X_{n+1}\in\cdot\mid X_1,\ldots,X_n)\overset{w}{\longrightarrow}\tilde{P}(\cdot)\qquad\mbox{a.s.},
\end{equation}

so that the parameters in (<ref>) directly influence the shape of $\tilde{P}$.

On the other hand, the condition of exchangeability itself imposes some restrictions on the form of the predictive distributions and hence on the reinforcement $R$. The following result from [Fortini et al., 2000] provides necessary and sufficient conditions for a system of predictive distributions to be consistent with exchangeability.

A sequence of random variables $(X_n)_{n\geq1}$ is exchangeable if and only if $(X_1,X_2)\sim(X_2,X_1)$ and, for all $n\geq1$,

\begin{equation}\label{model:exchangeable:predictive}
\mathbb{P}\bigl((X_{n+1},X_{n+2})\in\cdot\mid X_1,\ldots,X_n\bigr)=\mathbb{P}\bigl((X_{n+2},X_{n+1})\in\cdot\mid X_1,\ldots,X_n\bigr)\qquad\mbox{a.s.},
\end{equation}

\begin{equation}\label{model:exchangeable:predictive_perm}
\mathbb{P}(X_{n+1}\in\cdot\mid X_1=x_1,\ldots,X_n=x_n)=\mathbb{P}(X_{n+1}\in\cdot\mid X_1=x_{\sigma(1)},\ldots,X_n=x_{\sigma(n)}),
\end{equation}

for every permutation $\sigma$ of $\{1,\ldots,n\}$ and almost every $(x_1,\ldots,x_n)\in\mathbb{X}^n$ with respect to the marginal distribution of $(X_1,\ldots,X_n)$.

Based on Theorem <ref>, the reinforcement $R$ of any exchangeable MVPS must satisfy certain conditions that make it admissible under exchangeability. Notice, however, that equation (<ref>) is always true for MVPSs. Hence, information about $R$ can only be retrieved from the invariance property of the two-step ahead predictive distributions. In Section <ref>, using the identity in (<ref>), we show that $R$ must be a regular version of the conditional distribution for $\nu$ given some sub-$\sigma$-algebra of $\mathcal{X}$, up to a normalizing constant. This representation turns out to have major consequences for the shape and distribution of the directing random measure of any exchangeable MVPS, including the fact that the convergence in (<ref>) is in total variation. In addition, it also implies that exchangeable MVPSs are necessarily balanced. We give one important example next.

Let $(X_n)_{n\geq1}$ be an MVPS with reinforcement kernel

\begin{equation}\label{example:model:polya_sequence:reinforcement}
R_x=\delta_x\qquad\mbox{for }x\in\mathbb{X},
\end{equation}

i.e., we only reinforce the observed color with one additional ball. This process, also known as a sequence, was first studied by Blackwell and MacQueen [Blackwell and MacQueen, 1973] as an extension of the urn scheme (<ref>) to general Polish spaces $\mathbb{X}$. In [Blackwell and MacQueen, 1973], the authors prove that $(X_n)_{n\geq1}$ is exchangeable and its directing random measure $\tilde{P}$ has a Dirichlet process distribution with parameters $(\theta,\nu)$, denoted $\DP(\theta,\nu)$. We recall that $\tilde{P}\sim\DP(\theta,\nu)$ if, for every finite partition $B_1,\ldots,B_n\in\mathcal{X}$ of $\mathbb{X}$, the random vector $(\tilde{P}(B_1),\ldots,\tilde{P}(B_n))$ has a Dirichlet distribution with parameters $(\theta\nu(B_1),\ldots,\theta\nu(B_n))$,

\[\bigl(\tilde{P}(B_1),\ldots,\tilde{P}(B_n)\bigr)\sim\Dir\bigl(\theta\nu(B_1),\ldots,\theta\nu(B_n)\bigr).\]

An equivalent statement (see, e.g. <cit.>) is that $\tilde{P}$ is equal in law to

\begin{equation}\label{example:model:polya_sequence:limit}
\tilde{P}(\cdot)\overset{w}{=}\sum_{j=1}^\infty V_j\delta_{Z_j}(\cdot),
\end{equation}

where $V_j=W_j\prod_{i=1}^{j-1}(1-W_i)$ and $(W_j)_{j\geq1}$ are i.i.d. with $W_1\sim\Be(1,\theta)$, and $(Z_j)_{j\geq1}$ are i.i.d.$(\nu)$, independent of $(V_j)_{j\geq1}$.

In this case, it is easy to see that the reinforcement (<ref>) has the anticipated structure, since

\[R_x(\cdot)=\nu(\cdot\mid\mathcal{X})(x)\qquad\mbox{for }x\in\mathbb{X}.\]

§ MAIN RESULTS

§.§ General case

We begin this section with a simple characterization result for i.i.d. MVPSs.

An MVPS$(\theta,\nu,R)$ is i.i.d. if and only if, for $\nu$-a.e. $x$,

\begin{equation}\label{result:i.i.d.:equation:form}
\frac{R_x(\cdot)}{R_x(\mathbb{X})}=\nu(\cdot).
\end{equation}

The next theorem states that MVPSs that are exchangeable, but not i.i.d. are necessarily balanced. This fact is essential since it implies that the predictive distributions (<ref>) are a linear combination of reinforcement kernels and leads via (<ref>) to certain identities for $R$.

An exchangeable MVPS that is unbalanced is necessarily i.i.d.

If an MVPS$(\theta,\nu,R)$ is balanced, then $R_x(\mathbb{X})=m$ for $\nu$-a.e. $x$ and some constant $m>0$. It then follows from the particular form of the predictive distributions (<ref>) that we can equivalently reformulate the process as an MVPS with parameters $(\tilde{\theta},\nu,\tilde{R})$, where $\tilde{\theta}=\frac{\theta}{m}$ and $\tilde{R}_x=\frac{R_x}{m}$ with $\tilde{R}_x(\mathbb{X})=1$ for $\nu$-a.e. $x$. On the other hand, it is not hard to see that any i.i.d. MVPS$(\theta,\nu,R)$ with unbalanced $R$ is also an i.i.d. MVPS with parameters $(\theta,\nu,\nu)$. Thus, when studying i.i.d. MVPSs, we will implicitly assume that they are balanced. The above remarks are combined in the next corollary.

Any exchangeable MVPS$(\theta,\nu,R)$ is an exchangeable MVPS$(\tilde{\theta},\nu,\tilde{R})$ for some probability kernel $\tilde{R}$ and constant $\tilde{\theta}$.

We now consider the issue of representing $R$ as in (<ref>). For completeness, we first state the converse result, which is found in [Berti et al., 2023].

A balanced MVPS$(\theta,\nu,R)$ with reinforcement $\frac{R_x(\cdot)}{R_x(\mathbb{X})}=\nu(\cdot\mid\mathcal{G})(x)$ for $\nu$-a.e. $x$, where $\mathcal{G}$ is a sub-$\sigma$-algebra of $\mathcal{X}$, is exchangeable.

Although Berti et al. [Berti et al., 2023] focus only on the balanced case, it follows from Theorem <ref> that no unbalanced MVPS, having replacement kernel as in Theorem <ref>, can be exchangeable unless $\mathcal{G}=\{\emptyset,\mathbb{X}\}$, in which case the observation sequence is i.i.d.

Our main result studies the necessity of such a representation under exchangeability. It also has the added implication that $\mathcal{G}$ is countably generated (c.g.) under $\nu$, in the sense that there exists $C\in\mathcal{G}:\nu(C)=1$ and $\mathcal{G}\cap C$ is c.g., which holds if and only if $\nu(\cdot\mid\mathcal{G})$ is a.e. proper, that is,

\[\nu(A|\mathcal{G})(x)=\delta_x(A)\qquad\mbox{for all }A\in\mathcal{G}\mbox{ and }x\in F,\]

where $F\in\mathcal{X}$ is such that $\nu(F)=1$ (see [Berti and Rigo, 2007] and [Blackwell and Dubins, 1975] for more information).

Let $(X_n)_{n\geq1}$ be an exchangeable MVPS$(\theta,\nu,R)$. Then there exists a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{X}$ such that, for $\nu$-a.e. $x$,

\begin{equation}\label{result:general:equation:form}
\frac{R_x(\cdot)}{R_x(\mathbb{X})}=\nu(\cdot\mid\mathcal{G})(x).
\end{equation}

Moreover, $\mathcal{G}$ is c.g. under $\nu$.

A major consequence of Theorem <ref> is that one can characterize the directing random measure $\tilde{P}$ of any exchangeable MVPS using results developed in [Berti et al., 2023] under the assumption that $R$ is of the form (<ref>). In this case, Berti et al. <cit.> have shown that the reinforcement process $(R_{X_n})_{n\geq1}$ is in fact a sequence (see Example <ref>). In addition, the same results ensure that the predictive distributions (<ref>) converge almost surely in total variation to $\tilde{P}$. The following theorem is based on Theorem <ref> and Theorem <ref> here, and Theorems 9 and 10 in [Berti et al., 2023].

Let $(X_n)_{n\geq1}$ be an exchangeable MVPS$(\theta,\nu,R)$ with directing random measure $\tilde{P}$. Then, as $n\rightarrow\infty$,

\[\sup_{B\in\mathcal{X}}\bigl|\mathbb{P}(X_{n+1}\in B|X_1,\ldots,X_n)-\tilde{P}(B)\bigr|\overset{a.s.}{\longrightarrow}0.\]

Moreover, $\tilde{P}$ is equal in law to

\begin{equation}\label{result:general:limit:prior}
\tilde{P}(\cdot)\overset{w}{=}\sum_{j=1}^\infty V_jR_{Z_j}(\cdot),
\end{equation}

where $(V_j)_{j\geq1}$ and $(Z_j)_{j\geq1}$ are as in (<ref>), using as parameters $(\frac{\theta}{m},\nu)$, and $m>0$ is some constant such that $m=R_x(\mathbb{X})$ for $\nu$-a.e. $x$.

§.§ Absolutely continuous case

In this section, we consider MVPSs such that, for $\nu$-a.e. $x$, the replacement $R_x$ is absolutely continuous with respect to the base measure $\nu$, denoted $R_x\ll\nu$. Special cases include $k$-color urn models and MVPSs with discrete $\nu$. Moreover, the additional assumptions on $R$ induce further structure on the $\sigma$-algebra $\mathcal{G}$ in (<ref>).

Let $(X_n)_{n\geq1}$ be an exchangeable MVPS$(\theta,\nu,R)$. If $R_x\ll\nu$ for $\nu$-a.e. $x$, then there exists a partition $D_1,D_2,\ldots$ of $\mathbb{X}$ in $\mathcal{X}$ such that, for $\nu$-a.e. $x$,

\begin{equation}\label{result:abs_cont:equation:form}
\frac{R_x(\cdot)}{R_x(\mathbb{X})}=\sum_k\nu(\cdot\mid D_k)\cdot\mathbbm{1}_{D_k}(x).
\end{equation}

The converse result is immediate.

Note that the existence of the countable partition in (<ref>) does not depend on the enumerability of the color space $\mathbb{X}$ or on the form of the base measure $\nu$. The implications of Theorem <ref> are illustrated in the next Example <ref>, which describes $k$-color urn models with positive time-homogeneous reinforcement.

[$k$-color urn]
When the space of colors $\mathbb{X}=\{x_1,\ldots,x_k\}$ is finite, then $\nu(\cdot)=\sum_{i=1}^kp_i\delta_{x_i}(\cdot)$, where $p_i\geq0$ represents the initial fraction of balls with color $x_i$ in the urn. Furthermore, it is customary to state $R$ in terms of a reinforcement matrix $[a_{ij}]_{1\leq i,j\leq k}$, where $a_{ij}\geq0$ denotes the number of balls of color $x_j$ that will be added to the urn after color $x_i$ has been observed.

If the observation process $(X_n)_{n\geq1}$ is exchangeable, then we have from (<ref>) that

\[p_i=\mathbb{P}(X_1=x_i)=\sum_{j=1}^k\mathbb{P}(X_1=x_j)\mathbb{P}(X_2=x_i|X_1=x_j)=\sum_{j=1}^kp_j\frac{\theta p_i+a_{j,i}}{\theta+\sum_{l=1}^ka_{j,l}};\]

thus, if $p_i=0$ (i.e., there are initially no $x_i$ balls in the urn), then $a_{ji}=0$ for all $j:p_j>0$, so the color $x_i$ never appears in the urn. We can therefore work, for all practical reasons, on $\{x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_k\}$, in which case $R_{x_j}\ll\nu$, $j\neq i$, and $(X_n)_{n\geq 1}$ satisfies the conditions of Theorem <ref>.

Assuming $R_x(\mathbb{X})=1$, it follows from the particular form (<ref>) of $R$ that the reinforcement matrix of an exchangeable urn scheme with a finite number of colors has a block-diagonal design, i.e. $\mathbb{X}$ can be partitioned into subsets of colors $D_1,\ldots,D_n$, where a color reinforces and is reinforced only by the colors belonging to the same subset. Furthermore, the rows in each block are identical and equal to the conditional probability of $\nu$ given that a color from the same block has been observed. In other words, each $D_j$ can be viewed as representing different nuances of the same color, so that the reinforcement $1)$ will be identical for, say, light or dark red, and $2)$ will be equal to the initial number of balls for each nuance of red, normalized by the total number of red balls.

To illustrate the above points, let $\mathbb{X}=\{x_1,x_2,x_3\}$ and suppose that the urn has initial composition $(w_1,w_2,w_3)$, so that $\nu(\cdot)=\sum_{i=1}^3\frac{w_i}{\bar{w}}\delta_{x_i}(\cdot)$ with $\bar{w}=\sum_{i=1}^3w_i$. Put $\bar{w}_i=\sum_{j\neq i}w_j$, $i=1,2,3$. Then any measure-valued urn sequence $(X_n)_{n\geq1}$ on $\mathbb{X}$ with initial composition $(w_1,w_2,w_3)$ is exchangeable if and only if its associated reinforcement matrix $R$ has one of the following forms, up to some multiplicative constants $m,m_1,m_2,m_3>0$,

\begin{gather*}
\begin{blockarray}{cccc}
& x_1 & x_2 & x_3 \\
\begin{block}{c[ccc]}
x_1 & m_1\frac{w_1}{\bar{w}} & m_1\frac{w_2}{\bar{w}} & m_1\frac{w_3}{\bar{w}} \\
x_2 & m_2\frac{w_1}{\bar{w}} & m_2\frac{w_2}{\bar{w}} & m_2\frac{w_3}{\bar{w}} \\
x_3 & m_3\frac{w_1}{\bar{w}} & m_3\frac{w_2}{\bar{w}} & m_3\frac{w_3}{\bar{w}} \\
\end{block}\end{blockarray}
\qquad
\begin{blockarray}{ccc}
& & \\
\begin{block}{[ccc]}
m\frac{w_1}{\bar{w}_3} & m\frac{w_2}{\bar{w}_3} & 0 \\
m\frac{w_1}{\bar{w}_3} & m\frac{w_2}{\bar{w}_3} & 0 \\
0 & 0 & m \\
\end{block}\end{blockarray}
\qquad
\begin{blockarray}{ccc}
& & \\
\begin{block}{[ccc]}
m\frac{w_1}{\bar{w}_2} & 0 & m\frac{w_3}{\bar{w}_2} \\
0 & m & 0 \\
m\frac{w_1}{\bar{w}_2} & 0 & m\frac{w_3}{\bar{w}_2} \\
\end{block}\end{blockarray}\\
\begin{blockarray}{ccc}
\begin{block}{[ccc]}
m & 0 & 0 \\
0 & m\frac{w_2}{\bar{w}_1} & m\frac{w_3}{\bar{w}_1} \\
0 & m\frac{w_2}{\bar{w}_1} & m\frac{w_3}{\bar{w}_1} \\
\end{block}\end{blockarray}
\qquad
\begin{blockarray}{ccc}
\begin{block}{[ccc]}
m & 0 & 0 \\
0 & m & 0 \\
0 & 0 & m\\
\end{block}\end{blockarray}
\end{gather*}

Note that the first matrix corresponds to the case of an i.i.d. sequence of random variables with marginal distribution $\nu$, and the last one to the three-color urn model with a Dirichlet prior distribution with parameters $(\frac{w_1}{m},\frac{w_2}{m},\frac{w_3}{m})$.

[Discrete base measure]
The previous example has an immediate extension to more general spaces $\mathbb{X}$ when the base measure $\nu$ is discrete, which is true especially if $\mathbb{X}$ is countable. In this case, $\nu(\cdot)=\sum_{x\in\mathbb{X}_0} p(x)\delta_x(\cdot)$ for some countable subset $\mathbb{X}_0\subseteq\mathbb{X}$ and positive weights $p(x)>0$ such that $\sum_{x\in\mathbb{X}_0} p(x)=1$. If $(X_n)_{n\geq 1}$ is an exchangeable MVPS with discrete $\nu$, then it satisfies, by virtue of (<ref>),

\[1=\sum_{x\in\mathbb{X}_0}p(x)=\mathbb{P}(X_1\in\mathbb{X}_0)=\mathbb{P}(X_1\in\mathbb{X},X_2\in\mathbb{X}_0)=\sum_{x\in\mathbb{X}_0}p(x)\frac{\theta+R_x(\mathbb{X}_0)}{\theta+R_x(\mathbb{X})};\]

thus, $R_x(\mathbb{X}_0^c)=0$, and so $R_x\ll\nu$ for every $x\in\mathbb{X}$. Theorem <ref> then implies the existence of a partition such that $\frac{R_x}{R_x(\mathbb{X})}$ is a version of the conditional probability of $\nu$ given the $\sigma$-algebra generated by said partition.

The form of the replacement kernel in (<ref>) has further implications on the directing random measure $\tilde{P}$ of the exchangeable process. In particular, it follows from standard results in exchangeability theory that, for every $A\in\mathcal{X}$, on a set of probability one,

\begin{align*}
\tilde{P}(A)&=\limN\mathbb{E}[\tilde{P}(A)|X_1,\ldots,X_n]\\
&=\limN\mathbb{P}(X_{n+1}\in A|X_1,\ldots,X_n)=\limN\sum_k\mathbb{E}[\tilde{P}(D_k)|X_1,\ldots,X_n]\nu(A|D_k)=\sum_k\tilde{P}(D_k)\nu(A|D_k);
\end{align*}

\begin{equation}\label{result:abs_cont:equation:a.s.limit}
\tilde{P}(\cdot)\overset{a.s.}{=}\sum_k\tilde{P}(D_k)\,\nu(\cdot\mid D_k),
\end{equation}

as $\mathcal{X}$ is countably generated. The next theorem is a direct consequence of this fact and the distribution results in Theorem <ref> (see also Example 15 in [Berti et al., 2023] and Section 4.1.2 in [Bandyopadhyay and Thacker, 2022]).

Let $(X_n)_{n\geq1}$ be an exchangeable MVPS$(\theta,\nu,R)$ with directing random measure $\tilde{P}$. If $R_x\ll\nu$ for $\nu$-a.e. $x$, then $\tilde{P}$ has the form (<ref>) and, as $n\rightarrow\infty$,

\[\sup_{A\in\mathcal{X}}\bigl|\mathbb{P}(X_{n+1}\in A|X_1,\ldots X_n)-\tilde{P}(A)\bigr|\overset{a.s.}{\longrightarrow}0.\]

Moreover, for every choice of indices $k_1,\ldots,k_j$,

\[\bigl(\tilde{P}(D_{k_1}),\ldots,\tilde{P}(D_{k_j}),\tilde{P}(\cap_{i=1}^jD_{k_i}^c)\bigr)\sim\Dir\Bigl(\frac{\theta}{m}\nu(D_{k_1}),\ldots,\frac{\theta}{m}\nu(D_{k_j}),\frac{\theta}{m}\nu(\cap_{i=1}^jD_{k_i}^c)\Bigr),\]

using the notation in Example <ref>, where $m>0$ is such that $R_x(\mathbb{X})=m$ for $\nu$-a.e. $x$.

The directing random measure (<ref>) implies that the observations $X_1,X_2,\ldots$ will form clusters on the level of the sets $D_1,D_2,\ldots$ In fact, if we let $\tilde{P}^*(\cdot)=\sum_k\tilde{P}(D_k)\delta_k(\cdot)$ and $\nu^*(\cdot)=\sum_k\nu(D_k)\delta_k(\cdot)$, then, by Theorem <ref>, any exchangeable MVPS with absolutely continuous reinforcement is a Dirichlet process mixture model,

\begin{eqnarray}
X_n\mid\xi_n,\tilde{P}^*&\overset{ind.}{\sim}&\nu(\cdot\mid D_{\xi_n}) \nonumber\\
\xi_n\mid\tilde{P}^*&\overset{i.i.d.}{\sim}&\tilde{P}^* \nonumber\\
\tilde{P}^*&\sim&\mbox{DP}\Bigl(\frac{\theta}{m},\nu^*\Bigr) \nonumber
\end{eqnarray}

using the notation in Example <ref>, where $(\xi_n)_{n\geq1}$ is a sequence of exchangeable random labels such that $\xi_n=k$ if and only if $X_n\in D_k$. We refer to <cit.> for more details.

§.§ Singular case

In general, we have the decomposition

\[R_x=R_x^\perp+R_x^a,\]

where $R_x^\perp$ and $\nu$ are mutually singular, denoted by $R_x^\perp\independent\nu$, and $R_x^a\ll\nu$. The next theorem shows that, for fixed $x$, the replacement $R_x$ of any exchangeable MVPS is either mutually singular or absolutely continuous with respect to $\nu$. In addition, the support of any mutually singular component, $R_x^\perp$, does not intersect the support of any absolutely continuous one, $R_y^a$, implying a complete separation of the two regimes.

Let $(X_n)_{n\geq1}$ be an exchangeable MVPS$(\theta,\nu,R)$. Then there exists a set $S\in\mathcal{X}$ such that

\[R_x=R_x^\perp\quad\mbox{and}\quad R_x^\perp(S^c)=0,\qquad\mbox{for }\nu\mbox{-a.e. }x\mbox{ in }S,\]

\[R_x=R_x^a\quad\mbox{and}\quad R_x^a(S)=0,\qquad\mbox{for }\nu\mbox{-a.e. }x\mbox{ in }S^c.\]

It follows from Theorem <ref> and (<ref>) that, for any MVPS such that $\nu(S)>0$, its directing random measure $\tilde{P}$ can be decomposed with respect to $\nu$ into

\[\tilde{P}(\cdot)=\tilde{P}^\perp(\cdot)+\tilde{P}^a(\cdot)\overset{w}{=}\sum_jV_jR_{Z_j}^\perp(\cdot)\cdot\mathbbm{1}_S(Z_j)+\sum_jV_jR_{Z_j}^a(\cdot)\cdot\mathbbm{1}_{S^c}(Z_j),\]

so that $\tilde{P}^\perp(S^c)=0=\tilde{P}^a(S)$ a.s., where $S$ does not depend on the particular instance of $\tilde{P}$.

The following simple example demonstrates one such case.

Let $\nu$ be a diffuse probability measure. Fix $\theta=1$. Take $S\in\mathcal{X}$ such that $\nu(S)>0$. Let us define, for $x\in\mathbb{X}$,

\[R_x(\cdot)=\biggl\{\begin{array}{cc} \delta_x(\cdot), & x\in S; \\ \nu(\cdot\mid S^c), & x\in S^c.\end{array}\]

Then $R_x=R_x^\perp+R_x^a$, where $R_x=R_x^\perp$ for $x\in S$ with $R_x^\perp(S^c)=0$, and $R_x=R_x^a$ for $x\in S^c$ with $R_x^a(S)=0$.

Let $A,B\in\mathcal{X}$ and $x\in\mathbb{X}$. It is an easy exercise to check that

\[\int_AR_x(B)\nu(dx)=\int_BR_x(A)\nu(dx)\qquad\mbox{and}\qquad\int_AR_y(B)R_x(dy)=\int_BR_y(A)R_x(dy).\]

Therefore, by <cit.>, the MVPS $(X_n)_{n\geq1}$ with parameters $(\theta,\nu,R)$ is exchangeable.

\[\mathcal{G}:=\bigl\{A\in\mathcal{X}:\nu(A|S^c)=0\mbox{ or }\nu(A|S^c)=1\bigr\}.\]

Then $\mathcal{G}$ is a $\sigma$-algebra. Let $A\in\mathcal{G}$ and $B\in\mathcal{X}$. If $\nu(A|S^c)=0$, then $\nu(A\cap S^c)\nu(B|S^c)=0=\nu(A\cap B\cap S^c)$; else, if $\nu(A|S^c)=1$, then $\nu(A^c|S^c)=0$ and $\nu(A^c\cap B\cap S^c)=0$, so $\nu(A\cap S^c)\nu(B|S^c)=\nu(B\cap S^c)=\nu(A\cap B\cap S^c)$. In either case, we have that

\[\int_AR_x(B)\nu(dx)=\nu(A\cap B\cap S)+\nu(A\cap S^c)\nu(B|S^c)=\nu(A\cap B\cap S)+\nu(A\cap B\cap S^c)=\nu(A\cap B).\]

On the other hand, for all $t\in[0,1]$ and $B\in\mathcal{X}$, the set $\{x\in\mathbb{X}:R_x(B)\leq t\}$ is equal to one of $B\cap S$, $B^c\cap S$, $B\cup S^c$, $B^c\cup S^c\in\mathcal{G}$. Therefore, $x\mapsto R_x(B)$ is $\mathcal{G}$-measurable and, as a result, for $\nu$-a.e. $x$,

\[R_x(\cdot)=\nu(\cdot\mid\mathcal{G})(x).\]

It then follows from Theorem <ref> and (<ref>) that the directing random measure of $(X_n)_{n\geq1}$ has the form

\[\tilde{P}(\cdot)\overset{a.s.}{=}\tilde{P}(\cdot\cap S)+\nu(\cdot\mid S^c)\tilde{P}(S^c)\overset{w}{=}\sum_j V_j\delta_{Z_j}(\cdot\cap S)+\nu(\cdot\mid S^c)\sum_j V_j\delta_{Z_j}(S^c).\]

The decomposition implied in Theorem <ref> can be used to analyze separately the singular and the absolutely continuous parts of $R$. Concentrating on $R_x^\perp$, let us assume that $R_x\independent\nu$ for all $x\in\mathbb{X}$. By Theorem <ref>, such an MVPS is necessarily balanced. Moreover, $\nu$ is diffuse; else, if $\nu(\{z\})>0$ for some $z\in\mathbb{X}$, then $R_x(\{z\})=0$, and, using (<ref>),

\[\nu(\{z\})=\mathbb{P}(X_1=z)=\int_\mathbb{X}\frac{\theta\nu(\{z\})+R_x(\{z\})}{\theta+1}\nu(dx)=\frac{\theta}{\theta+1}\nu(\{z\}),\]

absurd. We show in Proposition <ref> that if $R_x$ is further discrete, then $x$ is an atom of $R_x$ for $\nu$-a.e. $x$. Examples include the sequence from Example <ref> with diffuse $\nu$, and Example 2 in [Berti et al., 2023] where the authors consider an MVPS with reinforcement $R_x=\frac{1}{2}(\delta_x+\delta_{-x})$. In addition, Proposition <ref> states for discrete $R$ that either $R_x=R_y$ or $R_x\independent R_y$, which is reminiscent of the block-diagonal design in Section <ref>.

Let $(X_n)_{n\geq1}$ be an exchangeable MVPS$(\theta,\nu,R)$. If $R_x\independent\nu$ and $R_x$ is discrete, then $R_x(\{x\})>0$ for $\nu$-a.e. $x$. Moreover,

\[\mbox{for }\nu\mbox{-a.e. }x\mbox{ and }y,\mbox{ either }R_x=R_y\mbox{ or }R_x\independent R_y.\]

§ PROOFS

Let $(X_n)_{n\geq1}$ be an i.i.d. MVPS$(\theta,\nu,R)$. Take $A,B\in\mathcal{X}$. It follows from (<ref>) that

\[\mathbb{P}(X_1\in A,X_2\in B)=\int_A\mathbb{P}(X_2\in B|X_1=x)\mathbb{P}(X_1\in dx)=\int_A\frac{\theta\nu(B)+R_x(B)}{\theta+R_x(\mathbb{X})}\nu(dx).\]

On the other hand,

\[\mathbb{P}(X_1\in A,X_2\in B)=\mathbb{P}(X_1\in A)\mathbb{P}(X_2\in B)=\nu(A)\nu(B)=\int_A\nu(B)\nu(dx).\]

Since $A$ is arbitrary, then $\frac{\theta\nu(B)+R_x(B)}{\theta+R_x(\mathbb{X})}=\nu(B)$, and so $\frac{R_x(B)}{R_x(\mathbb{X})}=\nu(B)$ for $\nu$-a.e. $x$. But $\mathcal{X}$ is countably generated, hence $\frac{R_x(dy)}{R_x(\mathbb{X})}=\nu(dy)$ for $\nu$-a.e. $x$.

Conversely, if $R$ is of the form (<ref>), then, for any $A\in\mathcal{X}$,

\[\mathbb{P}(X_{n+1}\in A|X_1,\ldots,X_n)=\frac{\theta\nu(A)+\sumN R_{X_i}(\mathbb{X})\nu(A)}{\theta+\sumN R_{X_i}(\mathbb{X})}=\nu(A),\]

implying that $(X_n)_{n\geq1}$ is i.i.d. with marginal distribution $\nu$.

Let $(X_n)_{n\geq1}$ be an MVPS$(\theta,\nu,R)$. Denote

\[\tilde{R}_x(dy)=\frac{R_x(dy)}{R_x(\mathbb{X})}\qquad\mbox{and}\qquad f(x)=R_x(\mathbb{X}),\qquad\mbox{for }x\in\mathbb{X}.\]

Define $H:=\bigl\{(x,y)\in\mathbb{X}^2:f(x)=f(y)\bigr\}$ and let $H_x$ be the $x$-section of $H$.

It follows from exchangeability and the form of the predictive distributions (<ref>) that, for every $A,B\in\mathcal{X}$,

\[\int_{A\times B}\frac{\theta\nu(dy)+R_x(dy)}{\theta+f(x)}\nu(dx)=\mathbb{P}(X_1\in A,X_2\in B)=\mathbb{P}(X_1\in B,X_2\in A)=\int_{A\times B}\frac{\theta\nu(dx)+R_y(dx)}{\theta+f(y)}\nu(dy),\]

implying the equivalence of the measures

\begin{equation}\label{proof:general:balance:identity1}
\frac{\theta\nu(dy)+R_x(dy)}{\theta+f(x)}\nu(dx)=\frac{\theta\nu(dx)+R_y(dx)}{\theta+f(y)}\nu(dy).
\end{equation}

Let $g(x,y)=(\theta+f(x))(\theta+f(y))$, for $x,y\in\mathbb{X}$. Then the following two measures, defined as the indefinite integrals of $g$ with respect to $\frac{\theta\nu(dy)+R_x(dy)}{\theta+f(x)}\nu(dx)$ and $\frac{\theta\nu(dx)+R_y(dx)}{\theta+f(y)}\nu(dy)$, are equal

\[(\theta+f(y))\bigl(\theta\nu(dy)+R_x(dy)\bigr)\nu(dx)=(\theta+f(x))\bigl(\theta\nu(dx)+R_y(dx)\bigr)\nu(dy),\]

or, after linearizing the measures,

\begin{align}\label{proof:general:balance:polynomial1}
\begin{aligned}
\theta^2\nu(&dx)\nu(dy)+\theta\bigl(f(y)\nu(dx)\nu(dy)+R_x(dy)\nu(dx)\bigr)+f(y)R_x(dy)\nu(dx)\\
\end{aligned}
\end{align}

Note that the parameter $\theta$ of an MVPS is chosen independently of $\nu$ and $R$. Therefore, by varying $\theta$, each side of (<ref>) becomes a 2-nd degree polynomial in $\theta$, implying that the "coefficients" of the corresponding terms must be equal, with one important caveat. Namely, the measures in (<ref>) need not be finite. They are, however, $\sigma$-finite with respect to the sets $F_n:=\{(x,y)\in\mathbb{X}^2:\frac{1}{n}\leq f(x),f(y)\leq n\}$, $n\geq1$; hence, if one works with the measures in (<ref>) restricted to $F_n$, all quantities from there on will be finite. The general result then follows from monotone convergence, letting $\mathbbm{1}_{F_n}(x,y)\uparrow 1$ as $n\rightarrow\infty$. For exposition reasons, we will omit this procedure and work as if all quantities were finite to begin with.

Having this in mind, as $\theta\downarrow 0$, we obtain

\[f(y)R_x(dy)\nu(dx)=f(x)R_y(dx)\nu(dy),\]

and thus, using $\frac{1}{f(x)f(y)}$ as density,

\begin{equation}\label{proof:general:balance:condition1a}
\tilde{R}_x(dy)\nu(dx)=\tilde{R}_y(dx)\nu(dy).
\end{equation}

On the other hand, differentiating (<ref>) with respect to $\theta$ and letting $\theta\downarrow0$, we get using (<ref>) that

\[f(y)\nu(dx)\nu(dy)+R_x(dy)\nu(dx)=f(x)\nu(dx)\nu(dy)+R_y(dx)\nu(dy)=f(x)\nu(dx)\nu(dy)+f(y)\tilde{R}_x(dy)\nu(dx),\]

which implies

\[(f(y)-f(x))\nu(dx)\nu(dy)=(f(y)-f(x))\tilde{R}_x(dy)\nu(dx),\]

and, analogously,

\[(f(y)-f(x))\nu(dx)\nu(dy)=(f(y)-f(x))\tilde{R}_y(dx)\nu(dy).\]

Thus, restricting the above (signed) measures to $H^c$,

\begin{equation}\label{proof:general:balance:condition1b}
\mathbbm{1}_{H^c}(x,y)\tilde{R}_x(dy)\nu(dx)=\mathbbm{1}_{H^c}(x,y)\nu(dx)\nu(dy)=\mathbbm{1}_{H^c}(x,y)\tilde{R}_y(dx)\nu(dy),
\end{equation}

and so, for $\nu$-a.e. $x$,

\begin{equation}\label{proof:general:balance:equation:i.i.d.1}
\mathbbm{1}_{H_x^c}(y)\tilde{R}_x(dy)=\mathbbm{1}_{H_x^c}(y)\nu(dy).
\end{equation}

Following (<ref>), we now consider the implications of the two-step ahead predictive distributions on $R$. Let $A,B\in\mathcal{X}$. It follows from (<ref>) that, for $\nu$-a.e. $x$,

\begin{align*}
&\int_{A\times B}\frac{\theta\nu(dz)+R_x(dz)+R_y(dz)}{\theta+f(x)+f(y)}\frac{\theta\nu(dy)+R_x(dy)}{\theta+f(x)}=\mathbb{P}(X_2\in A,X_3\in B|X_1=x)\\
&\hspace{10em}=\mathbb{P}(X_2\in B,X_3\in A|X_1=x)=\int_{A\times B}\frac{\theta\nu(dy)+R_x(dy)+R_z(dy)}{\theta+f(x)+f(z)}\frac{\theta\nu(dz)+R_x(dz)}{\theta+f(x)},
\end{align*}

implying the equivalence of the measures

\[\frac{\theta\nu(dz)+R_x(dz)+R_y(dz)}{(\theta+f(x)+f(y))(\theta+f(x))}\bigl(\theta\nu(dy)+R_x(dy)\bigr)=\frac{\theta\nu(dy)+R_x(dy)+R_z(dy)}{(\theta+f(x)+f(z))(\theta+f(x))}\bigl(\theta\nu(dz)+R_x(dz)\bigr).\]

Arguing as in (<ref>) and working again as if all measures were finite, we get

\begin{align*}
\theta^3\nu(dy)\nu(dz)&+\theta^2\Bigl\{\bigl(f(x)+f(z)\bigr)\nu(dy)\nu(dz)+R_x(dy)\nu(dz)+R_x(dz)\nu(dy)+R_y(dz)\nu(dy)\Bigr\}\\
\end{aligned}\\
\end{aligned}
\end{align*}

Equating the "coefficients" of the free terms, we have

\[f(z)R_x(dy)R_x(dz)+\bigl(f(x)+f(z)\bigr)R_y(dz)R_x(dy)=f(y)R_x(dy)R_x(dz)+\bigl(f(x)+f(y)\bigr)R_z(dy)R_x(dz);\]

thus, restricting the above measures to $H$, where $f(y)=f(z)$, yields

\begin{equation}\label{proof:general:balance:condition2a}
\mathbbm{1}_H(y,z)\tilde{R}_y(dz)\tilde{R}_x(dy)=\mathbbm{1}_H(y,z)\tilde{R}_z(dy)\tilde{R}_x(dz).
\end{equation}

In turn, the "coefficients" of the first degree terms imply that

\begin{align*}
\end{align*}

which on $H^c$ becomes, after cancelling $f(z)R_y(dz)\nu(dy)$ with $f(y)R_z(dy)\nu(dz)$ because of (<ref>), and integrating the expressions with respect to $\nu(dx)$,

\begin{align}
\begin{aligned}\label{proof:general:balance:condition2b}
\mathbbm{1}_{H^c}(y,z)f(z)R_x(dy)\nu(dz)\nu(dx)&+\mathbbm{1}_{H^c}(y,z)f(z)R_x(dz)\nu(dy)\nu(dx)\\
\end{aligned}
\end{align}

Observe now that, by (<ref>) and (<ref>),

\begin{align*}
\mathbbm{1}_{H^c}(y,z)R_y(dz)R_x(dy)\nu(dx)&=\mathbbm{1}_{H^c}(y,z)\frac{f(x)}{f(y)}R_y(dz)R_y(dx)\nu(dy)\\
\end{align*}

and, similarly,

\[\mathbbm{1}_{H^c}(y,z)R_z(dy)R_x(dz)\nu(dx)=\mathbbm{1}_{H^c}(y,z)f(z)R_x(dz)\nu(dy)\nu(dx).\]

This allows us to cancel the second and the forth terms on both sides of (<ref>). Further applying (<ref>) to $R_y(dz)\nu(dy)$ and $R_z(dy)\nu(dz)$ in the third terms of (<ref>), we obtain

\begin{align*}
\mathbbm{1}_{H^c}(y,z)f(z)R_x(dy)\nu(dz)\nu(dx)&+\mathbbm{1}_{H^c}(y,z)f(x)f(y)\nu(dx)\nu(dy)\nu(dz)\\
\end{align*}

hence, after “normalizing” by $f(x)$,

\begin{align}\label{proof:general:balance:condition2c}
\begin{aligned}
\mathbbm{1}_{H^c}(y,z)f(z)\tilde{R}_x(dy)\nu(dz)\nu(dx)&+\mathbbm{1}_{H^c}(y,z)f(y)\nu(dx)\nu(dy)\nu(dz)=\\
\end{aligned}
\end{align}

We now focus on the pairs $(x,y)$ such that $f(x)=f(y)$, in which case $\mathbbm{1}_H(x,y)\mathbbm{1}_{H^c}(y,z)=\mathbbm{1}_H(x,y)\mathbbm{1}_{H^c}(x,z)$. It follows from the first equation in (<ref>) that

\begin{align*}
\mathbbm{1}_H(x,y)\mathbbm{1}_{H^c}(y,z)f(y)\tilde{R}_x(dz)\nu(dy)\nu(dx)&=\mathbbm{1}_H(x,y)\mathbbm{1}_{H^c}(x,z)f(y)\tilde{R}_x(dz)\nu(dy)\nu(dx)\\
\end{align*}

thus, subtracting the corresponding terms in (<ref>), yields

\[\mathbbm{1}_H(x,y)\mathbbm{1}_{H^c}(x,z)f(z)\tilde{R}_x(dy)\nu(dz)\nu(dx)=\mathbbm{1}_H(x,y)\mathbbm{1}_{H^c}(x,z)f(z)\nu(dx)\nu(dy)\nu(dz),\]

which, after normalization becomes,

\[\mathbbm{1}_H(x,y)\mathbbm{1}_{H^c}(x,z)\tilde{R}_x(dy)\nu(dz)\nu(dx)=\mathbbm{1}_H(x,y)\mathbbm{1}_{H^c}(x,z)\nu(dx)\nu(dy)\nu(dz).\]

Integrating $z$ out, we obtain

\[\mathbbm{1}_H(x,y)\nu(H_x^c)\tilde{R}_x(dy)\nu(dx)=\mathbbm{1}_H(x,y)\nu(H_x^c)\nu(dx)\nu(dy).\]

Therefore, for $\nu$-a.e. $x$,

\begin{equation}\label{proof:general:balance:equation:i.i.d.2}
\mathbbm{1}_{H_x}(y)\nu(H_x^c)\tilde{R}_x(dy)=\mathbbm{1}_{H_x}(y)\nu(H_x^c)\nu(dy).
\end{equation}

Let us define $D^*:=\{x\in\mathbb{X}:\nu(H_x^c)>0\}$. Suppose $\nu(D^*)>0$. Take $y\in\mathbb{X}$. If $f(y)=f(x)$ for some $x\in D^*$, then $H_y^c=H_x^c$, and hence $\nu(H_y^c)=\nu(H_x^c)>0$; otherwise, if $f(y)\neq f(x)$ for all $x\in D^*$, then $D^*\subseteq H_y^c$ and $\nu(H_y^c)\geq\nu(D^*)>0$. As a result, $\nu(D^*)=1$. Therefore, (<ref>) leads to

\[\mathbbm{1}_{H_x}(y)\tilde{R}_x(dy)=\mathbbm{1}_{H_x}(y)\nu(dy)\]

for $\nu$-a.e. $x$. Summing this with (<ref>), we get for $\nu$-a.e. $x$ that

\[\tilde{R}_x(dy)=\nu(dy);\]

thus, $(X_n)_{n\geq1}$ is i.i.d. by Proposition <ref>.

If instead $\nu(D^*)=0$, then $\nu(H_x^c)=0$ for $\nu$-a.e. $x$. It follows for some fixed $x'$ with $\nu(H_{x'})=1$ that

\[f(x)=f(x')\qquad\mbox{for }\nu\mbox{-a.e. }x,\]

implying that the model is balanced.

If $(X_n)_{n\geq 1}$ is i.i.d., then Proposition <ref> implies (<ref>) with $\mathcal{G}=\{\emptyset,\mathbb{X}\}$. Suppose that $(X_n)_{n\geq1}$ is exchangeable, but not i.i.d. It follows from Theorem <ref> that the model is balanced, so we may assume, without loss of generality, that $R_x(\mathbb{X})=1$, for $x\in\mathbb{X}$; otherwise, we can normalize $\theta$ and $R$ (see Remark <ref>) and proceed from there. In this case, (<ref>) reduces to

\[R_x(dy)\nu(dx)=R_y(dx)\nu(dy),\]

and (<ref>) implies, after integrating with respect to $\nu(dx)$, that

\[R_y(dz)R_x(dy)\nu(dx)=R_z(dy)R_x(dz)\nu(dx).\]

Applying the two identities repeatedly, we obtain that

\begin{align*}
\end{align*}

Since $\mathcal{X}$ is countably generated, there exists $C\in\mathcal{X}$ such that $\nu(C)=1$ and, for all $x\in C$,

\[R_y(dz)R_x(dy)=R_x(dz)R_x(dy),\]

which itself implies that, for $R_x$-a.e. $y$,

\begin{equation}\label{proof:general:equality}
\end{equation}

Let us define

\[\mathcal{G}:=\{A\in\mathcal{X}:R_x(A)=\delta_x(A)\mbox{ for all }x\in C\}.\]

It is not hard to see that $\mathcal{G}$ is a $\sigma$-algebra. Fix $t\in[0,1]$ and $B\in\mathcal{X}$. Denote

\[E_{t,B}=\{x\in\mathbb{X}:R_x(B)<t\}.\]

Let $x\in C$. If $x\in E_{t,B}$, then $R_y(B)=R_x(B)<t$ for $R_x$-a.e. $y$ by (<ref>), so $R_x(E_{t,B})=1$. If instead $x\in E_{t,B}^c$, then again $R_y(B)=R_x(B)\geq t$ for $R_x$-a.e. $y$, and so $R_x(E_{t,B})=0$. As a result, $R_x(E_{t,B})=\delta_x(E_{t,B})$, so $E_{t,B}\in\mathcal{G}$ and, ultimately, $x\mapsto R_x(B)$ is $\mathcal{G}$-measurable.

Let $A\in\mathcal{G}$ and $B\in\mathcal{X}$. Then

\[\int_AR_x(B)\nu(dx)=\int_BR_x(A)\nu(dx)=\int_B\delta_x(A)\nu(dx)=\nu(A\cap B).\]

Therefore, $R_x(\cdot)=\nu(\cdot\mid\mathcal{G})(x)$ for $\nu$-a.e. $x$.

For the last part, define $C_0:=\{x\in\mathbb{X}:R_x(C)=1\}$ and $C_n:=\{x\in\mathbb{X}:R_x(C_{n-1})=1\}$, for $n\geq1$. From (<ref>).

\[1=\int_\mathbb{X}R_x(\mathbb{X})\nu(dx)=\int_CR_x(\mathbb{X})\nu(dx)=\int_\mathbb{X}R_x(C)\nu(dx),\]

so $\nu(C_0)=1$. By induction, $\nu(C_n)=1$ for all $n\geq1$; thus, letting $C^*=\bigcap_{n=0}^\infty C_n$, we have $C^*\subseteq C$, $\nu(C^*)=1$, and $R_x(C_n)=1$ for every $x\in C^*$ and $n\geq1$. As a result, $R_x(C^*)=1=\delta_x(C^*)$, which implies that $C^*\in\mathcal{G}$. On the other hand, by construction,

\[R_x(A)=\delta_x(A)\qquad\mbox{for all }A\in\mathcal{G}\mbox{ and }x\in C^*;\]

thus, $\mathcal{G}$ is countably generated under $\nu$ from <cit.> and <cit.>.

Assume, without loss of generality, that $R_x(\mathbb{X})=1$, $x\in\mathbb{X}$. By hypothesis $R_x\ll\nu$ for $\nu$-a.e. $x$, so $R_x(dy)\nu(dx)\ll\nu(dx)\nu(dy)$ and there exists a jointly measurable function $r:\mathbb{X}^2\rightarrow\mathbb{R}_+$ such that

\[R_x(dy)\nu(dx)=r(x,y)\nu(dx)\nu(dy).\]

It follows for each $B\in\mathcal{X}$ that $R_x(B)=\int_Br(x,y)\nu(dy)$ for $\nu$-a.e. $x$ and, since $\mathcal{X}$ is countably generated,

\begin{equation}\label{proof:abs_cont:density}
R_x(dy)=r(x,y)\nu(dy)\qquad\mbox{for }\nu\mbox{-a.e. }x.
\end{equation}

Let us define, for $x\in\mathbb{X}$,

\[D_x:=\{y\in\mathbb{X}:r(x,y)>0\},\qquad\mbox{and}\qquad D:=\{x\in\mathbb{X}:\nu(D_x)>0\}.\]

Then $D_x\in\mathcal{X}$ and $R_x(D_x)=1$ for $\nu$-a.e. $x$. Moreover,

\[1=\int_\mathbb{X}R_x(D_x)\nu(dx)=\int_D\int_{D_x}r(x,y)\nu(dy)\nu(dx)+\int_{D^c}\int_{D_x}r(x,y)\nu(dy)\nu(dx)=\nu(D).\]

On the other hand, by (<ref>), (<ref>) and (<ref>), we have the following identities

\begin{gather}
\end{gather}

Equations (<ref>) and (<ref>) imply that

\begin{align}
\mathbbm{1}_{D_x}(z)r(x,y)\nu(dx)\nu(dy)\nu(dz)&\overset{(a)}{=}\mathbbm{1}_{D_x}(z)\mathbbm{1}_{D_y}(z)r(x,y)\nu(dx)\nu(dy)\nu(dz) \nonumber\\
&=\mathbbm{1}_{D_x}(z)\mathbbm{1}_{D_y}(z)r(x,y)\frac{r(y,z)}{r(y,z)}\nu(dx)\nu(dy)\nu(dz) \nonumber\\
&=\mathbbm{1}_{D_x}(z)\mathbbm{1}_{D_y}(z)r(x,z)\frac{r(z,y)}{r(y,z)}\nu(dx)\nu(dy)\nu(dz) \nonumber\\
&=\mathbbm{1}_{D_x}(z)\mathbbm{1}_{D_y}(z)r(x,z)\nu(dx)\nu(dy)\nu(dz)\overset{(b)}{=}\mathbbm{1}_{D_x}(z)r(x,z)\nu(dx)\nu(dy)\nu(dz), \label{proof:abs_cont:equation1}
\end{align}

where $(a)$ (and, similarly, $(b)$) follows from (<ref>) by noticing that

\begin{align*}
\mathbbm{1}_{D_x}(z)\mathbbm{1}_{D_y^c}(z)r(x,y)\nu(dx)\nu(dy)\nu(dz)&=\mathbbm{1}_{D_x}(z)\mathbbm{1}_{D_y^c}(z)r(x,y)\frac{r(x,z)}{r(x,z)}\nu(dx)\nu(dy)\nu(dz) \nonumber\\
\end{align*}

Therefore, for every $A,B\in\mathcal{X}$, by (<ref>) and as $\nu(D)=1$,

\begin{align}
\int_AR_x(B)\nu(dx)&=\int_A\int_B\mathbbm{1}_{D_x}(y)\cdot r(x,y)\nu(dy)\nu(dx) \nonumber\\
&=\int_A\mathbbm{1}_D(x)\biggl(\int_B\mathbbm{1}_{D_x}(y)\cdot r(x,y)\frac{1}{\nu(D_x)}\Bigl(\int_\mathbb{X}\mathbbm{1}_{D_x}(z)\nu(dz)\Bigr)\nu(dy)\biggr)\nu(dx) \nonumber\\
&=\int_A\mathbbm{1}_D(x)\frac{1}{\nu(D_x)}\biggl(\int_B\mathbbm{1}_{D_x}(y)\Bigl(\int_\mathbb{X}\mathbbm{1}_{D_x}(z)\cdot r(x,z)\nu(dz)\Bigr)\nu(dy)\biggr)\nu(dx) \nonumber\\
\end{align}

On the other hand, we have

\begin{equation}\label{proof:abs_cont:partition:identity1}
1=\int_\mathbb{X}\int_{D_x}R_y(D_y)R_x(dy)\nu(dx)=\int_\mathbb{X}\int_{D_x}r(x,y)\Bigl(\int_{D_x\cap D_y}r(y,z)\nu(dz)\Bigr)\nu(dy)\nu(dx),
\end{equation}

where the second equality holds from (<ref>) since $r(x,y)r(y,z)=r(x,y)r(x,z)>0$ implies that $\mathbbm{1}_{D_x}(y)\mathbbm{1}_{D_y}(z)=\mathbbm{1}_{D_x}(y)\mathbbm{1}_{D_y\cap D_x}(z)$ for $\nu^3$-a.e. $(x,y,z)$. In addition, equation (<ref>) implies

\begin{align}
0=\int_\mathbb{X}\int_{D_x}R_y(D_y^c)R_x(dy)\nu(dx)&=\int_\mathbb{X}\int_{D_x}\int_{D_y^c}r(x,y)r(y,z)\nu(dz)\nu(dy)\nu(dx) \nonumber\\
&=\int_\mathbb{X}\int_{D_x}\int_{D_y^c}r(x,y)r(x,z)\nu(dz)\nu(dy)\nu(dx) \nonumber\\
&=\int_\mathbb{X}\int_{D_x}r(x,y)\Bigl(\int_{D_x\cap D_y^c}r(x,z)\nu(dz)\Bigr)\nu(dy)\nu(dx). \label{proof:abs_cont:partition:identity2}
\end{align}

But $\int_{D_x}r(x,y)\nu(dy)=1$ for $\nu$-a.e. $x$, and $r(x,z)>0$ when $z\in D_x$, so it follows from both (<ref>) and (<ref>) that, for some $E\in\mathcal{X}:\nu(E)=1$ and every $x\in E$, there exists a further set $E_x\in\mathcal{X}$ such that $\nu(E_x)=1$ and, for all $y\in E_x\cap D_x$, it holds $\nu(D_x^c\cap D_y)=0$ and $\nu(D_x\cap D_y^c)=0$; thus,

\begin{equation}\label{proof:abs_cont:partition:equivalence}
\nu(D_x)=\nu(D_x\cap D_y)=\nu(D_y).
\end{equation}

Let us define, for $x\in\mathbb{X}$,

\[\bar{D}_x:=E_x\cap D_x,\qquad\mbox{and}\qquad\mathbb{X}_0:=D\cap E\cap\{x\in\mathbb{X}:R_x(dy)=r(x,y)\nu(dy)\}.\]

Then $\nu(\bar{D}_x)=\nu(D_x)>0$ for all $x\in D$, and $\nu(\mathbb{X}_0)=1$.

Fix $x,y\in\mathbb{X}_0$ such that $\bar{D}_x\cap\bar{D}_y\neq\emptyset$. Pick $z\in\bar{D}_x\cap\bar{D}_y$. Since $\nu(D_x\cap D_z^c)=0$ and $\nu(D_y^c\cap D_z)=0$, it holds

\begin{align*}
\nu(\bar{D}_x)&=\nu(D_x)=\nu(D_x\cap D_y)+\nu(D_x\cap D_y^c)\\
&=\nu(D_x\cap D_y)+\nu(D_x\cap D_y^c\cap D_z)+\nu(D_x\cap D_y^c\cap D_z^c)=\nu(D_x\cap D_y)=\nu(\bar{D}_x\cap\bar{D}_y),
\end{align*}

and, similarly, $\nu(\bar{D}_y)=\nu(\bar{D}_x\cap\bar{D}_y)$. Hence, using that $R_x\ll\nu$ for $x\in\mathbb{X}_0$, we get

\begin{equation}\label{proof:abs_cont:partition_reminder1}
\end{equation}

If there exists, in addition, $y'\in\mathbb{X}_0$ such that $\bar{D}_y\cap\bar{D}_{y'}\neq\emptyset$, then $\nu(\bar{D}_{y'})=\nu(\bar{D}_y\cap\bar{D}_{y'})$ and

\[\nu(\bar{D}_x\cap\bar{D}_{y'})=\nu(\bar{D}_x\cap\bar{D}_y\cap\bar{D}_{y'})=\nu(\bar{D}_x\cap\bar{D}_y\cap\bar{D}_{y'})+\nu(\bar{D}_x^c\cap\bar{D}_y\cap\bar{D}_{y'})=\nu(\bar{D}_y\cap\bar{D}_{y'})=\nu(\bar{D}_{y'})>0,\]

which implies $\bar{D}_x\cap\bar{D}_{y'}\neq\emptyset$. If, on the other hand, $\bar{D}_x\cap\bar{D}_y=\emptyset$, then $\nu(\bar{D}_x\cap\bar{D}_y)=0$ and

\begin{equation}\label{proof:abs_cont:partition_reminder2}
\end{equation}

Therefore, we can partition $\mathbb{X}_0$ into equivalent classes according to whether $\bar{D}_x\cap\bar{D}_y\neq\emptyset$ or not, and we can pick one element from each class to create a family of subsets $\{\bar{D}_x\}_{x\in G}$, for some $G\subseteq\mathbb{X}_0$, such that $\bar{D}_x\cap \bar{D}_y=\emptyset$ whenever $x,y\in G$, and $\nu(\bar{D}_x)>0$ for all $x\in G$. In fact, $G$ is at most countable as $\nu$ is finite. Furthermore, (<ref>) and (<ref>) imply for each $x\in\mathbb{X}_0$ that $\sum_{x'\in G}R_x(\bar{D}_{x'})=1$, so

\[\nu\Bigl(\bigcup_{x'\in G}\bar{D}_{x'}\Bigr)=\int_{\mathbb{X}_0}\sum_{x'\in G}R_x(\bar{D}_{x'})\nu(dx)=\nu(\mathbb{X}_0)=1.\]

Define $\mathcal{G}:=\sigma(\{\bar{D}_x,x\in G\})$. Let $A,B\in\mathcal{X}$. Then

\begin{align*}
\int_AR_x(B)\nu(dx)&=\sum_{x'\in G}\int_{A\cap\bar{D}_{x'}}R_x(B)\nu(dx)\overset{(a)}{=}\sum_{x'\in G}\int_{A\cap\mathbb{X}_0}\mathbbm{1}_{\bar{D}_{x'}}(x)\nu(B|D_x)\nu(dx)\\
&\overset{(b)}{=}\int_A\sum_{x'\in G}\mathbbm{1}_{\bar{D}_{x'}}(x)\nu(B|\bar{D}_{x'})\nu(dx)=\int_A\nu(B|\mathcal{G})(x)\nu(dx)=\nu(A\cap B),
\end{align*}

where $(a)$ follows from (<ref>), and $(b)$ from (<ref>) as $\nu(\bar{D}_x^c\cap\bar{D}_{x'})=\nu(\bar{D}_x\cap\bar{D}_{x'}^c)=0$ for $x\in\bar{D}_{x'}\cap\mathbb{X}_0$. Therefore, $R_x(B)=\nu(B|\mathcal{G})(x)$ for $\nu$-a.e. $x$ and, as $\mathcal{X}$ is countably generated,

\[R_x(dy)=\nu(dy|\mathcal{G})(x).\]

Assume, without loss of generalization, that $R_x(\mathbb{X})=1$, for $x\in\mathbb{X}$. By Lebesgue decomposition,

\[R_x=R_x^\perp+R_x^a,\]

where $R_x^\perp\independent\nu$ for some $S_x\in\mathcal{X}$ such that $\nu(S_x)=0=R_x^\perp(S_x^c)$, and $R_x^a\ll\nu$. Then

\[R_x^\perp(\mathbb{X})=R_x^\perp(S_x)=R_x(S_x)\qquad\mbox{and}\qquad R_x^a(\mathbb{X})=R_x^a(S_x^c)=R_x(S_x^c).\]

Similarly to (<ref>), $R_x^a(dy)=r(x,y)\nu(dy)$ for some measurable function $r:\mathbb{X}^2\rightarrow\mathbb{R}_+$.

It follows from (<ref>) that

\begin{equation}\label{proof:general:decomposition:condition1}
\end{equation}

and, by (<ref>), for $\nu$-a.e. $x$,

\begin{equation}\label{proof:general:decomposition:condition2}
\begin{aligned}
\end{aligned}
\end{equation}

Fix one such $x$. Since $\nu(S_x)=0$, then $R_y^a(S_x)=0$ for all $y\in\mathbb{X}$, and so, using (<ref>),

\begin{equation}\label{proof:general:decomposition:result}
\int_\mathbb{X}R_y^\perp(S_x)R_x^\perp(dy)+\int_\mathbb{X}R_y^\perp(S_x)R_x^a(dy)=\int_\mathbb{X}R_z^\perp(\mathbb{X})R_x^\perp(dz)+\int_\mathbb{X}R_z^a(\mathbb{X})R_x^\perp(dz)=R_x^\perp(\mathbb{X}).
\end{equation}

On the other hand, by (<ref>),

\begin{align*}
\int_\mathbb{X}R_y^\perp(S_x)R_x^a(dy)&=\int_{S_x^c}R_y^\perp(S_x)R_x^a(dy)\\
&=\int_{y\in S_x^c}\int_{z\in S_x}r(x,y)R_y^\perp(dz)\nu(dy)\\
&\begin{aligned}=\int_{z\in S_x}\int_{y\in S_x^c}r(x,y)R_z^\perp(dy)\nu(dz)&+\int_{z\in S_x}\int_{y\in S_x^c}r(x,y)R_z^a(dy)\nu(dz)\\
&-\int_{y\in S_x^c}\int_{z\in S_x}r(x,y)R_y^a(dz)\nu(dy)\end{aligned}\\
\end{align*}

Plugging this in (<ref>), we get $\int_\mathbb{X}R_y^\perp(S_x)R_x^\perp(dy)=R_x^\perp(\mathbb{X})$, and thus

\begin{equation}\label{proof:decomposition:singular:support}
R_y^\perp(S_x)=1\qquad\mbox{for }R_x^\perp\mbox{-a.e. }y.
\end{equation}

But $1=R_y^\perp(S_x)\leq R_y^\perp(\mathbb{X})$, so $R_y^a(\mathbb{X})=0$ and $R_y^a(dz)R_x^\perp(dy)=0$, which implies, in particular, that

\[R_y(dz)R_x(dy)\nu(dx)=R_y^\perp(dz)R_x^\perp(dy)\nu(dx)+R_y(dz)R_x^a(dy)\nu(dx).\]

On the other hand, using the identity in (<ref>), we obtain

\[R_y(dz)R_x(dy)\nu(dx)=R_x(dz)R_x(dy)\nu(dx)=R_x(dz)R_x^\perp(dy)\nu(dx)+R_x(dz)R_x^a(dy)\nu(dx).\]

Combining the two result, we get, for $\nu$-a.e. $x$,

\[R_x(dz)R_x^\perp(dy)+R_x(dz)R_x^a(dy)=R_y^\perp(dz)R_x^\perp(dy)+R_y(dz)R_x^a(dy).\]

But $R_x^\perp(dy)\independent R_x^a(dy)$, so it must be

\[R_x(dz)R_x^\perp(dy)=R_y^\perp(dz)R_x^\perp(dy).\]

It follows from this result and (<ref>) that, for $\nu$-a.e. $x$,

\[R_x(S_x)=\int_{S_x}R_y^\perp(S_x)R_x(dy)=\int_{S_x}R_x(S_x)R_x(dy)=R_x(S_x)R_x(S_x);\]

thus, either $R_x(S_x)=1$, in which case $R_x(dy)=R_x^\perp(dy)$, or $R_x(S_x)=0$, in which case $R_x(dy)=R_x^a(dy)$.

Define $S:=\{x\in\mathbb{X}:R_x(S_x)=1\}$. It follows from above that $R_x^\perp(dy)=0$ for $\nu$-a.e. $x\in S^c$, and $R_x^a(dy)=0$ for $\nu$-a.e. $x\in S$. Moreover, by (<ref>), $R_x^\perp(S)=R_x^\perp(\mathbb{X})$ for $\nu$-a.e. $x$. Finally, using (<ref>), we get

\[\int_SR_x^\perp(\mathbb{X})\nu(dx)=\int_\mathbb{X}R_x^\perp(S)\nu(dx)+\int_\mathbb{X}R_x^a(S)\nu(dx)=\int_SR_x^\perp(\mathbb{X})\nu(dx)+\int_\mathbb{X}R_x^a(S)\nu(dx),\]

which implies that $R_x^a(S)=0$ for $\nu$-a.e. $x$.

As $R_x\independent\nu$, then $(X_n)_{n\geq1}$ is not i.i.d., and hence not unbalanced, so from Corollary <ref> we may assume that $R_x(\mathbb{X})=1$, without loss of generality.

Let us define, for $x\in\mathbb{X}$,

\[D_x:=\bigl\{y\in\mathbb{X}:R_x(\{y\})>0\bigr\},\qquad\mbox{and}\qquad D:=\{x\in\mathbb{X}:x\in D_x\}.\]

Then $R_x(\cdot)=\sum_{y\in D_x}p_x(y)\delta_y(\cdot)$ for some $p_x(y)>0$ such that $\sum_{y\in D_x}p_x(y)=1$.

It follows from the identity in (<ref>) and the fact that $R_x$ is discrete that there exists $C\in\mathcal{X}$ such that $\nu(C)=1$ and, for all $x\in C$,

\[R_x(dz)=R_y(dz)\qquad\mbox{for }y\in D_x;\]

thus, $D_x=D_y$ for all $y\in D_x$, so $y\in D_y$ or, equivalently, $y\in D$. Therefore, by (<ref>),

\[\nu(D)=\int_\mathbb{X}R_x(D)\nu(dx)=\int_{C^*}\Bigl(\sum_{y\in D_x}p_x(y)\delta_y(D)\Bigr)\nu(dx)=\int_{C^*}\Bigl(\sum_{y\in D_x}p_x(y)\Bigr)\nu(dx)=1.\]

Let $x,y\in C\cap D$ be such that $y\notin D_x$. Assume $D_x\cap D_y\neq\emptyset$. Take $z\in D_x\cap D_y$. It follows from above that $D_x=D_z=D_y$. But $y\in D$, and so $y\in D_y=D_x$, absurd. As a result, for $\nu$-a.e. $x$ and $y$, either $R_x(dz)=R_y(dz)$ or $D_x\cap D_y=\emptyset$.

§ DISCUSSION

The results in this paper allow us to state some universal facts about exchangeable MVPSs that were previously unknown: 1) all models are necessarily balanced; 2) the normalized reinforcement kernels are a.e. proper conditional distributions; 3) the prior distributions have the stick-breaking construction of a Dirichlet process. When $\mathbb{X}$ is countable or the reinforcement is dominated by $\nu$, then every MVPS is a Dirichlet process mixture model over a family of probability distributions with disjoint supports, derived from $\nu$. Relaxing parts of this structure while retaining exchangeability would lead to a very different sampling scheme from the one that (<ref>) implies. On the other hand, the fact that $R_x$ can be either absolutely continuous or mutually singular with respect to $\nu$, though surprising at first, seems very natural under exchangeability. Therefore, we expect that $R$ can be further decomposed along other lines, e.g. $R_x^\perp$ might be discrete only for some $x$ and diffuse for the rest.

§.§ Acknowledgments

This study is financed by the European Union-NextGenerationEU, through the National Recovery and Resilience Plan of the Republic of Bulgaria, project No. BG-RRP-2.004-0008

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# Relativistic generalised uncertainty and the corrected vacuum energy

Akshat<EMAIL_ADDRESS>
Department of Physics, Shiv Nadar Institution of Eminence
Greater Noida, Uttar Pradesh-201314, India

###### Abstract

In this short paper we make use of the recent covariant extension of the
generalised uncertainty principle to study the corrections to the vacuum
energy of the simplest scalar quantum field theory. We then calculate the
modifications to the Casimir effect that such a correction to the vacuum
energy would bring about. We emphasise that these corrections are indeed
physical.

##  1 Introduction

Predictions of a minimum measurable length have been a common feature of
theories of Quantum gravity (QG) [1, 2, 3, 4, 5]. Within QG phenomenology, the
Generalised Uncertainty Principle (GUP) which extends the Heisenberg
Uncertainty Principle to include high momenta cut-offs has been an important
approach at incorporating the idea of the smallest measurable length. [6, 7,
8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]. Recently, there have been
attempts to put GUP into a covariant framework [20, 21, 22, 23]. In
particular, Torodinnov et al extended GUP to relativistic scenarios [21] with
the modified commutation relation taking the form

$[x^{\mu},p^{\nu}]=i\hbar(1-\gamma
p^{\alpha}p_{\alpha})\eta^{\mu\nu}+i\hbar{\gamma}p^{\mu}p^{\nu}$ (1)

Here $c=1$, and all Greek indices $\in\\{0,1,2,3\\}$. $\gamma$ is the GUP
parameter with dimensions of inverse momentum squared,
$\gamma=\frac{\gamma_{0}}{M_{Pl}^{2}}$, $\gamma_{0}$ is a dimensionless that
fixes the GUP scale. The Relativistic GUP (RGUP) thus points at a minimum
position uncertainty that is Lorentz invariant. This ensures that Generalised
uncertainty can be formulated in the context of Quantum Field theory, making
it possible to calculate minimal length induced corrections to well-known
effects including, for example, scatterring amplitudes and electron-muon
cross-sections [24, 25, 26]. However, there seems to be no work exploring the
measurable effects of RGUP to free field theories.

The purpose of this present work is twofold. Firstly we show how even for the
simplest case of a free scalar field, imposing RGUP leads to modification in
the vacuum energy of the field. Secondly, by calculating the vacuum
fluctuations and the corrected Casimir force, we emphasise that this shift in
the vacuum energy is indeed physical and can have measurable consequences.

This paper is organized as follows. In section 2 we review RGUP modifications
to a scalar field and then go on to calculate the vacuum energy. In section 3,
using the results for the vacuum energy we study fluctuations and calculated
the corrected Casimir force. We end with a summary in section 4. For the rest
of the paper we assume natural units, that is, $\hbar=c=1$.

##  2 Modified Scalar field theory and the Vacuum

To start, we notice that the commutation relations (1) imply that the
variables $x^{\mu}$ and $p^{\nu}$ are not canonically conjugate. We can thus
define additional variables $x_{0}^{\mu}$ and $p_{0}^{\nu}$ that obey the
standard commutation rules

$[x_{0}^{\mu},p_{0}^{\nu}]=i\eta^{\mu\nu}$ (2)

Further, we can see that $p_{0}^{\mu}$ are related to $p^{\mu}$ via

$p_{\mu}=p_{0\mu}(1+\gamma p_{0}^{\mu}p_{0\mu})$ (3)

Thus a field $\phi$ that obeys the standard Klein-Gordon equation for
$p^{\mu}$ could thus be rewritten as

$(\Box+2\gamma\Box^{2}+m^{2})\phi=0$ (4)

where $\Box=\partial_{\mu}\partial^{\mu}$. Starting with (3), we restrict to
terms of the order $\gamma$ and neglect terms $\mathcal{O}(\gamma^{2})$ .The
$\gamma$ term above and in all subsequent equations is interpreted as RGUP
based Quantum gravity corrections to Quantum field theories [7]. For the above
equation the Lagrangian density as obtained by Bosso et al [24] is

$\mathcal{L}=\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}+\gamma\partial^{\nu}\partial_{\nu}\partial^{\mu}\phi\partial_{\mu}\phi$
(5)

This additional term in the Lagrangian density can be thought of as another
degree of freedom $\partial_{\mu}\partial^{\mu}\phi$ at each point in space in
addition to $\phi$

Defining the conjugate momenta

$\displaystyle\pi_{\alpha}^{\alpha}$
$\displaystyle=\frac{\partial\mathcal{L}}{\partial\left(\partial_{\alpha}\partial^{\alpha}\dot{\phi}\right)}=2\gamma\dot{\phi}$
(6) $\displaystyle\pi$
$\displaystyle=\frac{\partial\mathcal{L}}{\partial(\dot{\phi})}-\partial_{\beta}\partial^{\beta}\pi_{\alpha}^{\alpha}=\dot{\phi}$

Using (5) and (6), the Hamiltonian density can be calculated

$\mathcal{H}=\frac{1}{2}\pi^{2}+\frac{1}{2}m\phi^{2}+\frac{1}{2}(\nabla\phi)^{2}+\gamma(\nabla\phi)\cdot\left(\partial_{\sigma}\partial^{\sigma}\nabla\phi\right).$
(7)

The Hamiltonian density is that of a free scalar field with an extra $\gamma$
term. Here $\gamma\sim 10^{-38}$ ensuring that the theory is Ostragradsky
stable [27]. Alternatively the De Donder Weyl formulation [28] could have been
used to yield a much simpler looking Hamiltonian density where the $\gamma$
terms would cancel out and the RGUP corrections be encoded in the disperson
relation and the field definition. However, we shall stick to equation (7) as
our Hamiltonian density. This makes calculations simpler as written this way,
we can use the standard relativistic dispersion relations and the following
standard definition for a real scalar field

$\phi(\vec{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{2\omega_{\vec{k}}}}\left(a_{\vec{k}}e^{-i(\omega
t-\vec{k}\cdot\vec{x})}+a_{\vec{k}}^{\dagger}e^{+i(\omega
t-\vec{k}\cdot\vec{x})}\right).$ (8)

Note, $\vec{k}$ is simply $\vec{p}_{0}$ in natural units and thus obeys the
standard dispersion relations. Computing terms in the Hamiltonian density
using the above field definition is straightforward.

Now we wish to calculate the vacuum energy which turns out to be

$\matrixelement{0}{H}{0}=\int
d^{3}x\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{2\omega_{\vec{k}}}\frac{1}{2}\left(\left(\omega_{\vec{k}}^{2}+\vec{k}^{2}+m^{2}\right)-\gamma\left(\omega_{\vec{k}}^{2}(\vec{k}^{2}+m^{2})\right)+(\vec{k}^{2}+m^{2})^{2}\right)$
(9)

Plugging in the dispersion relations, we end up with

$\matrixelement{0}{H}{0}=\int
d^{3}x\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{2}(\omega_{\vec{k}}-\gamma\omega_{\vec{k}}^{3})$
(10)

The first term in the vacuum density is simply the ground state energy of the
harmonic oscillator integrated over all momentum modes, which is the standard
result. The second is the correction due to RGUP. While the modifications to
the commutation relations and the Hamiltonian were fairly involved, the
$\gamma$ modification to the vacuum energy has a rather simple form. This
simplicity lets us sketch some predictions for measurements of these Quantum
gravity corrections. In the following section we see this in the context of
the Casimir effect.

##  3 Corrections to the Casimir Force

The Casimir force is a physical consequence of the non-trivial vacuum in QFT,
experienced by the bounding plates when a Quantum field is restricted to a
finite region [29]. GUP corrections to the Casimir force have been previously
worked out by [30]. Another interesting paper is by Blasone et al. where they
use GUP arguements to put forward a heuristic derivation of the Casimir effect
[31].

This section is however different from the above mentioned articles as we are
making use of RGUP and working in a field theoretic framework. Our goal is not
to give a careful solid state physics based account of the Casimir force in a
laboratory setup, but to simply illustrate that the modifications to the
vacuum that are brought about by Generalised Uncertainty can have measurable
consequences. Further, for simplicity of calculations we will restrict our
analysis to 1+1 dimensions and to a massless scalar field [32].

The vacuum energy in 1+1 dimensions is

$\matrixelement{0}{H}{0}=\int
dx\int\frac{dk}{(2\pi)}\frac{1}{2}(\omega_{k}-\gamma\omega_{k}^{3})$ (11)

We impose periodic boundary conditions by restricting the field in a box of
length $L$. Further, within this box, we calculate the vacuum energy $E(l)$
between two thin metal sheets seperated by a distance $l$ such that the total
energy would be $E(l)+E(L-l)$. Following (7) $E(l)$ this turns out to be

$E(l)=\sum^{k}\frac{1}{2}\omega_{k}-\sum^{k}\frac{1}{2}\gamma\omega_{k}^{3}$
(12)

Making use of the mode expansion for the energy $\omega=\frac{n\pi}{l}$ in the
box We replace this infinite sum over $k$ with sum over all $n$. Thus, the
energy becomes

$E(l)=\frac{\pi}{2l}\sum^{n}n-\frac{\gamma\pi^{3}}{2l^{3}}\sum^{n}n^{3}$ (13)

We make use of exponential regularisation to obtain finite results from these
infinite sums. Physically this corresponds to introducing a UV cutoff in terms
of a length scale $\alpha$, that is to say that the higher frequency modes
contribute less and less to the sum.

Equation (10) thus becomes

$\displaystyle E(l)$
$\displaystyle=\frac{\pi}{2l}\sum_{n=1}^{\infty}ne^{-\alpha
n/l}-\frac{\gamma\pi^{3}}{2l^{3}}\sum_{n=1}^{\infty}n^{3}e^{-\alpha n/l}$ (14)
$\displaystyle=-\frac{\pi}{2}\frac{\partial}{\partial\alpha}\sum_{n=1}^{\infty}e^{-\alpha
n/l}+\frac{\gamma\pi^{3}}{2}\frac{\partial^{3}}{\partial\alpha^{3}}\sum_{n=1}^{\infty}e^{-\alpha
n/l}$
$\displaystyle=\frac{\pi}{2}\frac{\partial}{\partial\alpha}\left(\frac{e^{-\alpha/l}}{e^{-\alpha/l}-1}\right)-\frac{\gamma\pi^{3}}{2}\frac{\partial^{2}}{\partial\alpha^{2}}\left(\frac{\partial}{\partial\alpha}\left(\frac{e^{-\alpha/l}}{e^{-\alpha/l}-1}\right)\right)$
$\displaystyle=\frac{\pi}{2l}\frac{e^{-\alpha/l}}{(e^{-\alpha/l}-1)^{2}}-\frac{\gamma\pi^{3}}{2}\frac{\partial^{2}}{\partial\alpha^{2}}\left(\frac{e^{-\alpha/l}}{(e^{-\alpha/l}-1)^{2}}\right)$

The exponential suppression enables us to Taylor expand about
$\frac{\alpha}{l}\approx 0$. We get

$\displaystyle E(l)$
$\displaystyle=\frac{\pi}{2l}\left(\frac{l^{2}}{\alpha^{2}}-\frac{1}{12}+\frac{\alpha^{2}}{240l^{2}}+...\right)-\frac{\gamma\pi^{3}}{2l}\frac{\partial^{2}}{\partial\alpha^{2}}\left(\frac{l^{2}}{\alpha^{2}}-\frac{1}{12}+\frac{\alpha^{2}}{240l^{2}}+...\right)$
(15) $\displaystyle=\frac{\pi
l}{2\alpha^{2}}-\frac{\pi}{24l}+\frac{\gamma\pi^{3}l}{12\alpha^{4}}-\frac{\gamma\pi^{3}}{240l^{3}}+\mathcal{O}(\alpha^{2})$

The energy for the rest of the system can be calculated by substituting
$l\xrightarrow{}L-l$. The total energy turns out to be

$\displaystyle E$ $\displaystyle=E(l)+E(L-l)$ (16) $\displaystyle=\frac{\pi
L}{2\alpha^{2}}-\frac{\pi}{24}\left(\frac{1}{l}+\frac{1}{L-l}\right)+\frac{\gamma\pi^{3}L}{12\alpha^{4}}-\frac{\gamma\pi^{3}}{240}\left(\frac{1}{l^{3}}+\frac{1}{(L-l)^{3}}\right)+\mathcal{O}(\alpha^{2})$

Now, we can write down the Casimir force

$F_{Casimir}=-\frac{dE}{dl}=-\frac{\pi}{24l^{2}}-\frac{\gamma\pi^{3}}{720l^{4}}+\mathcal{F}(\alpha/l,l/L)$
(17)

$\mathcal{F}$ represents other terms that show up in the force. These terms
are dependent on the IR and UV regulators $L$ and $\alpha$, respectively.
These terms vanish as we take $L\xrightarrow{}\infty$ and
$\alpha\xrightarrow{}0$, and can simply be ignored.

We thus end up with an expression for the corrected Casimir force. Note that
upon taking the RGUP parameter $\gamma\xrightarrow{}0$, we recover the
standard expression for the Casimir force. The sign of the $\gamma$ term would
determine whether RGUP strengthens or weakens the attractive force between the
bounding plates. Further, the correction term in the force goes as $l^{-4}$
implying that the correction is relevant only for very small $l$ and is
quickly suppressed when the distance between the plates is increased. Note
also that this description cannot be valid for arbitrary small $l$ as the very
premise of the calculation involved a built-in minimal length. Thus, we assume
$l$ to be much larger than the minimum length scale.

Alternatively, starting from equation (13), we could have crudely made the
identification $\sum_{n}n\sim\frac{-1}{12}$ and
$\sum_{n}n^{3}\sim\frac{1}{120}$. However, it is instructive to use
exponential regularization as it also offers physical explanation for the
finiteness of the final answer, namely the UV cutoff [32]. Also, though we
chose exponential regularization for its usefulness in analytic calculations,
other regulators like the zeta-function regulator could have also been used.
For an interesting discussion on regulator choices in QFT and its connections
to number theory, see [33].

Note also that our analysis was only for a toy scalar model. To extend this to
a situation which is physically testable would require a similar analysis for
the electromagnetic field in 3+1 dimensions. In addition to this, the
infinitesimal thickness condition on the conducting plates will have to be
relaxed and the response of these plates to the electromagnetic field will
also have to be taken into account. Apart from this, we believe, a
quantitative analysis of how large the corrections would be given the
smallness of the parameters $\gamma$ and $l$, also deserves further
investigation.

##  4 Summary

In this paper we studied some consequences of the Generalised Uncertainty
Principle applied to the simplest Relativistic Quantum field Theory. In
particular, we worked out how the vacuum energy gets modified by imposing a
Covariant Generalised Uncertainty Relation on the field. We emphasised that
these modification to the vacuum and vacuum fluctuations are indeed real and
to this effect calculated the corrections to the Casimir force that such a
modification of the vacuum energy would lead to.

## Acknowledgements

I thank the anonymous referee for their comments on an earlier version of this
manuscript. I also wish to thank Noel Jobu and Rahul Ghosh for useful
discussions.

## References

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# The Goldman bracket characterizes homeomorphisms

Siddhartha Gadgil  Department of Mathematics,
Indian Institute of Science,
Bangalore 560012, India<EMAIL_ADDRESS>
###### Abstract.

We show that a homotopy equivalence between compact, connected, oriented
surfaces with non-empty boundary is homotopic to a homeomorphism if and only
if it commutes with the Goldman bracket.

Nous montrons que d’une équivalence d’homotopie entre les compacte, con-
connect s, les surfaces orient es avec bord non vide est homotope à un homéo-
morphisme si et seulement si elle commute avec le Goldman bracket.

###### 1991 Mathematics Subject Classification:

Primary 57M99; Secondary 17B99

##  1\. Introduction

Given a homotopy equivalence between manifolds of the same dimension, a
fundamental question in topology is whether it is homotopic to a
homeomorphism. Basic examples of exotic homotopy equivalences – ones that are
not homotopic to the homeomorphisms, are maps between surfaces with boundary.
For example, the three-holed sphere and the one-holed torus are homotopy
equivalent but not homeomorphic.

We assume throughout that all surfaces we consider are oriented. In this note,
we show that there is a simple and natural characterisation of when a homotopy
equivalence $f\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ between compact
surfaces is homotopic to a homeomorphism in terms of the _Goldman bracket_ ,
which is a Lie Algebra structure associated to a surface. More precisely, if
$\hat{\pi}(\Sigma)$ denotes the set of free homotopy classes of closed curves
in $\Sigma$, then the Goldman bracket (whose definition we recall below) is a
bilinear map

$[\cdot,\cdot]\colon\thinspace\mathbb{Z}[\hat{\pi}(\Sigma)]\times\mathbb{Z}[\hat{\pi}(\Sigma)]\to\mathbb{Z}[\hat{\pi}(\Sigma)],$

which is skew-symmetric and satisfies the Jacobi identity.

######  Theorem 1.1.

A homotopy equivalence $f\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ between
compact, connected, oriented surfaces with non-empty boundary is homotopic to
a homeomorphism if and only if it commutes with the Goldman bracket, i.e., for
all $x,y\in\mathbb{Z}[\hat{\pi}(\Sigma_{1})]$, we have,

(1) $[f_{*}(x),f_{*}(y)]=f_{*}([x,y]),$

where
$f_{*}\colon\thinspace\mathbb{Z}[\hat{\pi}(\Sigma_{1})]\to\mathbb{Z}[\hat{\pi}(\Sigma_{1})]$
is the function induced by $f$.

For closed surfaces, every homotopy equivalence is homotopic to a
homeomorphism. So the corresponding result holds trivially in this case.

Homotopy equivalences that are homotopic to homeomorphisms can be
characterised as those that preserve the so called _peripheral structure_.
However, our result has the advantage that the characterization is in terms of
a structure defined without reference to the boundary. Furthermore, the
Goldman bracket is the simplest instance of so called _string topology_ [2],
which in turn is related to the Floer homology of the cotangent bundle [1].

##  2\. Preliminaries

###  2.1. The Goldman bracket

We recall the definition of the Goldman bracket [4] on a connected surface
$\Sigma$. Let $\hat{\pi}(\Sigma)$ denote the set of free homotopy classes of
curves in $\Sigma$. For a closed curve $\alpha$, let
$\langle\alpha\rangle\in\hat{\pi}(\Sigma)$ denote its homotopy class.

Given homotopy classes $x$ and $y$ of closed curves, we consider smooth,
oriented representatives $\alpha$ and $\beta$ that intersect transversally in
double points. The bracket of $x$ and $y$ is then defined as the sum

(2) $[x,y]=\sum_{p\in\alpha\cap\beta}\varepsilon_{p}\
\langle\alpha*_{p}\beta\rangle,$

where, for $p\in\alpha\cap\beta$, $\varepsilon_{p}$ is the sign of the
intersection between $\alpha$ and $\beta$ at $p$ and $\alpha*_{p}\beta$ is the
product of the loops $\alpha$ and $\beta$ viewed as based at $p$.

This expression is well defined by the first part of the following remarkable
theorem of Goldman. A completely topological proof of this has been given by
Chas [3].

######  Theorem 2.1 (Goldman [4]).

The bracket defined by Equation 2 has the following properties.

  1. (1)

Equation 2 gives a well-defined bilinear map
$\mathbb{Z}[\hat{\pi}(\Sigma)]\times\mathbb{Z}[\hat{\pi}(\Sigma)]\to\mathbb{Z}[\hat{\pi}(\Sigma)]$,
i.e., the right hand side of the equation depends only on the homotopy classes
of $\alpha$ and $\beta$.

  2. (2)

The bracket is skew-symmetric and satisfies the Jacobi identity.

  3. (3)

If $\alpha$ is a simple curve and $x=\langle\alpha\rangle$, then, for
$y\in\hat{\pi}(\Sigma)$, $[x,y]=0$ if and only if $y=\langle\beta\rangle$ for
a closed curve $\beta$ such that $\beta$ is disjoint from $\alpha$.

###  2.2. Peripheral classes

For a connected surface $\Sigma$, if $p\in\Sigma$ is a point,
$\hat{\pi}(\Sigma)$ is the set of conjugacy classes in $\pi_{1}(\Sigma,p)$.
The power operations $\pi_{1}(\Sigma,p)\to\pi_{1}(\Sigma,p)$,
$\alpha\mapsto\alpha^{n}$, $n\in\mathbb{Z}$, and the inverse function
$\alpha\mapsto\alpha^{-1}$ on $\pi_{1}(\Sigma,p)$ induce well-defined
functions on $\hat{\pi}(\Sigma)$. A class $x\in\hat{\pi}$ that is not a non-
trivial power is called _primitive_.

Suppose henceforth that $\Sigma$ is a compact, connected surface with non-
empty boundary. An element of $\hat{\pi}$ is said to be _peripheral_ if it can
be represented by a curve $\alpha\subset\partial\Sigma$.

Assume further that $\Sigma$ has negative Euler characteristic, i.e., $\Sigma$
is not a disc or annulus. Then each component of $\partial\Sigma$ corresponds
to a pair of primitive peripheral classes (one for each orientation of the
boundary curve) which are inverses of each other. Further, the primitive
peripheral classes corresponding to different boundary components are
different.

##  3\. Proof of Theorem 1.1

We now turn to the proof of Theorem 1.1. It is clear from the definition that
a homeomorphism commutes with the Goldman bracket. As homotopic maps induce
the same function on $\hat{\pi}$, it follows that if a homotopy equivalence
$f\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ is homotopic to a homeomorphism,
then $f_{*}$ commutes with the Goldman bracket.

The rest of the paper is devoted to proving the converse. Assume henceforth
that $f\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ is a map between connected,
compact, oriented surfaces with non-empty boundary so that for all
$x,y\in\mathbb{Z}[\hat{\pi}(\Sigma_{1})]$, we have,

$[f_{*}(x),f_{*}(y)]=f_{*}([x,y]).$

######  Lemma 3.1.

For a compact surface $\Sigma$ with boundary, a non-trivial class
$x\in\hat{\pi}(\Sigma)$ is peripheral if and only if for all
$y\in\hat{\pi}(\Sigma)$, we have $[x,y]=0.$

###### Proof.

Assume $x$ is peripheral, i.e., $x=\langle\alpha\rangle$ with
$\alpha\subset\partial\Sigma$. As any closed curve in $\Sigma$ is freely
homotopic to one in the interior of $\Sigma$, every element
$y\in\hat{\pi}(\Sigma)$ can be represented by a curve $\beta\subset
int(\Sigma)$. Hence $\alpha\cap\beta=\phi$, which implies by Equation 2 that
$[x,y]=0$.

Conversely, let $x=\langle\alpha\rangle$ be such that $[x,y]=0$ for all
$y\in\hat{\pi}(\Sigma)$. If $x$ is not peripheral, it is well known that there
is a simple closed curve $\beta$ so that the geometric intersection number of
$\alpha$ and $\beta$ is non-zero, i.e., $\alpha$ is not homotopic to a curve
that is disjoint from $\beta$. This follows from the fact that there is a pair
of simple curves in $\Sigma$ that _fill_ $\Sigma$ (i.e., all curves with zero
geometric intersection with each of them is peripheral), or alternatively by
considering the geodesic representative for $x$ with respect to a hyperbolic
metric and using the result that geodesics intersect minimally.

Now, by Goldman’s theorem (Theorem 2.1, part (3)), as the simple closed curve
$\beta$ has non-zero geometric intersection number with $\alpha$ and
$y=\langle\beta\rangle$, we have $[x,y]\neq 0$, a contradiction.

∎

Recall that we assume that we have a map
$f\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ such that $f_{*}$ commutes with the
Goldman bracket. In case $\Sigma_{1}$ (hence $\Sigma_{2}$) is a disc or an
annulus, it is easy to see that any homotopy equivalence is homotopic to a
homeomorphism. We can hence assume henceforth that $\Sigma_{1}$ and
$\Sigma_{2}$ have negative Euler characteristic.

######  Lemma 3.2.

Suppose $f\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ induces a surjection
$f_{*}\colon\thinspace\hat{\pi}(\Sigma_{1})\to\hat{\pi}(\Sigma_{2})$. Then if
$x\in\hat{\pi}(\Sigma_{1})$ is a peripheral class, so is $f_{*}(x)$.

###### Proof.

As $x$ is peripheral, by Lemma 3.1 $[x,y]=0$ for all
$y\in\hat{\pi}(\Sigma_{1})$. As $f_{*}$ commutes with the Goldman bracket, it
follows that $[f_{*}(x),f_{*}(y)]=0$ for all $y\in\hat{\pi}(\Sigma_{1})$. As
$f_{*}$ is surjective, $[f_{*}(x),z]=0$ for all $z\in\hat{\pi}(\Sigma_{2})$.
It follows by Lemma 3.1 that $f_{*}(x)$ is peripheral. ∎

######  Lemma 3.3.

Suppose $f\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ induces an injection
$f_{*}\colon\thinspace\hat{\pi}(\Sigma_{1})\to\hat{\pi}(\Sigma_{2})$. Then,
for $x\in\hat{\pi}(\Sigma_{1})$, if $f_{*}(x)$ is peripheral then $x$ is
peripheral.

###### Proof.

As $f_{*}(x)$ is peripheral, by Lemma 3.1 $[f_{*}(x),f_{*}(y)]=0$ for all
$y\in\hat{\pi}(\Sigma_{1})$. As $f_{*}$ is injective and commutes with the
Goldman bracket, it follows that $[x,y]=0$ for all
$y\in\hat{\pi}(\Sigma_{1})$. Thus, $x$ is peripheral by Lemma 3.1. ∎

######  Lemma 3.4.

Suppose $f\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ is a homotopy equivalence
and $C_{1}$, $C_{2}$, …$C_{n}$ are the boundary components of $\Sigma_{1}$.
Then there are boundary components $C^{\prime}_{1}$, $C_{2}^{\prime}$,…,
$C^{\prime}_{n}$ of $\Sigma_{2}$ such that $f(C_{i})$ is homotopic to
$C_{i}^{\prime}$ for all $i$, $1\leq i\leq n$. Further the boundary components
$C_{i}^{\prime}$ are all distinct and $\partial\Sigma_{2}=C_{1}^{\prime}\cup
C_{2}^{\prime}\cup\dots\cup C_{n}^{\prime}$.

###### Proof.

As $f$ is a homotopy equivalence,
$f_{*}\colon\thinspace\hat{\pi}(\Sigma_{1})\to\hat{\pi}(\Sigma_{2})$ is both
surjective and injective and maps primitive classes to primitive classes. For
$1\leq i\leq n$, let $\alpha_{i}$ be a closed curve parametrizing $C_{i}$. By
Lemma 3.2, $\alpha_{i}^{\prime}=f(\alpha_{i})$ represents a primitive
peripheral class for $1\leq i\leq n$, and hence represents a boundary
component $C^{\prime}_{i}$. Furthermore, as $\Sigma_{1}$ has negative Euler
characteristic, for $i\neq j$, $\alpha_{i}\neq\alpha_{j}^{\pm 1}$, and hence
$\alpha^{\prime}_{i}\neq{\alpha^{\prime}}_{j}^{\pm 1}$. It follows that the
curves $\alpha_{i}^{\prime}$ are homotopic to curves representing distinct
boundary components $C_{i}^{\prime}$, $1\leq i\leq n$.

Finally, we see that $\partial\Sigma_{2}=C_{1}^{\prime}\cup
C_{2}^{\prime}\cup\dots\cup C_{n}^{\prime}$. Namely, if $\gamma$ is a curve
parametrizing a boundary component $C^{\prime\prime}$ of $\partial\Sigma_{2}$,
$\langle\gamma\rangle=f_{*}(x)$ for some class $x$ (as $f_{*}$ is surjective).
The class $x$ must be peripheral by Lemma 3.3 and primitive as $\gamma$ is
primitive. Hence $x$ is represented by either $\alpha_{i}$ or
$\alpha_{i}^{-1}$ for some $i$, $1\leq i\leq n$. In either case,
$C^{\prime\prime}=C_{i}^{\prime}$. ∎

######  Lemma 3.5.

Suppose $f\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ is a homotopy equivalence.
Then $f$ is homotopic to a map $g\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ so
that $g(\partial\Sigma_{1})=\partial\Sigma_{2}$ and
$g|_{\partial\Sigma_{1}}\colon\thinspace\partial\Sigma_{1}\to\partial\Sigma_{2}$
is a homeomorphism.

###### Proof.

By Lemma 3.4, the restriction of $f$ to $\partial\Sigma_{1}$ is homotopic in
$\Sigma_{2}$ to a homeomorphism
$\varphi\colon\thinspace\partial\Sigma_{1}\to\partial\Sigma_{2}$. Let
$h\colon\thinspace\partial\Sigma_{1}\times[0,1]\to\partial\Sigma_{2}$ be such
a homotopy, i.e., $h$ is a map so that $h(\cdot,0)=f(\cdot)$ and
$h(\cdot,1)=\varphi(\cdot)$.

Let
$\Sigma_{1}^{\prime}=\Sigma_{1}\coprod_{\partial\Sigma_{1}=(\partial\Sigma_{1}\times\\{0\\})}\partial\Sigma_{1}\times[0,1]$
be the space obtained from the disjoint union of $\Sigma_{1}$ and
$\partial\Sigma_{1}\times[0,1]$ by identifying points
$x\in\partial\Sigma_{1}\subset\Sigma_{1}$ with the corresponding points
$(x,0)\in\partial\Sigma_{1}\times[0,1]$. Define a map
$g^{\prime}\colon\thinspace\Sigma_{1}^{\prime}\to\Sigma_{2}$ by
$g^{\prime}(x)=f(x)$ for $x\in\Sigma_{1}$ and $g^{\prime}(x,t)=h(x,t)$ for
$(x,t)\in\partial\Sigma_{1}\times[0,1]$.

Observe that $\partial\Sigma_{1}^{\prime}=\partial\Sigma_{1}\times\\{1\\}$.
The inclusion map $\Sigma_{1}\to\Sigma_{1}^{\prime}$ is homotopic to a
homeomorphism $\psi$ so that, for $x\in\partial\Sigma_{1}$, $\psi(x)=(x,1)$.
Let $H^{\prime}\colon\thinspace\Sigma_{1}\to\Sigma_{1}^{\prime}$ be such a
homotopy, i.e., $H^{\prime}$ is a map so that $H^{\prime}(\cdot,0)$ is the
inclusion map and $H^{\prime}(\cdot,1)=\psi(\cdot)$. Let
$g\colon\thinspace\Sigma_{1}\to\Sigma_{2}$ be given by

$g(x)=g^{\prime}(H^{\prime}(x,1))=g^{\prime}(\psi(x)).$

Then, for $x\in\partial\Sigma_{1}$,
$g(x)=g^{\prime}(\psi(x))=g^{\prime}((x,1))=h(x,1)=\varphi(x)$, i.e.
$g|_{\partial\Sigma_{1}}=\varphi$. Hence $g$ is a map from
$\Sigma_{1}\to\Sigma_{2}$ so that $g(\partial\Sigma_{1})=\partial\Sigma_{2}$
and
$g|_{\partial\Sigma_{1}}\colon\thinspace\partial\Sigma_{1}\to\partial\Sigma_{2}$
is a homeomorphism. Finally, for $x\in\Sigma_{1}$,
$g^{\prime}(H^{\prime}(x,0))=g^{\prime}(x)=f(x)$. Hence, a homotopy from $f$
to $g$ is given by

$H(x,t)=g^{\prime}(H(x,t)).$

∎

By a theorem of Nielsen, $g$, and hence $f$, is homotopic to a homeomorphism
(see Lemma 1.4.3 of [5] – as
$g\colon(\Sigma_{1},\partial\Sigma_{1})\to(\Sigma_{2},\partial\Sigma_{2})$
induces an isomorphism of fundamental groups, it follows that $g$ is homotopic
to a homeomorphism). This completes the proof of Theorem 1.1.∎

## References

  * [1] Abbondandolo, A. and Schwarz, M., Floer homology of cotangent bundles and the loop product, Geom. Topol. 14 (2010), 1569 1722.
  * [2] M. Chas and D. Sullivan, String Topology, Annals of Mathematics (to appear).
  * [3] Chas, M., Combinatorial Lie bialgebras of curves on surfaces, Topology 43, (2004), 543-568.
  * [4] Goldman, W. M., Invariant functions on Lie groups and Hamiltonian flows of surface group representations, Invent. Math. 85, No.2, (1986), 263-302.
  * [5] Waldhausen, F., On irreducible 3-manifolds which are sufficiently large. Ann. of Math. (2) 87 (1968) 56 88.


When $p=0$, (A.1) is a trivial equality. When $v=0$ then (A.1) is also
straightforward. Thus, we consider $p>0$ and we can assume, without loss of
generality, that $0<\|v\|\leqslant\|u\|$. Observe that (A.1) is equivalent to

$\|u\|^{2p+2}-2\|u\|^{p}\|v\|^{p}\langle
u,v\rangle+\|v\|^{2p+2}\leqslant(p+1)^{2}R^{2p}(\|u\|^{2}-2\langle
u,v\rangle+\|v\|^{2}).$ (A.2)

Also note that $\|u\|^{2}-2\langle u,v\rangle+\|v\|^{2}=0$ iff $\|u-v\|=0$ iff
$u=v$, in which case the inequality is also trivial. Thus we can also assume
that $u\neq v$. Setting $x=\|u\|$, $y=\|v\|$ and $z=\langle u,v\rangle$, we
have $0<y\leqslant x\leqslant R$, $-xy\leqslant z\leqslant xy$,
$x^{2}-2z+y^{2}\neq 0$ and (A.2) becomes

$x^{2p+2}-2x^{p}y^{p}z+y^{2p+2}\leqslant(p+1)^{2}R^{2p}(x^{2}-2z+y^{2}).$
(A.3)

Case I. $-xy\leqslant z\leqslant 0$.

In this case, since $z\leqslant 0$, $x^{p}\leqslant(p+1)R^{p}$, and
$y^{p}\leqslant(p+1)R^{p}$, we have

$\displaystyle x^{2p+2}-2x^{p}y^{p}z+y^{2p+2}-(p+1)^{2}R^{2p}(x^{2}-2z+y^{2})$
$\displaystyle\quad=x^{2}[x^{p}-(p+1)R^{p}][x^{p}+(p+1)R^{p}]+y^{2}[y^{p}-(p+1)R^{p}][y^{p}+(p+1)R^{p}]$
$\displaystyle\qquad-2z(x^{p}y^{p}-(p+1)^{2}R^{2p})\leqslant 0,$

which proves (A.3).

Case II. $0<z\leqslant xy$.

Define

$f(x,y,z)=\frac{x^{2p+2}-2x^{p}y^{p}z+y^{2p+2}}{x^{2}-2z+y^{2}}.$

We have

$\frac{\partial f}{\partial
z}(x,y,z)=\frac{(-2x^{p}y^{p})(x^{2}-2z+y^{2})+2(x^{2p+2}-2x^{p}y^{p}z+y^{2p+2})}{(x^{2}-2z+y^{2})^{2}},$

and since

$\displaystyle(-2x^{p}y^{p})(x^{2}-2z+y^{2})+2(x^{2p+2}-2x^{p}y^{p}z+y^{2p+2})$
$\displaystyle\quad=2x^{p+2}(x^{p}-y^{p})-2y^{p+2}(x^{p}-y^{p})=2(x^{p}-y^{p})(x^{p+2}-y^{p+2})\geqslant
0,$

it follows that $\frac{\partial f}{\partial z}(x,y,z)\geqslant 0$. Thus, for
each pair $(x,y)$ satisfying the conditions mentioned above, the function
$(0,xy]\ni z\mapsto f(x,y,z)$ is increasing and, consequently,
$f(x,y,z)\leqslant f(x,y,xy)$. Set

$g(x,y):=f(x,y,xy)=\frac{x^{2p+2}-2x^{p+1}y^{p+1}+y^{2p+2}}{x^{2}-2xy+y^{2}}=\frac{(x^{p+1}-y^{p+1})^{2}}{(x-y)^{2}}.$

A direct application of the Mean Value Theorem to the function $h(t)=t^{p+1}$
shows that there exists $\xi\in(y,x)$ such that

$x^{p+1}-y^{p+1}=(p+1)\xi^{p}(x-y)\leqslant(p+1)x^{p}(x-y)\leqslant(p+1)R^{p}(x-y),$

which implies $f(x,y,z)\leqslant g(x,y)\leqslant(p+1)^{2}R^{2p}$, and
completes the proof. ∎

The next result is taken from [17, Lemma 2.2], and we present a slightly
adapted proof.

######  Proposition A.4.

If $H$ is a real Hilbert space with inner product $\langle\cdot,\cdot\rangle$
and associate norm $\|\cdot\|$, then for $p\geqslant 0$ and $x,y\in H$ we have

$\langle\|x\|^{p}x-\|y\|^{p}y,x-y\rangle\geqslant 2^{-p}\|x-y\|^{p+2}.$ (A.4)

###### Proof.

When $p=0$, the inequality is trivial and $C_{0}=1$. Hence we assume $p>0$.
Clearly (A.4) is trivial when $x=0$. We can assume, without loss of
generality, that $\|y\|\leqslant\|x\|$ and $x\neq 0$. Note that (A.4) is
equivalent to

$\|x\|^{p+2}+\|y\|^{p+2}-(\|x\|^{p}+\|y\|^{p})\langle x,y\rangle\geqslant
2^{-p}(\|x\|^{2}+\|y\|^{2}-2\langle x,y\rangle)^{\frac{p+2}{2}}.$

Dividing both sides by $\|x\|^{p+2}$ we obtain

$1+\frac{\|y\|^{p+2}}{\|x\|^{p+2}}-\left(1+\frac{\|y\|^{p}}{\|x\|^{p}}\right)\frac{\|y\|}{\|x\|}\frac{\langle
x,y\rangle}{\|x\|\|y\|}\geqslant
2^{-p}\left(1+\frac{\|y\|^{2}}{\|x\|^{2}}-2\frac{\|y\|}{\|x\|}\frac{\langle
x,y\rangle}{\|x\|\|y\|}\right)^{\frac{p+2}{2}}$

Setting $t=\frac{\|y\|}{\|x\|}\in[0,1]$ and $s=\frac{\langle
x,y\rangle}{\|x\|\|y\|}\in[-1,1]$ we obtain

$1+t^{p+2}-(t+t^{p+1})s\geqslant 2^{-p}(1+t^{2}-2ts)^{\frac{p+2}{2}}.$ (A.5)

Since $1+t^{2}-2ts=(t-s)^{2}+1-s^{2}=0$ if and only if $t=s=1$, we can prove
(A.5) for $t\in[0,1)$ and $s\in[-1,1)$, where $1+t^{2}-2ts\neq 0$. Define
$h\colon[0,1)\times[-1,1)\to\mathbb{R}$ by

$h(t,s)=\frac{1+t^{p+2}-(t+t^{p+1})s}{(1+t^{2}-2ts)^{\frac{p+2}{2}}}.$

We have

$\frac{\partial h}{\partial
s}(t,s)=\frac{t[(p+2)(1+t^{p+2})-(1+t^{p})(1+t^{2})-p(t+t^{p+1})s]}{(1+t^{2}-2ts)^{\frac{p+4}{2}}},$

and, for $s<1$,

$\displaystyle(p+2)(1+t^{p+2})$
$\displaystyle-(1+t^{p})(1+t^{2})-p(t+t^{p+1})s$
$\displaystyle\geqslant(p+2)(1+t^{p+2})-(1+t^{p})(1+t^{2})-p(t+t^{p+1}).$

For all $t\in[0,1]$ note that

$\displaystyle(p+2)$
$\displaystyle(1+t^{p+2})-(1+t^{p})(1+t^{2})-p(t+t^{p+1})$
$\displaystyle=p(1-t^{p+1})(1-t)+(1-t^{p})(1-t^{2})\geqslant 0$

thus $\frac{\partial h}{\partial s}(t,s)\geqslant 0$ for $s\in[-1,1)$ and
$t\in[0,1)$. Hence, for each fixed $t\in[0,1)$, $s\mapsto h(t,s)$ is
increasing, which implies that $h(t,s)\geqslant h(t,-1)$ for each $s\in[-1,1)$
and $t\in[0,1)$. Set

$g(t)=h(t,-1)=\frac{t^{p+2}+t^{p+1}+t+1}{(1+t)^{p+2}},$

and note that for $t\in[0,1]$ we have

$g^{\prime}(t)=\frac{(p+1)(t+1)(t^{p}-1)}{(1+t)^{p+3}}\leqslant 0.$

Thus $g$ is decreasing and hence

$h(t,s)\geqslant h(t,-1)=g(t)\geqslant g(1)=2^{-p},$

which proves (A.5) and completes the result. ∎

Data Availability. Data sharing is not applicable to this article as no new
data were created or analyzed in this study.

## Declarations

Competing Interests. The authors have not disclosed any competing interests.

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