# Polarization and coherence in mean field games driven by private and social
utility

Paolo Dai Pra111University of Verona, Department of Computer Science, 15,
Strada Le Grazie, I - 37134 Verona, Italy; contact<EMAIL_ADDRESS>Elena Sartori222University of Padova, Department of Mathematics “Tullio Levi-
Civita”, 63, Via Trieste, I - 35121, Padova, Italy; contact:
<EMAIL_ADDRESS>Marco Tolotti333Ca’ Foscari University of Venice,
Department of Management, Cannaregio 873, I - 31121, Venice, Italy; contact:
<EMAIL_ADDRESS>
###### Abstract

We study a mean field game in continuous time over a finite horizon, T, where
the state of each agent is binary and where players base their strategic
decisions on two, possibly competing, factors: the willingness to align with
the majority (conformism) and the aspiration of sticking with the own type
(stubbornness). We also consider a quadratic cost related to the rate at which
a change in the state happens: changing opinion may be a costly operation.
Depending on the parameters of the model, the game may have more than one Nash
equilibrium, even though the corresponding N-player game does not. Moreover,
it exhibits a very rich phase diagram, where polarized/unpolarized,
coherent/incoherent equilibria may coexist, except for T small, where the
equilibrium is always unique. We fully describe such phase diagram in closed
form and provide a detailed numerical analysis of the $N$-player counterpart
of the mean field game. In this finite dimensional setting, the equilibrium
selected by the population of players is always coherent (favoring the
subpopulation whose type is aligned with the initial condition), but it does
not necessarily minimize the cost functional. Rather, it seems that, among the
coherent ones, the equilibrium prevailing is the one that most benefits the
underdog subpopulation forced to change opinion.

Keywords: Finite Population Dynamics; Mean field games; Multiple Nash
Equilibria; Phase transition; Social Interaction

JEL Classification: C61; C73; D91

MSC Classification: 91A16; 91B14

##  1 Introduction

In this paper we analyze a simple continuous-time dynamic multi-agent model,
and study the limit as the number of agents goes to infinity. We consider a
group of $N$ interacting agents (players), who are allowed to control their
binary state choosing the probability rate of “flipping” them. This rate is a
feedback control, that may depend on the state of all players and (measurably)
on time. Each player aims at minimizing an individual cost, which is comprised
by a running cost and a final reward. We consider a standard quadratic running
cost. At some final time $T>0$, each player gets a reward given as the sum of
two different terms:

  * •

the first mimics a _social driver_ and favors imitation: the player gets a
higher reward if she conforms with the majority, and if the majority becomes
polarized (close to consensus); the majority is over the whole population,
making the interaction between players of mean field type;

  * •

the second models the private (individual) desire to align the state with the
sign of a static and predetermined random variable denoting her personal type.

These two terms are possibly competing and represent a classical social
dilemma: the former mimics _conformism_ , i.e., the adherence to social norms
and is often referred to as social utility; the latter models _stubbornness_ ,
namely, the aspiration of the agent to stay as close as possible to the
prescription of personal traits, hence mimicking a private (or individual)
utility. As notion of optimality we adopt that of Nash equilibrium, and our
aim is to understand the system’s behavior in the limit as
$N\rightarrow+\infty$. This falls into the realm of mean field games,
introduced by J.-M. Lasry and P.-L. Lions and, independently, by M. Huang,
R.P. Malhamé and P.E. Caines (cf. [14], [13]), as limit models for symmetric
many player dynamic games as the number of players tends to infinity; see, for
instance, the lecture notes [6] and the two-volume work [7].

The variable representing the type is introduced in the model as a random
field and is treated as an observable static component of player’ state.
Therefore, this term introduces random disorder and, to the best of our
knowledge, it is one of the first attempts to do it in mean field games.

In literature, this dilemma has been analyzed from different perspectives: [5]
is usually considered as a pioneering study on the trade-off between private
and social drivers in (static) binary choice models. In [4], a generalization
to a dynamic continuous-time setting is proposed, which is not a mean field
game, as agents play static games at random times. [9] proposes a model of
consensus formation similar to our, where only the social component is present
and individual preferences are not considered. [2] is one of the rare examples
studying the interplay between stubbornness and imitation drivers in the realm
of mean field games. However, their mathematical setting is rather different
from ours: in that paper state variables are real with Gaussian initial
distribution; moreover, their linear quadratic optimization problem is solved
by affine controls which preserve Gaussianity; as a consequence, their optimal
control is always unique. Models close to the one proposed here, but without
individual preferences, have also been introduced as examples of non-
uniqueness of equilibria in mean field games (see, e.g., [3, 1, 8, 15, 10,
12]).

Similarly to what is contained in the last cited references, the model that we
are proposing here, has the following remarkable feature: for each $N$, there
is a unique Nash equilibrium for the $N$-player game; however, the
corresponding mean field game may have multiple equilibria. This is
reminiscent of a common paradigm in Statistical Physics: finite volume Gibbs
states are uniquely defined, but thermodynamic limit may be non unique,
indicating a phase transition. The analogy with models in Statistical Physics
can be carried on further: the model that we propose corresponds to the mean
field Ising model (or Curie-Weiss one), when there are no private
signals/types, and to the random field Curie-Weiss model, when disorder is
introduced. The time horizon $T$ plays a role similar to the inverse
temperature in the models cited above: the higher $T$, the smaller the
contribution of the running cost. The $N$-player game as well as its mean
field limit are presented in Section 2. In remarkable analogy with the Curie-
Weiss model, we show that the mean field game has a unique equilibrium for
small $T$, whereas several equilibria emerge as $T$ increases.
A detailed study of the different equilibria of the mean field limit is
proposed in Section 3. In particular, we will see that a number of different
types of equilibria can be identified: polarized/unpolarized (related to the
size of the majority), coherent/incoherent (alignment of the final population
state with the initial state). It is, therefore, natural to ask which one of
these equilibria is selected by taking the limit of the unique equilibrium for
the N-player game. In [8], in absence of individual preferences and with a
much simpler phase diagram, this question was rigorously answered, while a
rigorous analysis is presently out of reach here. We, therefore, run numerical
simulations to capture this selection. We see that there is, indeed, a unique
equilibrium emerging from the N-player approximation: it is always coherent,
but it could be polarized or not, depending on some parameters of the model.
Notably, the prevailing equilibrium is not necessarily the one that minimizes
the aggregate cost suffered by the population of interacting agents. Some
remarks about the rationale behind the selection of the equilibrium in the
case of a finite population are collected in Section 4.2. Section 5 contains
some concluding remarks. The Appendix contains all technical proofs of the
results stated in Section 3.

##  2 A continuous-time binary strategic game

###  2.1 The $N$-player game

We consider $N$ players whose binary state vector is denoted by ${\bf
x}:=(x_{1},\ldots,x_{N})\in\\{-1,1\\}^{N}$. To each player is also assigned a
variable $y_{i}\in\\{-\epsilon,\epsilon\\}$, where $\epsilon>0$ is a given
constant, and we set ${\bf y}:=(y_{1},\ldots,y_{N})$; the components of ${\bf
y}$ will be referred to as local fields. The vector state ${\bf x}={\bf x}(t)$
evolves in continuous time, while ${\bf y}$ is static. Each player is allowed
to control her state with a feedback control $u_{i}(t,{\bf x},{\bf y})$ which
may depend on time, and on the values of ${\bf x}$ and ${\bf y}$. We assume
each $u_{i}$, as function of $t$, to be nonnegative, measurable and locally
integrable. Thus, for a given control ${\bf u}=(u_{1},\ldots,u_{N})$, the
state of the system, ${\bf x}(t)$, evolves as a Markov process, whose law is
uniquely determined as the solution of the martingale problem for the time-
dependent generator

$\mathcal{L}_{t}f({\bf x}):=\sum_{i=1}^{N}u_{i}(t,{\bf x},{\bf y})\left[f({\bf
x}^{i})-f({\bf x})\right]=:\sum_{i=1}^{N}u_{i}(t,{\bf x},{\bf
y})\nabla^{i}f({\bf x}),$

where ${\bf x}^{i}$ is the vector state obtained from ${\bf x}$ by replacing
the component $x_{i}$ with $-x_{i}$. In order to fully define the dynamics, we
prescribe the joint distribution of the initial states ${\bf x}(0)$ and of the
local fields ${\bf y}$. For simplicity we assume all these variables are
independent: all $x_{i}(0)$ have mean $m_{0}\in[-1,1]$, whereas all $y_{i}$
have mean $0$. Each player aims at minimizing her own cost, depending on the
controls of all players, which is given by

$J_{i}({\bf u}):={\mathbb{E}}\left[\frac{1}{2}\int_{0}^{T}u_{i}^{2}(t,{\bf
x}(t),{\bf y})dt-x_{i}(T)(m_{N}(T)+y_{i})\right],$ (1)

where $m_{N}(t):=\frac{1}{N}\sum_{i=1}^{N}x_{i}(t)$, is the mean state of the
population. Here, $T>0$ is the time horizon of the game. Besides the standard
quadratic running cost in the control, two other terms contribute to the cost:

  * •

the term $-x_{i}(T)m_{N}(T)$ favors _polarization_ : each agent profits from
being aligned with the majority at the final time $T$;

  * •

the term $-x_{i}(T)\,y_{i}$ incentivizes each agent to align with her own
local field. As ${\bf y}$ is uniformly distributed on
$\\{-\epsilon,\epsilon\\}^{N}$, this term inhibits alignment of behaviors,
hence polarization.

From a technical viewpoint, we note that, by rescaling time, one could
normalize to $1$ the time horizon and multiplying by $T$ the reward. Thus, the
time horizon $T$ may be seen as tuning the relevance of the final reward as
compared to the “natural inertia” expressed by the running cost.

Given a control vector ${\bf u}$ and a measurable and locally integrable
function

$\beta:[0,T]\times\\{-1,1\\}^{N}\times\\{-\epsilon,\epsilon\\}^{N}\rightarrow[0,+\infty),$

we define the control vector $\left[{\bf u}^{i},\beta\right]$ by

$\left[{\bf u}^{i},\beta\right]_{j}=\left\\{\begin{array}[]{ll}u_{j}&\mbox{for
}j\neq i\\\ \beta&\mbox{for }j=i\end{array}\right..$

######  Definition 2.1.

A control vector ${\bf u}$ is a Nash equilibrium if for each $\beta$ as above,

$J_{i}({\bf u})\leq J_{i}(\left[{\bf u}^{i},\beta\right]),\quad\forall
i=1,\ldots,N.$

Nash equilibria may be obtained via the Hamilton-Jacobi-Bellman equation (see
for details [11])

$\begin{cases}\frac{\partial{v_{i}}}{\partial t}(t,{\bf x},{\bf
y})+\sum_{j=1}^{N}a_{*}(\nabla^{j}v_{j}(t,{\bf x},{\bf
y}))\nabla^{j}v_{i}(t,{\bf x},{\bf
y})+\frac{1}{2}a_{*}^{2}(\nabla^{i}v_{i}(t,{\bf x},{\bf y}))=0\\\ v_{i}(T,{\bf
x},{\bf y})=-x_{i}(m_{N}+y_{i}),\end{cases}$ (2)

where $\nabla^{j}v_{j}(t,{\bf x},{\bf y})=v_{j}(t,{\bf x}^{j},{\bf
y})-v_{j}(t,{\bf x},{\bf y})$, $m_{N}=\frac{1}{N}\sum_{i=1}^{N}x_{i}$, and

$a_{*}(p):=\arg\min_{a\geq 0}\left[ap-\frac{1}{2}a^{2}\right]=p^{-}$

with $p^{-}$ denoting the negative part of $p\in{\mathbb{R}}$. Note that (2)
is a system of $N\times 2^{N}\times 2^{N}$ ordinary differential equations
with locally Lipschitz vector field and global solutions. Therefore, it admits
a unique solution ${\bf v}:=(v_{1},\ldots,v_{N})$; moreover, there exists a
unique Nash equilibrium ${\bf u}$ given by

$u_{i}(t,{\bf x},{\bf y}):=a_{*}(\nabla^{i}v_{i}(t,{\bf x},{\bf y})),\quad
i=1,\ldots,N.$

###  2.2 The mean field game

The mean field game is the formal limit of the above $N$-player game, as seen
from a representative player. Denote respectively by $x\in\\{-1,1\\}$ and
$y\in\\{-\epsilon,\epsilon\\}$ the state and the local field of a
representative player. Given a (deterministic) $m\in[-1,1]$, the player aims
at minimizing the cost

$J(u):={\mathbb{E}}\left[\frac{1}{2}\int_{0}^{T}u^{2}(t,x(t),y)dt-x(T)(m+y)\right]$
(3)

under the Markovian dynamics with infinitesimal generator

$L^{u}_{t}f(x,y):=u(t,x,y)[f(t,-x,y)-f(t,x,y)].$ (4)

As above, admissible controls are measurable and locally integrable functions.
Consistently with the $N$-player game, the initial state $x(0)$ and the local
field $y$ are independent, with means $m_{0}$ and $0$, respectively. As we
will see below, the convexity of the running cost implies uniqueness of the
optimal control $u_{*}^{m}$, which actually depends on the choice of $m$.
Denoting by $x^{m}_{*}(t)$ the evolution of the state for the optimal control,
the solution of the mean field game is completed by finding the solutions of
the Consistency Equation

$m={\mathbb{E}}\left(x^{m}_{*}(T)\right).$ (5)

Therefore, we handle the mean field game first by solving, for $m$ fixed, the
optimal control problem (3)-(4) via Dynamic Programming. This leads to the
Hamilton-Jacobi-Bellman equation

$\begin{cases}\frac{\partial V(t,x,y)}{\partial t}+\min_{u\geq 0}\left[u\nabla
V(t,x,y)+\frac{1}{2}u^{2}\right]=0\\\ V(T,x,y)=-x(m+y)\end{cases}$ (6)

with $\nabla V(t,x,y):=V(t,-x,y)-V(t,x,y)$. The optimal feedback control is
given by

$u_{*}^{m}(t,x,y)=\left(\nabla V(t,x,y)\right)^{-}.$

Then we solve (5) using (4) to obtain the evolution of
$m(t,y):={\mathbb{E}}_{y}\left(x^{m}_{*}(t)\right)$, where ${\mathbb{E}}_{y}$
denotes expectation conditioned to the local field
$y\in\\{-\epsilon,\epsilon\\}$. It follows that
${\mathbb{E}}\left(x^{m}_{*}(t)\right)=\frac{1}{2}\left[m(t,\epsilon)+m(t,-\epsilon)\right]$.
By (4), we obtain the Kolmogorov forward equation

$\frac{d}{dt}m(t,y)={\mathbb{E}}_{y}\left[-2u_{*}(t,x^{m}_{*}(t),y)x^{m}_{*}(t)\right].$
(7)

Now we proceed with the explicit solutions of the steps just described. Next,
we discuss the solutions of the Consistency Equation (5) in terms of the three
parameters of the model: $T$, $\epsilon$ and $m_{0}$.

###  2.3 Solving the Hamilton-Jacobi-Bellman equation

Our aim here is to determine the value function $V(t,x,y)$ which solves (6).
Note that this value function also depends on $m$. It is convenient to set
$z(t,y):=V(t,-1,y)-V(t,1,y).$ Note also that $\nabla V(t,x,y)=xz(t,y).$ Using
(6), we can subtract the two equations for $V(t,-1,y)$ and $V(t,1,y)$,
obtaining a closed equation for $z(t,y)$:

$\begin{cases}\frac{d}{dt}z(t,y)=\frac{1}{2}z(t,y)|z(t,y)|\\\
z(T,y)=2(m+y).\end{cases}$ (8)

It is a key fact that the equations for $z(t,\epsilon)$ and $z(t,-\epsilon)$
are decoupled, so they can be solved separately by separation of variables.
Indeed, observing that, by uniqueness, the sign of $z(t,y)$ is constant in
$t\in[0,T]$, we can rewrite (8) as

$\frac{d}{dt}\left(\frac{1}{z(t,y)}\right)=\frac{1}{2}\text{sign}(m+y)$

that, integrated over $[t,T]$, yields

$z(t,y)=\frac{2(m+y)}{|m+y|(T-t)+1}.$ (9)

At this point we can also compute the value function $V(t,x,y)$. Plugging the
optimal control in (6) we get

$\frac{\partial}{\partial
t}V(t,1,y)=\frac{1}{2}\left(z^{-}(t,y)\right)^{2}=\left\\{\begin{array}[]{ll}0&\quad\mbox{
if }m+y\geq 0\\\ \left(\frac{2(m+y)}{|m+y|(T-t)+1}\right)^{2}&\quad\mbox{ if
}m+y<0.\end{array}\right.$

Integrating this last identity from $t$ to $T$ we get $V(t,1,y)$.
Having $V(t,1,y)$ and $z(t,y)=V(t,-1,y)-V(t,1,y)$ we can also obtain
$V(t,-1,y)$. The final result is

$V(t,x,y)=\left\\{\begin{array}[]{ll}-|m+y|&\quad\mbox{ if
}\text{sign}(m+y)\in\\{0,x\\}\\\
-|m+y|+\frac{2|m+y|}{|m+y|(T-t)+1}&\quad\mbox{ if
}\text{sign}(m+y)=-x.\end{array}\right.$ (10)

###  2.4 Solving the Kolmogorov forward equation

We begin by observing that

$u_{*}(t,x,y)=\left(\nabla
V(t,x,y)\right)^{-}=\left(xz(t,y)\right)^{-}=\frac{1+x}{2}z^{-}(t,y)+\frac{1-x}{2}z^{+}(t,x)=\frac{1}{2}|z(t,y)|-\frac{x}{2}z(t,y).$

Plugging this into (7) we obtain

$\begin{cases}\frac{d}{dt}m(t,y)=-m(t,y)|z(t,y)|+z(t,y)\\\
m(0,y)=m_{0},\end{cases}$

where we used the fact that $x^{2}=1$. Recalling that the sign of $z(t,y)$ is
constant and equals $\rho:=\text{sign}(m+y)$ we can rewrite this last equation
as

$-\rho\frac{d}{dt}\log(1-\rho m(t,y))=z(t,y)=\frac{2(m+y)}{\rho(m+y)(T-t)+1},$

which, integrated from $0$ to $t$, yields

$m(t,y)=\rho\left[1-(1-\rho
m_{0})\left(\frac{|m+y|(T-t)+1}{|m+y|T+1}\right)^{2}\right].$ (11)

Hence,

${\mathbb{E}}\left(x_{*}^{m}(t)\right)=\frac{1}{2}\left(m(t,\epsilon)+m(t,-\epsilon)\right).$
(12)

##  3 Equilibria and phase diagram

This section is entirely devoted to the analysis of the solutions to the
Consistency Equation (5). As said, $m$ corresponds to a solution of the MFG if
and only if it solves such equation. Relying on (11) and (12), the Consistency
Equation can be rewritten as

$\frac{1}{2}\text{sign}(m+\epsilon)\left[1-\frac{(1-\text{sign}(m+\epsilon)m_{0})}{(|m+\epsilon|T+1)^{2}}\right]+\frac{1}{2}\text{sign}(m-\epsilon)\left[1-\frac{(1-\text{sign}(m-\epsilon)m_{0})}{(|m-\epsilon|T+1)^{2}}\right]=m.$
(13)

Solutions of (13) will be called equilibria. We are now going to identify all
such equilibria. Moreover, depending on the values of the parameters of the
model, we classify them emphasizing the presence or the absence of two
different features:

  * -

_polarization_ , which expresses the fact that agent alignment outscores
individual preference,

  * -

_coherence_ , which indicates the fact that the majority of the agents aligns
with the sign of the initial condition $m_{0}$.

We begin by pointing out a symmetry property: if we denote by
$F(m,\epsilon,T,m_{0})$ the l.h.s. of (13), then

$F(-m,\epsilon,T,-m_{0})=F(m,\epsilon,T,m_{0}).$ (14)

Therefore, without loss of generality, in the remainder of this article we
study (13) in the case $m_{0}\geq 0$. We can thus specify precisely four
classes of equilibria:

  * •

equilibria $m\in(\epsilon,1]$ will be called polarized coherent;

  * •

equilibria $m\in[-1,-\epsilon]$ will be called polarized incoherent;

  * •

equilibria $m\in(0,\epsilon]$ will be called unpolarized coherent;

  * •

equilibria $m\in[-\epsilon,0]$ will be called unpolarized incoherent.

Before stating the formal results describing in detail all the solutions to
(13), we provide a visual example, where all the possible situations are
depicted. To this aim, in Figure 1 we plot the full phase diagram in the
parameters $\epsilon,T$, having fixed $m_{0}=0.25$. The right picture is a
zoom of the region within the dashed lines. We can identify nine regions, each
of them corresponding to a specific typology of solutions to (13). For
example, _region 1_ , characterized by an intermediate value of $T$ and
$\epsilon$ small, shows the presence of three equilibria, one
polarized/coherent, one polarized/incoherent, and one unpolarized/incoherent.

Figure 1: Full phase diagram for $m_{0}=0.25$. Different regions identify different properties of the solutions to (13), both in terms of numerosity and classification.  Region | Number of | Polarized | Polarized | Unpolarized | Unpolarized
---|---|---|---|---|---
| equilibria | Coherent | Incoherent | Coherent | Incoherent
1 | 3 | 1 | 1 | 0 | 1
2 | 5 | 1 | 1 | 2 | 1
3 | 5 | 1 | 2 | 2 | 0
4 | 5 | 2 | 2 | 1 | 0
5 | 3 | 1 | 2 | 0 | 0
6 | 1 | 1 | 0 | 0 | 0
7 | 3 | 1 | 0 | 2 | 0
8 | 3 | 2 | 0 | 1 | 0
9 | 1 | 0 | 0 | 1 | 0
Table 1: Total number of equilibria for each region of the phase diagram as in
Figure 1, and number of equilibria disentangled for their typology
(coherent/incoherent; polarized/unpolarized).

In Table 1, we summarize the results in terms of number of equilibria and
their typology for all the regions on the phase diagram as depicted in Figure
1. We note that in regions 6 and 9, for $T$ small, there is a unique
equilibrium which is always coherent, and it is polarized in 6, for $\epsilon$
small, whereas it is unpolarized in 9, for $\epsilon$ large. On the opposite,
in zones 2, 3 and 4, for $T$ large, there are five equilibria, and three of
them are always coherent. Finally, in zones 1, 5, 7 and 8 (for $T$
intermediate), there are three equilibria. In this situation, we see two
different zones: in regions 1 and 5 (for $\epsilon$ small), only one
equilibrium is coherent; in regions 7 and 8 (for $\epsilon$ large), they are
all coherent.

Note that the way the number of equilibria depends of the parameters
$\epsilon$ and $T$ is far from obvious. For instance, fixing $\epsilon\simeq
0.5$, the number of equilibria is not monotonic in $T$. Similarly, fixing
$T\simeq 3$, there is no monotonicity in $\epsilon$.

In the remainder of this section, we state the results describing the phase
diagram, specifying at what times the phase transitions occur. This will also
serve to specify the algebraic form of all the curves separating the regions
depicted in Figure 1. To ease readability, we organize them in four
propositions, one for each type of equilibrium of the MFG, i.e., one for each
class identified by the possible polarization or coherence of the equilibria,
which are listed in Table 1. As already mentioned, we restrict to the case
$m_{0}\geq 0$.

######  Proposition 3.1 (Polarized coherent equilibria: $m>\epsilon$).

…

  * (i)

Suppose $\epsilon\leq m_{0}$. Then, $\forall T>0$, there is a unique
equilibrium $m=M(T,\epsilon,m_{0})$ in $(\epsilon,1]$, [regions 6, 5, 1, 2 in
Fig. 1]. Moreover, $\lim_{T\rightarrow+\infty}M(T,\epsilon,m_{0})=1.$

  * (ii)

Suppose $\epsilon\geq\frac{1+m_{0}}{2}$. Then there exists a unique
$T^{(1)}_{c}=T^{(1)}_{c}(\epsilon,m_{0})>0$ such that the graph of the curve
in the plane $(z,m)$ of equation $z=F(m,\epsilon,T^{(1)}_{c},m_{0})$ is
tangent to the line of equation $z=0$. Moreover,

    * –

for $T<T^{(1)}_{c}(\epsilon,m_{0})$, there is no equilibrium in
$(\epsilon,1]$, [region 9];

    * –

for $T=T^{(1)}_{c}(\epsilon,m_{0})$, there is a unique equilibrium in
$(\epsilon,1]$, [separatrix of regions 9 and 8];

    * –

for $T>T^{(1)}_{c}(\epsilon,m_{0})$, there are two equilibria in
$(\epsilon,1]$, [regions 8, 4].

  * (iii)

Suppose $m_{0}<\epsilon<\frac{1+m_{0}}{2}$. Define

$T^{*}(\epsilon,m_{0}):=-\frac{1}{2\epsilon}+\frac{1}{2\epsilon}\sqrt{\frac{1-m_{0}}{1+m_{0}-2\epsilon}}>0$
(15)

Then there exists a unique
$\epsilon_{*}^{(1)}\in\left(m_{0},\frac{1+m_{0}}{2}\right)$ such that
$\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T^{*}(\epsilon,m_{0}),m_{0})=0.$
Moreover, if $m_{0}<\epsilon\leq\epsilon_{*}^{(1)}\\!,$

    * –

for $T\leq T^{*}(\epsilon,m_{0})$, there is no equilibrium in $(\epsilon,1]$,
[region 9];

    * –

for $T>T^{*}(\epsilon,m_{0})$, there is a unique equilibrium in
$(\epsilon,1]$, [regions 6, 7, 5, 3, 1, 2].

If $\epsilon_{*}^{(1)}<\epsilon<\frac{1+m_{0}}{2}$, there exists a unique
$T^{(1)}_{c}(\epsilon,m_{0})>0$ defined as for the case
$\epsilon\geq\frac{1+m_{0}}{2}$. Furthermore,
$T^{(1)}_{c}(\epsilon,m_{0})<T^{*}(\epsilon,m_{0})$, for each $m_{0}$ the map
$\epsilon\mapsto T_{c}^{(1)}(\epsilon,m_{0})$ is continuous,
$T_{c}^{(1)}(\epsilon,m_{0})\rightarrow T^{*}(\epsilon_{*}^{(1)},m_{0})$ as
$\epsilon\downarrow\epsilon_{*}^{(1)}$, $T_{c}^{(1)}$ defined here for
$\epsilon\in(\epsilon_{*}^{(1)},\frac{1+m_{0}}{2})$ connects continuously at
$\epsilon=\frac{1+m_{0}}{2}$ with $T_{c}^{(1)}$ defined for
$\epsilon\geq\frac{1+m_{0}}{2}$, and

    * –

for $T<T_{c}^{(1)}(\epsilon,m_{0})$, there is no equilibrium in
$(\epsilon,1]$, [region 9];

    * –

for $T=T_{c}^{(1)}(\epsilon,m_{0})$, there is a unique equilibrium in
$(\epsilon,1]$, [separatrix of 9 and 8];

    * –

for $T_{c}^{(1)}(\epsilon,m_{0})<T<T^{*}(\epsilon,m_{0})$, there are two
equilibria in $(\epsilon,1]$, [regions 8, 4];

    * –

for $T\geq T^{*}(\epsilon,m_{0})$, there is a unique equilibrium in
$(\epsilon,1]$, [regions 7, 3].

######  Proposition 3.2 (Polarized incoherent equilibria: $m<-\epsilon$).

…

  * (i)

Suppose $\epsilon\geq\frac{1-m_{0}}{2}$. Then,

    * –

for $T<T^{(1)}_{c}(\epsilon,-m_{0})$, there is no equilibrium in
$[-1,-\epsilon)$, [regions 9, 6, 7, 8];

    * –

for $T=T^{(1)}_{c}(\epsilon,-m_{0})$, there is a unique equilibrium in
$[-1,-\epsilon)$, [separatrix of 6, 7, 8 from 5, 3, 4];

    * –

for $T>T^{(1)}_{c}(\epsilon,-m_{0}))$, there are two equilibria in
$[-1,-\epsilon)$, [regions 5, 3, 4].

  * (ii)

Suppose $0<\epsilon<\frac{1-m_{0}}{2}$. Then
$T^{(1)}_{c}(\epsilon,-m_{0})<T^{(1)}_{*}(\epsilon,-m_{0})$ , and,

    * –

for $T<T^{(1)}_{c}(\epsilon,-m_{0})$, there is no equilibrium in
$[-1,-\epsilon)$, [region 6];

    * –

for $T=T^{(1)}_{c}(\epsilon,-m_{0})$, there is a unique equilibrium in
$[-1,-\epsilon)$, [separatrix of 6 and 5];

    * –

for $T^{(1)}_{c}(\epsilon,-m_{0})<T<T^{(1)}_{*}(\epsilon,-m_{0})$, there are
two equilibria in $[-1,-\epsilon)$, [region 5];

    * –

for $T\geq T^{*}(\epsilon,-m_{0})$, there is a unique equilibrium in
$[-1,-\epsilon)$, [regions 1, 2].

######  Proposition 3.3 (Unpolarized coherent equilibria: $0\leq
m\leq\epsilon$).

…

  * (i)

Suppose $\epsilon\leq m_{0}$. Then there exists a unique
$T^{(2)}_{c}=T^{(2)}_{c}(\epsilon,m_{0})>0$ such that the graph of the curve
in the plane $(z,m)$ of equation $z=F(m,\epsilon,T^{(2)}_{c},m_{0})$ is
tangent to the line of equation $z=0$. Moreover,

    * –

for $T<T^{(2)}_{c}(\epsilon,m_{0})$, there is no equilibrium in
$(0,\epsilon)$, [regions 6, 5, 1];

    * –

for $T=T^{(2)}_{c}(\epsilon,m_{0})$, there is a unique equilibrium in
$(0,\epsilon)$, [separatrix of 1 and 2];

    * –

for $T>T^{(2)}_{c}(\epsilon,m_{0})$, there are two equilibria in
$(0,\epsilon)$, [region 2].

  * (ii)

Suppose $\epsilon\geq\frac{1+m_{0}}{2}$. Then, $\forall T>0$, there is a
unique equilibrium $m=M(T,\epsilon,m_{0})$ in $(0,\epsilon)$, [regions 9, 8,
4]. Moreover, $\lim_{T\rightarrow+\infty}M(T,\epsilon,m_{0})=0.$

  * (iii)

Suppose $m_{0}<\epsilon<\frac{1+m_{0}}{2}$ and let $T^{*}(\epsilon,m_{0})$
given by (15).
For $T\leq T^{*}(\epsilon,m_{0})$ , there is a unique solution in
$(0,\epsilon]$, [regions 9, 8, 4]; the solution is $\epsilon$ if and only if
$T=T^{*}(\epsilon,m_{0})$. Consider now the function

$V(s,\epsilon,m_{0}):=\frac{\sqrt{s}}{8}\left\\{16+\sqrt{s}\left[16+3m_{0}\left(\frac{2m_{0}}{\epsilon
s}\right)^{1/3}\right]\right\\}\\\
+\frac{\epsilon}{2}(1+\sqrt{s})^{2}\left\\{-2-4\sqrt{s}+s\left[-2+3\left(\frac{2m_{0}}{\epsilon
s}\right)^{1/3}\right]\right\\}.$

Then,

    * –

the equation for the unknown $\epsilon,$
$V\left(\frac{m_{0}}{4\epsilon},\epsilon,m_{0}\right)=0$ has a unique solution
$\epsilon_{*}^{(2)}(m_{0})\in\left(m_{0},\frac{1+m_{0}}{2}\right)$ (unless
$m_{0}=0$ for which $\epsilon_{*}^{(2)}(m_{0})=0$);

    * –

there is a unique
$\epsilon_{*}^{(3)}(m_{0})\in\left(\epsilon_{*}^{(2)}(m_{0}),\frac{1+m_{0}}{2}\right)$
such that the curve in the plane $(s,z)$ of equation
$z=V(s,\epsilon_{*}^{(3)}(m_{0}),m_{0})$ is tangent to the line of equation
$z=0$.

Moreover,

    * (a)

if $m_{0}<\epsilon\leq\epsilon_{*}^{(2)}(m_{0})$, the critical time
$T^{(2)}_{c}=T^{(2)}_{c}(\epsilon,m_{0})>0$ defined in point (i) is well
defined, and $T^{*}(\epsilon,m_{0})<T^{(2)}_{c}(\epsilon,m_{0})$;

      * -

for $T^{*}(\epsilon,m_{0})<T\leq T^{(2)}_{c}(\epsilon,m_{0})$, there is no
equilibrium in $(0,\epsilon]$, [regions 6, 5, 1];

      * -

for $T=T^{(2)}_{c}(\epsilon,m_{0})$, there is a unique equilibrium in
$(0,\epsilon)$, [separatrix of 1 from 2, 3];

      * -

for $T>T^{(2)}_{c}(\epsilon,m_{0})$, there are two equilibria in
$(0,\epsilon)$, [regions 3, 2];

    * (b)

if $\epsilon_{*}^{(2)}(m_{0})<\epsilon<\epsilon_{*}^{(3)}(m_{0})$,
$T^{(2)}_{c}$ is not well defined. This implies that there are two times
$T^{(2)}_{c}(\epsilon,m_{0})<T^{(3)}_{c}(\epsilon,m_{0})$ such that the graph
of the curve in the plane $(z,m)$ of equation
$z=F(m,\epsilon,T^{(2)}_{c},m_{0})$ is tangent to the line of equation $z=0$
and

      * -

for $T^{*}(\epsilon,m_{0})<T\leq T^{(2)}_{c}(\epsilon,m_{0})$, there are two
equilibria in $(0,\epsilon]$, [region 7 in Fig. 1];

      * -

for $T=T^{(2)}_{c}(\epsilon,m_{0})$, there is a unique equilibrium in
$(0,\epsilon)$, [separatrix of 7 and 6];

      * -

for $T^{(2)}_{c}(\epsilon,m_{0})<T<T^{(3)}_{c}(\epsilon,m_{0})$, there is no
equilibrium in $(0,\epsilon)$, [region 6];

      * -

for $T=T^{(3)}_{c}(\epsilon,m_{0})$, there is a unique equilibrium in
$(0,\epsilon)$, [separatrix of 6 and 7];

      * -

for $T>T^{(3)}_{c}(\epsilon,m_{0})$, there are two equilibria in
$(0,\epsilon])$, [region 3];

    * (c)

if $\epsilon_{*}^{(3)}(m_{0})\leq\epsilon<\frac{1+m_{0}}{2}$, there are two
equilibria in $(0,\epsilon])$, $\forall T>T^{*}(\epsilon,m_{0})$, [regions 7,
3].

######  Proposition 3.4 (Unpolarized incoherent equilibria: $-\epsilon\leq
m<0$).

…

  * (i)

Suppose $\epsilon\geq\frac{1-m_{0}}{2}$. Then there is no equilibrium in
$[-\epsilon,0)$, $\forall T>0$, [regions 9, 6, 7, 8, 5, 3, 4].

  * (ii)

Suppose $\epsilon<\frac{1-m_{0}}{2}$ and let $T^{*}(\epsilon,m_{0})$ be given
by (15). Then,

    * –

for $T\leq T^{*}(\epsilon,-m_{0})$, there is no equilibrium in $(-\epsilon,0)$
and $m=-\epsilon$ is an equilibrium if and only if $T=T^{*}(\epsilon,-m_{0})$,
[regions 9, 6, 5, 3];

    * –

for $T>T^{*}(\epsilon,-m_{0})$, there is a unique equilibrium in
$[-\epsilon,0)$, [regions 1, 2].

##  4 The $N$-player game: HJB and numerical results

In this section, we provide a different glance to the problem and consider
again the representative agent in a setting where she is best responding to a
population of $N$ opponents (note that, in doing this, the population is thus
formed by $N+1$ players). We now derive a new _large dimensional_ HJB equation
used to run simulations of a finite population in order to identify the
emerging (unique) equilibrium in the finite dimensional model. This approach
is inspired by [12].

All parameters and variables are as described in the previous sections. Recall
that each agent $j$ is characterized by a predetermined _local field_
$y_{j}\in\\{-\epsilon,\,\epsilon\\}$, where $\epsilon\in[0,1]$ and by a time-
varying state variable $x_{j}(t)\in\\{-1,\,1\\}$. We take agent $i$ playing
the role of the representative agent. Concerning the remaining population of
$N$ players, we introduce two summary statistics as the number of “ones” in
the two subpopulations with different local fields (i.e., different
$\epsilon$). To this aim, we define

$n^{+}_{N}=\sum_{j\neq
i}\mathbb{I}_{\\{x_{j}=1\\}}\mathbb{I}_{\\{y_{j}=\epsilon\\}};\quad
n^{-}_{N}=\sum_{j\neq
i}\mathbb{I}_{\\{x_{j}=1\\}}\mathbb{I}_{\\{y_{j}=-\epsilon\\}};\quad
n^{\epsilon}_{N}=\sum_{j\neq i}\mathbb{I}_{\\{y_{j}=\epsilon\\}}.$

Note that $n^{\epsilon}_{N}$ is a static variable, whereas $n^{+}_{N}$ and
$n^{-}_{N}$ change in time, and take values, respectively, in
$\\{0,\,1,\ldots,n_{N}^{\epsilon}\\}$ and
$\\{0,\,1,\ldots,N-n_{N}^{\epsilon}\\}$. By taking advantage of the symmetries
of the model we search equilibrium controls that, for the representative
player $i$, are feedback depending on the state $x_{i}$, on the local field
$y_{i}$ and on the aggregate variables $n^{+}_{N},n^{-}_{N}$ and
$n^{\epsilon}_{N}$, and symmetrically for all other players. We denote by
$\alpha(x_{i},y_{i},n^{+},n^{-},n^{\epsilon},t)$ the feedback control strategy
of player $i$, while each other player $j\neq i$ uses the feedback control

$\beta(x_{j},y_{j},n^{+}\\!-\mathbb{I}_{\\{x_{j}=1\\}}\mathbb{I}_{\\{y_{j}=\epsilon\\}}+\mathbb{I}_{\\{x_{i}=1\\}}\mathbb{I}_{\\{y_{i}=\epsilon\\}},n^{-}\\!-\mathbb{I}_{\\{x_{j}=1\\}}\mathbb{I}_{\\{y_{j}=-\epsilon\\}}+\mathbb{I}_{\\{x_{i}=1\\}}\mathbb{I}_{\\{y_{i}=-\epsilon\\}},n^{\epsilon}\\!-\mathbb{I}_{\\{y_{j}=\epsilon\\}}+\mathbb{I}_{\\{y_{i}=\epsilon\\}},t)$

where, for instance, we have used the fact that

$n^{+}-\mathbb{I}_{\\{x_{j}=1\\}}\mathbb{I}_{\\{y_{j}=\epsilon\\}}+\mathbb{I}_{\\{x_{i}=1\\}}\mathbb{I}_{\\{y_{i}=\epsilon\\}}=\sum_{k\neq
j}\mathbb{I}_{\\{x_{k}=1\\}}\mathbb{I}_{\\{y_{k}=\epsilon\\}}\,.$

Under these assumptions, the triple $(x_{i},n^{+},n^{-})$ evolves as a
continuous-time Markov chain, with the following transitions

$\begin{split}(x,n^{+},n^{-})&\mapsto(-x,n^{+},n^{-})\text{ with rate
}u(t)=\alpha(x,y,n^{+},n^{-},n^{\epsilon},t);\\\
(x,n^{+},n^{-})&\mapsto(x,n^{+}+1,n^{-})\text{ with rate
}\gamma^{+}(x,n^{+},n^{-},n^{\epsilon},t)=\Big{(}n^{\epsilon}-n^{+}\Big{)}\\!\cdot\beta\Big{(}\\!\\!-\\!1,+\epsilon,n^{+}\\!+\mathbb{I}_{\\{x=1\\}},n^{-},n^{\epsilon},t\Big{)};\\\
(x,n^{+},n^{-})&\mapsto(x,n^{+}-1,n^{-})\text{ with rate
}\delta^{+}(x,n^{+},n^{-},n^{\epsilon},t)=n^{+}\cdot\beta\Big{(}1,+\epsilon,n^{+}\\!-\mathbb{I}_{\\{x=-1\\}},n^{-},n^{\epsilon},t\Big{)};\\\
(x,n^{+},n^{-})&\mapsto(x,n^{+},n^{-}+1)\text{ with rate
}\gamma^{-}(x,n^{+},n^{-},n^{\epsilon},t)=\Big{(}N\\!-n^{\epsilon}\\!-n^{-}\Big{)}\\!\cdot\beta\Big{(}\\!\\!-\\!1,-\epsilon,n^{+}\\!,n^{-}\\!+\mathbb{I}_{\\{x=1\\}},n^{\epsilon},t\Big{)};\\\
(x,n^{+},n^{-})&\mapsto(x,n^{+},n^{-}-1)\text{ with rate
}\delta^{-}(x,n^{+},n^{-},n^{\epsilon},t)=n^{-}\cdot\beta\Big{(}1,-\epsilon,n^{+},n^{-}\\!-\mathbb{I}){\\{x=-\\!1\\}},t\Big{)}.\end{split}$

The best response $u(t)=\alpha(x(t),y,n^{+}(t),n^{-}(t),n^{\epsilon},t)$ for
player $i$ is the one minimizing the cost

$\mathbb{E}\left[\int_{0}^{T}\frac{u^{2}(t)}{2}\,dt-x(T)\,(m_{N+1}(T)+y)\right],$

with

$m_{N+1}(T)=2\frac{(n^{+}(t)+n^{-}(t))+\mathbb{I}_{\\{x=1\\}}}{N+1}-1.$

By Dynamic Programming, the Value Function for this stochastic optimal control
problem solves

$\begin{split}&\frac{\partial V}{\partial
t}+\min_{u}\left[\frac{u^{2}}{2}+u\nabla_{x}V+\gamma^{+}\nabla^{+}_{\gamma}V+\delta^{+}\nabla^{+}_{\delta}V+\gamma^{-}\nabla^{-}_{\gamma}V+\delta^{-}\nabla^{-}_{\delta}V\right]=0\\\
&V(x,y,n^{+},n^{-},T)=-x\left(\frac{2(n^{+}+n^{-}+\mathbb{I}_{\\{x=1\\})}}{N+1}-1\right).\end{split}$

The minimisation over $u$ leads to the optimal feedback

$\alpha^{*}(x,y,n^{+},n^{-},t)=\left[V(-x,y,n^{+},n^{-},t)-V(x,y,n^{+},n^{-},t)\right]^{-}$
(16)

and, finally, to the HJB equation

$\begin{split}\dot{V}=&\
\frac{1}{2}\left(\left[V(-x,y,n^{+},n^{-},t)-V(x,y,n^{+},n^{-},t)\right]^{-}\right)^{2}\\\
&-\gamma^{+}(x,n^{+},n^{-},t)\cdot[V(x,+\epsilon,n^{+}+1,n^{-},t)-V(x,+\epsilon,n^{+},n^{-},t)]\\\
&-\delta^{+}(x,n^{+},n^{-},t)\cdot[V(x,+\epsilon,n^{+}-1,n^{-},t)-V(x,+\epsilon,n^{+},n^{-},t)]\\\
&-\gamma^{-}(x,n^{+},n^{-},t)\cdot[V(x,-\epsilon,n^{+},n^{-}+1,t)-V(x,-\epsilon,n^{+},n^{-},t)]\\\
&-\delta^{-}(x,n^{+},n^{-},t)\cdot[V(x,-\epsilon,n^{+},n^{-}-1,t)-V(x,-\epsilon,n^{+},n^{-},t)].\end{split}$
(17)

The unique Nash equilibrium of the game is obtained by setting
$\alpha=\beta=\alpha^{*}$, under which the HJB reduces to a system of
$4\,(n^{\epsilon}+1)(N-n^{\epsilon}+1)$ ordinary differential equations in the
state variables $V$, and can be solved numerically. Specifically, we use the
software Matlab to solve an ODE system backward in time, meaning that the
final conditions play the role of initial conditions and the variation is
inverted in time (the r.h.s of (17) is multiplied by $-1$).

###  4.1 Simulations of the $N$-player system

Having obtained the Nash equilibrium feedback control $\alpha^{*}$ for $N+1$
players, we rescale the problem to $N$ players, and simulate the evolution of
the sufficient statistics

$n^{+}(t)=\sum_{i=1}^{N}\mathbb{I}_{\\{x_{i}=1\\}}\mathbb{I}_{\\{y_{i}=\epsilon\\}},\
\
n^{-}(t)=\sum_{i=1}^{N}\mathbb{I}_{\\{x_{i}=1\\}}\mathbb{I}_{\\{y_{i}=-\epsilon\\}},$

which has the Markovian evolution

$\begin{split}(n^{+},n^{-})\mapsto(n^{+}+1,n^{-})&\text{ with rate
}\Big{(}n^{\epsilon}-n^{+}\Big{)}\cdot\alpha^{*}(-1,+\epsilon,n^{+},n^{-},n^{\epsilon},t);\\\
(n^{+},n^{-})\mapsto(n^{+}-1,n^{-})&\text{ with rate
}n^{+}\cdot\alpha^{*}(1,+\epsilon,n^{+}-1,n^{-},n^{\epsilon},t);\\\
(n^{+},n^{-})\mapsto(n^{+},n^{-}+1)&\text{ with rate
}\Big{(}N-n^{\epsilon}-n^{-}\Big{)}\cdot\alpha^{*}(-1,-\epsilon,n^{+},n^{-},n^{\epsilon},t);\\\
(n^{+},n^{-})\mapsto(n^{+},n^{-}-1)&\text{ with rate
}n^{-}\cdot\alpha^{*}(1,-\epsilon,n^{+},n^{-}-1,n^{\epsilon},t).\end{split}$

Initializing appropriately $(n^{+}(0),n^{-}(0))$, by independently assigning
to each player $x_{i}(0)=1$ with probability $\frac{1+m_{0}}{2}$ and
$y_{i}=\epsilon$ with probability $\frac{1}{2}$, we run simulations of the
above dynamics, eventually obtaining samples for
$m_{N}(T)=2\frac{n^{+}(T)+n^{-}(T)}{N}-1$. Averaging over $S$ independent
simulations we estimate the expectation $\mathbb{E}(m_{N}(T))$.

More in detail, in our first series of experiments we fix $m_{0}=0.25$, and we
take different values of $\epsilon$ and $T$. Concerning the number of
simulations, we set $S=100$, whereas the number of agents in the population is
$N=30$. This figure could appear too small to describe a _large population_.
However, what we see in our simulations is that the expected values of $m_{N}$
are approximating very well the asymptotic equilibrium of the mean field game,
except for some _transition windows_ that we will discuss in more detail.
Later, in a second series of experiments, we will also consider $N=60$. Note
that, by increasing $N$, the numerical problem becomes quickly intractable
because of the high dimension of the HJB associated with the $N$-dimensional
system.444For $N=30$, we have a system of 1024 equations. For $N=60$, the
number increases to 3844.

In Figure 2, we compare $\mathbb{E}(m_{N}(T))$ (red circles), the equilibrium
emerging in the finite dimensional model, with the mean field equilibria
described as the solutions to the Fixed Point Equation (13) (black lines).
Specifically, we consider four different values of $\epsilon\in\\{0.5,\ 0.52,\
0.6\\}$ and we let $T$ vary from $T=0$ to a large time, where the largest
equilibrium value of $m(T)$ is approaching the limit value of $1$. We expect
that $\mathbb{E}(m_{N}(T))$ converges in $N$ towards one of the mean field
equilibria solving (13). This is verified in our simulations; notably, for
certain values of the parameter $\epsilon$, we see a clear transition from a
polarized to a unpolarized equilibrium or vice versa. In fact, by looking at
the four panels of Figure 2, if $\epsilon$ is large enough (see Panel C), the
individual behavior prevails for all $T$ and the population sticks with the
smallest (unpolarized and coherent) equilibrium. When $\epsilon$ is small
enough (see Panel A), the selected equilibrium changes continuously in $T$, as
in the large $\epsilon$ case, but this equilibrium is unpolarized for small
$T$ and polarized for larger $T$. More interesting is the case of intermediate
values of $\epsilon$ (see Panel B). Here we see a continuous branch of
unpolarized and coherent equilibria existing for all $T>0$, while a branch of
polarized coherent equilibria emerges for $T$ sufficiently large. In this case
the $N$-player game agrees with the unique unpolarized equilibrium for small
$T$, jumps to the branch of polarized equilibria as it emerges, but for larger
$T$ it jumps back to the less polarized equilibrium. We actually see “smooth
transitions” rather than “jumps”, but this could be due to the small value of
$N$ in simulations.

Figure 2: Values of $m(T)$ (black dots) and $m_{N}(T)$ (red large circles).

Note that this switch from polarized to unpolarized is not seen for all values
of the initial condition $m_{0}$, as seen in Figure 3.

Figure 3: Values of $m(T)$ (black dots) and $m_{N}(T)$ (red large circles).
Here, $m_{0}=0.2$ and $\epsilon=0.48$.

In the next section, we provide a justification behind the emergence of one
selected equilibrium, in case that multiple equilibria are present in the mean
field limit.

###  4.2 A rationale behind the equilibrium selection

In all the numerical experiments we have performed about $m_{N}(T)$, the
equilibrium emerging in the $N$-dimensional system, we can recognize two
important properties:
Property 1. The equilibrium $m_{N}(T)$ is always coherent. Namely, if
$m_{0}>0$, then $m_{N}(T)>0$.
Property 2. For some values of $\epsilon$, $m_{N}(T)$ switches from a
polarized to an unpolarized equilibrium (or vice versa) depending on the
length of the time horizon $T$.
Concerning Property 1, when we look at the finite dimensional system, the
equilibrium $m_{N}(T)$ converges, for $N$ large, to one of the values $m(T)$
solving $\eqref{eq}$. Note that, among those values, there is at least one
coherent equilibrium (i.e., an equilibrium with the same sign as $m_{0}$). It
is then plausible to presume that the finite population will select one of the
coherent equilibria, in that conveying to an incoherent equilibrium would ask
for a (implausible) mobilization of the subpopulation ex-ante aligned with the
sign of $m_{0}$. Property 2., instead, deals with the eventual polarization of
the coherent equilibrium selected when playing the finite dimensional game.
Here the discussion is more subtle, since we do not have a clear and trivial
explanation of the evident phase transitions we see in the simulations. The
simplest explanation could be that the population chooses the equilibrium
which minimizes the _total cost_ , $J(u)$, defined in (3). Interestingly, this
functional, being a function of the control, can be rewritten in terms of
$m(T)$. We can take advantage of (10) to see that, depending on the value of
$x\in\\{-1,+1\\}$ and $y\in\\{-\epsilon,+\epsilon\\}$, we can specify the cost
needed to reach a certain equilibrium value $m(T)$:

$v_{m(T)}(x,y)=\left\\{\begin{array}[]{ll}-|m(T)+y|&\quad\mbox{ if
}\text{sign}(m(T)+y)\in\\{0,x\\}\\\
-|m(T)+y|+\frac{2|m(T)+y|}{T\,|m(T)+y|+1}&\quad\mbox{ if
}\text{sign}(m(T)+y)=-x,\end{array}\right.$ (18)

where $v_{m(T)}(x,y)=V(0,x,y)$ evaluated at $m=m(T)$. This functional
describes the total cost sustained by each subpopulation indexed by $x$ and
$y$ to reach the equilibrium $m(T)$. We can now derive the costs sustained by
the _underdog_ subpopulation (the one whose local filed is opposite in sign to
$m_{0}$) and by its opponent, namely the one whose local filed is aligned with
$m_{0}$. With a slight abuse of notation, we denote such quantities with
$J^{(-\epsilon)}(m(T))$ and $J^{(+\epsilon)}(m(T))$ to emphasize the
dependence on the prevailing equilibrium. Accordingly, we will also write
$J(m(T))$ as the total cost sustained by the entire system. We have

$J^{(-\epsilon)}(m(T))=\frac{1-m_{0}}{2}\,v_{m(T)}(-1,-\epsilon)+\frac{1+m_{0}}{2}\,v_{m(T)}(+1,-\epsilon)$
$J^{(+\epsilon)}(m(T))=\frac{1-m_{0}}{2}\,v_{m(T)}(-1,+\epsilon)+\frac{1+m_{0}}{2}\,v_{m(T)}(+1,+\epsilon)$

and

$J(m(T))=\frac{1}{2}J^{(-\epsilon)}(m(T))+\frac{1}{2}J^{(-\epsilon)}(m(T)).$

It is not difficult to see that, when considering only coherent equilibria
(i.e., $m(T)$ such that $\text{sign}(m(T))=\text{sign}(m_{0})$), $J(m(T))$
decreases in $m(T)$, in the sense that the more polarized the equilibrium is,
the lower the cost to reach it. Therefore, we could expect the polarized
equilibrium to prevail. However, as said, for certain values of the
parameters, this is not the case. In line with our simulations, we now shape a
different conjecture. We see that the prevailing equilibrium is the one that,
among the coherent ones, minimizes the functional $J^{(-\epsilon)}$, namely
the cost related to the _underdog_ subpopulation. We rephrase this conjecture
in the following fact, which embraces both the previous properties.
Property 3. The equilibrium $m_{N}(T)$ of the $N$-dimensional system converges
to the coherent solution of (13) that minimizes $J^{(-\epsilon)}$.

In some sense, abstracting a two-player game played between the favorite
player ($y=+\epsilon$) and the underdog one ($y=-\epsilon$), we can say that
the former _imposes_ that the equilibrium will be coherent (and this minimizes
her effort), whereas the latter decides about polarization (again, minimizing
effort given the previous selection).

To provide evidence about the goodness of Property 3., in Figure 4, we plot
the phase diagram of $J^{(-\epsilon)}$ for $m_{0}=0.25$ and for the same
values of $\epsilon$ and $T$ seen in Figure 2. As said before, for each
equilibrium value $m(T)$ of the mean field limit, we have the corresponding
value of $J^{(-\epsilon)}$.

Figure 4: Phase diagram for $m(T)$ (left panels) and $J^{(-\epsilon)}$ (right
panels) for two different values of $\epsilon$. Here, $\epsilon=0.5$ (top
panels) and $\epsilon=0.52$ (bottom panels). In the left panels, the red
points denote the solution $m(T)$ corresponding to the minimum value of
$J^{(-\epsilon)}$, (depicted in red in the corresponding right panel).

Note that, for $\epsilon=0.52$, we see two transitions corresponding to the
points where the two branches of $J^{(-\epsilon)}$ related to the polarized
and unpolarized coherent equilibria intersect themselves. In the lower panel
of Figure 4, we zoom on the right-bottom panel to better recognize such
intersection. We see that the two branches of $J^{(-\epsilon)}$ related to the
polarized and unpolarized coherent equilibria intersect themselves at
$T\approx 8.9$. Notably, this point lies in the time interval, where the
emerging equilibrium of the $N$-dimensional system jumps from the polarized
equilibrium to the unpolarized one (see Panel C of Figure 2). We do not report
all the figures related to the other values of $\epsilon$, but the same fact
still appears, thus corroborating Property 3.

Finally, we show that the solution $m(T)$ related to the prevailing solution
$m_{N}(T)$ does not necessarily minimize to total cost $J(m(T))$. In Figure 5
(left panel), for $\epsilon=0.52$, we plot the value of $m(T)$ that minimizes
$J^{(-\epsilon)}$ (bold blue circles) and $J$ (thin black line). It can be
seen that the two cases coincide as soon as $T$ exceed the value $T\approx
8.9$ discussed above. In the right panel of the same figure, we plot two
different branches of $J$, one corresponding to $m(T)$ which minimizes
$J^{(-\epsilon)}$, and the second one related to $m(T)$ which minimizes $J$
(thin black line). We see that the two curves differ exactly for $T$ larger
than the intersection value depicted above; this is the value of $T$ where we
know that the equilibrium $m_{N}(T)$ in the finite dimensional model jumps
from the polarized to the unpolarized one. This shows that the equilibrium
emerging in the population dynamics does not always minimize the total cost.
In some sense, the unpolarized equilibrium partially favors the underdog
subpopulation, in that $J^{(-\epsilon)}$ is minimized among the coherent
equilibria.

Figure 5: Left: The value of $m(T)$ minimizing the functional
$J^{(-\epsilon)}$ (bold blue circles) and minimizing $J$ (thin black line).
Right: The functional $J$ computed for $m(T)$ minimizing $J^{(-\epsilon)}$
(bold circles) and the minimum level of $J$ (thin black line). Here,
$\epsilon=0.52$.

We now run a second series of experiment; the aim is still to reinforce the
goodness of Property 3, showing that $m_{N}(T)$ fairly well approximates the
coherent mean field game equilibrium minimizing $J^{(-\epsilon)}$. For this
round of simulations, we increase the number of agents taking $N=60$.
Concerning the other parameters, we choose different values of $m_{0}$,
$\epsilon$ and $T$, in order to consider cases where there are one, three or
five solutions to (13). Specifically, in the first panel of simulations, we
fix $T=1$ and let $m_{0}$ and $\epsilon$ vary (cf. the first six experiments
reported in Table 2); in the second panel, we fix $m_{0}=0.2$ and let $T$ and
$\epsilon$ vary (cf. the second six experiments reported in Table 2). In Table
2, we report all results. The first three columns summarize the parameters of
each experiment. In the fourth, we report the value of $m(T)$ which matches
what prescribed by Property 3. Columns five and six pertain to the finite
dimensional model; we indicate by $\overline{m}_{N}(T)$ the average value of
$m_{N}(T)$ as resulting by running $S=100$ simulations, with an indication of
its standard deviation, $SD(m_{N}(T))$. Finally, in columns seven and eight,
we measure the goodness of the approximation by reporting
$|\overline{m}_{N}(T)-m(T)|$ and $|\overline{m}_{N}(T)-m(T)|/SD(m_{N}(T))$,
respectively. By looking at these latter columns, it is evident that the
equilibrium prevailing in the finite dimensional model aligns with what
prescribed by Property 3. This is testified by the fact that the difference
between $\overline{m}_{N}(T)$ and $m(T)$ is close to $0$ (the highest value is
$0.038$ in experiment n.7) and that such difference is always below one
standard deviation (the highest ratio is, again, in experiment n.7).

$T$ | $m_{0}$ | $\epsilon$ | $m(T)$ | $\overline{m}_{N}(T)$ | $SD(m_{N}(T))$ | $|\overline{m}_{N}(T)-m(T)|$ | $\frac{|\overline{m}_{N}(T)-m(T)|}{SD(m_{N}(T))}$
---|---|---|---|---|---|---|---
1 | 0.1 | 0.42 | 0.8261 | 0.8240 | 0.0794 | 0.0021 | 0.0264
1 | 0.1 | 0.45 | 0.8126 | 0.8020 | 0.0859 | 0.0106 | 0.1234
1 | 0.1 | 0.5 | 0.0506 | 0.0390 | 0.0765 | 0.0116 | 0.1516
1 | 0.5 | 0.55 | 0.8962 | 0.9171 | 0.0553 | 0.0209 | 0.3779
1 | 0.5 | 0.6 | 0.8818 | 0.8860 | 0.0569 | 0.0042 | 0.0738
1 | 0.5 | 0.7 | 0.1276 | 0.1080 | 0.0783 | 0.0196 | 0.2503
2.3 | 0.2 | 0.5 | 0.8552 | 0.8173 | 0.0788 | 0.0379 | 0.4810
2.3 | 0.2 | 0.58 | 0.0583 | 0.0460 | 0.0818 | 0.0123 | 0.1504
2.8 | 0.2 | 0.5 | 0.8925 | 0.8827 | 0.0715 | 0.0098 | 0.1371
3.5 | 0.2 | 0.7 | 0.0203 | 0.0173 | 0.0473 | 0.0030 | 0.0634
5.5 | 0.2 | 0.7 | 0.0094 | 0.0077 | 0.0307 | 0.0017 | 0.0554
9 | 0.2 | 0.7 | 0.0039 | 0.0030 | 0.0171 | 0.0009 | 0.0526
Table 2: Results of numerical experiments run to check the goodness of
Property 3.

##  5 Conclusions

We have studied a simple mean field game on a time interval $[0,T]$, where
players can control their binary state according to a functional made of a
quadratic cost and a final reward. This latter depends on two competing
drivers: (i) a social component rewarding conformism, namely, being part of
the majority (conformism) and (ii) a private signal favoring the coherence of
individuals with respect to a personal type (stubbornness). The trade-off
between these two factors, associated with the antimonotonicity of the
objective functional, leads to a fairly reach phase diagram. Specifically, the
presence of multiple Nash Equilibria for the mean field game has been
detected; moreover, when looking at the aggregate outcome of the game, several
different types of equilibria can emerge in terms of polarization (fraction of
conformists) and coherence (sign of the majority at the final time $T$
compared to the sign of the initial condition).

We have described and characterized the full phase diagram and discussed the
role of all the parameters of the model with respect to the aforementioned
classification of possible equilibria. We have also analyzed a $N$-player
version of the same mean field game. It is a well-known that in this latter
case, the Nash Equilibrium is necessarily unique. It becomes, therefore,
interesting to identify which equilibrium is selected by the $N$-finite
population, in case the corresponding mean field game exhibits multiple
equilibria. In this respect, we detected phase transitions, in the sense that,
depending on one or more parameters of the model, the equilibrium emerging in
the finite-dimensional game is always coherent, but it may turn from an
unpolarized one to a polarized and vice versa, depending on the length of the
time horizon, $T$. This fact seems to be new in the mean field game
literature. At a first glance, we could expect the finite dimensional
population to select the equilibrium minimizing the cost associated to the
equilibrium for the entire system (interpreted as the collective cost). By
contrast, what emerges from our simulations is that the equilibrium prevailing
in the finite dimensional game is the one converging, for $N$ large, to the
coherent equilibrium that minimizes the cost functional associated with the
_ex-ante underdog_ subpopulation, namely, the collection of such players whose
private signal opposites the sign of the majority at time zero. Put
differently, it seems that the _ex-ante_ favorite subpopulation (namely, the
one whose private signal is aligned with the initial condition) imposes the
selection of a coherent equilibrium, whereas the _ex-ante_ underdog
subpopulation (namely, the one whose private signal opposites the sign of the
initial condition) decides about polarization.

##  Appendix A Proof of results

### Proof of Proposition 3.1: polarized coherent equilibria ($m>\epsilon$)

For $m>\epsilon$ (13) can be rewritten as

$F(m,\epsilon,T,m_{0}):=1-m-\frac{1-m_{0}}{\epsilon^{2}T^{2}}\varphi\left(\frac{1+mT}{\epsilon\,T}\right)=0,$
(19)

where

$\varphi(y)=\frac{y^{2}+1}{(y^{2}-1)^{2}}.$

Before describing the solutions of the equation $F(m,\epsilon,T,m_{0})$ in
$(\epsilon,1]$ we give some simple facts that will be useful in the proof.

Fact 1. The map $m\mapsto F(m,\epsilon,T,m_{0})$ is strictly concave for
$m\in(\epsilon,1]$, for each $\epsilon,T,m_{0}$.

Proof. This comes immediately from the fact that $\varphi$ is strictly convex
in $(1,+\infty)$. Indeed,

$\varphi^{\prime}(y)=-\frac{2y(3+y^{2})}{(y^{2}-1)^{3}},\ \ \
\varphi^{\prime\prime}(y)=\frac{6(y^{4}+6y^{2}+1)}{(y^{2}-1)^{4}}.$ (20)

Note also that $\varphi$ is strictly decreasing.∎

As a consequence of Fact 1, (19) can have at most two solutions in
$(\epsilon,1]$.

Fact 2. The map $T\mapsto F(m,\epsilon,T,m_{0})$ is strictly increasing for
$T\in[0,+\infty)$, for each $m,\epsilon,m_{0}$, with $m>\epsilon$.

Proof. To see this, note that

$\frac{\partial F}{\partial
T}(m,\epsilon,T,m_{0})=\frac{2(1-m_{0})\left[m(1+Tm)^{3}-3\epsilon^{2}T(1+Tm)-\epsilon^{4}T^{3}\right]}{\left((1+Tm)^{3}-\epsilon^{2}T^{2}\right)^{3}},$

which has the same sign as
$h(m,\epsilon,T,m_{0}):=m(1+Tm)^{3}-3\epsilon^{2}T(1+Tm)-\epsilon^{4}T^{3}.$
Observing that $h(\epsilon,\epsilon,T,m_{0})=\epsilon$, that

$\frac{\partial h}{\partial
m}(m,\epsilon,T,m_{0})=(1+Tm)^{3}+3Tm(1+Tm)^{2}-3\epsilon^{2}T^{2}$

is increasing in $m$ and $\frac{\partial h}{\partial
m}(\epsilon,\epsilon,T,m_{0})=4\epsilon^{3}T^{3}+6\epsilon^{2}T^{2}+6\epsilon\,T+1>0,$
we conclude that $\frac{\partial F}{\partial T}(m,\epsilon,T,m_{0})>0$ for
$m\in(\epsilon,1]$. This proves Fact2.∎

Fact 3. $\forall m,\epsilon,m_{0}$, with $m>\epsilon,$

$F(m,\epsilon,0^{+},m_{0}):=\lim_{T\downarrow 0}F(m,\epsilon,T,m_{0})=m_{0}-m$

and

$\lim_{T\uparrow+\infty}F(\epsilon,\epsilon,T,m_{0})=\frac{1+m_{0}-2\epsilon}{2},$

whereas for $m\in(\epsilon,1]$,

$\lim_{T\uparrow+\infty}F(\epsilon,\epsilon,T,m_{0})=1-m.$

These are simple asymptotics, and the proof is omitted.

Now we can prove Proposition 3.1. We begin with the case $\epsilon\leq m_{0}$.
By Fact 3, $F(\epsilon,\epsilon,0^{+},m_{0})=m_{0}-\epsilon\geq 0;$ thus, by
Fact 2, $F(\epsilon,\epsilon,T,m_{0})>0$, $\forall T>0$. Since, clearly,
$F(1,\epsilon,T,m_{0})<0$ and $m\mapsto F(m,\epsilon,T,m_{0})$ is strictly
concave by Fact 1, the uniqueness of the solution to (19) follows readily and
case (i) is proved.

Now, consider the case $\epsilon\geq\frac{1+m_{0}}{2}$ (which implies
$\epsilon>m_{0}$) . By Facts 2 and 3, $F(\epsilon,\epsilon,T,m_{0})<0$,
$\forall T>0$. Since $F(m,\epsilon,0^{+},m_{0})=m_{0}-m$, by continuity (19)
has no solution in $(\epsilon,1]$ for $T$ small enough. Now we claim that
there exists a unique $T_{c}^{(1)}=T_{c}^{(1)}(\epsilon,m_{0})>0$ such that
the graph of $z=F(m,\epsilon,T_{c}^{(1)},m_{0})$ is tangent to the line $z=0$.
Since $F(m,\epsilon,T,m_{0})$ is strictly increasing in $T$ (Fact 2), such
$T_{c}^{(1)}$, if any, is unique. To prove the existence of $T_{c}^{(1)}$ by
continuity it is enough to show that there are values of $m\in(\epsilon,1]$
and $T>0$ such that $F(m,\epsilon,T,m_{0})>0$: this is true $\forall
m\in(\epsilon,1)$ as $F(m,\epsilon,T,m_{0})\rightarrow 1-m$ for
$T\uparrow+\infty$ (Fact 3). The conclusions in case (ii) are now obvious
consequences of concavity and $T$-monotonicity of $F$.

Consider, finally, the case $m_{0}<\epsilon<\frac{1+m_{0}}{2}$. We first
investigate the following equation in the unknown $T>0$:

$F(\epsilon,\epsilon,T,m_{0})=0,$ (21)

that can be rewritten as

$1-\epsilon-\frac{2\epsilon^{2}T^{2}+2\epsilon\,T+1}{4\epsilon^{2}T^{2}+4\epsilon\,T+1}(1-m_{0})=0.$

This can be cast as a quadratic equation in $T$: it has real solutions if and
only if $\epsilon<\frac{1+m_{0}}{2}$, and in this case the only positive
solution is $T^{*}(\epsilon,m_{0})$ given in (15). Note that $\epsilon>m_{0}$
implies that $T^{*}(\epsilon,m_{0})>0$. Now, $T$-monotonicity and
$m$-concavity of $F(m,\epsilon,T,m_{0})$ and the fact that

$\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T,m_{0})=-1+2(1-m_{0})\frac{T(\epsilon\,T+1)(4\epsilon^{2}T^{2}+2\epsilon\,T+1)}{2\epsilon\,T+1}$
(22)

is strictly increasing in $T$, show that there are two alternatives:

  1. (a)

if $\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T^{*}(\epsilon,m_{0}),m_{0})\leq 0$, then,

     * •

for $T\leq T^{*}(\epsilon,m_{0})$, the map $m\mapsto F(m,\epsilon,T,m_{0})$ is
less than or equal to $0$ at $m=\epsilon$ and it is decreasing: there are no
solutions to (19) in $(\epsilon,1]$;

     * •

for $T>T^{*}(\epsilon,m_{0})$, the map $m\mapsto F(m,\epsilon,T,m_{0})$ is
strictly positive at $m=\epsilon$, concave and negative at $m=1$, so (19) has
a unique solution in $(\epsilon,1]$;

  2. (b)

if $\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T^{*}(\epsilon,m_{0}),m_{0})>0$, then by continuity
there exists
$T_{c}^{(1)}=T_{c}^{(1)}(\epsilon,m_{0})\in(0,T^{*}(\epsilon,m_{0}))$ such
that the graph of $z=F(m,\epsilon,T_{c},m_{0})$ is tangent to the line $z=0$;
as above, this $T_{c}^{(1)}$ is unique. Thus,

     * •

for $T<T_{c}^{(1)}(\epsilon,m_{0})$, (19) has no solutions in $(\epsilon,1]$;

     * •

for $T=T_{c}^{(1)}(\epsilon,m_{0})$, (19) has a unique solution in
$(\epsilon,1]$;

     * •

for $T_{c}^{(1)}(\epsilon,m_{0})<T<T^{*}(\epsilon,m_{0})$, (19) has two
solutions in $(\epsilon,1]$;

     * •

for $T\geq T^{*}(\epsilon,m_{0})$, (19) has a unique solution in
$(\epsilon,1]$.

Note that $T_{c}^{(1)}$ is defined as in case (ii), and by the Implicit
Function Theorem it is continuous at $\epsilon$, $\forall\epsilon$ in its
domain. To complete the proof of case (iii) we are left to show that there
exists $\epsilon_{*}^{(1)}\in\left(m_{0},\frac{1+m_{0}}{2}\right)$ such that

$\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T^{*}(\epsilon,m_{0}),m_{0})>0\ \ \mbox{ if and only if
}\ \ \epsilon\in\left(\epsilon_{*}^{(1)},\frac{1+m_{0}}{2}\right),$

so that $T_{c}^{(1)}(\epsilon,m_{0})$ is defined for
$\epsilon\in(\epsilon_{*}^{(1)},1]$. Moreover, if this is the case, continuity
implies that

$\lim_{\epsilon\downarrow\epsilon_{*}^{(1)}}T_{c}^{(1)}(\epsilon,m_{0})=T^{*}(\epsilon^{*},m_{0}).$

To complete the proof we are left to show the existence of such
$\epsilon_{*}^{(1)}$. This is established by proving that the map
$\epsilon\mapsto\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T^{*}(\epsilon,m_{0}),m_{0})$ is strictly increasing,
negative at $\epsilon=m_{0}$ and diverging to $+\infty$ at
$\epsilon=\frac{1+m_{0}}{2}$. We use the expressions (22) and (15), and the
change of variable $y:=1+2\epsilon\,T^{*}(\epsilon).$ Note that, by (15),

$\epsilon\,T^{*}(\epsilon)=-\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1-m_{0}}{1+m_{0}-2\epsilon}},$

so $y=\sqrt{\frac{1-m_{0}}{1+m_{0}-2\epsilon}}$. It is easily seen that
$\frac{dy}{d\epsilon}>0$, $y(m_{0})=1$ and
$\lim_{\epsilon\uparrow\frac{1+m_{0}}{2}}y(\epsilon)=+\infty.$
We can write, using (22),

$G(\epsilon,m_{0}):=\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T^{*}(\epsilon,m_{0}),m_{0})=-1+(1-m_{0})\frac{(y^{2}-1)(y^{2}-y+1)}{2\epsilon
y}.$

This shows that $G(m_{0},m_{0})=-1$ and that $G(\epsilon,m_{0})$ diverges to
$+\infty$ as $\epsilon\uparrow\frac{1+m_{0}}{2}$. So the last step is to prove
that $G$ is strictly increasing in $\epsilon$. We have

$\frac{\partial
G}{\partial\epsilon}(\epsilon,m_{0})=\frac{1-m_{0}}{2\epsilon}\frac{d}{dy}\left(\frac{(y^{2}-1)(y^{2}-y+1)}{y}\right)\frac{dy}{d\epsilon}-\frac{1-m_{0}}{2\epsilon^{2}}\frac{(y^{2}-1)(y^{2}-y+1)}{y}.$

Using the facts that $\frac{dy}{d\epsilon}=\frac{y^{3}}{1-m_{0}}\mbox{ and
}\epsilon=\frac{(1+m_{0})y^{2}-(1-m_{0})}{y^{2}},$ it follows that
$\frac{\partial G}{\partial\epsilon}(\epsilon,m_{0})$ has the same sign as

$\frac{\epsilon
y^{3}}{1-m_{0}}\frac{3y^{4}-2y^{3}-y^{2}+1}{y^{2}}-\frac{(y^{2}-1)(y^{2}-y+1)}{y}\\\
=\frac{\frac{1+m_{0}}{1-m_{0}}y^{2}-1}{y}(y^{2}(y-1)(3y+1)+1)-\frac{(y^{2}-1)(y^{2}-y+1)}{y}\\\
\geq\frac{y^{2}-1}{y}(y^{2}(y-1)(3y+1)+1)-\frac{(y^{2}-1)(y^{2}-y+1)}{y}\\\
=(y^{2}-1)(y-1)(3y^{2}+y-1)>0,\ \ \forall\,y>1,$ (23)

where we have used the fact that $m_{0}\geq 0$ and the expression in the
second line of (23) is increasing in $m_{0}$. This completes the proof. ∎

### Proof of Proposition 3.2: polarized incoherent equilibria ($m<-\epsilon$)

The proof repeats some of the arguments seen in the proof of Proposition 3.1.
It is convenient to take advantage of the symmetry relation (14). This implies
that we can equivalently find the equilibria in $(\epsilon,1]$ after replacing
$m_{0}$ by $-m_{0}$. For the regime $\epsilon\geq\frac{1-m_{0}}{2}$ the proof
of part (ii) of Proposition 3.1 applies with no changes, as the assumption
$m_{0}\geq 0$ was not used.
For the case $0<\epsilon<\frac{1-m_{0}}{2}$ we can adapt the proof of part
(iii) of Proposition 3.1, where the assumption $m_{0}\geq 0$ was only used to
prove the existence of $\epsilon_{*}^{(1)}$. Here we obtain the same behavior
seen for $\epsilon_{*}^{(1)}<\epsilon<\frac{1+m_{0}}{2}$ in Proposition 3.1:
to repeat the same argument we need to show that

$\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T^{*}(\epsilon,-m_{0}),-m_{0})>0,\ \ \ \forall\epsilon:\
\ 0<\epsilon<\frac{1-m_{0}}{2}.$

Indeed, as seen in the proof of Proposition 3.1,

$\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T^{*}(\epsilon,-m_{0}),-m_{0})=-1+(1+m_{0})\frac{(y^{2}-1)(y^{2}-y+1)}{2\epsilon
y},$

with $y=\sqrt{\frac{1+m_{0}}{1-m_{0}-2\epsilon}}$. Note that, as $m_{0}\geq
0$, $y>1$ for all $0<\epsilon<\frac{1+m_{0}}{2}$. In particular, $y^{2}-y+1>1$
so that, with a further simple computation we get

$\frac{\partial}{\partial
m}F(\epsilon,\epsilon,T^{*}(\epsilon,-m_{0}),-m_{0})>-1+(1+m_{0})\frac{y^{2}-1}{2\epsilon
y}=-1+\frac{\epsilon+m_{0}}{\epsilon}y>0.$

Thus, the proof for the case $0<\epsilon<\frac{1-m_{0}}{2}$ can be carried out
in the same way as the proof of part (iii) of Proposition 3.1 (case
$\epsilon_{*}^{(1)}<\epsilon<\frac{1+m_{0}}{2}$). ∎

### Proof of Proposition 3.3: unpolarized coherent equilibria ($0\leq
m\leq\epsilon$)

For $m\in[0,\epsilon]$, equation (13) becomes:

$F(m,\epsilon,T,m_{0}):=\frac{2(\epsilon\,T+1)Tm+m_{0}\left[(\epsilon\,T+1)^{2}+m^{2}T^{2}\right]}{\left((\epsilon\,T+1)^{2}-m^{2}T^{2}\right)^{2}}-m=0.$
(24)

We begin by observing that

$F(m,\epsilon,T,m_{0})=\frac{1}{(\epsilon\,T+1)^{2}}\varphi\left(\frac{T}{1+\epsilon\,T}m,m_{0}\right)-m,$
(25)

where

$\varphi(z,m_{0}):=\frac{2z+m_{0}(1+z^{2})}{(1-z^{2})^{2}}.$ (26)

In particular, this allows to prove easily that, if $m_{0}\geq 0$, then
$m\mapsto F(m,\epsilon,T,m_{0})$ is convex.

  * (i)

Note first that $F(0,\epsilon,T,m_{0})=\frac{m_{0}}{(\epsilon\,T+1)^{2}}>0$
(recall that here $m_{0}\geq\epsilon>0$). Moreover,

$F(\epsilon,\epsilon,T,m_{0})=\frac{2(\epsilon\,T+1)\epsilon\,T+m_{0}\left[(\epsilon\,T+1)+\epsilon^{2}T^{2}\right]}{\left((\epsilon\,T+1)^{2}-\epsilon^{2}T^{2}\right)^{2}}-\epsilon=\psi(\epsilon\,T,m_{0})-\epsilon,$

with

$\psi(y):=\frac{2(y+1)y+m_{0}\left[(y+1)^{2}+y^{2}\right]}{(2y+1)^{2}}.$

As $\psi^{\prime}(y)=\frac{2(1-m_{0})}{(2y+1)^{3}},$ we deduce that
$F(\epsilon,\epsilon,T,m_{0})$ is strictly increasing in $T$, except for
$m_{0}=1$ (in this case it is constant). In all cases we have that, since
$F(\epsilon,\epsilon,0,m_{0})=m_{0}-\epsilon\geq 0$,
$F(\epsilon,\epsilon,T,m_{0})>0,$ $\forall T>0$. Thus, the map $m\mapsto
F(m,\epsilon,T,m_{0})$ is strictly positive at the endpoints of the interval
$(0,\epsilon)$. We also have $F(m,\epsilon,0,m_{0})=m_{0}-m>0$ $\forall
m\in(0,\epsilon)$. Moreover, it is easily seen that for each
$m\in(0,\epsilon)$,

$\lim_{T\rightarrow+\infty}F(m,\epsilon,T,m_{0})=-m<0.$ (27)

Then there must be a time $T_{c}^{(2)}(\epsilon,m_{0})$ such that for
$T<T_{c}^{(2)}(\epsilon,m_{0})$, $F(m,\epsilon,T,m_{0})>0$, $\forall
m\in(0,\epsilon)$, and the graph of the convex function
$y=F(m,\epsilon,T_{c}^{(2)}(\epsilon,m_{0}),m_{0})$ is tangent to the line
$y=0$. Set

$\mathcal{T}:=\left\\{T>0:\min_{m\in(0,\epsilon)}F(m,\epsilon,T,m_{0})\leq
0\right\\}.$

The proof of this point (i) is completed as we show that
$\mathcal{T}=[T_{c}^{(2)}(\epsilon,m_{0}),+\infty)$. If this is not the case,
by continuity, there must be a time $\hat{T}\geq T_{c}^{(2)}(\epsilon,m_{0})$,
$\hat{m}\in(0,\epsilon)$ and $\delta>0$ such that

$F(\hat{m},\epsilon,\hat{T},m_{0})=0,\ \
\frac{d}{dm}F(\hat{m},\epsilon,\hat{T},m_{0})=0,$ (28)

but $F(\hat{m},\epsilon,t,m_{0})>0,$ $\forall t\in(\hat{T},\hat{T}+\delta)$.
To show that this is impossible, it is enough to prove that

$\frac{d}{dT}F(\hat{m},\epsilon,\hat{T},m_{0})<0.$ (29)

To see this it is convenient to perform the following change of variables:
$u:=\frac{m}{\epsilon}\in(0,1)$ and $r:=\epsilon\,T$, so that

$F(m,\epsilon,T,m_{0})=G(u,\epsilon,r,m_{0}):=\frac{1}{(1+r)^{2}}\varphi\left(\frac{r}{1+r}u,m_{0}\right)-\epsilon
u,$ (30)

where $\varphi$ is given in (26). Note that (29) is equivalent to

$\frac{d}{dr}G(\hat{u},\epsilon,\hat{r},m_{0})<0,$ (31)

where $\hat{u}:=\frac{\hat{m}}{\epsilon}$ and $\hat{r}:=\epsilon\hat{T}$.

Claim: $\hat{r}>\frac{1}{2}$.

Proof. To see this, note that, being $\epsilon\leq m_{0}$,

$G(u,\epsilon,r,m_{0})\geq
H(u,r,m_{0}):=\frac{1}{(1+r)^{2}}\varphi\left(\frac{r}{1+r}u,m_{0}\right)-m_{0}u.$

The claim follows if we show that, $\forall r\leq\frac{1}{2}$,

$H(u,r,m_{0})>0,\ \ \forall u\in(0,1).$ (32)

Since $H(u,r,m_{0})$ is linear in $m_{0}$, it is enough to prove (32) for
$m_{0}\in\\{0,1\\}$. For $m_{0}=0$ this is obvious, so we show it for
$m_{0}=1$:

$H(u,r,1)=\frac{\left(1+\frac{r}{1+r}u\right)^{2}}{(1+r)^{2}\left(1-\frac{r^{2}u^{2}}{(1+r)^{2}}\right)^{2}}-u=\frac{(1-u)(r^{2}u^{2}-(r^{2}+2r)u+1)}{(1+r(1-u))^{2}},$

which has the same sign as $p(u):=r^{2}u^{2}-(r^{2}+2r)u+1).$ If
$r\leq\frac{1}{2}$ (indeed here $r<2$ would suffice),
$p^{\prime}(u)=2r^{2}u-r^{2}-2r<0$, $\forall u\in(0,1)$, so, $p(u)\geq
p(1)=1-2r\geq 0,$ $\forall u\in(0,1)$ and for $r\leq\frac{1}{2}$. This
completes the proof of the Claim.∎

We are now left with the proof of (31). Note that

$\frac{d}{dr}G(u,\epsilon,r,m_{0})=-\frac{2}{(1+r)^{3}}\varphi\left(\frac{r}{1+r}u,m_{0}\right)+\frac{u}{(1+r)^{4}}\varphi^{\prime}\left(\frac{r}{1+r}u,m_{0}\right),$
(33)

where $\varphi^{\prime}(z,m_{0})=\frac{d}{dz}\varphi(z,m_{0}).$ By (28),

$G(\hat{u},\epsilon,\hat{r},m_{0})=0\ \Rightarrow\
\varphi\left(\frac{\hat{r}}{1+\hat{r}}\hat{u},m_{0}\right)=\epsilon\hat{u},$

and

$\frac{d}{du}G(\hat{u},\epsilon,\hat{r},m_{0})=0\ \Rightarrow\
\frac{r}{(1+r)^{3}}\varphi^{\prime}\left(\frac{\hat{r}}{1+\hat{r}}\hat{u},m_{0}\right)=\epsilon.$

Inserting these identities in (34) we obtain

$\frac{d}{dr}G(\hat{u},\epsilon,\hat{r},m_{0})=\frac{\epsilon\hat{u}}{1+\hat{r}}\left(-2+\frac{1}{\hat{r}}\right)<0,\quad\text{
as }\hat{r}>\frac{1}{2}.$ (34)

  * (ii)

Note that $F(0,T,\epsilon,m_{0})=\frac{m_{0}}{(\epsilon\,T+1)^{2}}>0$.
Moreover,
$\lim_{T\rightarrow+\infty}F(\epsilon,T,\epsilon,m_{0})=\frac{1+m_{0}}{2}-\epsilon\leq
0.$ Since, as shown in point (i), the map $T\mapsto
F(\epsilon,T,\epsilon,m_{0})$ is strictly increasing, then
$F(\epsilon,T,\epsilon,m_{0})<0,$ $\forall T>0$. Thus, $m\mapsto
F(m,T,\epsilon,m_{0})$ has opposite sign at the endpoints of $(0,\epsilon)$:
by convexity there is a unique solution $m=M(T,\epsilon,m_{0})$ to
$F(m,T,\epsilon,m_{0})=0$. The fact that
$\lim_{T\rightarrow+\infty}M(T,\epsilon,m_{0})=0$ follows from the fact that
$\lim_{T\rightarrow+\infty}F(0,T,\epsilon,m_{0})=0$ and that
$\lim_{T\rightarrow+\infty}\frac{d}{dm}F(0,T,\epsilon,m_{0})=-1.$

  * (iii)

In this case $F(0,T,\epsilon,m_{0})=\frac{m_{0}}{(\epsilon\,T+1)^{2}}\geq 0$.
Moreover, $F(\epsilon,T,\epsilon,m_{0})<0\ \iff\ T<T^{*}(\epsilon,m_{0}),$ as
already seen in the proof of Proposition 3.1, point (iii). By convexity of
$m\mapsto F(m,T,\epsilon,m_{0})$, for $T\leq T^{*}(\epsilon,m_{0})$ Equation
(24) has a unique solution in $[0,\epsilon)$. The fact that $m=0$ is a
solution if and only if $m_{0}=0$ is easily verified. Moreover, by (27), for
$T$ sufficiently large $F(m,T,\epsilon,m_{0})$ attains negative values, it is
positive at $m=0$ and $m=\epsilon$, so (24) has two solutions in
$[0,\epsilon)$. We need, however, a sharper analysis for
$T>T^{*}(\epsilon,m_{0})$. Note that, in this case $F(0,T,\epsilon,m_{0})>0$
and $F(\epsilon,T,\epsilon,m_{0})>0$. Thus, by convexity of $m\mapsto
F(m,T,\epsilon,m_{0})$, (24) has zero or two solutions, except for the
“special times” $T$ for which the graph of the map $m\mapsto
F(m,T,\epsilon,m_{0})$ is tangent to the horizontal axis. Note that these
special times are identified as ($T$-component of the) solutions in
$(0,\epsilon)\times(T^{*}(\epsilon,m_{0}),+\infty)$ of the system

$\left\\{\begin{split}F(m,T,\epsilon,m_{0})&=0\\\
\frac{d}{dm}F(m,T,\epsilon,m_{0})&=0.\end{split}\right.$ (35)

Note that (35) has no solutions with $T\leq T^{*}(\epsilon,m_{0})$, so we may
look for solutions of (35) in $(0,\epsilon)\times(0,+\infty)$. The remaining
part of the proof is based on the following Lemma.

######  Lemma A.1.

Let $\epsilon_{*}^{(2)}(m_{0})$ and $\epsilon_{*}^{(3)}(m_{0})$ be as in the
statement of Proposition 3.3. Then,

$m_{0}<\epsilon_{*}^{(2)}(m_{0})<\epsilon_{*}^{(3)}(m_{0})<\frac{1+m_{0}}{2},$

(unless for $m_{0}=0$, where
$0=\epsilon_{*}^{(2)}(0)<\epsilon_{*}^{(3)}(0)<\frac{1}{2}$) such that

    * (a)

for $m_{0}<\epsilon\leq\epsilon_{*}^{(2)}(m_{0})$, (35) has a unique solution
$(\tilde{m}(\epsilon,m_{0}),\tilde{T}(\epsilon,m_{0}))$;

    * (b)

for $\epsilon_{*}^{(2)}(m_{0})<\epsilon<\epsilon_{*}^{(3)}(m_{0})$, (35) has
two solutions $(\hat{m}(\epsilon,m_{0}),T_{c}^{(2)}(\epsilon,m_{0}))$ and
$(\tilde{m}(\epsilon,m_{0}),T_{c}^{(3)}(\epsilon,m_{0}))$ with
$T_{c}^{(2)}(\epsilon,m_{0})<T_{c}^{(3)}(\epsilon,m_{0})$;

    * (c)

for $\epsilon_{*}^{(3)}(m_{0})\leq\epsilon<\frac{1+m_{0}}{2}$, (35) has no
solutions.

The proof of this lemma is postponed after the end of this section. The
desired result of the solutions of (24) readily follows from this Lemma.
Indeed, in case (a), there is a unique special time
$T_{c}^{(2)}(\epsilon,m_{0})$. Since, for large $T$, (24) has two solutions,
necessarily for $T^{*}(\epsilon,m_{0})<T\leq T_{c}^{(2)}(\epsilon,m_{0})$ we
must have $F(m,T,\epsilon,m_{0})>0$, $\forall m\in(0,\epsilon)$, so (24) has
no solution.
In case (b) there are two special times
$T_{c}^{(2)}(\epsilon,m_{0})<T_{c}^{(3)}(\epsilon,m_{0})$: the only
possibility is that, for
$T^{*}(\epsilon,m_{0})<T<T_{c}^{(2)}(\epsilon,m_{0})$, the graph of $m\mapsto
F(m,T,\epsilon,m_{0})$ crosses twice the horizontal axis (two solutions for
(24)), for $T_{c}^{(2)}(\epsilon,m_{0})<T<T_{c}^{(3)}(\epsilon,m_{0})$ it
stays above the horizontal axis (no solutions for (24)) and it crosses again
twice the horizontal axis for $T>T_{c}^{(3)}(\epsilon,m_{0})$. The proof is
therefore completed.∎

### Proof of Lemma A.1

By the change of variables $u:=\frac{m}{\epsilon}\in(0,1)$ and
$r:=\epsilon\,T$ as in (30), we may replace $F(m,\epsilon,T,m_{0})$ by

$G(u,\epsilon,r,m_{0}):=\frac{1}{(1+r+ru)^{2}}\left[m_{0}+\frac{2(1+m_{0})(1+r)ru}{(1+r-ru)^{2}}\right]-\epsilon
u=0,$

$u\in(0,1)$, $r>0$, and (35) is equivalent to

$\left\\{\begin{split}G(u,\epsilon,r,m_{0})&=0\\\
\frac{d}{du}G(u,\epsilon,r,m_{0})&=0.\end{split}\right.$ (36)

Letting, as above,
$U(u,\epsilon,r,m_{0}):=m_{0}(1+r-ru)^{2}+2(1+m_{0})r(1+r)u-\epsilon
u\left[(1+r)^{2}-r^{2}u^{2}\right]^{2}=0,$ we have that
$G(u,\epsilon,r,m_{0})=\frac{1}{\left[(1+r)^{2}-r^{2}u^{2}\right]^{2}}U(u,\epsilon,r,m_{0}).$
It is immediately seen that (36) is equivalent to

$\left\\{\begin{split}U(u,\epsilon,r,m_{0})&=0\\\
\frac{d}{du}U(u,\epsilon,r,m_{0})&=0,\end{split}\right.$ (37)

We use again the identity
$U(u,\epsilon,r,m_{0})=u\frac{d}{du}U(u,\epsilon,r,m_{0})+\left(m_{0}-4\epsilon
r^{2}u^{3}\right)\left((1+r)^{2}-u^{2}r^{2}\right),$ which implies that (37)
is equivalent to

$\left\\{\begin{split}\left(m_{0}-4\epsilon r^{2}u^{3}\right)&=0\\\
\frac{d}{du}U(u,\epsilon,r,m_{0})&=0.\end{split}\right.$ (38)

Summing up, the number of pairs $(m,T)\in(0,\epsilon)\times(0,+\infty)$
solving (35) equals the number of solutions $r$ to the equation

$W(r,\epsilon,m_{0}):=\frac{d}{du}U\left(\left(\frac{m_{0}}{4\epsilon
r^{2}}\right)^{1/3},\epsilon,r,m_{0}\right)=0$ (39)

such that $\frac{m_{0}}{4\epsilon r^{2}}<1$, i.e.,
$r>\sqrt{\frac{m_{0}}{4\epsilon}}$. With the help of a symbolic calculator we
obtain

$8W(r,\epsilon,m_{0})=r\left\\{16+\left[16+3m_{0}\left(\frac{2m_{0}}{\epsilon
r^{2}}\right)^{1/3}\right]r\right\\}+4\epsilon(1+r)^{2}\left\\{-2-4r+\left[-2+3\left(\frac{2m_{0}}{\epsilon
r^{2}}\right)^{1/3}\right]r^{2}\right\\}.$

Now set $V(s,\epsilon,m_{0}):=W(\sqrt{s},\epsilon,m_{0}).$ Thus we are left
with the problem of finding the solutions of

$V(s,\epsilon,m_{0})=0$ (40)

with $s>\frac{m_{0}}{4\epsilon}$. This last change of variable is a trick to
get the following claim.

Claim The map $s\mapsto V(s,\epsilon,m_{0})$ is strictly concave in
$\left(\frac{m_{0}}{4\epsilon},+\infty\right)$.

Proof. Using again a symbolic calculator we get, letting
$k:=\left(\frac{2m_{0}}{\epsilon}\right)^{1/3}$,

$-12s^{3/2}\frac{d^{2}}{ds^{2}}V(s,\epsilon,m_{0})=6+m_{0}ks^{1/6}+\epsilon\left[-12+4k^{2}s^{-1/6}+36s+5k^{2}s^{1/3}+24s^{3/2}-8k^{2}s^{5/6}\right].$

To prove the claim we need to show that this last expression is positive. Note
that $s>\frac{m_{0}}{4\epsilon}=\left(\frac{k}{2}\right)^{3}$. Moreover, as
$m_{0}<\epsilon<\frac{1+m_{0}}{2}$, then $\frac{4m_{0}}{1+m_{0}}<k^{3}<2.$
Again from $\epsilon<\frac{1+m_{0}}{2}$, the inequality
$\frac{d^{2}}{ds^{2}}V(s,\epsilon,m_{0})<0$ follows if we show that

$6+m_{0}ks^{1/6}+\frac{1+m_{0}}{2}\left[-12+4k^{2}s^{-1/6}+36s+5k^{2}s^{1/3}+24s^{3/2}-8k^{2}s^{5/6}\right]>0.$
(41)

Using $s>\left(\frac{k}{2}\right)^{3}$ and $k^{3}>\frac{4m_{0}}{1+m_{0}}$ we
have that $\frac{1+m_{0}}{2}5k^{2}s^{1/3}>\frac{1+m_{0}}{4}5k^{3}>5m_{0},$ so
the l.h.s. of (41) is bounded from below by

$m_{0}ks^{1/6}-m_{0}+(1+m_{0})\left[2k^{2}s^{-1/6}+18s+12s^{3/2}-4k^{2}s^{5/6}\right].$
(42)

Now, for $s\leq\frac{1}{4}$ the expression in (42) is bounded from below by

$\begin{split}-m_{0}+(1+m_{0})\left[2k^{2}s^{-1/6}-4k^{2}s^{5/6}\right]&=-m_{0}+\frac{1+m_{0}}{s^{1/6}}\left(2k^{2}-4k^{2}s\right)\\\
&\geq-m_{0}+(1+m_{0})k^{2}s^{-1/6}\\\ &\geq-
m_{0}+(1+m_{0})\left(\frac{4m_{0}}{1+m_{0}}\right)^{2/3}\\\ &\geq-
m_{0}+(1+m_{0})4^{2/3}\frac{m_{0}}{1+m_{0}}>0.\end{split}$

For $s>\frac{1}{4}$, the expression in (42) is bounded from below by

$\begin{split}-m_{0}+18s+12s^{3/2}-4k^{2}s^{5/6}&\geq-
m_{0}+s^{5/6}\left[18\left(\frac{1}{4}\right)^{1/6}+12\left(\frac{1}{4}\right)^{2/3}-4k^{2}\right]\\\
&\geq-
m_{0}+s^{5/6}\left[\left(\frac{1}{4}\right)^{1/6}+12\left(\frac{1}{4}\right)^{2/3}-4\,2^{2/3}\right]\\\
&\geq-m_{0}+8s^{5/6}\geq-m_{0}+2>0.\end{split}$

This proves (41) and therefore the claim. ∎

Now, using concavity of $V(s,\epsilon,m_{0})$ and the fact that

$\lim_{s\rightarrow+\infty}V(s,\epsilon,m_{0})=\lim_{r\rightarrow+\infty}W(r,\epsilon,m_{0})=-\infty,$

we have that (40) has a unique solution whenever
$V\left(\frac{m_{0}}{4\epsilon},\epsilon,m_{0}\right)>0.$ We get

$V\left(\frac{m_{0}}{4\epsilon},\epsilon,m_{0}\right)=\frac{m_{0}(1+m_{0})}{2\epsilon}+(1+m_{0})\sqrt{\frac{m_{0}}{\epsilon}}-\epsilon\left(1+e\sqrt{\frac{m_{0}}{\epsilon}}\right),$

so

$V\left(\frac{m_{0}}{4\epsilon},\epsilon,m_{0}\right)\Big{|}_{\epsilon=m_{0}}=\frac{3}{2}(1-m_{0})>0,$
$V\left(\frac{m_{0}}{4\epsilon},\epsilon,m_{0}\right)\Big{|}_{\epsilon=\frac{(1+m_{0})}{2}}=-\frac{1}{2}(1-m_{0})<0$
(43)

and

$\frac{d}{d\epsilon}V\left(\frac{m_{0}}{4\epsilon},\epsilon,m_{0}\right)=-1-\sqrt{\frac{m_{0}}{\epsilon}}-\frac{m_{0}(1+m_{0})}{2\epsilon^{2}}-\frac{\sqrt{\frac{m_{0}}{\epsilon}}(1+m_{0})}{2\epsilon}<0.$

Therefore, there is a unique $\epsilon_{*}^{(2)}(m_{0})$, with
$m_{0}\\!<\\!\epsilon_{*}^{(2)}(m_{0})\\!<\\!\frac{1+m_{0}}{2}$, such that
$V\\!\left(\frac{m_{0}}{4\epsilon_{*}^{(2)}},\epsilon_{*}^{(2)},m_{0}\right)\\!\\!=\\!0$.
Moreover, for $m_{0}<\epsilon<\epsilon_{*}^{(2)}(m_{0})$, (40) has a unique
solution and, since

$\frac{d}{ds}V(s,\epsilon,m_{0})\Big{|}_{s=\frac{m_{0}}{4\epsilon}}=\frac{2(1+m_{0}-2\epsilon)\left(1+\sqrt{\frac{m_{0}}{\alpha}}\right)}{\sqrt{\frac{m_{0}}{\alpha}}}>0,$

it follows that (40) has two solutions as $\epsilon$ crosses
$\epsilon_{*}^{(2)}(m_{0})$. This actually occurs until $\epsilon$ reaches
$\epsilon_{*}^{(3)}(m_{0})$, where $\epsilon_{*}^{(3)}(m_{0})$ is
characterized by the fact that the graph of $s\mapsto
V(s,\epsilon_{*}^{(3)}(m_{0}),m_{0})$ is tangent to the horizontal axis. This
is a consequence of the following monotonicity property:

$\frac{d}{d\epsilon}V(s,\epsilon,m_{0})<0,\quad\forall
s\geq\frac{m_{0}}{4\epsilon}.$ (44)

This suffices to characterize $\epsilon_{*}^{(3)}\\!(m_{0})$; to complete the
proof we need to show that
$\epsilon_{*}^{(3)}\\!(m_{0})\\!<\\!\frac{1+m_{0}}{2}$ which is equivalent to

$V\left(s,\frac{1+m_{0}}{2},m_{0}\right)<0,\quad\forall
s\geq\frac{m_{0}}{2(1+m_{0})}.$ (45)

We are therefore left to prove (44) and (45). We begin with (44).

$\frac{d}{d\epsilon}V(s,\epsilon,m_{0})=-\frac{1}{8\epsilon}\left[m_{0}\left(\frac{2m_{0}}{\epsilon}\right)^{1/3}s^{2/3}+4\epsilon(1+\sqrt{s})^{2}\left(2+4\sqrt{s}+4s-\left(\frac{2m_{0}}{\epsilon}\right)^{2/3}s^{1/3}\right)\right].$

To show that this expression is negative we just observe that, being
$\frac{2m_{0}}{\epsilon}\leq 2$,

$2+4\sqrt{s}+4s-\left(\frac{2m_{0}}{\epsilon}\right)^{2/3}s^{1/3}>2+4\sqrt{s}+4s-2s^{1/3}>\left\\{\begin{array}[]{ll}2-2s^{1/3}\geq
0&\mbox{for }s\leq 1\\\ 4s-2s^{1/3}>0&\mbox{for }s>1.\end{array}\right.$

This establishes (44). Now we show (45). We recall that $s\mapsto
V\left(s,\frac{1+m_{0}}{2},m_{0}\right)$ is strictly concave for
$s\geq\frac{m_{0}}{2(1+m_{0})}$. Moreover, again with the help of a symbolic
calculator

$\frac{d}{ds}V\left(s,\frac{1+m_{0}}{2},m_{0}\right)\Big{|}_{s=\frac{m_{0}}{2(1+m_{0})}}=0.$

So it is enough to show that

$V\left(\frac{m_{0}}{2(1+m_{0})},\frac{1+m_{0}}{2},m_{0}\right)<0,$

that has been seen already in (43). The proof is now complete.∎

### Proof of Proposition 3.4: unpolarized incoherent equilibria
($-\epsilon\leq m<0$)

As done in Proposition 3.2, we use the symmetry (14). So we look for solutions
$m\in(0,\epsilon]$ of the equation $F(m,\epsilon,T,-m_{0})=0;$ this allows to
reuse some of the ideas in Proposition 3.3.

  * (i)

We employ here the change of variables seen in (30), so

$\begin{split}F(m,\epsilon,T,-m_{0})&=G(u,\epsilon,r,-m_{0}):=\frac{1}{(1+r)^{2}}\varphi\left(\frac{r}{1+r}u,-m_{0}\right)-\epsilon
u\\\ &=\frac{2(1+r)ru-
m_{0}\left[(1+r)^{2}+r^{2}u^{2}\right]}{\left((1+r)^{2}-r^{2}u^{2}\right)^{2}}-\epsilon
u\\\
&=\frac{1}{(1+r+ru)^{2}}\left[-m_{0}+\frac{2(1-m_{0})(1+r)ru}{(1+r-ru)^{2}}\right]-\epsilon
u.\end{split}$

We need to show that this last expression is strictly negative $\forall
u\in(0,1)$; it is enough to show this for $\epsilon=\frac{1-m_{0}}{2}$. This
amounts to prove that

$-m_{0}+\frac{2(1-m_{0})(1+r)ru}{(1+r-ru)^{2}}<\frac{1-m_{0}}{2}u(1+r+ru)^{2},$

which follows from

$\frac{2(1-m_{0})(1+r)ru}{(1+r-ru)^{2}}<\frac{1-m_{0}}{2}u(1+r+ru)^{2}\ \iff\
4r(1+r)<\left[(1+r)^{2}-r^{2}u^{2}\right]^{2},$

$\forall u\in(0,1)$. This last inequality follows if we show it holds for
$u=1$, i.e., $4r(1+r)<(1+2r)^{2},$ that is clearly true $\forall r>0$.

  * (ii)

As seen in the proof of point (i) of Proposition 3.3,
$F(\epsilon,\epsilon,T,-m_{0})$ is strictly increasing in $T$. Moreover, the
same computation done after (21) shows that $F(\epsilon,\epsilon,T,-m_{0})>0$
if and only if $T>T^{*}(\epsilon,-m_{0})$. By the same change of variables
used in point (i), (24) is equivalent to the equation

$\frac{1}{(1+r+ru)^{2}}\left[-m_{0}+\frac{2(1-m_{0})(1+r)ru}{(1+r-ru)^{2}}\right]-\epsilon
u=0,$

with $u\in(0,1)$, which is also equivalent to

$U(u,\epsilon,r,m_{0}):=-m_{0}(1+r-ru)^{2}+2(1-m_{0})r(1+r)u-\epsilon
u\left[(1+r)^{2}-r^{2}u^{2}\right]^{2}=0.$

Our proof is based on the following claim.

Claim. Let $u^{*}\in(0,1)$ be such that $U(u^{*},\epsilon,r,m_{0})=0$. Then,
$\frac{d}{du}U(u^{*},\epsilon,r,m_{0})>0.$

Proof. We use the identity

$U(u,\epsilon,r,m_{0})=u\frac{d}{du}U(u,\epsilon,r,m_{0})+\left(-m_{0}-4\epsilon
r^{2}u^{3}\right)\left((1+r)^{2}-u^{2}r^{2}\right)<u\frac{d}{du}U(u,\epsilon,r,m_{0})$

which proves the claim. ∎

This clearly implies that such $u^{*}$ exists if and only if
$U(1,\epsilon,r,m_{0})>0$ (as $U(0,\epsilon,r,m_{0})=-m_{0}(1+r)^{2}\leq 0$),
i.e., if and only if $r>\epsilon\,T^{*}(\epsilon,-m_{0})$ and in this case it
is unique. This completes the proof.∎

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