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1

Antiferromagnetic Potts model

on the Erdos-Renyi random graph∗

Pierluigi Contucci† Sander Dommers‡ Cristian Giardina§ Shannon Starr¶

July 25, 2011

Abstract

We study the antiferromagnetic Potts model on the Erdos-Renyi random graph. By identifying

a suitable interpolation structure and proving an extended variational principle we show that the

replica symmetric solution is an upper bound for the limiting pressure which can be recovered in

the framework of Derrida-Ruelle probability cascades. A comparison theorem with a mixed model

made of a mean field Potts-antiferromagnet plus a Potts-Sherrington-Kirkpatrick model allows to

show that the replica symmetric solution is exact at high temperatures.

Keywords: Mean field, dilute antiferromagnet, q-state Potts model, interpolation, extended vari-

ational principle, spin glass, replica symmetry breaking.

MCS numbers: Primary 60B10, 60G57, 82B20; Secondary 60K35.

1 Introduction and main results

In this paper we give some rigorous results on the antiferromagnetic Potts model on the Erdos-Renyirandom graph.

It is well known that antiferromagnetic Potts models on graphs are related, at zero temperature,to the graph coloring problem which consists in placing colors on the graph vertices in such a waythat two of them connected by an edge have different color. In recent time an algorithmic approachwas developed to study their colorability [7] based on methods and ideas from disordered system, inparticular on the replica symmetry breaking scheme introduced within the mean field theory of spinglasses [21].

Since the Erdos-Renyi random graph has a locally tree-like structure and large loops, the statisticalmechanics model with antiferromagnetic interactions has been reported to display some spin glassbehavior in the physics literature [25]. In particular it has been argued that the one-step replicasymmetry breaking solution does not get improved by a higher number of steps [28].

The rigorous theory of spin glasses has, on the other hand, made important progresses in the lastdecade by means of the Guerra-Toninelli [16] interpolation scheme, their thermodynamic limit control,the Guerra-Talagrand [23] theorem for the free energy of the Sherrington-Kirkpatrick model and withthe general scheme introduced by Aizenman-Sims-Starr [1] which includes an extended variationalprinciple to obtain the exact solution for a large class of models.

This paper is a first attempt to derive within mathematical rigor some properties of the antiferro-magnetic Potts model on the Erdos-Renyi random graph. A full treatment of the ferromagnetic Ising

∗pierluigi.contucci@unibo.it, s.dommers@tue.nl, cristian.giardina@unimore.it, sstarr@math.rochester.edu†Universita di Bologna, Piazza di Porta S.Donato 5, 40127 Bologna, Italy‡Eindhoven University of Technology, Department of Mathematics and Computer Science, P.O. Box 513, 5600 MB

Eindhoven, The Netherlands§Universita di Modena e Reggio E., viale A. Allegri, 9 - 42121 Reggio Emilia , Italy¶University of Rochester, Department of Mathematics, Rochester, NY, 14627, USA

1

case has been given in [10] for locally tree-like random graphs and extended in [11]. The techniquesused there are heavily based on the use of ferromagnetic Griffiths-Kelly-Sherman and Griffiths-Hurst-Sherman inequalities and do not apply to our case. The extension to Potts random variables is anopen problem.

For the antiferromagnetic model we introduce here an interpolation scheme and prove its mono-tonicity. It had been previously derived [5] a new discrete version of the Guerra-Toninelli interpolationargument to prove existence of the thermodynamic limit for this model for all T > 0, as well as directlyfor T = 0 . This gave a positive solution of a conjecture due to Aldous [3].

We consider the actual value of the pressure. We show that the continuous interpolation method ofGuerra-Toninelli [16] applies in a suitably extended form (see [12] and [18]). We then prove an extendedvariational principle for the considered model and we use it to obtain a rigorous control of free energybounds. In particular we show that an upper bound is obtained by restricting the optimization schemeto hierarchical structures (Ruelle probability cascades) which recovers the replica symmetric solutionheuristically introduced by the physicists. We moreover show that such an upper bound is exact forhigh enough temperatures. Our analysis applies to an arbitrary number q ∈ N of values for the spinvariables.

The paper is organized in three parts. In the first part, dealing with existence and simple bounds,the antiferromagnetic q-states Potts model on the Erdos-Renyi random graph is defined (Section 2)and the thermodynamic limit of its pressure is proved to exist (Section 3). By an estimate using theCauchy-Schwarz inequality, it is also shown in Section 4 that the pressure admits an elementary upperbound.

The part on existence and simple bounds is then followed by the part on the extended variationalprinciple. The cavity functional is constructed in Section 5 and it is shown to lead to a variationalexpression for the pressure. A class of optimizers, given by the Derrida-Ruelle random probabilitycascades, is discussed in Section 6. In Section 7 it is shown how this class leads to an upper boundfor the pressure which coincides with the replica symmetric solution (in particular the trivial replicasymmetric solution, i.e., the one involving a uniform overlap probability measure, reproduce the simpleestimate of Section 4). This part on the extended variational principle is comparable to the analogousresults for the Sherrington-Kirkpatrick model ([14, 1]). Note that the SK model is defined on thecomplete graph and the spin glass behavor has its origin in the random couplings of random signs.Here frustration is produced by deterministic anti-ferromagnetic couplings, while the randomness isgiven by the underlying spatial structure of the Erdos-Renyi random graph.

The last part deals with the full control of the high temperature region. In Section 8 it is shownthat for temperatures high enough the quenched pressure admits the trivial replica symmetric expres-sion as a lower bound. This result, combined with the result of Section 4, proves that in the hightemperature region the pressure is given by the trivial replica symmetric formula. Since the replicasymmetric solution is known to have a negative entropy at low temperature this proves the existenceof a temperature below which the solution is not valid anymore. We discuss these implications andother points in Section 9, wherein we summarize our results and present some further questions forconsideration. Interestingly, the lower bound in the high temperature region leads to the study of amodel defined on the complete graph with a Hamiltonian which is the one of the SK model plus theantiferromagnetic Potts-Curie-Weiss model. This last model is studied in Appendix A by a suitableextension of the techniques developed in [15].

2 The model

We study the antiferromagnetic q-states Potts model on the Erdos-Renyi random graph. We considera set of N vertices, each vertex has attached a spin variables which can take q ∈ N values: σi ∈1, 2, . . . , q. The edges of the graph are constructed using a set of N2 independent and identical

2

random variables Ji,j (with i, j = 1, . . . , N) having a Poisson distribution with

E[Ji,j ] =c

2N. (1)

In the following we will denote by E the expectation with respect to the randomness of the graph.Because of the double counting (we allow Ji,j to differ from Jj,i) we obtain an average vertex degreec. Strictly speaking, in the construction of the Erdos-Renyi random graph we should use randomvariables distributed as Binomial with parameter c/(2N). However the two choices are equivalent inthe thermodynamic limit N → ∞ and we find it more convenient to work with the Poisson setting.

The Hamiltonian of the model is given by

HN(σ) =

N∑

i,j=1

Ji,jδ(σi, σj) (2)

where δ(x, y) denotes the Kronecker delta function. Note that since the Ji,j are non-negative, themodel is anti-ferromagnetic. For a given inverse temperature β, we consider the random partitionfunction

ZN (c, β) =∑

σ∈1,...,qN

e−βHN (σ) , (3)

the expectation of a spin function f : 1, . . . , qN 7→ R with respect to the random Boltzmann-Gibbsstate

ω(f(σ)) =1

ZN(c, β)

∑

σ∈1,...,qN

f(σ)e−βHN (σ) (4)

and the quenched expectation〈f(σ)〉 = E[ω(f(σ))] . (5)

As usual in disordered systems, it will be convenient to define the quenched state for a function ofthe spins of an arbitrary number n ∈ N of real copies. This is obtained by considering the productBoltmann-Gibbs state and then averaging over the disorder. Namely, for a function f of n spinconfigurations σ(1), . . . , σ(n) we define

Ω(f(σ(1), . . . , σ(n))) =1

ZnN(β)

∑

σ(1),...,σ(n)

f(σ(1), . . . , σ(n))e−β(HN (σ(1))+...+HN (σ(n))) (6)

and then〈f(σ(1), . . . , σ(n))〉 = E[Ω(f(σ(1), . . . , σ(n)))] . (7)

The main thermodynamic quantity we will study is the quenched pressure per particle

pN(c, β) =1

NE[lnZN (c, β)] . (8)

and its thermodynamic limitp(c, β) = lim

N→∞pN (c, β) . (9)

An important observable that will appear later is the sequence (for n ≥ 1) of arrays qN (r1, r2, . . . , rn)with (r1, r2, . . . , rn) ∈ 1, . . . qn, which represents the generalized multi-overlap between n spin con-figurations σ(1), . . . , σ(n), and is defined as

qN (r1, r2, . . . , rn) =1

N

N∑

i=1

δ(σ(1)i , r1)δ(σ

(2)i , r2) · · · δ(σ

(n)i , rn) . (10)

3

3 Thermodynamic limit of the pressure

Theorem 3.1 The pressure per particle defined in (8) is a superadditive sequence, i.e. for all N,N1, N2

with N1 + N2 = NNpN(β) ≥ N1pN1(β) + N2pN2(β) . (11)

Therefore the infinite volume limit of the pressure per particle defined in (9) exists and is equal to

p(β) = supN

pN (β) . (12)

Remark 3.2 The fact that pN (c, β) has a finite upper bound, uniform in N , can be trivially provedby the Jensen inequality N−1

E(lnZN ) ≤ N−1 lnE(ZN ), where the right hand side is the pressure ofthe antiferromagnetic Potts-Curie-Weiss model. (See, for example, (82).)

Proof: The proof is obtained by interpolation. For a partition of the systems of size N into twosubsystems of sizes N1 and N2 and for a t ∈ [0, 1] we consider the following independent Poissonrandom variables:

J′

i,j ∼ Poisson

(ct

2N

)

J′′

i,j ∼ Poisson

(c(1 − t)

2N1

)(13)

J′′′

i,j ∼ Poisson

(c(1 − t)

2N2

).

We define the interpolating Hamiltonian

HN (σ, t) =

N∑

i,j=1

J′

i,jδ(σi, σj) +

N1∑

i,j=1

J′′

i,jδ(σi, σj) +

N∑

i,j=N1+1

J′′′

i,jδ(σi, σj), (14)

which induces an interpolating random partition function

ZN (c, β, t) =∑

σ∈1,...,qN

e−βHN (σ,t) , (15)

the expectation with respect to an interpolating random Boltzmann-Gibbs

ωt(f(σ)) =1

ZN(c, β, t)

∑

σ∈1,...,qN

f(σ)e−βHN (σ,t) , (16)

and an interpolating quenched pressure

pN (c, β, t) =1

NE [ZN (c, β, t)] . (17)

Since pN (β, 1) = pN (β) and pN (β, 0) = pN1(β) + pN2(β) the first statement of the theorem (Eq. (11))follows from the fundamental theorem of calculus if one can show that the interpolating pressure ismonotonically non-decreasing in t. The derivative of the interpolating pressure reads

dpN (c, β, t)

dt=

c

2E

1

N2

N∑

i,j=1

lnωt(e−βδ(σi,σj))

− 1

NN1

N1∑

i,j=1

lnωt(e−βδ(σi,σj)) − 1

NN2

N∑

i,j=N1+1

lnωt(e−βδ(σi,σj))

, (18)

4

where the following identity has been used: for a vector X = (X1, . . . , Xm) of m independent Poissonrandom variables Xi with parameter λi(t) and a function f : Nm → R

d

dtE[f(X)] = E

[m∑

i=1

dλi(t)

dt(f(X1, . . . , Xi + 1, . . . , Xm) − f(X1, . . . , Xi, . . . , Xm))

]. (19)

Expression (18) can be further simplified using the identity

e−βδ(σi,σj) = 1 − (1 − e−β)δ(σi, σj), (20)

and the Taylor expansion

ln(1 − x) = −∞∑

n=1

xn

n, ∀|x| < 1 . (21)

One obtains

dpN (c, β, t)

dt= − c

2E

∞∑

n=1

(1 − e−β)n

n

1

N2

N∑

i,j=1

ωnt (δ(σi, σj))

− 1

NN1

N1∑

i,j=1

ωnt (δ(σi, σj)) −

1

NN2

N∑

i,j=N1+1

ωnt (δ(σi, σj))

, (22)

which can be rewritten, using the definition of the sequence of generalized multi-overlap arrays in (10),as

dpN (c, β, t)

dt= − c

2E

[∞∑

n=1

(1 − e−β)n

n(23)

q∑

r1,...,rn=1

Ωt

(q2N (r1, . . . , rn) − N1

Nq2N1

(r1, . . . , rn) − N2

Nq2N2

(r1, . . . , rn)

)].

From this expression one sees by inspection that the derivative is non-negative, employing a standardconvexity argument as in [16]. Therefore monotonicity of the interpolating pressure is established andsuper-additivity (Eq.(11)) follows. The second claim of the theorem, i.e., the existence of the thermo-dynamic limit of the pressure and its realization as a supremum (Eq.(12)) is a standard consequenceof super-additivity (see [16]).

Remark 3.3 The same computation goes through also for the ferromagnetic model with the changeβ 7→ −β. However 1 − eβ < 0 for β > 0 and therefore the series in (23) has alternating signs andmonotonicity can not be derived anymore by inspection. We believe however that the interpolationis monotone also in the ferromagnetic case, though in the opposite direction. This belief is based ontwo facts. Firstly, pressure sub-additivity for the ferro-magnetic model on the Erdos-Renyi randomgraph has been checked numerically for small system sizes [2]. This is in agreement with a monotonicbehavior of the interpolating pressure. Secondly, and more importantly, the numerical checks ([2])for the Ising case (q = 2) show that for 0 ≤ t ≤ 1 the series in (23) is dominated by the first termand therefore one would be left with the same interpolating pressure of the Curie-Weiss model whichis know to be sub-additive. This is indeed rigorously shown at zero temperature in [9]. Although theferromagnetic model has been fully solved in [10], it would be interesting to extend the monotonicityresult to all temperature.

5

4 “Trivial” pressure bound

As a preliminary result, we show that the quenched pressure is bounded from above by the so-called“trivial replica symmetric pressure”, which improves the Jensen bound of the previous section.

Lemma 4.1 For all β ≥ 0 and N ≥ 1,

pN (c, β) ≤ pTRS(c, β) (24)

with

pTRS(c, β) =c

2ln

(1 − 1 − e−β

q

)+ ln q. (25)

Proof: Using (19), we have that

d

dcpN (c, β) =

1

2N2E

N∑

i,j=1

lnω(e−βδ(σi,σj)). (26)

By Jensen’s inequality and the concavity of the logarithm, this is bounded from above by

1

2E ln

1

N2

N∑

i,j=1

ω(e−βδ(σi,σj))

≤ 1

2lnE

1 − (1 − e−β)ω

(1

N2

N∑

i,j=1

δ(σi, σj)

)

, (27)

where also (20) was used. Observe that

1

N2

N∑

i,j=1

δ(σi, σj) =

q∑

r1=1

(1

N

N∑

i=1

δ(σi, r1)

)2

=

q∑

r1=1

q2N (r1) . (28)

Since all the qN (r1) are nonnegative and add up to 1, we can use Holder’s inequality to get

1 =

q∑

r1=1

qN (r1) ≤

√√√√q∑

r1=1

q2N (r1) · √q, (29)

and thus

1

N2

N∑

i,j=1

δ(σi, σj) ≥1

q. (30)

Hence,d

dcpN (c, β) ≤ 1

2ln

(1 − 1 − e−β

q

). (31)

The bound in the lemma is then obtained by using the fundamental theorem of calculus and observingthat

pN (0, β) = ln q. (32)

5 Extended Variational Principle

We begin with a proposition that is useful in the ultimate definition of the extended variationalprinciple.

6

Proposition 5.1 Suppose that N,M ∈ N are chosen and let µM be any measure on 1, . . . , qM .Then

pN (c, β) ≤ GN,M (c, β, µM ) , (33)

whereGN,M (c, β, µM ) = G

(1)N,M (c, β, µM ) −G

(2)N,M (c, β, µM )

with

G(1)N,M (c, β, µM ) =

1

NE

ln∑

τ∈1,...,qM

µM (τ)∑

σ∈1,...,qN

exp

−β

N∑

i=1

M∑

j=1

Kijδ(σi, τj)

, (34)

where the Kij’s are i.i.d. Poisson random variables with parameter c/M , and

G(2)N,M (c, β, µM ) =

1

NE

ln

∑

τ∈1,...,qM

µM (τ) exp

−β

M∑

i,j=1

Lijδ(τi, τj)

, (35)

where the Lij’s are i.i.d. Poisson random variables with parameters cN/(2M2).

Proof: For a t ∈ [0, 1] we consider the following independent Poisson random variables:

(Jij)Ni,j=1 , (Kij : i = 1, . . . , N ; j = 1, . . . ,M) , (Lij)

Mi,j=1 , (36)

such that

E[Jij ] =(1 − t)c

2N, E[Kij ] =

ct

M, E[Lij ] =

(1 − t)cN

2M2, (37)

for all appropriate indices i, j. We define

HN,M(σ, τ, t) =

N∑

i,j=1

Jijδ(σi, σj) +

N∑

i=1

M∑

j=1

Kijδ(σi, τj) +

M∑

i,j=1

Lijδ(τi, τj) , (38)

and we also define

ZN,M (c, β, µM , t) =∑

τ∈1,...,qM

µM (τ)∑

σ∈1,...,qN

e−βHN,M(σ,τ,t) , (39)

and

ωt(f(σ, τ)) =1

ZN,M (c, β, µM , t)

∑

τ∈1,...,qM

µM (τ)∑

σ∈1,...,qN

f(σ, τ)e−βHN,M (σ,τ,t). (40)

Then1

NE[lnZN,M(c, β, µM , 0)] = pN (c, β) + G

(2)N,M (c, β, µM ) , (41)

since if t = 0 then HN,M (σ, τ, 0) splits into a summand only depending on σ and one only dependingon τ . Furthermore,

1

NE[lnZN,M (c, β, µM , 1)] = G

(1)N,M (c, β, µM ) . (42)

Moreover, as in the proof of Theorem 3.1, one can show that

d

dt

( 1

NE[lnZN,M (c, β, µM , t)]

)

= − c

2E

1

N2

N∑

i,j=1

lnωt(e−βδ(σi,σj)) − 2

NM

N∑

i=1

M∑

j=1

lnωt(e−βδ(σi,τj)) +

1

M2

M∑

i,j=1

lnωt(e−βδ(τi,τj))

=c

2E

∞∑

n=1

(1 − e−β)n

n

1

N2

N∑

i,j=1

ωnt (δ(σi, σj)) −

2

NM

N∑

i=1

M∑

j=1

ωnt (δ(σi, τj)) +

1

M2

M∑

i,j=1

ωnt (δ(τi, τj))

.

(43)

7

This can again be rewritten in terms of generalized multi-overlaps as

c

2E

[∞∑

n=1

(1 − e−β)n

n

q∑

r1,...,rn=1

Ωt

(q2N (r1, . . . , rn) − 2qN (r1, . . . , rn)qM (r1, . . . , rn) + q2M (r1, . . . , rn)

)]

=c

2E

[∞∑

n=1

(1 − e−β)n

n

q∑

r1,...,rn=1

Ωt

((qN (r1, . . . , rn) − qM (r1, . . . , rn)

)2)], (44)

which is obviously nonnegative for all t. This proves that

pN (c, β) + G(2)N,M (c, β, µM ) =

1

NE[lnZN,M (c, β, µM , 0)]

≤ 1

NE[lnZN,M (c, β, µM , 1)] = G

(1)N,M (c, β, µM ) . (45)

In other words,

pN(c, β) ≤ G(1)N,M (c, β, µM ) −G

(2)N,M (c, β, µM ) . (46)

In order to accommodate a limit where M approaches ∞, we mention an equivalent version of thefunction GN,M (c, β, µM ):

Lemma 5.2 Let I(1), I(2), . . . be i.i.d. random variables uniformly chosen from 1, . . . , N, and letJ(1), J(2), . . . be i.i.d. random variables uniformly chosen from 1, . . . ,M. Independently of this,let K be a Poisson random variable with parameter cN and let L be a Poisson random variable withparameter cN/2. Then

G(1)N,M(c, β, µM ) =

1

NE

ln

∑

τ∈1,...,qM

µM (τ)∑

σ∈1,...,qN

exp

(−β

K∑

k=1

δ(σI(k), τJ(k))

) ,

and

G(2)N,M (c, β, µM ) =

1

NE

ln

∑

τ∈1,...,qM

µM (τ) exp

(−β

L∑

k=1

δ(τJ(2k−1), τJ(2k))

) .

Proof: This follows from a property of Poisson random variables known as Poisson thinning (orsometimes called Bernoulli thinning). Previously we had i.i.d., Poisson random variables Kij and Lij .Now we have just two Poisson random variables K and L, but we have i.i.d., uniform random variablesI(k), J(k). The Poisson thinning property refers to the fact that the families

Kij = #k ≤ K : (I(k), J(k)) = (i, j) and Lij = #k ≤ L : (J(2k − 1), J(2k)) = (i, j) ,

are distributed identically to Kij and Lij . The reader may also check this as an exercise.

Corollary 5.3 Suppose that µM is a random measure on 1, . . . , qM . Then

pN (c, β) ≤ E [GN,M(c, β, µM )] ,

where the symbol E denotes the expectation with respect to µM .

Proof: For each random realization of µM , we have

pN (c, β) ≤ GN,M (c, β, µM ) ,

8

almost surely, according to Proposition 5.1. So the corollary follows by elementary properties of theexpectation.

The following Proposition may be viewed as the M → ∞ limit of Proposition 5.1.

Proposition 5.4 Let (µα)∞α=1 be a random sequence such that each µα is positive, and the seriesconverges, almost surely. Also let (τα,k : α ∈ N , k ∈ N) be a random array of elements of 1, . . . , q.We write L for the measure which describes the joint distribution of

((µα)α∈N , (τα,k : α ∈ N , k ∈ N)) .

We say that L is “exchangeable” if

((µα)α∈N , (τα,k : α ∈ N , k ∈ N))D= ((µα)α∈N , (τα,π(k) : α ∈ N , k ∈ N)) ,

for every non-random permutation π of N which moves only finitely many k’s. HereD= indicates

equality in distribution.For an exchangeable L, we define

GN (c, β,L) = G(1)N (c, β,L) −G

(2)N (c, β,L) , (47)

where

G(1)N =

1

NE

ln

∞∑

α=1

µα

∑

σ∈1,...,qN

exp

(−β

K∑

k=1

δ(σI(k), τα,k)

) , (48)

and

G(2)N =

1

NE

[ln

∞∑

α=1

µα exp

(−β

L∑

k=1

δ(τα,2k−1, τα,2k)

)], (49)

where E is the expectation over L as well as the random variables (I(k))∞k=1, K and L, where weassume that (I(k)),K, L are all independent of one another and of ((µα), (τα,k)), and I(1), I(2), . . .are i.i.d, uniform on 1, . . . , N, K is Poisson with mean cN and L which is Poisson with mean cN/2.Then we have

pN (c, β) ≤ GN (c, β,L) ,

for every exchangeable L.

Remark 5.5 The condition of exchangeability is not very restrictive. Given any non-exchangeableL, we may obtain an exchangeable law by a standard symmetrization procedure. The advantage ofassuming that L is symmetric is that it allows us to simplify the expression of GN . Otherwise it wouldbe more complicated.

Proof: For any fixed M , consider a random realization of the sequence (µα)α∈N and (τα,k : α ∈N , k ∈ 1, . . . ,M). Define the random measure µM on 1, . . . , qM defined from this as

µM (τ) =

∞∑

α=1

µα1(τα,1, . . . , τα,M ) = τ ,

for each τ ∈ 1, . . . , qM , where 1· · · represents the indicator function of the condition given by· · · . This is the empirical measure, but where we merely truncate the full sequence (τα,1, τα,2, . . . )to the first M components of the spin. Another useful way to state the same thing is to notice thatfor any non-random function f : 1, . . . , qM → R, we have

∑

τ∈1,...,qM

f(τ)µM (τ) =

∞∑

α=1

µαf(τα,1, . . . , τα,M ) . (50)

9

In fact, it is not necessary that f is non-random, merely that it is independent of (µα)α∈N and(τα,k : α ∈ N , k ∈ N). Note that µM is a random measure, but according to Corollary 5.3 this stillgives an upper bound. Specifically,

pN (c, β) ≤ E [GN,M(c, β, µM )] ,

where the expectation is over the law L for (µα)α∈N and (τα,k : α ∈ N , k ∈ N), from which µM isderived as a measurable function.

According to Lemma 5.2, we may write

G(1)N,M(c, β, µM ) =

1

NE

ln

∑

τ∈1,...,qM

µM (τ)∑

σ∈1,...,qN

exp

(−β

K∑

k=1

δ(σI(k), τJ(k))

) ,

where I(1), I(2), . . . are i.i.d. random variables uniformly chosen from 1, . . . , N, and J(1), J(2), . . .are i.i.d random variables uniformly chosen from 1, . . . ,M, and independently of this, K is a Poissonrandom variable with parameter cN . Note that, according to (50), we may rewrite this as

1

Nln

∑

τ∈1,...,qM

µM (τ)∑

σ∈1,...,qN

exp

(−β

K∑

k=1

δ(σI(k), τJ(k))

)

=1

Nln

∞∑

α=1

µα

∑

σ∈1,...,qN

exp

(−β

K∑

k=1

δ(σI(k), τα,J(k))

).

But, conditional on the event that J(1), . . . , J(K) are all distinct elements of 1, . . . ,M, we haveequality in distribution,

1

Nln

∞∑

α=1

µα

∑

σ∈1,...,qN

exp

(−β

K∑

k=1

δ(σI(k), τα,J(k))

)

D=

1

Nln

∞∑

α=1

µα

∑

σ∈1,...,qN

exp

(−β

K∑

k=1

δ(σI(k), τα,k)

)(51)

where we have replaced the random indices J(1), . . . , J(K) by the non-random indices 1, . . . ,K, be-cause we assumed that (τα,1, . . . , τα,M ) are exchangeable, meaning equal in distribution under finitepermutations. Note that here we use the fact that K and J(1), J(2), . . . are independent of (µα)α∈N and(τα,k : α ∈ N , k ∈ N). Moreover, conditional on the value of K, the probability that J(1), . . . , J(K)are all distinct is

M(M − 1) · · · (M −K + 1)

MK.

If we take a single realization of K for all M ’s, then we see that this conditional probability convergesto 1, pointwise, almost surely. So we are justified in making the rearrangement in (51), with highprobability. Moreover, conditioning on K, we see that the function on the left hand side of (51) isbounded in the interval [log(q) − β(K/N), log(q)]. This is summable against the distribution of K.Therefore, by the dominated convergence theorem, we have

limM→∞

E

[G

(1)N,M(c, β, µM )

]= G

(1)N (c, β,L) .

A similar argument holds for the second term G(2)N (c, β,L).

Theorem 5.6 For any c and β, we have the “extended variational principle,”

p(c, β) = limN→∞

infL

GN (c, β,L) (52)

10

Proof: We view the equality as a concatenation of an upper bound and a lower bound. The upperbound is obtained by optimizing the upper bound in Proposition 5.4 and taking the limit N → ∞.The lower bound is proved combining sub-additivity with Fekete’s lemma [22],

p(c, β) = limN→∞

limM→∞

1

NE

[ln

ZN+M (c, β)

ZM (c, β)

](53)

using a particular choice of L∗ such that

limM→∞

1

NE

[ln

ZN+M (c, β)

ZM (c, β)

]= GN (c, β,L∗) . (54)

This is obtained as a weak limit of Boltzmann-Gibbs measures. By definition,

1

NE

[ln

ZM+N (c, β)

ZM (c, β)

]=

1

NE

[ln

∑τ∈1,...,qM

∑σ∈1,...,qN e−βHM,N (τ,σ)

∑τ∈1,...,qM e−βHM(τ)

],

where, with Jij ∼ Poiss(

c2(M+N)

)and Kij ∼ Poiss

(c

M+N

),

HM,N (τ, σ) =

M∑

i,j=1

Jijδ(τi, τj) +

M∑

j=1

N∑

i=1

Kijδ(τj , σi) +

N∑

i,j=1

Jijδ(σi, σj) (55)

and, with Jij ∼ Poiss(

c2M

),

HM (τ) =

M∑

i,j=1

Jijδ(τi, τj). (56)

We split the latter Hamiltonian as follows.

HM (τ) =M∑

i,j=1

Jijδ(τi, τj) +M∑

i,j=1

Lijδ(τi, τj), (57)

where Lij ∼ Poiss(

cN2M(M+N)

). We are interested in the limit M → ∞, so we will ignore the term

∑Ni,j=1 Jijδ(σi, σj) in HM,N (τ, σ), because this converges to 0 almost surely. Hence,

1

NE

[ln

ZM+N (c, β)

ZN(c, β)

]=

1

NE

ln

∑τ∈1,...,qM

∑σ∈1,...,qN µ∗

M (τ) exp(−β∑M

j=1

∑Ni=1 Kijδ(τj , σi)

)

∑τ∈1,...,qM µ∗

M (τ) exp(−β∑M

i,j=1 Lijδ(τi, τj))

,

where

µ∗M (τ) = exp

−β

M∑

i,j=1

Jijδ(τi, τj)

(58)

Using Poisson thinning we can rewrite this as

1

NE

[ln

ZM+N (c, β)

ZN (c, β)

]=

1

NE

ln

∑τ∈1,...,qM

∑σ∈1,...,qN µ∗

M (τ) exp(−β∑KM

k=1 δ(τJ(k), σI(k)))

∑τ∈1,...,qM µ∗

M (τ) exp(−β∑LM

k=1 δ(τJ(2k−1), τJ(2k)))

,

where KM ∼ Poiss(MN cM+N ) and LM ∼ Poiss(M2 cN

2M(M+N) ), and where I(1), I(2), . . . and J(1), J(2), . . .

are i.i.d. random variables uniformly chosen on 1, . . . , N and 1, . . . ,M respectively. Since

limM→∞

KM = K ∼ Poiss(cN) , limM→∞

LM = L ∼ Poiss

(cN

2

)

we obtain formula (54) with L∗ = limM→∞ µ∗M .

11

6 Derrida-Ruelle Random Probability Cascade

Theorem (5.6) shows that the cavity functional in eq. (47) needs to be optimized over measures L. Theoptimal choice of the measure has been conjectured to be related to Derrida-Ruelle random probabilitycascades (RPC). In the following we will show how one can obtain increasing level of Replica SymmetryBreaking by considering RPC with an increasing number of levels. We will concentrate on the simplestrealization of that which produces the Replica Symmetric solution.

6.1 One level RPC

Given m ∈ (0, 1), let Λm be the measure on (0,∞): dΛm(x) = mx−m−1 dx. Suppose that ξ1, ξ2, . . . are a Poisson point process with intensity measure Λm. This has the important property that the set

λαξα∞α=1

has the same distribution as the setE[λm

α ]1/mξα∞α=1 ,

for i.i.d., positive random variables λ1, λ2, . . . . Let

ξα =ξα∑∞α=1 ξα

.

Then using properties of the logarithm

E

[ln

∞∑

α=1

ξαλα

]=

1

mlnE [λm

α ] .

As a preliminary step, we will calculate the upper bound from Proposition 5.4 for µα = ξα, withall the τ ’s being i.i.d., uniform on 1, . . . , q.

We start with the formula for G(2)N , which is easier. The important thing is to condition on the value

of L. This is because L is random, but it is not independent for the different values of α. Conditioningon L, the multipliers

λα = exp

(−β

L∑

k=1

δ(τα,2k−1, τα,2k)

)

are conditionally independent. We may calculate

E[λmα |L] =

L∏

k=1

E

[e−mβδ(τα,2k−1, τα,2k)

]=

(1 − 1

q+

1

qe−mβ

)L

.

So

E

[ln

∞∑

α=1

ξα exp

(−β

L∑

k=1

δ(τα,2k−1, τα,2k)

) ∣∣∣L]

= L ln

(1 − 1 − e−mβ

q

).

Since L has mean cN/2, this gives

G(2)N =

c

2mln

(1 − 1 − e−mβ

q

). (59)

For G(1)N we can rewrite this as

G(1)N =

1

NE

ln

∞∑

α=1

µα

∑

σ∈1,...,qN

exp

−β

N∑

i=1

K(i)∑

k=1

δ(σi, τα,i,k)

,

12

where each K(i) = |k ≤ K : I(k) = i| is an independent Poisson random variable with parameterc. This holds because of the special property for Poisson random variables, that the sum of twoindependent Poisson random variables is Poisson.

Once again the random variables K(1), . . . ,K(N) are the same for all α’s, not independent. There-fore, we condition on K(1), . . . ,K(N). In this case

λα =∑

σ∈ΣNr

exp

−β

N∑

i=1

K(i)∑

k=1

δ(σi, τα,i,k)

=N∏

i=1

q∑

σi=1

exp

−β

K(i)∑

k=1

δ(σi, τα,i,k)

.

The τα,i,k’s are all independent. In particular they are independent for different values of i. Therefore,

E[λmα |K(1), . . . ,K(N)] =

N∏

i=1

Eτ1,...,τK(i)

[(q∑

σi=1

e−β∑K(i)

k=1 δ(σi,τk)

)m].

Using this, if we let κ be a Poisson random variable with mean c, then

G(1)N =

1

mEκ lnE

τ1,...,τκ

[(q∑

σ=1

e−β∑κ

k=1 δ(σ,τk)

)m]. (60)

6.2 Trivial bound

A special case arises in the limit m ↑ 1. We get

G(2)N =

c

2ln

(1 − 1 − e−β

q

), (61)

directly. For the more complicated term, we get

G(1)N = E

κ lnEτ1,...,τκ

[q∑

σ=1

e−β∑

κk=1 δ(σ,τk)

]

= Eκ ln

q∑

σ=1

Eτ1,...,τκ

[e−β

∑κk=1 δ(σ,τk)

]

= Eκ ln

q∑

σ=1

κ∏

k=1

Eτk[e−βδ(σ,τk)

]

= Eκ ln

[q∑

σ=1

(1 − 1 − e−β

q

)κ]

= Eκ ln

[q

(1 − 1 − e−β

q

)κ]

= ln q + c ln

(1 − 1 − e−β

q

).

Combining this with (61) we obtain

pN(c, β) ≤ c

2ln

(1 − 1 − e−β

q

)+ ln q .

13

which matches the upper bound from equation (24). This is not the output of the full replica symmetricansatz. This is the output of the “trivial replica symmetric” ansatz: the one with overlap equal to 0,as we will see.

For the full replica symmetric ansatz, we must also allow for the possibility of a nontrivial overlap,which we parametrize by a number p ∈ [0, 1]. This is what we do next. Then the trivial RS ansatz isthe one with p = 0.

6.3 Two level RPC

In order to derive the full replica symmetric ansatz, we need to take a two-level RPC, and then take alimiting case. For Proposition 5.4, all that is required of the µα’s is that the index set α is countable.Therefore, we may consider instead of a single countable index, a pair α = (α1, α2).

We let ξ1α1 be a Poisson point process with intensity measure Λm1 . For each α1, we let ξ2α2

(α1)be a Poisson point process with intensity measure Λm2 . These are independent of each other fordifferent values of α1. They are also independent of the first point process ξ1α1

. Then we define

ξα = ξ1α1· ξ2α2

(α1) .

We also define

ξα =ξα∑

α∈N2 ξα.

We assume that m1 < m2. The reason for this is that∑

α1ξ1α1

has a moment only up to (and not

including) m1, and∑

α2ξ2α2

(α1) has a moment only up to (and not including) m2. But the distributionof

∑

α2

ξ(α1,α2)

α1∈N

is the same as the distribution ofE

[(∑

α2

ξ2α2(α1)

)m1]1/m1

ξα1

α1∈N

.

Since we need the sum to be finite almost surely, in order to define ξα, this means we must havem2 > m1.

In addition to the choice µα = ξα, we also choose an ansatz for the spins τα. Actually, we haveτα,k’s, but we will assume that the distribution is i.i.d., for different values of k. So we just focus ona single value of k, and suppress that index from the notation momentarily. Let τ1α1

be i.i.d., uniformon 1, . . . , q. Let τ2α also be i.i.d., uniform on 1, . . . , q, independent for every different choice ofα = (α1, α2), even if the α1’s coincide (as long as the α2’s are different, and vice-versa). Let us alsochoose i.i.d., Bernoulli random variables Xα, with probability p to be 1, and probability 1− p to be 0.Let

τα = Xατ1α1

+ (1 −Xα)τ2α2(α1) .

In other words, the probability that τα is just τ1α1is p. This is a common value for the ”cluster”

consisting of all α’s with the first part α1.

To calculate G(2)N , the easier of the two parts of the cavity field functional, we condition on ξ1α1

and on τ1α1

. Since the τ2α’s and Xα’s are i.i.d., we do not condition on them. Then we note that

P(τα,2k−1 = τα,2k | τ1α1,k) = P(X1 = X2 = 1)1τ1α1,2k−1 = τ1α1,2k + [1 −P(X1 = X2 = 1)] · 1

q.

14

This implies that

E

[e−m2βδ(τα,2k−1,τα,2k)

∣∣ τ1α1]

= p2e−m2βδ(τ1α1,2k−1,τ

1α1,2k) + (1 − p2)

(1 − 1 − e−m2β

q

).

Forλα = e−βδ(τα,2k−1,τα,2k) ,

we have thatλαξα d

= λ0ξα ,where

λ0 = E

[E

[e−m2βδ(τα,2k−1,τα,2k)

∣∣ τ1α1]m1/m2

]1/m1

.

So we get in general

lnλ0 =1

m1ln

[(1 − 1

q

)(1 − (1 − p2)(1 − e−m2β)

q

)m1/m2

+1

q

(1 −

[p2 +

1 − p2

q

] (1 − e−m2β

))m1/m2]

Because the spin fields are independent for different values of k, the effect of the L is just to multiplythis final answer. Therefore, taking the expectation of that, and dividing by N gives

G(2)N =

c

m1ln

[(1 − 1

q

)(1 − (1 − p2)(1 − e−m2β)

q

)m1/m2

+1

q

(1 −

[p2 +

1 − p2

q

] (1 − e−m2β

))m1/m2]

To get the replica symmetric ansatz, we use the 2-level RPC and take the limits m2 ↑ 1 and m1 ↓ 0.Taking m2 ↑ 1, gives

lnλ0 =1

m1ln

[(1 − 1

q

)(1 − (1 − p2)(1 − e−β)

q

)m1

+1

q

(1 −

[p2 +

1 − p2

q

] (1 − e−β

))m1].

Then taking m1 ↓ 0 gives

lnλ0 =

(1 − 1

q

)ln

(1 − (1 − p2)(1 − e−β)

q

)+

1

qln

(1 −

[p2 +

1 − p2

q

] (1 − e−β

)).

Thus the replica symmetric value of the first term is

G(2)N =

c

2

(1 − 1

q

)ln

(1 − (1 − p2)(1 − e−β)

q

)+

c

2qln

(1 −

[p2 +

1 − p2

q

] (1 − e−β

)). (62)

The more complicated term is G(1)N . Conditioning on the spins at the first level τ1α, and all the

K(i) values, we get (using a notation which is clear from the context)

E[λm2α | K(i)Ni=1, τ1α,i,kα,i,k] =

N∏

i=1

EXi,k,τ

2i,k

q∑

σi=1

K(i)∏

k=1

[e−βδ(σi,τi,k)

]

m2 ,

where the τ2i,k are all i.i.d., uniform on 1, . . . , q, and

τi,k = Xi,kτ1i,k + (1 −Xi,k)τ2i,k .

The Xi,k’s are i.i.d., Bernoulli-p random variables. Since the formulas are identically distributed fordifferent i’s and since there are N such i’s (canceling the division by N), we get the formula

G(1)N =

1

m1Eκ lnE

τ1k

EXk,τ2k

[(q∑

σ=1

κ∏

k=1

[e−βδ(σ,τk)

])m2]m1/m2

,

15

where once again κ is a Poisson random variable with mean c, and now τ1k and τ2k are all i.i.d.,uniform random variables on 1, . . . , q, and Xk are all i.i.d., Bernoulli random variables with meanp, and τk = Xkτ

1k + (1 −Xk)τ2k for each k.

We now take m2 ↑ 1 and m1 ↓ 0. Taking m2 ↑ 1 gives

1

m1Eκ lnE

τ1k

[EXk,τ

2k

[q∑

σ=1

κ∏

k=1

[e−βδ(σ,τk)

]]m1],

which can be rewritten

1

m1Eκ lnE

τ1k

[(q∑

σ=1

κ∏

k=1

EXk,τ

2k

[e−βδ(σ,τk)

])m1].

But

EXk,τ

2k

[e−βδ(σ,τk)

]= pe−βδ(σ,τ1

k) + (1 − p)

(1 − 1 − e−β

q

).

Using this and taking the limit m1 ↓ 0 gives

G(1)N = E

κ,τ1k

[ln

(q∑

σ=1

κ∏

k=1

(pe−βδ(σ,τ1

k) + (1 − p)

(1 − 1 − e−β

q

)))].

Let us now rewrite this in a manner which is appropriate for taking derivatives at p = 0. Wecan write e−βδ(σ,τ1

k) = 1 − (1 − e−β)δ(σ, τ1k ). Since the average value of δ(σ, τ1k ) is 1/q, we may alsoincorporate that:

e−βδ(σ,τ1k) = 1 − 1 − e−β

q− (1 − e−β)(δ(σ, τ1k ) − q−1) .

Then we may rewrite EXk,τ

2k

[e−βδ(σ,τk)

]as

(1 − 1 − e−β

q

)− p(1 − e−β)(δ(σ, τ1k ) − q−1) .

The formula for G(1)N is simpler if we introduce a new variable, x = (1 − e−β)/(q − e−β). Therefore,

we obtain

G(1)N = ln q + c ln

(1 − 1 − e−β

q

)+ E

κ,τ1k

[lnE

σ

[κ∏

k=1

(1 − px (qδ(σ, τ1k ) − 1)

)]] ∣∣∣∣∣

x= 1−e−β

q−(1−e−β )

. (63)

6.4 Local Replica Symmetric stability analysis of p = 0 and the critical

temperature

Now we want to consider this formula as a function of p perturbatively near 0. This allows us to derivesome results of Zdeborova and Krzaka la [28] in the present context. We say that the p = 0 RS ansatzis “stable to RS perturbations” if it is a local minimizer of the extended variational principle in theset of RS ansatze.

Starting from the simpler term, (62), we rewrite G(2)N as

c

2ln

(1 − 1 − e−β

q

)+

c(q − 1)

2qln

(1 +

p2(1 − e−β)

q − (1 − e−β)

)+

c

2qln

(1 − (q − 1)(1 − e−β)p2

q − (1 − e−β)

).

Using x = (1 − e−β)/(q − e−β) this is simpler:

G(2)N =

c

2ln

(1 − 1 − e−β

q

)+

c

2q

[(q − 1) ln(1 + p2x) + ln(1 − (q − 1)p2x)

] ∣∣∣∣x= 1−e−β

q−(1−e−β )

. (64)

16

This is an even function of p, so only even powers will appear. Taylor expansion shows that

(q − 1) ln(1 + p2x) + ln(1 − (q − 1)p2x) = −1

2q(q − 1)p4x2 .

Therefore,

G(2)N =

c

2ln

(1 − 1 − e−β

q

)− c(q − 1)p4x2

4+ O(p6)

∣∣∣x= 1−e−β

q−(1−e−β )

.

This givesd4

dp4G

(2)N

∣∣∣∣p=0

= −6c(q − 1)x2∣∣∣x= 1−e−β

q−(1−e−β )

. (65)

Now turning to the more difficult term, let us start with (63). Let us write f(σ, τ) = (qδ(σ, τ)− 1).Then we have

G(1)N = ln q + c ln

(1 − 1 − e−β

q

)+ E

κ,τ1k

[lnE

σ

[κ∏

k=1

(1 − pxf(σ, τ1k )

)]] ∣∣∣∣∣

x= 1−e−β

q−(1−e−β )

. (66)

As usual, we may interpret the function

Eσ

[κ∏

k=1

(1 − pxf(σ, τ1k )

)]

as a cumulant generating function. But the random variable is multi-linear in p. Therefore, whenexpanding in p, we have to take account of these terms. Also, notice that Eσ[f(σ, τ)] = E

τ [f(σ, τ)] = 0as long as the expectations are with respect to the uniform measure. Because of this, various termsvanish either in the expectation over E

σ or in the expectation over Eτ1

k.For instance, using the fact that E

σ[f(σ, τ)] = 0, we see that the first derivative in p equals0. Moreover, since each factor is linear in p, in taking multiple derivatives (of a single copy of theproduct) means we cannot repeat the derivative of any factor. So we obtain

d2

dp2Eσ

[κ∏

k=1

(1 − pxf(σ, τ1k )

)]

= x2κ∑

j,k=1j 6=k

Eσ[f(σ, τ1j )f(σ, τ1k )] .

But then taking the expectation over Eτ1

k gives 0 because since j 6= k, we have

Eτ1

kEσ[f(σ, τ1j )f(σ, τ1k )] = E

σ[Eτ1j [f(σ, τ1j )] · Eτ1

k [f(σ, τ1k )]]

= 0 .

Continuing, we may easily see that the third derivative is again 0 since Eσ[f(σ, τ)] = 0. Then, the

next simplest term arises from

d4

dp4Eσ

[κ∏

k=1

(1 − pxf(σ, τ1k )

)]

= x4κ∑

j,k,ℓ,m=1j 6=k 6=ℓ 6=m

Eσ[f(σ, τ1j )f(σ, τ1k )f(σ, τ1ℓ )f(σ, τ1m)]

− 3x4

κ∑

j,k=1j 6=k

Eσ [f(σ, τ1j )f(σ, τ1k )]

2

.

17

We can rewrite this by expanding the square of the sum, and using replicated spin variables for productsof expectations:

d4

dp4Eσ

[κ∏

k=1

(1 − pxf(σ, τ1k )

)]

= x4κ∑

j,k,ℓ,m=1j 6=k 6=ℓ 6=m

Eσ[f(σ, τ1j )f(σ, τ1k )f(σ, τ1ℓ )f(σ, τ1m)]

− 3x4κ∑

j,k=1j 6=k

κ∑

ℓ,m=1ℓ 6=m

Eσ,σ′

[f(σ, τ1j )f(σ, τ1k )f(σ′, τ1ℓ )f(σ′, τ1m)] .

Any distinct terms for j, k, ℓ,m vanish in the expectation over Eτ1k. Therefore all must be paired. That

means that the first summand vanishes entirely. In the second summand, we require (ℓ,m) = (j, k) or(ℓ,m) = (k, j). These two possibilities give an extra factor of 2. Hence, we obtain

d4

dp4G

(1)N

∣∣∣∣p=0

= −6x4Eκ,τ1

kκ∑

j,k=1j 6=k

(Eσ[f(σ, τ1j )f(σ, τ1k )]

)2∣∣∣∣∣x= 1−e−β

q−(1−e−β )

.

A calculation gives

Eσ [f(σ, τ1j )f(σ, τ1k )] =

−1 if τ1j 6= τ1k ,

(q − 1) if τ1j = τ1k .

Using the i.i.d., uniform distribution on τ1k gives Pτ1j = τ1k = 1/q. Therefore,

Eτ1

kEσ[f(σ, τ1j )f(σ, τ1k )] = 0 ,

as we claimed before. But now we also have

(Eσ[f(σ, τ1j )f(σ, τ1k )]

)2=

1 if τ1j 6= τ1k ,

(q − 1)2 if τ1j = τ1k ,

which gives

Eτ1

k[(Eσ[f(σ, τ1j )f(σ, τ1k )]

)2]= q − 1 .

Therefore, also using the fact that Eκ[#(j, k) ∈ 1, . . . , κ2 : j 6= k] equals E

κ[κ(κ − 1)] = c2, weobtain

d4

dp4G

(1)N

∣∣∣∣p=0

= −6c2(q − 1)x4∣∣∣x= 1−e−β

q−(1−e−β )

. (67)

Since the RS ansatz is G(1)N − G

(2)N this means that the local stability of the trivial RS ansatz is

determined by the sign of the difference between the terms in (67) and (65)

d4

dp4

[G

(1)N −G

(2)N

]= −6c2(q − 1)x4 + 6c(q − 1)x2 = 6c(q − 1)x2[1 − cx2] .

We assume q > 1 otherwise the model is trivial (the 1-state Potts model). So the p = 0 solution islocally stable if and only if

1 − cx2 > 0 for x =1 − e−β

q − (1 − e−β).

Note that (β = 0) ⇒ (x = 0) while (β → ∞) ⇒ (x → 1q−1 ). Therefore, we see that the p = 0 solution

is stable at all temperatures if c < c∗, where

c∗ = (q − 1)2 . (68)

18

For instance, for the Ising model which has q = 2, this merely states that c∗ = 1, which is thepercolation threshold. For c > c∗ there is a giant component, and for c < c∗ there is not.

If c > c∗ then the critical temperature for stability is given by the condition

1 − cx2 = 0 ⇔ x =1√c

⇔ 1

x=

√c .

Since

x =1 − e−β

q − (1 − e−β)⇔ 1

x=

q

1 − e−β− 1 ,

this means that the critical value of β satisfies

q

1 − e−β− 1 =

√c ⇔ 1 − e−β =

q

1 +√c

⇔ e−β = 1 − q

1 +√c.

So

β∗ = − ln

(1 − q

1 +√c

). (69)

7 Replica Symmetric Solution

We would now like to compare to Dembo and Montanari’s results [10], as well as those of [11]. So farwe have imposed the condition that all the τ1k ’s and τ2k ’s are i.i.d., uniform on 1, . . . , q, and

τk =

τ1k if Xk = 1,

τ2k if Xk = 0.

This is a very simple special case of the replica symmetric ansatz. The more general case yields thefollowing result.

Theorem 7.1 The quenched pressure is bounded from above by the Replica Symmetric pressure.Namely,

p(c, β) ≤ E ln

[q∑

s=1

κ∏

k=1

(1 − (1 − e−β)Pk(s)

)]− c

2E ln

[1 − (1 − e−β)

q∑

s=1

P1(s)P2(s)

], (70)

where κ is a Poisson random variables with mean c, the Pk = (Pk(s))qs=1 are i.i.d. random probabilityvectors, i.e. Pk(s) ≥ 0 for each s and

∑qs=1 Pk(s) = 1 a.s., satisfying the following equality in

distribution

(P1(s))qs=1

d=

( ∏κk=1

(1 − (1 − e−β)Pk(s)

)∑q

s=1

∏κk=1 (1 − (1 − e−β)Pk(s))

)q

s=1

. (71)

Proof: Suppose that (Ω,F ,P) is a probability space. Suppose that F1 is a sub-σ-algebra such that(ξ1α1

)∞α1=1 are F1-measurable, and have the appropriate Poisson point process distribution with inten-sity ξm1dξ, as stated before. Also, suppose that (ξ2α1,α2

)∞α2=1 are independent of F1, and independentof each other for different α1, and distributed according to the Poisson point process distributionwith intensity ξm2dξ, for 0 < m1 < m2 < 1. Finally, suppose that the conditional distributionτ(α1,α2),k∞α2=1, conditioned on F1 are independent:

P(τ(α1,1),k ∈ A1, τ(α1,2),k ∈ A2, · · · | F1) =∞∏

α2=1

P(τ(α1,α2),k ∈ Aα2 | F1) ,

assuming that only finitely many An’s are different than 1, . . . , q. Also, assume that the τ(α1,α2),k’sare independent for different values of α1.

19

Finally, assume that K and L are independent of all the ξα’s and τα,k’s as before. For notationalsimplicity, suppose that there is a F0 such that K and L are F0-measurable and such that all the ξα’sand ταk

’s are independent of F0.Then the extended variational principle works as before with

G(1)N =

1

m1E lnE

[E

[(q∑

s=1

e−∑

κk=1 βδ(s,τk)

)m2 ∣∣∣F1

] ∣∣∣F0

]m1/m2

,

where we write τk for a generic τα,k, and κ is a Poisson-c random variable, measurable with respect toF0. Similarly,

G(2)N =

c

2m1E lnE

[E

[e−m2βδ(τ1,τ2)

∣∣F1

]m1/m2∣∣∣F0

].

Taking the limit m1 ↓ 0 and m2 ↑ 1 gives

G(1)N = E lnE

[q∑

s=1

e−∑

κk=1 βδ(s,τk)

∣∣∣F1

],

andG

(2)N =

c

2E lnE

[e−βδ(τ1,τ2)

∣∣F1

].

The most general choice is to have F1-measurable i.i.d., random vectors Pk = (Pk(s))qs=1 chosenaccording to a distribution with support in the set such that Pk(s) ≥ 0 for each s, and

∑qs=1 Pk(s) = 1,

a.s. Then each τk is Pk-distributed. In this case,

G(2)N =

c

2E ln

[1 − (1 − e−β)

q∑

s=1

P1(s)P2(s)

],

and

G(1)N = E ln

[q∑

s=1

κ∏

k=1

(1 − (1 − e−β)Pk(s)

)],

Assuming an underlying distribution dρ(P ),

G(2)N =

c

2

∫ln

[1 − (1 − e−β)

q∑

s=1

P1(s)P2(s)

]dρ(P1) dρ(P2) ,

and

G(1)N = E

κ

∫ln

[q∑

s=1

κ∏

k=1

(1 − (1 − e−β)Pk(s)

)]

κ∏

k=1

dρ(Pk) .

Taking a Frechet derivative of ρ in the direction of δρ gives a condition for criticality/extremality:

c

∫ln

[1 − (1 − e−β)

q∑

s=1

P1(s)P2(s)

]dδρ(P1) dρ(P2)

= Eκ

[κ

∫ln

[q∑

s=1

κ∏

k=1

(1 − (1 − e−β)Pk(s)

)]dδρ(P1)

κ∏

k=2

dρ(Pk)

].

Here we have used the product rule (Leibniz rule) and the symmetry of the formulas with respectto permutations of the indices on the Pk’s. In the first integral, both ρ(dP1) and ρ(dP2) must bedifferentiated which gives a factor of 2. In the second one, we get an extra factor of κ. Note that

20

E[κf(κ)] = cE[f(κ + 1)]. Note that because of the differentiation, P1 had distribution δρ, and it isonly the κ− 1 last Pk’s that had distribution ρ. Therefore, rewriting P1 as P0 and shifting, we have

∫ln

[1 − (1 − e−β)

q∑

s=1

P0(s)P1(s)

]dδρ(P0) dρ(P1)

= Eκ

[∫ln

[q∑

s=1

(1 − (1 − e−β)P0(s)

) κ∏

k=1

(1 − (1 − e−β)Pk(s)

)]dδρ(P0)

κ∏

k=1

dρ(Pk)

].

Note that, in order for ρ + δρ to be a probability measure, given that ρ is already a probabilitymeasure, it is necessary that δρ is a signed measure, with total measure 0. Therefore, if we integrateany function against δρ(dP0), which is constant with respect to P0, then the integral is zero. Usingthis, and properties of the logarithm, we see that

∫ln

[1 − (1 − e−β)

q∑

s=1

P0(s)P1(s)

]dδρ(P0) dρ(P1)

= Eκ

[∫ln

[q∑

s=1

(1 − (1 − e−β)P0(s)

) ∏κk=1

(1 − (1 − e−β)Pk(s)

)∑q

s=1

∏κk=1 (1 − (1 − e−β)Pk(s))

]dδρ(P0)

κ∏

k=1

dρ(Pk)

].

We have normalized the multiplier, so that summed over s it just gives 0. Therefore, we may rewritethis by distributing the two terms in the factor (1 − (1 − e−β)P0(s)), to get

∫ln

[1 − (1 − e−β)

q∑

s=1

P0(s)P1(s)

]dδρ(P0) dρ(P1)

= Eκ

[∫ln

[1 − (1 − e−β)

q∑

s=1

P0(s)

∏κk=1

(1 − (1 − e−β)Pk(s)

)∑q

s=1

∏κk=1 (1 − (1 − e−β)Pk(s))

]dδρ(P0)

κ∏

k=1

dρ(Pk)

].

Since δρ is general, this implies that the condition for extremality is

(P1(s))qs=1d=

( ∏κk=1

(1 − (1 − e−β)Pk(s)

)∑q

s=1

∏κk=1 (1 − (1 − e−β)Pk(s))

)q

s=1

.

8 High temperature region

A guess at high temperature would be that ωN (δ(σi, σj)) is about 1/q generically (for i 6= j) since thatis the exact value at infinite temperature. A calculation, using (25) and (26), shows that

pN (c, β) − pTRS(c, β) =c

2

∫ 1

0

E[ln(1 − aq(β)ωN,ct,β

(δ(σi, σj) − q−1

))]dt ,

where ωN,c,β is the Boltzmann-Gibbs state at inverse temperature β, associated to the HamiltonianHN (c, σ) (here we write explicitly the dependence on the parameter c) and

aq(β) =1 − e−β

1 − (1 − e−β)q−1.

An important observation is that

aq(β)[1 − q−1] = 1 − e−β

1 − (1 − e−β)q−1∈ [0, 1) for q ≥ 1 and β ≥ 0.

21

By Taylor’s theorem, we may write

ln(1 − ax) = −n∑

k=1

akxk

k+ Rk+1(a, x) ,

where Rk+1(a, x) is the remainder term, of order (ax)k+1:

Rk+1(a, x) = −(ax)k+1

∫ 1

0

(1 − t)k

(1 − ax)k+1dx .

We want to consider this expansion with a = aq(β) and x = δ(σi, σj) − q−1. In particular, we noticethat for ax ≤ µ < 1, we can bound

Rk+1(a, x) ≥ − 1

k + 1

(ax

1 − µ

)k+1

.

If also k is odd, then we know that Rk+1(a, x) ≥ 0. This leads to the family of bounds: for any c, β ≥ 0,and any integer ℓ ∈ 1, 2, . . .,

0 ≥ pN (c, β) − pTRS(c, β) ≥ − c

2N2

N∑

i,j=1

2ℓ−1∑

k=1

akq(β)

k

∫ 1

0

E[ωkN,ct,β

(δ(σi, σj) − q−1

)]dt

− cb2ℓ(q, β)

2N2

N∑

i,j=1

∫ 1

0

E[ω2ℓN,ct,β

(δ(σi, σj) − q−1

)]dt ,

where

b2ℓ(q, β) = a2ℓq (β)

∫ 1

0

(1 − t)2ℓ−1

(1 − aq(β)[1 − q−1]t)2ℓdt .

We will do a second order approximation, so we take ℓ = 1, and note that

b2(q, β) = a2q(β)

∫ 1

0

(1 − t)

(1 − aq(β)[1 − q−1]t)2dt .

In terms of doing the second order approximation, since the mean-field antiferromagnet has the correctfirst order behavior and the SK mean-field spin glass has the correct second order behavior, we introducethis as an intermediate step. Let us define

H(2)N (σ) =

1

2N

N∑

i,j=1

δ(σi, σj) ,

which is the mean-field antiferromagnetic term, and

H(3)N (σ) = − 1√

2N

N∑

i,j=1

Yijδ(σi, σj) ,

where the Yij ’s are i.i.d., standard, normal random variables. All the random variables are supposedto be independent. We will henceforth write the original model with a superscript,

H(1)N (c, σ) = HN (c, σ) ,

in order to look like the other two terms.

22

Let us define

ZN = ZN (c, λ, κ) =∑

σ∈1,...,qN

exp[−βH

(1)N (c, σ) − λH

(2)N (σ) −√

κH(3)N (σ)

].

The important parameter β will be implicit in this notation, since it will stay constant. The parametersc, λ and κ will vary. We write ωN for the expectation with respect to the probability measure on1, . . . , qN with weights

exp[−βH

(1)N (c, σ) − λH

(2)N (σ) −√

κH(3)N (σ)

]

ZN

. (72)

Then the easiest calculation is

∂

∂λE

[1

Nln ZN

]= − 1

2N2

N∑

i,j=1

E [ωN(δ(σi, σj))] . (73)

By Gaussian integration by parts, we have

∂

∂κE

[1

Nln ZN

]=

1

4N2

N∑

i,j=1

E[ωN

(δ2(σi, σj)

)− ω2

N (δ(σi, σj))].

But, of course, δ2(σi, σj) = δ(σi, σj). So this gives

∂

∂κE

[1

Nln ZN

]=

1

4N2

N∑

i,j=1

E [ωN (δ(σi, σj))] −1

4N2

N∑

i,j=1

E[ω2N (δ(σi, σj))

]. (74)

For later reference, we note that this implies

1

2N2

N∑

i,j=1

E [ωN(δ(σi, σj))] = − ∂

∂λE

[1

Nln ZN

], (75)

and

1

2N2

N∑

i,j=1

E[ω2N (δ(σi, σj))

]= −2

∂

∂κE

[1

Nln ZN

]+

1

2N2

N∑

i,j=1

E [ωN (δ(σi, σj))]

= −2∂

∂κE

[1

Nln ZN

]− ∂

∂λE

[1

Nln ZN

].

(76)

Finally, by the Poisson differentiation formula (cfr (26)),

∂

∂cE

[1

Nln ZN

]=

1

2N2

N∑

i,j=1

E[ln(1 − [1 − e−β]ωN (δ(σi, σj))

)].

But the power series and the bounds for the logarithm are still valid:

∂

∂cE

[1

Nln ZN

]≥ 1

2ln

(1 − 1 − e−β

q

)− aq(β)

2N2

N∑

i,j=1

E[ωN(δ(σi, σj) − q−1)

]

− b2(q, β)

2N2

N∑

i,j=1

E[ω2N (δ(σi, σj) − q−1)

].

23

Expanding out the terms just involving q−1, we see that

∂

∂cE

[1

Nln ZN

]≥ 1

2ln

(1 − 1 − e−β

q

)+

aq(β)

2q− b2(q, β)

2q2

− aq(β) − 2b2(q, β)q−1

2N2

N∑

i,j=1

E [ωN(δ(σi, σj))]

− b2(q, β)

2N2

N∑

i,j=1

E[ω2N (δ(σi, σj))

].

(77)

Now we can use the formulas (75) and (76) to obtain

∂

∂cE

[1

Nln ZN

]≥ 1

2ln

(1 − 1 − e−β

q

)+

a(β)

2q− b(β)

2q2

+ [a(β) − 2b(β)q−1 + b(β)]∂

∂λE

[1

Nln ZN

]

+ 2b(β)∂

∂κE

[1

Nln ZN

].

Therefore, for a fixed β, we take a curve in (c, λ, κ) space:

c(t) = c0 − t , λ(t) = [a(β) − 2b(β)q−1 + b(β)]t and κ(t) = 2b(β)t .

Then the differential inequality above and the fundamental theorem of calculus gives the bound

E

[1

Nln ZN

] ∣∣∣∣∣

t=c0

t=0

≤ −c02

ln

(1 − 1 − e−β

q

)− c0a(β)

2q+

c0b(β)

2q2.

(This is reminiscent of Gronwall’s inequality, in that one integrates a differential inequality to get atypical bound.) In other words,

E

[1

Nln ZN (c, 0, 0)

]≥ E

[1

Nln ZN(0, [a(β) − 2b(β)q−1 + b(β)]c, 2b(β)c)

]

+c

2ln

(1 − 1 − e−β

q

)+ c

(a(β)

2q− b(β)

2q2

).

The previous inequality tell us that to study the high temperature of the antiferromagnetic modelon the Erdos-Renyi we are led to consider ZN (0, λ, κ) for sufficiently small λ and κ. Interestingly,this is a model we have not seen studied before: it is a mean-field antiferromagnet in conjunctionwith a mean-field spin glass. The two models are conducive to study together since they both satisfyinequalities going in the same direction. However, as the mean-field antiferromagnet never has a phasetransition, it is not surprising to learn that it does not make any significant change to the behavior ofthe spin glass at high temperature.

Lemma 8.1 For λ and κ in some open neighborhood of the origin, we have

limN→∞

1

NE

[ln ZN (0, λ, κ)

]= ln q − λ

2q+

κ(q − 1)

4q2.

We prove this result in the appendix. For now, we mention how it results in the high temperatureregion for the diluted antiferromagnet.

24

Corollary 8.2 There exist β∗ > 0 such that for all β < β∗ we have the asymptotic lower bound,

p(c, β) = limN→∞

E

[1

Nln ZN(c, 0, 0)

]≥ ln q +

c

2ln

(1 − 1 − e−β

q

)= pTRS(c, β) .

Combined with Lemma 4.1 this implies, for all β < β∗,

p(c, β) = pTRS(c, β) . (78)

Proof: (Proof of the corollary.) We already know

E

[1

Nln ZN (c, 0, 0)

]≥ E

[1

Nln ZN(0, [a(β) − 2b(β)q−1 + b(β)]c, 2b(β)c)

]

+c

2ln

(1 − 1 − e−β

q

)+ c

(a(β)

2q− b(β)

2q2

).

(79)

Settingλ = [a(β) − 2b(β)q−1 + b(β)]c ,

andκ = 2b(β)c ,

we see thatZN (0, [a(β) − 2b(β)q−1 + b(β)]c, 2b(β)c) = ZN (0, λ, κ) .

But by the lemma, we know that

limN→∞

1

NE

[ln ZN (0, λ, κ)

]= ln q − λ

2q+

κ(q − 1)

4q2.

In this case we see that

− λ

2q+

κ(q − 1)

4q2= − [a(β) − 2b(β)q−1 + b(β)]c

2q+

2b(β)c(q − 1)

4q2.

Simplifying, this is

−a(β)c

2q+

b(β)c

2q2,

thus we see that

limN→∞

E

[1

Nln ZN(0, [a(β) − 2b(β)q−1 + b(β)]c, 2b(β)c)

]= ln q − c

(a(β)

2q− b(β)

2q2

).

Putting this together with (79) gives the corollary.

9 Summary and Outlook

We have considered the antiferromagnetic Potts model on an Erdos-Renyi graph. While this modelhas some important distinctions from a true spin glass model, it also has many qualitative similarities.It is a disordered model in statistical mechanics, and there is a high degree of frustration. We havedemonstrated that the interpolation method for mean-field and diluted mean-field spin glasses alsoapplies to this model. This gives an extended variational principle, analogous to the known resultsfor the Sherrington-Kirkpatrick and Vianna-Bray model. In particular, this allows us to interpret thereplica symmetric ansatz of theoretical physics as a rigorous bound on the free energy. By adapting theinterpolation method of Guerra and Toninelli, we were also able to prove rigorously that the replicasymmetric solution is exact for sufficiently high temperatures.

We now make some remarks about the context of our results, further implications, and importantopen problems.

25

1. It should be mentioned that the second moment method does not work to control the hightemperature region, because ZN does not concentrate around its mean. To show this, note thatfor a Poisson random variable J with mean λ, we have

Ee−βJ =

∞∑

k=1

e−βke−λλk

k!= e−λeλe

−β

= e−λ(1−e−β). (80)

Using this, a simple computation shows that the second moment of ZN equals

E[Z2N ] =

∑σ,σ′∈1,...,qN exp

−cN

2(1 − e−β)

q∑

r1=1

(q2N (r1) + q

′2N (r1)

)

+cN

2(1 − e−β)2

q∑

r1,r2=1

q2N (r1, r2)

, (81)

Compared to

E[ZN ]2 =∑

σ,σ′∈1,...,qN

exp

−cN

2(1 − e−β)

q∑

r1=1

(q2N (r1) + q

′2N (r1)

), (82)

this grows exponentially faster.

2. It is known that for low connectivity there is no phase transition. It has been proved indeed [4]that the ground state entropy is rigorously equal to the replica symmetric prediction when theconnectivity is low enough. Because we prove here that for sufficiently high temperatures thepressure is given exactly by the trivial replica symmetric formula, this also proves as a subsidiaryresult that the model must have a phase transition for sufficiently large values of the connectivityparameter. This is because the entropy of the trivial replica symmetric ansatz

sTRS(c, β) =c

2ln

(1 − 1 − e−β

q

)+

βc

2q

e−β

1 − 1−e−β

q

+ ln q

becomes negative (and physically unrealizable) at sufficiently low temperatures and large enoughc.

3. Additionally, due to the extended variational principle, one may prove upper bounds associatedto any level of replica symmetry breaking, as in [14, 12]. In particular it has been observed[20, 28] that there are various transition points, such as a temperature at which the replicasymmetric ansatz is unstable to a 1-level RSB solution. In our notation, this means that thereplica symmetry breaking solution is a better optimizer of the extended variational principle.This gives a second proof of a phase transition, since their calculations are entirely rigorous forthe purposes of showing this instability, and one can use an argument as in [24]. We remarkthat this approach implies a phase transition for a temperature at least as low as that requiredto obtain local instability of the replica symmetric ansatz, although, as claimed in [20, 28] forvarious parameter ranges there is a first order transition which happens prior to that point.

4. The open neighborhood of the origin for the parameters (λ, κ) referred to in Lemma 8.1 is shownin Figure 2, where we have relabelled y = λ, z = κ. Combining this with Corollary 8.2, one mayin principle determine explicit bounds for the high temperature region. But these bounds willnot be sharp.

5. As an example, we mention that it is known that for each value of q, there is a critical value ofthe connectivity c∗(q) such that if c ≤ c∗(q) then the replica symmetric ansatz is correct all the

26

way down to zero temperature. We cannot re-derive this low connectivity result. Technically,the obstruction is the appearance of the constants a(β) and b(β) multiplying c in equation (79)for the parameters λ and κ. While the constant a(β) remains bounded as β → ∞, the constantb(β) diverges, which effectively rescales c to ∞. It would be interesting to find a solution to thisproblem.

A Proof of Lemma 8.1

In this appendix, we will carry out the high temperature analysis to prove Lemma 8.1. Our approachis reminiscent of the one carried out for the Sherrington-Kirkpatrick model in [15].

A.1 Spin glass monotonicity principle

Suppose that we have two Hamiltonians: a non-random Hamiltonian H0 : ΩN → R, and a Gaussianspin glass Hamiltonian, which we write as

H(σ,J ) =

MN∑

k=1

Jkhk(σ) ,

where each hk : ΩN → R is a non-random interaction, and J = (J1, . . . , JMN) are i.i.d., N (0, 1)

random variables. What is important is that

E[H(σ,J )] = 0 ,

for each σ ∈ ΩN , and there is a covariance which we may write explicitly in this case as

E[H(σ,J )H(σ′,J )] =

MN∑

k=1

hk(σ)hk(σ′) .

For each t ≥ 0, we define an interpolating Hamiltonian:

HN (J , σ, t) = H(0)N (σ) +

√tHN (σ,J ) +

t

2E[H2

N (σ,J )]. (83)

We define the random partition function

ZN (J , t) =∑

σ∈ΩN

e−HN (J ,σ,t) ,

and the “quenched” pressure

pN(t) =1

NE [lnZN (J , t)] . (84)

We also define the random Boltzmann-Gibbs measure

〈f(σ)〉J ,N,t =∑

σ∈ΩN

f(σ)e−HN (J ,σ,t)

ZN (J , t).

Lemma A.1 For t ≥ 0,dpNdt

= − 1

2NE[〈HN (J ′, σ)〉2J ,N,t

], (85)

where J and J ′ are independent realizations of the i.i.d., N (0, 1) noise. In particular pN (t) is mono-tone non-increasing.

27

Proof: Suppose that we have a general Hamiltonian

HN (J , σ, t, u) = H(0)N (σ) + tH

(1)N (σ) +

√uHN (J , σ) ,

where H(1)N (σ) is another non-random Hamiltonian. Then

∂

∂tE

[1

Nln∑

σ∈ΩN

e−HN (J ,σ,t,u)

]= − 1

NE

[⟨H

(1)N (σ)

⟩

J ,N,t,u

], (86)

where

〈f(σ)〉J ,N,t,u =∑

s∈ΩN

f(σ)e−HN (J ,σ,t,u)

ZN (J , t, u)and ZN (J , t, u) =

∑

σ∈ΩN

e−HN (J ,σ,t,u) .

Similarly,

∂

∂uE

[1

Nln∑

σ∈ΩN

e−HN (J ,σ,t,u)

]= − 1

2N√uE

[〈HN (J , σ)〉N,t,u

].

But now we use Gaussian-integration-by parts to determine that

− 1

2N√uE

[〈HN (J , σ)〉N,t,u

]= − 1

2N√u

MN∑

k=1

E

[Jk

∑

σ∈ΩN

hk(σ)e−HN (J ,σ,t,u)

∑σ′∈ΩN

e−HN (J ,σ′,t,u)

]

= − 1

2N√u

MN∑

k=1

∑

σ∈ΩN

hk(σ)E

[∂

∂Jk

(e−HN (J ,σ,t,u)

∑σ′∈ΩN

e−HN (J ,σ′,t,u)

)]

=1

2N

MN∑

k=1

∑

σ∈ΩN

hk(σ)E

[hk(σ)e−HN (J ,σ,t,u)

∑σ′∈ΩN

e−HN (J ,σ′,t,u)−∑

σ′∈ΩN

hk(σ′)e−HN (J ,σ,t,u)−HN (J ,σ′,t,u)

Z2N(J , t, u)

].

This can be written more concisely as

∂

∂uE

[1

Nln∑

σ∈ΩN

e−HN (J ,σ,t,u)

]=

1

2NE

[⟨E[H2

N (J , σ)]⟩

J ,N,t,u− 〈E [HN (J , σ)HN (J , σ′)]〉J ,N,t,u

],

where we write

〈f(σ, σ′)〉J ,N,t,u =∑

σ,σ′∈ΩN

f(σ, σ′)e−HN (J ,σ,t,u)−HN (J ,σ′,t,u)

Z2N (J , t, u)

.

Putting this together with (86) and specializing to

H(1)N (σ) = E

[H2

N (J , σ)],

and setting u = t, we have

∂

∂tE

[1

Nln∑

σ∈ΩN

e−HN (J ,σ,t,t)

]= − 1

2NE

[〈E [HN (J , σ)HN (J , σ′)]〉J ,N,t,u

].

Finally, we can make the inner expectation and outer expectations independent by using independentrealizations of the noise J . Doing this, and using the fact that conditional on J , the probabilitymeasure associated to 〈f(σ, σ′)〉J ,N,t,u makes σ and σ′ independent and identically distributed, weobtain the desired result.

28

A.2 Upper and lower bounds for the spin glass plus mean-field interactions

We will need to introduce various new terms to the Hamiltonian. Therefore, we take the opportunityto redefine some terms and introduce some new ones. In completing the proof we will relate these backto the original definitions from the last section. Here are the new terms in the Hamiltonian.

• The mean-field Potts-type Hamiltonian,

HPottsMF (σ) =

1

2N

N∑

i,j=1

[δ(σi, σj) − q−1

]. (87)

• The mean-field Potts spin-glass. This has Hamiltonian

HSGPotts(σ) = −

N∑

i,j=1

Jij√2N

[δ(σi, σj) − q−1

], (88)

for random coupling constants Jij , which are i.i.d., standard, normal random variables, meaningE[Jij ] = 0, E[J2

ij ] = 1.

• A spin-spin overlap term. This has configuration space ΩNq × ΩN

q . We denote the pair as (σ, τ)

for σ, τ ∈ 1, . . . , qN . The Hamiltonian is

H(2)over(σ, τ) =

1

2N

N∑

i,j=1

[δ(σi, σj) − q−1

] [δ(τi, τj) − q−1

]. (89)

We observe some simple calculations in the following lemma.

Lemma A.2 There are the following relations among these Hamiltonians:

E[HSG

Potts(σ)HSGPotts(τ)

]= H(2)

over(σ, τ) and

H(2)over(σ, σ) = (1 − 2q−1)HPotts

MF (σ) +N

2q2.

We leave the proof of Lemma A.2 as an exercise for the reader. The most important step is theidentity [

δ(σi, σj) − q−1]2

= (1 − 2q−1)δ(σi, σj) + q−2 .

We are interested in the following quantities:

HN (y, z, σ) = yHPottsMF (σ) +

√z HSG

Potts(σ) − Nz

2q2, (90)

ZN(y, z) =∑

σ∈ΩNq

e−HN (y,z,σ) , (91)

pN (y, z) =1

NE [lnZN (y, z)] . (92)

Guerra and Toninelli proved that the thermodynamic limit of this pressure exists in [17]. It is re-markable that they are able to prove the existence of the thermodynamic limit even with mean-fieldferromagnetic interactions, not just antiferromagnetic terms. More recently, other researchers haveconsidered this model in certain regimes outside the high-temperature region, for the Ising case, q = 2,to try to establish the Ghirlanda-Guerra identities [8]. We will not need such results, since we restrictattention to a high temperature region.

Guerra and Toninelli’s general spin glass monotonicity principle implies the following.

29

Corollary A.3 For any y ∈ R, z ≥ 0

pN (y, z) ≤ pN

(y − 1 − 2q−1

2z, 0

). (93)

More precisely,

d

dtpN

(y +

1 − 2q−1

2t, z + t

) ∣∣∣∣t=0

= − 1

2NE

[⟨H(2)

over(σ, σ′)⟩

y,z

]

= − 1

4N2

N∑

i,j=1

E[〈δ(σi, σj) − q−1〉2y,z

],

(94)

where 〈·〉y,z denotes the random Boltzmann-Gibbs average,

〈f(σ)〉y,z =∑

σ∈ΩN

f(σ)e−HN (y,z,σ)

ZN (y, z)and 〈f(σ, σ′)〉y,z =

∑

σ,σ′∈ΩN

f(σ, σ′)e−HN (y,z,σ)−HN (y,z,σ′)

Z2N (y, z)

.

Proof: The proof is an application of Lemma A.1. Suppose that y and z are fixed, and, let usdefine the interpolating Hamiltonians for the definition of (83). Let

HN (J , σ) = HSGPotts(σ) .

According to Lemma A.2, this has variance

H(1)N (σ) = E

[H2

N (J , σ)]

= (1 − 2q−1)HPottsMF (σ) +

N

2q2,

We also define the non-random Hamiltonian

H(0)N (σ) =

(y − 1 − 2q−1

2z

)HPotts

MF (σ) .

Then we see that the interpolating Hamiltonian,

HN (σ, t) := H(0)N (σ) +

√tHN (J , σ) +

t

2E[H2

N (J , σ)],

may actually be written in terms of the Hamiltonian in (90) as

HN (σ, t) = HN (y(t), z(t), σ) for y(t) = y − 1 − 2q−1

2(z − t) , z(t) = t .

Defining pN(t) relative to the interpolating Hamiltonian in (84), we see that it related to (92) as

pN (t) = pN(y(t), z(t)) .

Therefore, we apply (85), applied to t = z. Again using Lemma A.2 to get the covariance, this givesprecisely (94).

The inequality (93) is a simple application of Lemma A.1. But that lemma will always only leadto upper bounds. In order to get lower bounds, we use the explicit form of the right hand side of (94)and introduce a term to the Hamiltonian corresponding to this. We define

H(2)N (x, y, z, σ, τ) = −xH(2)

over(σ, τ)+y[HPottsMF (σ)+HPotts

MF (τ)]+√z [HSG

Potts(σ)+HSGPotts(τ)]−Nz

q2, (95)

30

for x, y ∈ R and z ≥ 0. The reason we take −x is that we want the two spin configurations σ and τ tobe ferromagnetically coupled. We will define the general partition function for all three of these termsas

Z(2)N (x, y, z) =

∑

(σ,τ)∈ΩNq ×ΩN

q

e−H(2)N

(x,y,z,σ,τ) , (96)

and the quenched finite-volume approximation to the “pressure,”

p(2)N (x, y, z) =

1

2NE

[lnZ

(2)N (x, y, z)

]. (97)

Then, by a calculation like (86), we obtain

∂p(2)N

∂x(x, y, z) =

1

2NE

[〈H(2)

over(σ, τ)〉x,y,z], (98)

where now we write 〈· · · 〉x,y,z for the random Gibbs measure associated to the partition function

Z(2)N (x, y, z), which is consistent with 〈·〉y,z in the special case x = 0.

Because of this, one may rewrite (94) as

d

dtpN

(y +

1 − 2q−1

2t, z + t

) ∣∣∣∣t=0

= −∂p(2)N

∂x(x, y, z)

∣∣∣∣x=0

. (99)

Guerra and Toninelli’s next key step is to replace the derivative by a finite difference approximation.A general property guarantees that pN(x, y, z) is convex in x. Therefore

∂p(2)N

∂x(x, y, z)

∣∣∣∣x=0

≤ p(2)N (x, y, z) − p

(2)N (0, y, z)

x=

1

xp(2)N (x, y, z) − 1

xpN (y, z) ,

for any x > 0. So we may obtain an inequality from (99):

d

dtpN

(y +

1 − 2q−1

2t, z + t

)≥ 1

xpN

(y +

1 − 2q−1

2t, z + t

)− 1

xp(2)N

(x, y +

1 − 2q−1

2t, z + t

).

This leads to the following:

Corollary A.4 For any x > 0,

d

dt

(e−t/xpN

(y +

1 − 2q−1

2t, z + t

))≥ −e−t/x

xp(2)N

(x, y +

1 − 2q−1

2t, z + t

). (100)

This corollary is a direct consequence of the previous inequality. The key fact is that now, turningthe inequality around,

− d

dt

(e−t/xpN

(y +

1 − 2q−1

2t, z + t

))≤ e−t/x

xp(2)N

(x, y +

1 − 2q−1

2t, z + t

),

and now Guerra and Toninelli’s monotonicity principle in the form of Lemma A.1 may be applied toyield and upper bound for the right hand side.

Corollary A.5 For x, y ∈ R, z ≥ 0, and t ∈ [0, z],

p(2)N (x, y, z) ≤ p

(2)N

(x + z, y − 1 − 2q−1

2z, 0

), (101)

31

This is proved similarly to Corollary A.3, except that now we use the pair, (σ, τ) ∈ Ω2N , as thebasic spin configuration.

Proof: Now we replace N by 2N and consider the pair (σ, τ) ∈ Ω2N . Let

H2N (J , σ, τ) = HSGPotts(σ) + HSG

Potts(τ) .

Using Lemma A.2, this has variance

H(1)2N (σ, τ) = E

[H2

2N (J , σ, τ)]

= (1 − 2q−1)(HPotts

MF (σ) + HPottsMF (τ)

)+ 2H(2)

over(σ, τ) +N

q2.

This time, we define the non-random Hamiltonian

H(0)2N (σ, τ) = −(x + z)H(2)

over(σ, τ) +

(y − 1 − 2q−1

2z

)(HPotts

MF (σ) + HPottsMF (τ)

).

The interpolating Hamiltonian is

H2N (σ, τ, t) := H(0)2N (σ, τ) +

√tH2N (J , σ, τ) +

t

2H

(1)2N (σ, τ) .

In terms of (95) this isH2N (σ, τ, t) = HN (x(t), y(t), z(t), σ) ,

for

x(t) = x + z − t , y(t) = y − 1 − 2q−1

2(z − t) , z(t) = t .

Defining p2N (t) relative to the interpolating Hamiltonian H2N (σ, τ, t) as in (84), but with N replacedby 2N , it is related to (97) as

p2N (t) = p(2)N (x(t), y(t), z(t)) .

Therefore, we apply (85), applied to t = 0 and t = z. This gives the inequality that was claimed. Thistime, we do not calculate the remainder, so we do not need to use the general covariance formula.

A.3 Conclusion of Lemma 8.1.

We may now establish Lemma 8.1 by completing the proof. By combining (93), (100) and (101), wehave

pN

(y − 1 − 2q−1

2z, 0

)≤ pN (y, z)

≤ ez/xpN

(y − 1 − 2q−1

2z, 0

)−∫ z

0

p(2)N

(x + z − t, y − 1 − 2q−1

2z, 0

)et/x

xdt . (102)

Note that we may rewrite this as

p(2)N

(0, y − 1 − 2q−1

2z, 0

)≤ pN (y, z)

≤ ez/xp(2)N

(0, y − 1 − 2q−1

2z, 0

)−∫ z

0

p(2)N

(x + z − t, y − 1 − 2q−1

2z, 0

)et/x

xdt .

But p(2)N (x, y, 0) represents the pressure from a true mean-field classical spin system. This may be

calculated by large deviation techniques. Thus both the lower and upper bounds may be calculatedthis way. When they match, we will be guaranteed to be in the high temperature region.

32

Lemma A.6 For each u, v ∈ R, let us define

L(u, v; ρ) =u

2

q∑

a,b=1

(ρab − q−2

)2+v

2

q∑

a=1

[q∑

b=1

(ρab − q−2)

]2+

q∑

b=1

[q∑

a=1

(ρab − q−2)

]2−

q∑

a,b=1

ρab ln ρab .

where we assume ρ is in M2q, the set of all probability distributions ρ = (ρab)

qa,b=1 on 1, . . . , q2. For

any x, y ∈ R, we have

limN→∞

ln p(2)N (x, y, 0) =

1

2maxρ∈M2

q

L(x, xq−1 − y; ρ) .

Proof: For a pair (σ, τ) ∈ ΩNq × ΩN

q , let us define ρ(σ, t) = (ρab(σ, t))qa,b=1 ∈ M2

q, where

ρa,b(σ, τ) =1

N

N∑

i=1

δ(σi, a)δ(τi, b) ,

for a, b = 1, . . . , q. Let us also define

ρ(1)a (σ) =1

N

N∑

i=1

δ(σi, a) and ρ(2)b (τ) =

1

N

N∑

i=1

δ(τi, b) .

We note that

ρ(1)a (σ) =

q∑

b=1

ρa,b(σ, τ) and ρ(2)b (τ) =

q∑

a=1

ρa,b(σ, τ) .

Therefore, we may write

H(2)over(σ, τ) =

1

2N

N∑

i,j=1

[δ(σi, σj) − q−1

]·[δ(τi, τj) − q−1

]

=1

2N

N∑

i,j=1

q∑

a,b=1

δ(σi, a)δ(σj , a)δ(τi, b)δ(tj , b) − q−1

q∑

a=1

δ(σi, a)δ(σj , a) − q−1

q∑

b=1

δ(τi, b)δ(τi, b) + q−2

=N

2

q∑

a,b=1

[ρa,b − q−2]2 − q−1

q∑

a=1

[ρ(1)a − q−1

]2− q−1

q∑

b=1

[ρ(2)b − q−1

]2 .

Similarly,

HPottsMF (σ) + HPotts

MF (τ) =N

2

(q∑

a=1

[ρ(1)a − q−1

]2+

q∑

b=1

[ρ(2)b − q−1

]2)

.

Therefore, using (96), we have the non-random partition function when z = 0:

Z(2)(x, y, 0) =∑

σ,τ∈ΩNq

eNE(x,xq−1−y;ρ(σ,τ)) ,

where

E(u, v; ρ) =u

2

q∑

a,b=1

[ρab − q−2

]2+

v

2

q∑

a=1

[q∑

b=1

(ρab − q−2)

]2+

q∑

b=1

[q∑

a=1

(ρab − q−2)

]2 .

But standard large deviation calculations give

limN→∞

ln #(σ, τ) ∈ ΩNq : ρ(σ, τ) ∈ AN

= −maxρ∈A

q∑

a,b=1

ρab ln ρab

33

u

v

β(q)

β(q2)

Figure 1: By Lemma A.7, we can deduce that the region in grey is a subset of the region of pairs (u, v)such that L(u, v; ρ) has the symmetric state ρ = ρ∗ as an optimizer.

for any set A ⊆ M2q which is the closure of its interior. (See for instance, [26].) Then Varadhan’s

lemma implies that

limN→∞

lnZ(2)(x, y, 0)

N= max

ρ∈M2q

E(x, xq−1 − y; ρ) −q∑

a,b=1

ρab ln ρab

.

This is the desired result because L(u, v; ρ) = E(u, v; ρ) −∑qa,b=1 ρab ln ρab.

We define ρ∗ ∈ M2q to be the symmetric state:

ρ∗ab = q−2 for each a, b = 1, . . . , q.

We say that (u, v) is a “trivial point” of L if ρ∗ is an optimizer of L(u, v; ρ). In other words, (u, v) isa trivial point of L if

maxρ∈M2

q

L(u, v; ρ) = 2 ln q .

Note that in the case that there are several optimizers, we still say that (x, y) is a trivial point if oneof the optimizers is ρ∗.

Lemma A.7 (i) If (u, v) is a trivial point of L, then so is (u′, v′) for all pairs such that u′ ≤ u andv′ ≤ v.(ii) If (u1, v1) and (u2, v2) are trivial points, then (u, v) is also a trivial point for any convex combination

u = tu1 + (1 − t)u2 and v = tv1 + (1 − t)v2 , for t ∈ [0, 1] .

(iii) For q ≥ 2, let

β(q) =2(q − 1)

q − 2ln(q − 1) ,

which is the critical temperature for the q-state Potts ferromagnet. Then (β(q2), 0) and (0, β(q)) areboth trivial points.

In Figure 1, we have sketched a region of pairs (u, v) which are trivial, in the sense defined above.Proof: Suppose u′ ≤ u and v′ ≥ v. Then for any ρ ∈ M2

q, we have

L(u, v; ρ)−L(u′, v′; ρ) =u− u′

2

q∑

a,b=1

(ρab − q−2

)2+v − v′

2

q∑

a=1

[q∑

b=1

(ρab − q−2

)]2

+

q∑

b=1

[q∑

a=1

(ρab − q−2

)]2 .

34

Hence, relative to the symmetric measure ρ∗ such that ρ∗ab ≡ q−2, we have

L(u, v; ρ) − L(u, v; ρ∗) = L(u′, v′; ρ) − L(u′, v′; ρ∗) +u− u′

2

q∑

a,b=1

(ρab − q−2

)2

+v − v′

2

q∑

a=1

[q∑

b=1

(ρab − q−2

)]2

+

q∑

b=1

[q∑

a=1

(ρab − q−2

)]2 .

The last two terms are nonnegative. Therefore if

L(u′, v′, ρ) > L(u′, v′; ρ∗) ,

for some ρ ∈ M2q, then it follows that

L(u, v; ρ) > L(u, v; ρ∗) ,

for that same ρ. Part (i) of the lemma is the statement of the contrapositive of this result.To prove part (ii), note that for (u, v) = t(u1, v1) + (1 − t)(u2, v2), we have

L(u, v; ρ) = tL(u1, v1; ρ) + (1 − t)L(u2, v2; ρ) ,

for every ρ ∈ M2q. Therefore, if L(u, v; ρ) > L(u, v; ρ∗), then either

L(u1, v1; ρ) > L(u1, v1; ρ∗) ,

orL(u2, v2; ρ) > L(u2, v2; ρ∗) ,

or both. So, once again, the result follows by taking the contrapositive statement.For part (iii) first note that when v = 0, then L(u, 0; ρ) is precisely the large-deviation optimization

problem for the mean-field r-state Potts model with r = q2. This proves that (u = β(q2), v = 0) is atrivial point. This follows from the rigorous analysis of the mean-field Potts model. See for instance,[19] or [6].

When u = 0 the model is not equivalent to the q-state Potts model. But we may show that theoptimizer is still the same. Given any ρ ∈ M2

q, let us define two measures in Mq, which are measures

on just the set 1, . . . , q, as ρ(1) and ρ(2), the marginals

ρ(1)a =

q∑

b=1

ρab and ρ(2)b =

q∑

a=1

ρab .

Then we may define a new “product measure,” ρ ∈ M2q, such that

ρab = ρ(1)a ρ(2)b .

Note that the term attached to v/2 in the partition function is

q∑

a=1

(ρ(1)a − q−1)2 +

q∑

b=1

(ρ(2)b − q−1)2 .

But the process of taking marginals and product measures stabilizes in the sense that the marginals ofρ are: ρ(1) = ρ(1) and ρ(2) = ρ(2). So, if u = 0 then the only difference between the function evaluatedat ρ and ρ comes from the entropy term, and we obtain:

L(0, v; ρ) − L(0, v; ρ) = −q∑

a,b=1

[ρab ln(ρab) − ρab ln(ρab)] .

35

But a standard calculation with entropy (see for example, chapter 1 of [27]) shows that

−q∑

a,b=1

[ρab ln(ρab) − ρab ln(ρab)] =

q∑

a,b=1

ρab ln

(ρabρab

),

and this is always nonnegative. So, given any ρ ∈ M2q, one can always replace ρ by ρ, and L(0, v; ·)

will not decrease.But also,

L(0, v; ρ) = L(v; ρ(1)) + L(v; ρ(2)) ,

where for any ρ ∈ Mq, we have

L(β; ρ) =β

2

q∑

a=1

(ρa − q−1)2 −q∑

a=1

ρa ln(ρa) .

This is the large-deviation optimization problem for the mean-field q-state Potts model, at inversetemperature β. Also note that there is not restriction on ρ(1) or ρ(2): given any two such measures,we may have started out by choosing ρ to be the their product measure. Therefore, the optimizationproblem for L(0, v; ·) has the same solution as twice the optimization problem for the q-state Pottsmodel at inverse temperature v. (If v > 0 then saying the Potts model has a negative inverse tem-perature just means we consider the antiferromagnet instead of the ferromagnet, which has no phasetransition because of “convexity.”) In particular, this proves that the critical value of v is v = β(q). Ifv ≥ −β(q) then ρ∗ is still an optimizer for L(0, v; ·). Again, this follows from the rigorous analysis ofthe mean-field q-state Potts model, as in [19] or [6].

Proof of Lemma 8.1: Combining (102), Lemma A.6 and Lemma A.7, we see that

ln q ≤ lim infN→∞

pN (y, z) ≤ lim supN→∞

pN(y, z) ≤ ez/x ln q −∫ z

0

(ln q)et/x

xdx = ln q ,

as long as (u, v) is a trivial point of L for

u = x + z and v = (x + z)q−1 − y +1 − 2q−1

2z = xq−1 − y +

1

2z .

Here it is required that x > 0. But the various bounds we established for the partial derivatives of pNapply to show that it is Lipschitz. Therefore, even if (u, v) is only a trivial point of L for

u = z and v =1

2z − y , (103)

it is still the case that pN (y, z) → ln q as N → ∞. From Lemma A.7 once again, (u, v) is a trivialpoint if it lies in the generalized triangle with vertices (0, β(q)), (β(q2), 0) and (0,−∞). But (103) isa linear mapping. So this means the condition for (z, y) is that it lies in the generalized triangle withvertices (0,−β(q)), (β(q2), 1

2β(q2)) and (0,∞). This is shown in Figure 2. In this region, we know thatpN(y, z) has the limiting value ln q. Note that this does include a neighborhood of (0, 0): 0 ≤ z ≤ ǫand −δ < y < δ for some ǫ, δ > 0, as desired.

Now, combining this with (94), since pN (y, z) is constant in this region, we see that

1

N2

N∑

i,j=1

E[〈δ(σi, σj) − q−1〉2y,z

]

converges to zero, as N → ∞, almost everywhere in this region. This also means that for almost every(y, z) in this region,

1

N2

N∑

i,j=1

〈δ(σi, σj)〉y,z P→ q−1 as N → ∞, ,

36

z

y

β(q)

(β(q2), 12β(q2))

Figure 2: The region in grey is a subset of the region of pairs (y, z) (drawn with the coordinate pairsreversed as (z, y) on the ordinate and abscissa) such that pN (y, z) converges to its trivial value ln q.

whereP→ indicates convergence in probability. We can use this to determine the desired formula for

N−1E

[ln ZN (0, λ, κ)

].

When λ = κ = 0 we have the trivial formula

1

NE

[ln ZN (0, 0, 0)

]= ln q .

(Recall that we will assume c = 0, throughout.) That serves as a starting point. We may calculate

1

NE

[ln ZN (0, λ, κ)

]= ln q +

∫ 1

0

d

dtE

[1

Nln ZN (0, λ(t), κ(t))

]dt ,

for any differentiable path (λ(t), κ(t)) such that: λ(0) = 0, λ(1) = λ, κ(0) = 0, κ(1) = κ.We can relate the partial derivatives of this to the partial derivatives of pN (y, z). Recall that in

our definition of HN (y, z) we did not exactly match the definition needed to obtain ZN (0, λ, κ) fromLemma 8.1. We subtracted some trivial terms to obtain pN (y, z) that made the analysis proceed moreeasily. Specifically, we subtracted q−1 from δ(σi, σj) in (87) and (88), and we subtracted Nz/(2q2) in(90). But these are all constant shifts, not depending on the spin configurations. Therefore, none ofthese changes affects the random Boltzmann-Gibbs measures, because they cancel in the numeratorand denominator in the definition of the Boltzmann-Gibbs weights. In other words, considering thedefinition of ωN (· · · ) from (72), we do have

ωN(· · · ) = 〈· · · 〉y,z if c = 0 , λ = y and κ = z .

The partial derivatives of N−1E[ln ZN (0, λ, κ)] are calculated in (73) and (74):

∂

∂λE

[1

Nln ZN

]= − 1

2N2

N∑

i,j=1

E [ωN (δ(σi, σj))] and

∂

∂κE

[1

Nln ZN

]=

1

4N2

N∑

i,j=1

E [ωN (δ(σi, σj))] −1

4N2

N∑

i,j=1

E[ω2N (δ(σi, σj))

].

But since we know that

1

N2

N∑

i,j=1

〈δ(σi, σj)〉y,z P→ q−1 as N → ∞, ,

for almost every (y, z) in the specified region, this means that

∂

∂λE

[1

Nln ZN

]N→∞−→ − 1

2qand

∂

∂κE

[1

Nln ZN

]N→∞−→ 1

4q− 1

4q2,

37

for almost every pair (λ, κ) in the same region.Now there seems to be a small problem. We have established convergence of the derivatives for

Lebesque-a.e. pair (λ, κ) in the desired region. But we have to integrate over a 1-dimensional curve,which has Lebesgue measure zero. (In fact, we could have concluded convergence a.e. with respect toa 1-dimensional Lebesgue measure, but it would have been for a curve, with a specification for theslope between y and z, which may not have matched our desired curve.) However, we also know that

the derivatives in question are all uniformly bounded. That implies continuity of E[

1N ln ZN (0, λ, κ)

]

with respect to λ and κ. So we can average a bit over these points, to turn the contour integral intoan area integral. Then we can reduce the averaging window, and use continuity to conclude that weobtain the original pressure. Therefore, using this, we do conclude that for (λ, κ) in this region wehave

limN→∞

1

NE

[ln ZN (0, λ, κ)

]= ln q − λ

2q+

(q − 1)κ

4q2,

as claimed.

Acknowledgements

We thank Alessandra Bianchi and Anton Bovier for useful discussion on the ferromagnetic version of themodel. We thank F. Krzakala for several useful observations. PC thanks STRATEGIC RESEARCHGRANT (University of Bologna). The work of SD is supported in part by The Netherlands Organisa-tion for Scientific Research (NWO). CG acknowledges INTERNATIONAL RESEARCH PROJECTS(Fondazione Cassa di Risparmio and University of Modena) for partial financial support.

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