Journal of Theoretical Probabilityhttps://doi.org/10.1007/s10959-020-01051-8

Color Representations of Ising Models

Malin P. Forsström1

Received: 5 August 2020 / Revised: 6 October 2020 / Accepted: 22 October 2020© The Author(s) 2020

AbstractIn Steif and Tykesson (J Prob 16:899–955, 2019), the authors introduced the so-calledgeneral divide and color models. One of the best-known examples of such a model isthe Ising model with external field h = 0, which has a color representation given bythe random cluster model. In this paper, we give necessary and sufficient conditionsfor this color representation to be unique. We also show that if one considers the Isingmodel on a complete graph, then for many h > 0, there is no color representation.This shows, in particular, that any generalization of the random cluster model whichprovides color representations of Ising models with external fields cannot, in general,be a generalized divide and color model. Furthermore, we show that there can be atmost finitely many β > 0 at which the random cluster model can be continuouslyextended to a color representation for h �= 0.

Mathematics Subject Classification (2020) 60G99 · 60K35

1 Introduction

A simple mechanism for constructing random variables with a positive dependencystructure is the so-called generalized divide and color model. This model was firstintroduced in [12], but similar constructions had already arisen in many differentcontexts.

Definition 1.1 Let S be a finite set. A {0, 1}S-valued random variable X := (Xi )i∈Sis called a generalized divide and color model if X can be generated as follows.

1. Choose a random partition π of S according to some arbitrary distribution μ.2. Let π1, . . . , πm be the partition elements of π . Independently for each i ∈

{1, 2, . . . ,m}, pick a “color” ci ∼ (1− p)δ0 + pδ1, and assign all the elements inπi the color ci by letting X j = ci for all j ∈ πi .

B Malin P. Forsströmmalinpf@kth.se

1 Department of Mathematics, KTH Royal Institute of Technology, 100 44 Stockholm, Sweden

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The final {0, 1}-valued process X is called the generalized divide and color modelassociated to μ and p, and we say that μ is a color representation of X .

As detailed in [12], many processes in probability theory are generalized divideand color models, one of the most prominent examples being the Ising model with noexternal field. To define this model, let G = (V , E) be a finite connected graph withvertex set V and edge set E . We say that a random vector X = (σi )i∈V ∈ {0, 1}V isan Ising model on G with interaction parameter β > 0 and external field h ∈ R if Xhas probability density function νG,β,h proportional to

exp(β

∑{i, j}∈E

(1σi=σ j − 1σi �=σ j

) + h∑i∈V

(1σi=1 − 1σi=0

)).

The parameter β will be referred to as the interaction parameter and the parameter has the strength of the external field. We will write XG,β,h to denote a random variablewith XG,β,h ∼ νG,β,h . It is well known that the Ising model has a color representationwhen h = 0 given by the random cluster model. To define the random cluster modelassociated with the Ising model we first, for G = (V , E) and w ∈ {0, 1}E , defineEw := {e ∈ E : we = 1} and note that this defines a partition π [w] of V , wherev, v′ ∈ V are in the same partition element of π [w] if and only if they are in the sameconnected component of the graph (V , Ew). Let ‖π [w]‖ be the number of partitionelements of π [w] and let BV denote the set of partitions of V . For r ∈ (0, 1) andq ≥ 0, the random cluster model μG,r ,q is defined by

μG,r ,q(π′) = 1

Z ′G,r ,q

∑

w∈{0,1}E :π [w]=π ′

[∏e∈E

rw(e)(1 − r)1−w(e)]q‖π [w]‖, π ′ ∈ BV .

where Z ′G,r ,q is a normalizing constant ensuring that this is a probability measure.

It is well known (see, e.g., [10]) that if one sets r = 1 − exp(−2β), q = 2 andp = 1/2, then μG,r ,q is a color representation of Xβ,0. To simplify notation, we willwrite μG,r := μG,r ,2.

Since many properties of the Ising model with h = 0 have been understood byusing a color representation (given by the random cluster model, see, e.g., [1,6,10]),it is natural to ask if there is a color representation also when h > 0. Moreover,Theorems 1.2 and 1.4 in [8], which state that a random coloring of a set can havemore than one color representation, motivates asking whether there are any colorrepresentationwhen h = 0which is different from the randomclustermodel. Themainobjective of this paper is to provide partial answers to these questions by investigatinghow generalized divide and color models relate to some Ising models, both in thepresence and absence of an external field.

In order to be able to present these results, we will need some additional notation.Let S be a finite set. For any measurable space (S, σ (S)), we let P(S) denote theset of probability measures on (S, σ (S)). When S = {0, 1}T for some finite set T ,then we always consider the discrete σ -algebra, i.e., we let σ({0, 1}T ) be the set ofall subsets of {0, 1}T . Recall the we let BS denote the set of partitions of S. If π ∈ BS

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and T ⊆ S, we let π |T denote the partition of T induced from π in the natural way.On BS we consider the σ -algebra σ(BS) generated by {π |T }T⊆S, π∈BS . In analogywith [12], we let RERS denote the set of all probability measures on (BS, σ (BS))

(RER stands for random equivalence relation). For a graph G with vertex set V andedge set E , we let RERG

V denote the set of probability measures μ ∈ RERV whichhas support only on partitions π ∈ BV whose partition elements induce connectedsubgraphs of G. For each p ∈ (0, 1), we now introduce the mapping �p from RERS

to the set of probability measures on {0, 1}S as follows. Let μ ∈ RERS . Pick π

according to μ. Let π1, π2, . . . , πm be the partition elements of π . Independently foreach i ∈ {1, 2, . . . ,m}, pick ci ∼ (1− p)δ0 + pδ1 and let X j = ci for all j ∈ πi . Thisyields a random vector X = (Xi )i∈S whose distribution will be denoted by �p(μ).The random vector X will be referred to as a generalized divide and color model,and the measure μ will be referred to as a color representation of X or �p(μ). Notethat �p : BS → P({0, 1}S). We will say that a probability measure ν ∈ P({0, 1}S)has a color representation if there is a measure μ ∈ RERS and p ∈ (0, 1) such thatν = �p(μ). For ν ∈ P(S), we let �−1(ν) := {μ ∈ RERS : �p(μ) = ν}. Then,ν has a color representation if and only iff there is p ∈ (0, 1) such that �−1

p (ν) isnon-empty. By Theorems 1.2 and 1.4 in [8], �−1

p (ν) can be non-empty if and onlyif the one-dimensional marginals of ν are all equal to (1 − p)δ0 + pδ1. From this itimmediately follows that for any graph G and any β > 0, we have �−1

p (νG,β,0) = ∅whenever p �= 1 − e−2β .

Our first result is the following theorem, which states that for any finite graph Gand any β > 0, XG,β,0 has at least two distinct color representations.

Theorem 1.2 Let n ∈ N and let G be a connected graph with n ≥ 3 vertices. Further,let β > 0. Then, there are at least two distinct probability measuresμ,μ′ ∈ RERV (G)

such that �1/2(μ) = �1/2(μ′) = νG,β,0. Furthermore, if G is not a tree, then there

are at least two distinct probability measures μ,μ′ ∈ RERGV (G) such that �1/2(μ) =

�1/2(μ′) = νG,β,0.

We remark that if a graphG has only one or two vertices and β > 0, then it is knownfrom Theorem 2.1 in [12] (see also Theorems 1.2 and 1.4 in [8]) that �−1

1/2(νG,β,0) ={μG,1−exp(−2β)}. In other words, when h = 0, the Ising model XG,β,h has a uniquecolor representation, given by the random cluster model μG,1−exp(−2β). To get anintuition for what should happen when h > 0, we first look at a few toy examples.One of the simplest such examples is the Ising model on a complete graph with threevertices. The following result was also included as Remark 7.8(iii) in [12] and in [8]as Corollary 1.8.

Proposition 1.3 Let G be the complete graph on three vertices. Let β > 0 be fixed.For each h > 0, let ph ∈ (0, 1) be such that the marginal distributions of XG,β,h aregiven by (1 − ph)δ0 + phδ1. Then, the following holds.

(i) For each h > 0, we have∣∣�−1

ph (νG,β,h)∣∣ = 1, i.e., XG,β,h has a unique color

representation for any h > 0.(ii) For each h > 0, let μh be defined by {μh} = �−1

p(β,h)(νG,β,h). Then, μ0(π) :=limh→0 μh(π) exists for all π ∈ BV (G), and �1/2(μ0) = νG,β,0. However, μ0 �=μG,1−e−2β .

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Interestingly, if we increase the number of vertices in the underlying graph by one,the picture immediately becomes more complicated.

Proposition 1.4 Let G be the complete graph on four vertices. Let β > 0 be fixed.For each h > 0, let ph ∈ (0, 1) be such that the marginal distributions of XG,β,h aregiven by (1 − ph)δ0 + phδ1. To simplify notation, set x := e2β and yh := e2h. Then,XG,β,h has a color representation if and only if

x5 + 3x2yh + 4xy2h − 2x3y2h + x5y2h − 3y3h + 7x2y3h − x4y3h − x6y3h

+ 4xy4h − 2x3y4h + x5y4h + 3x2y5h + x5y6h ≥ 0.(1)

In particular, if we let β0 := log(2 + √3)/2, then the following holds.

(i) If β < β0, then XG,β,h has a color representation for all sufficiently small h > 0,(ii) Ifβ > β0, then XG,β,h has no color representation for any sufficiently small h > 0.

Moreover, there is no decreasing sequence h1, h2, . . .with limn→∞ hn = 0 andμhn ∈�−1

phn(νG,β,h) such that limn→∞ μhn (π) = μG,1−e−2β (π) for all π ∈ BV (G). In other

words, the random cluster model does not arise as a subsequential limit of colorrepresentations of XG,β,h as h → 0, for any β > 0.

Interestingly, this already shows that there are graphs G and parameters β, h > 0 suchthat the corresponding Ising model XG,β,h does not have any color representations.

The proof of Proposition 1.4 can in principle be extended directly to completegraphs with more than four vertices, but it quickly becomes computationally heavyand the analogues of (1) become quite involved (see Remark 4.1 for the analogueexpression for a complete graph on five vertices). In Fig. 1, we draw the set of all pairs(β, h) ∈ R

2+ which satisfies the inequality in (1) together with the corresponding setfor a complete graph on five vertices.

Figure 1, together with the previous two propositions, suggests that the followingconjectures should hold for all complete graphs G on at least four vertices.

I. If β > 0 is sufficiently small, then XG,β,h has a color representation for all h ∈ R.II. For each β > 0, XG,β,h has a color representations for all sufficiently large h ∈ R.III. Ifβ is sufficiently large, then XG,β,h has no color representation for any sufficiently

small h > 0.IV. Ifβ, h > 0 and XG,β,h has a color representation, then so does XG,β ′,h and XG,β,h′

for all h′ > h and β ′ ∈ (0, β).V. The random cluster model corresponding to XG,β,0 does not arise as a subsequen-

tial limit of color representations of XG,β,h as h → 0.

Our next result concerns the last of these conjectures.

Theorem 1.5 Let n ∈ N and let G be a connected and vertex-transitive graph withn vertices. For each β ≥ 0 and h > 0, let pβ,h ∈ (0, 1) be such that the marginaldistributions of XG,β,h are given by (1− pβ,h)δ0+ pβ,hδ1. Then, there is a set B ⊆ R+with |B| ≤ n(n − 1) such that for all β ∈ R+\B, all sequences h1 > h2 > . . . withlimm→∞ hm = 0 and all sequencesμm ∈ �−1

pβ,hm(νG,β,h), there is a partition π ∈ BV

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Fig. 1 The sets of all(β, h) ∈ R

2+ which are such that

the Ising model XG,β,h , for Gbeing a complete graph on fourvertices (red) and five vertices(black) respectively, has at leastone color representation (seeProposition 1.4 andRemark 4.1).

0 1 2 3 40

1

2

3

4

h

such that limm→∞ μm(π) �= μG,1−e−2β (π). In other words, the random cluster modelon G can arise as a subsequential limit when h → 0 of color representations of XG,β,h

for at most n(n − 1) different values of β.

Interestingly, Theorem 1.5 does not require n to be large. The set of exceptionalvalues for β where the random cluster model could arise as a limit is a consequence ofthe proof strategy used, and could possibly be shown to be empty by using a differentproof.

Our next result shows that the third of the conjectures above is true when n issufficiently large.

Theorem 1.6 Let n ∈ N and let G be the complete graph on n vertices. Further, letβ̂ > 0 and β := β̂/n. If β̂ ≥ β̂c = 1 and n is sufficiently large, then XG,β,h has nocolor representation for any sufficiently small h > 0.

As a consequence of this theorem, the fifth conjecture above is true when β > 1/nand n is sufficiently large.

Our last result gives a partial answer to the second conjecture.

Theorem 1.7 Let n ∈ N and let G be the complete graph on n vertices. Let h > 0 andβ = β(h) be such that (n − 1)β(h) < h. Then, XG,β(h),h has a color representationfor all sufficiently large h.

Simulations suggest that the previous result should be possible to extend toβ(h) < h. This is a much stronger statement, especially for large n, and wouldrequire a different proof strategy. The assumption that (n − 1)β(h) ≤ h, made inTheorem 1.7, is, however, a quite natural condition, since this exactly correspondsto that νG,β,h(1S0[n]\S) is decreasing in |S|, for S ⊆ [n] (here 1S0[n]\S denotes thebinary string x ∈ {0, 1}n with x(i) = 1 when i ∈ S and x(i) = 0 when i /∈ S).

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The rest of this paper will be structured as follows. In Sect. 2, we give the back-ground and definitions needed for the rest of the paper. In Sect. 3, we give a proof ofTheorem 1.2. In Sect. 4, we give proofs of Propositions 1.3 and 1.4 and also discusswhat happens when G is the complete graph on five vertices. Next, in Sect. 5, weprove Theorems 1.5 and 1.6, and in Sect. 6, we give a proof of Theorem 1.7. Finally,in Sect. 7, we state and prove a few technical lemmas which are used throughout thepaper.

2 Background and Notation

The main purpose of this section is to give definitions of the notation used throughoutthe paper, as well as some more background to the questions studied.

2.1 The Original Divide and Color Model

When the generalized divide and color model was introduced in [12], it was introducedas a generalization of the so-called divide and color model, first defined by Häggströmin [11]. To define this family of models, let G be a finite graph with vertex set V andedge set E . Let r ∈ (0, 1), and let λ be a finitely supported probability measure onR. The divide and color model X = (Xv)v∈V (associated to r and λ) is the randomcoloring of V obtained as follows.

1. Pick π ∼ μG,r ,1.2. Let π1, . . . , πm be the partition elements of π . Independently for each i ∈

{1, 2, . . . ,m}, pick ci ∼ λ, and assign all the vertices v ∈ πi the color ci byletting Xv = ci for all v ∈ πi

Note that if λ = (1 − p)δ0 + pδ1, then X ∼ �p(μG,r ,1).Since its introduction in [11], properties of the divide and color model have been

studied in several papers, including, e.g., [3,4,9]. Several closely relatedmodels, whichall in some way generalize the divide and color model by considering more generalmeasures λ, have also been considered (see, e.g., [2]).

We stress that this is not the model we discuss in this paper. To avoid confusionbetween the divide and color model and the generalized divide and color model, wewill usually talk about color representations rather than generalized divide and colormodels.

2.2 Generalizations of the Coupling Between the IsingModel and the RandomCluster Model

When h > 0, there is a generalization of the random cluster model (see, e.g., [5])from which the Ising model can be obtained by independently assigning colors todifferent partition elements. This model has been shown to have properties which canbe used in similar ways as analogue properties of the random cluster model. However,since this model uses different color probabilities for different partition elements,it is not a generalized divide and color models. On the other hand, Proposition 1.4

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shows that there are graphs G and parameters β, h > 0 such that XG,β,h has nocolor representation. This motivates considering less restrictive generalizations of therandom cluster model, such as the one given in [5].

2.3 General Notation

Let 1 denote the indicator function, and for each n ∈ N, define [n] := {1, 2, . . . , n}.When G is a graph, we will let V (G) denote the set of vertices of G and E(G)

denote its set of edges. For each graph G with |V (G)| = n, we assume that a bijectionfrom V (G) to [n] is fixed, and in this way identify binary strings σ ∈ {0, 1}V (G) withthe corresponding binary strings in {0, 1}n . The complete graph on n vertices will bedenoted by Kn .

For all finite sets S, disjoint sets T , T ′ ⊆ S and σ = (σi )i∈S ∈ {0, 1}S , wenow make the following definitions. Let σ |T denote the restriction of σ to T . Writeσ |T ≡ 1 if σi = 1 for all i ∈ T , and analogously write σ |T ≡ 0 if σi = 0 for alli ∈ T . We let 1T 0T

′denote the unique binary string σ ∈ {0, 1}T∪T ′

with σ |T ≡ 1and σ |T ′ ≡ 0. Whenever ν is a signed measure on ({0, 1}S, σ ({0, 1}S), we writeν(1T ) := ∑

T ′⊆[n]\T ν(1T∪T ′0S\(T∪T ′)).We let ‖σ‖ := ∑

i∈S σi and defineχT (σ ) :=∏i∈T (−1)1σi=0 = (−1)n−‖σ‖.We now give some notation for working with set partitions. To this end, recall that

when S is a finite set we let BS denote the set of partitions of S. If S = [n] for somen ∈ N, we let Bn := B[n]. If π ∈ BS has partition elements π1, π2, …, πm , we writeπ = (π1, . . . , πm). Now assume that a finite set S, a partition π ∈ BS and a binarystring σ ∈ {0, 1}S are given. We write π �σ if σ is constant on the partition elementsof π . If π � σ , πi is a partition element of π and j ∈ πi , we write σπi := σ j . Notethat this function is well defined exactly when π � σ . Next, we let ‖π‖ denote thenumber of partition elements of π . Combining these notations, if π � σ then we let‖σ‖π := ∑‖π‖

i=1 σπi . If T ⊆ S, we write π |T to denote the restriction of the π to theset T (so that π |T ∈ BT ). If μ is a signed measure on ({0, 1}S, σ ({0, 1}S)), T ⊆ Sand π ∈ BT , we let μ|T (π) := μ

({π ′ ∈ BS : π ′|T = π}). If π ′, π ′′ ∈ BS , then wewrite π ′ � π ′′ if for each partition element of π ′ is a subset of some partition elementof π ′′.

We let Sn denote the set of all permutations of [n]. Sn acts naturally on Bn bypermuting the elements in [n]. When τ ∈ Sn and π = (π1, . . . , πm) ∈ Bn , we letτ ◦ π := (τ (π1), . . . , τ (πm)). If μ is a signed measure on ({0, 1}n, σ ({0, 1}n)) whichis such that μ(τ ◦π) = μ(π) for all τ ∈ Sn and π ∈ Bn , we say that μ is permutationinvariant.

Finally, recall when G = (V , E) is a finite graph and w ∈ {0, 1}E , we defineEw := {e ∈ E : we = 1} and note that this defines a partition π [w] ∈ BV if we letv, v′ ∈ V be in the same partition element of π [w] if and only if they are in the sameconnected component of the graph (V , Ew). If T ⊆ V and |T | ≥ 2, then we let π [T ]be the unique partition in ∈ BV in which T is a partition element and all other partitionelements are singletons.

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2.4 The Associated Linear Operator

Let n ∈ N, ν ∈ P({0, 1}n) and p = ν(1{1}). It was observed in [12] that ifμ ∈ RER[n]is such that �p(μ) = ν, then μ and ν satisfy the following set of linear equations.

ν(σ ) =∑

π∈Bn : π�σ

p‖σ‖π (1 − p)‖π‖−‖σ‖π μ(π), σ ∈ {0, 1}n . (2)

Moreover, whenever a nonnegative measure μ on (Bn, σ (Bn)) satisfies these equa-tions, then μ ∈ RER[n] and �p(μ) = ν, i.e., then μ is a color representation of ν.A signed measure μ on (Bn, σ (Bn)) which satisfies (2), but which is not necessarilynonnegative, will be called a formal solution to (2). If we for a finite set S let RER∗

Sdenote the set of signed measures on (BS, σ (BS)) and P∗({0, 1}) denote the set ofsigned measures on ({0, 1}n, σ ({0, 1}n)), then for each p ∈ (0, 1) we can use (2) toextend �p : RERS → P({0, 1}S) to a mapping �∗

S : RER∗S → P∗({0, 1}), whose

restriction to RERS is equal to �.The matrix corresponding to the system of linear equations given in (2) is given by

An,p(σ, π) :={p‖σ‖π (1 − p)‖π‖−‖σ‖π if π � σ

0 else, σ ∈ {0, 1}n, π ∈ Bn . (3)

It was shown in [8] that An,1/2 has rank 2n−1, and that when p ∈ (0, 1)\{1/2}, thenAn,p has rank 2n − n. When we use the matrix An,p to think about (2) as a system oflinear equations, we will abuse notation slightly and let μ ∈ RER∗[n] denote both thesigned measure and the corresponding vector (μ(π))π∈Bn , given some unspecifiedand arbitrary ordering of Bn .

2.5 Subsequential Limits of Color Representations

Assume that n ∈ N and that a family N = (νp)p∈(0,1) of probability measures onP({0, 1}n) are given. Further, assume that for each p ∈ (0, 1), themarginal distributionof νp is given by (1 − p)δ0 + pδ1. We say that a measure μ ∈ RER[n] arise as asubsequential limit of color representations of measures in N as p → 1/2, if thereis a sequence p1, p2, . . . in (0, 1)\{1/2} with lim j→∞ p j = 1/2 and a measureμ j ∈ �−1

p j(ν j ) such that for all π ∈ Bn we have lim j→∞ μ j (π) = μ(π).

3 Color Representations of XG,ˇ,0

In this section, we give a proof of Theorem 1.2.

Proof of Theorem 1.2 When p = 1/2, σ ∈ {0, 1}n and π ∈ Bn , then

A(σ, π) := An,p(σ, π) ={2−‖π‖ if π � σ,

0 otherwise.(4)

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For S ⊆ [n] and π ∈ Bn , define

A′(S, π) :=∑

σ∈{0,1}n : σ |S≡1

A(σ, π) = 2−‖π |S‖. (5)

Since for any S ⊆ [n] and π ∈ Bn we have

∑T⊆[n] : S⊆T

A′(T , π)(−1)|T |−|S| =∑

T⊆[n] :S⊆T

∑σ∈{0,1}n :

σ |T ≡1

A(σ, π)(−1)|T |−|S|

=∑

σ∈{0,1}n :σ |S≡1

A(σ, π)∑

T⊆[n] :σ |T ≡1

(−1)|T |−|S| = A(1S0[n]\S, π)

it follows that A and A′ are row equivalent. Moreover, by Möbius inversion theorem,applied to the set of subsets of [n] ordered by inclusion, the matrix

A′′(S, π) :=∑

S′ : S′⊆S

2|S′|(−1)|S|−|S′|A′(S′, π), S ⊆ [n], π ∈ Bn (6)

is row equivalent to A′, and hence also to A. By Theorem 1.2 in [8], A has rank 2n−1,and hence the same is true for A′′.

Now note that if S ⊆ [n], π ∈ Bn , and we let T1, T2, …, T‖π |S‖ denote the partitionelements of π |S , then

∑S′ : S′⊆S

2|S′|(−1)|S|−|S′|A′(S′, π) = 2|S| ∑S′ : S′⊆S

(−2)|S′|−|S|A′(S′, π)

= 2|S| ∑S′ : S′⊆S

(−2−1)|S|−|S′| · 2−‖π |S′ ‖

= 2|S| ∑S1,...,Sm :

∀i∈[m] : Si⊆Ti

m∏i=1

(−2−1)|Ti |−|Si |2−1Si �=∅

= 2|S|m∏i=1

∑Si : Si⊆Ti

(−2−1)|Ti |−|Si | · 2−1Si �=∅

= 2|S|m∏i=1

(1 + (−1)|Ti |) · (2−1)|Ti |+1

= 1(π |S has only even-sized partition elements).

and hence

A′′(S, π) = 1(π |S has only even-sized partition elements). (7)

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Let T be a spanning tree of G. Let BTn ⊆ Bn denote the partitions of [n] whose

partition elements induce connected subgraphs of T . Note that the number of suchpartitions is equal to 2n−1. For S ⊆ [n] with |S| even and π ∈ BT

n , define

AT (S, π) := 1(π |S has only even-sized partition elements). (8)

Then, AT is a submatrix of A′′. We will show that AT has full rank. Since AT is a2n−1 by 2n−1 matrix, this is equivalent to having nonzero determinant. To see thatdet AT �= 0, note first that if S ⊆ [n], |S| is even and π ∈ BT

n , then

B(S, π) :=∑

π ′ : π ′�π

(−1)|π |−|π ′|AT (S, π ′)

= 1

⎛⎝

π has only even-sized partition elementsand any finer partition of S has at least

one odd-sized partition element

⎞⎠ .

Since all partition elements of π ∈ BTn induce connected subgraphs of G, B is a

permutation matrix. Since all permutation matrices have nonzero determinant, thisimplies that B, and hence also AT , has full rank.

Since AT has 2n−1 rows and columns, this implies, in particular, that AT has rank2n−1. On the other hand, AT is a submatrix of A′′, and A′′ is row equivalent to Awhich also has rank 2n−1. This implies, in particular, that when we solve (2), we canuse the columns corresponding to partitions in BT

n as dependent variables.Now recall that since XG,β,0 is the Isingmodel on some graphG, XG,β,0 has at least

one color representation given by μG,1−e−2β . The random cluster model μG,1−e−2β

gives strictly positive mass to all partitions π ∈ Bn whose partition elements induceconnected subgraphs of G. In particular, it gives strictly positive mass to all partitionsinBT

n . If we use the columns corresponding to partitions inBTn as dependent variables,

then all dependent variables are given positive mass by μG,1−e−2β . Since n ≥ 3, thereis at least one free variable. By continuity, it follows that we can find another colorrepresentation by increasing the value of this free variable a little. If G is not a tree,there will be at least of free variable corresponding to a partition π ∈ Bn\BT , whosepartition elements induce connected subgraphs of G (but not T ). From this the desiredconclusion follows. ��

Remark 3.1 It is not the case that all sets of 2n−1 columns of An,1/2 have full rank. Tosee this, note first that there are exactly |Bn|−|Bn−1| partitions inBn in which 1 is not asingleton. If |Bn|−|Bn−1| ≥ 2n−1, then there is a setB′ ⊆ Bn of such partitions of size2n−1. An easy calculation shows that this happens whenever n ≥ 4. Let μ ∈ RER∗[n]be a signed measure on Bn with support only on B′, and let ν := �∗

1/2(μ). Then, by

definition, if ν ∈ P({0, 1}n) then if ν(1{1}0[n]\{1}) = 0. In particular, this implies thatthe columns of An,1/2 corresponding to the partitions in B′ cannot have full rank.

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4 Color Representations of XKn,ˇ,h for n ∈ {3, 4, 5}In this section, we provide proofs of Propositions 1.3 and 1.4. In both cases, Math-ematica was used to simplify the formulas for the color representations for differentvalues of β and h.

Proof of Proposition 1.3 Fix h > 0 and let p = νK3,β,h(1{1}). Note that since|V (K3)| = 3, the relevant set of partitions is given by

B3 ={({1, 2, 3}), ({1, 2}, {3}), ({1}, {2, 3}), ({1, 3}, {2}), ({1}, {2}, {3})

}.

By Theorem 1.4 in [8], the linear equation system A3,pμ = νK3,β,h has a uniqueformal solution μ ∈ RER∗[3]. With some work, one verifies that this solution satisfies

μ(({1, 2}, {3})) = μ

(({1, 3}, {2})) = μ

(({1}, {2, 3})),

and

μ(({1, 2, 3}))

=(e4β − 1

)2e4h

(e4β + e4β+2h + e4β+4h + 5e2h + 2e4h + 2

)(e4β + 2e2h + e4h

) (e4β+4h + 2e2h + 1

) (e4β + e4β+2h + e4β+4h + e2h

)

μ(({1}, {2, 3}))

=(e4β − 1

)e2h

(e2h + 1

)2 (e4β − e4β+2h + e4β+4h + 3e2h

)(e4β + 2e2h + e4h

) (e4β+4h + 2e2h + 1

) (e4β + e4β+2h + e4β+4h + e2h

)

μ(({1}, {2}, {3})) =

(e2h + 1

)2 (e4β − e4β+2h + e4β+4h + 3e2h

)2(e4β + 2e2h + e4h

) (e4β+4h + 2e2h + 1

) (e4β + e4β+2h + e4β+4h + e2h

) .

Since e4β+4h ≥ e4β+2h , it is immediately clear that for all π ∈ B3, β > 0 and h > 0,μ(π) is nonnegative, and hence (i) holds.

By letting h → 0 in the expression for μ(({1}, {2}, {3})), while keeping β fixed,

one obtains

limh→0

μ(({1}, {2}, {3})) = 4

1 + 3e4β.

On the other hand, it is easy to check that

μK3,e−2β(({1}, {2}, {3})) = 4e−2β

3 + e4β.

Since these two expressions are not equal for any β > 0, (ii) holds. ��

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Journal of Theoretical Probability

Proof of Proposition 1.4 Fix h > 0 and let p := νK4,β,h(1{1}). Since νK4,β,h ispermutation invariant, it follows that if νK4,β,h has a color representation, then ithas at least one color representation which is invariant under the action of S4. Itis easy to check that there are exactly five different partitions in B4/S4, namely({1, 2, 3, 4}), ({1, 2, 3}, {4}), ({1, 2}, {3, 4}), ({1}, {2}, {3, 4}) and ({1}, {2}, {3}, {4}).By Theorem 1.5(i) in [8], the linear subspace spanned by all S4-invariant formal solu-tionsμ of A4,pμ = νK4,β,h has dimension one. By linearity, this implies, in particular,that if νK4,β,h has a color representation, then it has at least one color representationwhich is S4-invariant and gives zero mass to at least one of the partitions in B4/S4.This gives us one unique solution for each choice of partition in B4/S4.

Solving the corresponding linear equation systems inMathematica and studying thesolutions, after some work, one obtains the desired necessary and sufficient condition.

��Remark 4.1 With the same strategy as in the proof of Proposition 1.4, one can findanalogous necessary and sufficient conditions for G = Kn when n ≥ 5. However,already when n = 5 the analogue inequality is quite complicated; in this case, onecan show that a necessary and sufficient condition to have a color representation isthat

− x18y10 − x18y8 + x18y7 − x18y6 − x18y4 + x16y14 + x16y12 + x16y9

+ 3x16y8 + 2x16y7 + 3x16y6 + x16y5 + x16y2 + x16 − 3x14y11

− x14y10 − 12x14y9 + 6x14y8 − 11x14y7 + 6x14y6 − 12x14y5 − x14y4

− 3x14y3 + 9x12y13 − 3x12y12 + 6x12y11 + 7x12y10 + 15x12y9 − 4x12y8

+ 18x12y7 − 4x12y6 + 15x12y5 + 7x12y4 + 6x12y3 − 3x12y2 + 9x12y

+ 19x10y12 + 27x10y11 + 14x10y10 + 36x10y9 − 21x10y8 + 45x10y7 − 21x10y6

+ 36x10y5 + 14x10y4 + 27x10y3 + 19x10y2 + 20x8y12 − 18x8y11 + 15x8y10

− 11x8y9 + 50x8y8 − 102x8y7 + 50x8y6 − 11x8y5 + 15x8y4 − 18x8y3

+ 20x8y2 + 82x6y11 + 83x6y10 + 76x6y9 + 178x6y8 + 197x6y7 + 178x6y6

+ 76x6y5 + 83x6y4 + 82x6y3 + 54x4y10 + 226x4y9 + 152x4y8 + 386x4y7

+ 152x4y6 + 226x4y5 + 54x4y4 − 84x2y9 + 12x2y8 − 156x2y7 + 12x2y6

− 84x2y5 − 72y8 − 56y7 − 72y6 ≥ 0

where x = e2β and y = e2h .

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Journal of Theoretical Probability

5 Color Representations for small h > 0

5.1 Existence

The goal of this subsection is to provide a proof of Theorem 1.6. A main tool in theproof of this theorem will be the following result from [8].

Theorem 5.1 (Theorem 1.7 in [8]) Let n ∈ N and let (νp)p∈(0,1) be a family of prob-ability measures on {0, 1}n. For each p ∈ (0, 1), assume that νp has marginalspδ1 + (1 − p)δ0, and that for each S ⊆ [n], νp(1S) is two times differentiable inp at p = 1/2. Assume further that for any S ⊆ T ⊆ [n] and any p ∈ (0, 1), we havethat

νp(0S1T \S) = ν1−p(0

T \S1S).

Then, the set of solutions {(μ(π))π∈Bn } to (4) which arise as subsequential limits ofsolutions to (2) as p → 1/2 is exactly the set of solutions to the system of equations

{∑π∈Bn

2−‖π‖μ(π) = ν1/2(1S), S ⊆ [n], |S|even∑π∈Bn

‖π‖S2−‖π |S‖+1μ(π) = ν′1/2(1

S), S ⊆ [n], |S|odd. (9)

Remark 5.2 By applying elementary row operations to the system of linear equationsin (9), one obtains the following equivalent system of linear equations.

∑π∈Bn

1(π |S has at most one odd-sized partition element) μ(π)

={∑

S′⊆S(−2)|S′|+1ν′1/2(1

S′) if |S| is odd∑

S′⊆S(−2)|S′|ν1/2(1S′) if |S| is even. (10)

(See the last equation in the proof of Theorem 1.7 in [8].)

The proof of Theorem 1.6 will be divided into two parts, corresponding to the twocases β̂ > 1 and β̂ = 1.

Proof of Theorem 1.6 when β̂ > 1 (the supercritical regime). Let zβ̂be the unique

positive root to the equation z = tanh(β̂z). Then, it is well known (see, e.g., [7], p.181) that as n → ∞, we have

2‖XKn ,β,0‖ − n

nd→ 1

2δ−z

β̂+ 1

2δz

β̂

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Journal of Theoretical Probability

and hence, by Lemma 7.4, as n → ∞ we have that

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∑σ∈{0,1}n χ∅(σ ) νKn ,β,0(σ ) = 1∑σ∈{0,1}n (2‖σ‖ − n)χ{1}(σ ) νKn ,β,0(σ ) ∼ nz2

β̂∑σ∈{0,1}n χ[2](σ ) νKn ,β,0(σ ) ∼ z2

β̂∑σ∈{0,1}n (2‖σ‖ − n)χ[3](σ ) νKn ,β,0(σ ) ∼ nz4

β̂∑σ∈{0,1}n χ[4](σ ) νKn ,β,0(σ ) ∼ z4

β̂.

(11)

By Remark 5.2, any color representationμ of νKn ,β,0 which arise as a subsequentiallimit of color representations of νKn ,β,h as h → 0 must satisfy (10). By Lemma 7.3,for any S ⊆ [n] we have that

∑S′⊆S

(−2)|S|νKn ,β,0(1S′

) = (−1)|S| ∑σ∈{0,1}n

χS(σ ) νKn ,β,0(σ )

and

∑S′⊆S

(−2)|S|+1ν′Kn ,β,0(1

S′) = (−1)|S|+1

∑σ∈{0,1}n (2‖σ‖ − n)χS(σ ) νKn ,β,0(σ )∑σ∈{0,1}n (2‖σ‖ − n)χ{1}(σ ) νKn ,β,0(σ )

and hence (10) is equivalent to that

∑π∈Bn

1(π |S has at most one odd-sized partition element) μ(π)

= λ(n)S :=

⎧⎨⎩

∑σ∈{0,1}n (2‖σ‖−n)χS(σ ) νKn ,β,0(σ )∑σ∈{0,1}n (2‖σ‖−n)χ{1}(σ ) νKn ,β,0(σ )

if |S| is odd∑

σ∈{0,1}n χS(σ ) νKn ,β,0(σ ) if |S| is even.(12)

Note that since νKn ,β,0 is permutation invariant, we have λ(n)S = λ

(n)[|S|] for any

S ⊆ [n]. This implies, in particular, that if (15) has a nonnegative solution, then itmust have at least one permutation invariant nonnegative solution μ(n). Observe thatμ(n)|B4 satisfies (15) for any set S ⊆ {1, 2, 3, 4}. For this reason, for each n ≥ 4, wenow consider the system of linear equations given by

{∑π∈B4

1(π |S has at most one odd-sized partition element) μ(π) = λ(n)S

μ(π) = μ(τ ◦ π) for all τ ∈ Sn,

S ⊆ [4]. (13)

Let A(4) denote the corresponding matrix. By Theorem 1.5(i) in [8], the null space ofA(4) has dimension one. Since (13) is a set of linear equations, this implies that if anonnegative solution exists, then there will exist a nonnegative solution μ in whicheither μ

(({1, 2, 3, 4})), μ(({1, 2, 3}, {4})), μ(({1, 2}, {3, 4})), μ(({1, 2}, {3}, {4})) or

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Journal of Theoretical Probability

μ(({1}, {2}, {3}, {4})) is equal to zero. Next, note that when S ⊆ [n] satisfies |S| ≤ 4,

by (11), we have that λ(∞)S := limn→∞ λ

(n)S = z2�|S|/2�

β̂. Define λ(n) = (λ

(n)S )S⊆[4] and

λ(∞) = (λ(∞)S )S⊆[4]. Using these definitions, one verifies that the five (permutation

invariant) solutions μ(∞)1 , μ(∞)

2 , μ(∞)3 , μ(∞)

4 and μ(∞)5 to

A(4)μ = λ(∞) (14)

with at least one zero entry are given by

μ(∞)1

(({1, 2, 3, 4}), ({1}, {2, 3, 4}), ({1, 2}, {3, 4}), ({1}, {2}, {3, 4}), ({1}, {2}, {3}, {4}))

= (0, z2

β̂, z4

β̂/3, −z2

β̂(1 + z2

β̂/3), (1 + z2

β̂)2)

μ(∞)2

(({1, 2, 3, 4}), ({1}, {2, 3, 4}), ({1, 2}, {3, 4}), ({1}, {2}, {3, 4}), ({1}, {2}, {3}, {4}))

= (z2β̂, 0, z2

β̂(z2

β̂− 1)/3, −z2

β̂(z2

β̂− 1)/3, (z2

β̂− 1)2

)

μ(∞)3

(({1, 2, 3, 4}), ({1}, {2, 3, 4}), ({1, 2}, {3, 4}), ({1}, {2}, {3, 4}), ({1}, {2}, {3}, {4}))

= (z4β̂, −z2

β̂(z2

β̂− 1), 0, z2

β̂(z2

β̂− 1),−(z2

β̂− 1)(1 + 3z2

β̂))

μ(∞)4

(({1, 2, 3, 4}), ({1}, {2, 3, 4}), ({1, 2}, {3, 4}), ({1}, {2}, {3, 4}), ({1}, {2}, {3}, {4}))

= (z2β̂(3 + z2

β̂)/4, −z2

β̂(z2

β̂− 1)/4, z2

β̂(z2

β̂− 1)/4, 0, −(z2

β̂− 1)

)

and

μ(∞)5

(({1, 2, 3, 4}), ({1}, {2, 3, 4}), ({1, 2}, {3, 4}), ({1}, {2}, {3, 4}),

({1}, {2}, {3}, {4}))

= ((1 + z2

β̂)2/4, −(z2

β̂− 1)2/4, (z2

β̂− 1)(1 + 3z2

β̂)/12, −(z2

β̂− 1)/3, 0

).

Above, all the negative entries are displayed in red. Since all of these solutions have atleast one strictly negative entry, it follows that any solutionsμ(∞) to A(4)μ(∞) = λ(∞)

must have at least one strictly negative entry.Now letμ(n) be a permutation invariant solution to (15). Then,μ(n)|B4 must satisfy

A(4)μ(n)|B4 = λ(n). But this implies that for all S ⊆ [4] we have that

limn→∞(A(4)μ(n)|B4 − A(4)μ(∞))(S) = lim

n→∞ λ(n)(S) − λ(∞)(S) = 0.

Since

limn→∞(A(4)μ(n)|B4 − A(4)μ(∞))(S) = A(4)( lim

n→∞ μ(n)|B4 − μ(∞))(S)

this show that

limn→∞ μ(n)|B4 − μ(∞) ∈ Null A(4).

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Journal of Theoretical Probability

This, in particular, implies thatμ(n) must have negative entries for all sufficiently largen, and hence, the desired conclusion follows. ��

We now give a proof of Theorem 1.6 in the case β̂ = 1. This proof is very similarto the previous proof, but requires that we look at the distribution of the first fivecoordinates of XKn ,β,0 and are more careful with the asymptotics.

Proof of Theorem 1.6 when β̂ = 1 (the critical regime). Assume that n is large. ByTheorem V.9.5 in [7], as n → ∞, we have that

2‖XKn ,β,0‖ − n

n3/4d→ X ,

where X is a random variable with probability density function f (x) = 31/4�(1/4)√2

e−x4/12, x ∈ R.Let m(n)

1 , m(n)2 , etc., be the moments of XKn ,β,0. By Lemma 7.4, as n → ∞, we

have that

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

∑σ∈{0,1}n χ∅(σ ) νKn ,β,0(σ ) = 1∑σ∈{0,1}n (2‖σ‖ − n)χ{1}(σ ) νKn ,β,0(σ ) ∼ n1/2m(n)

2∑σ∈{0,1}n χ[2](σ ) νKn ,β,0(σ ) ∼ n−1/2m(n)

2∑σ∈{0,1}n (2‖σ‖ − n)χ[3](σ ) νKn ,β,0(σ ) ∼ m(n)

4∑σ∈{0,1}n χ[4](σ ) νKn ,β,0(σ ) ∼ n−1m(n)

4∑σ∈{0,1}n (2‖σ‖ − n)χ[5](σ ) νKn ,β,0(σ ) ∼ n−1/2m(n)

6 .

For each S ⊆ [n], let λ(n)S be defined by (15) and define λ(n) = (λ

(n)S )S⊆[n]. Further,

let λ̂(n) be the vector containing only the largest order term of each entry of λ(n), i.e.,let ⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

λ̂(n)(∅) = 1

λ̂(n)([1]) = 1

λ̂(n)([2]) = n−1/2m(n)2

λ̂(n)([3]) = n−1/2m(n)4 /m(n)

2

λ̂(n)([4]) = n−1m(n)4

λ̂(n)([5]) = n−1m(n)6 /m(n)

2 .

Then, we have that λ̂(n)(S) − λ(n)(S) = o(n−1/2) for all S ⊆ [n].Wenowproceed as in the supercritical case.ByRemark5.2, any color representation

μ of νKn ,β,0 which arise as a subsequential limit of color representations of νKn ,β,h ash → 0 must satisfy (10). By Lemma 7.3, for any S ⊆ [n] we have that

∑S′⊆S

(−2)|S|νKn ,β,0(1S′

) = (−1)|S| ∑σ∈{0,1}n

χS(σ ) νKn ,β,0(σ )

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Journal of Theoretical Probability

and

∑S′⊆S

(−2)|S|+1ν′Kn ,β,0(1

S′) = (−1)|S|+1

∑σ∈{0,1}n (2‖σ‖ − n)χS(σ ) νKn ,β,0(σ )∑σ∈{0,1}n (2‖σ‖ − n)χ{1}(σ ) νKn ,β,0(σ )

and hence (10) is equivalent to that

∑π∈Bn

1(π |S has at most one odd-sized partition element) μ(π)

= λ(n)S :=

⎧⎨⎩

∑σ∈{0,1}n (2‖σ‖−n)χS(σ ) νKn ,β,0(σ )∑σ∈{0,1}n (2‖σ‖−n)χ{1}(σ ) νKn ,β,0(σ )

if |S| is odd∑

σ∈{0,1}n χS(σ ) νKn ,β,0(σ ) if |S| is even.(15)

Note that νKn ,β,0 is permutation invariant, and hence λ(n)S = λ

(n)[|S|] for any S ⊆ [n].

This implies, in particular, that if (15) has a nonnegative solution, then it must haveat least one permutation invariant nonnegative solution μ(n). Observe that μ(n)|B5

satisfies (15) for any set S ⊆ {1, 2, 3, 4, 5}. For this reason, for each n ≥ 5, we nowconsider the system of linear equations given by

{∑π∈B5

1(π |S has at most one odd-sized partition element) μ(π) = λ(n)S

μ(π) = μ(τ ◦ π) for all τ ∈ Sn, S ⊆ [5].

(16)

Let A(5) denote the corresponding matrix and define

�5 :={({1, 2, 3, 4, 5}), ({1, 2, 3, 4}, {5}), ({1, 2, 3}, {4, 5}), ({1, 2, 3}, {4}, {5}),({1, 2}, {3, 4}, {5}), ({1, 2}, {3}, {4}, {5}), ({1}, {2}, {3}, {4}, {5})

}⊆ B5.

By Theorem 2.2(i) in [8], the null space of A(5) has dimension two. This implies thatif a nonnegative solution μ to (16) exists, then there will exist two district partitionsπ ′, π ′′ ∈ �5 and a nonnegative solution μ to (16) such that at μ(π ′) = μ(π ′′) = 0.

Let μ(n)1 , μ

(n)2 , . . . , μ

(n)

(72)∈ RER∗[5] be the solutions of (16) which are such

that at least two of μ(({1, 2, 3, 4, 5})), μ

(({1, 2, 3, 4}, {5})), μ

(({1, 2, 3}, {4, 5})),

μ(({1, 2, 3}, {4}, {5})), μ(({1, 2}, {3, 4}, {5})), μ(({1, 2}, {3}, {4}, {5})) and

μ(({1}, {2}, {3}, {4}, {5})) are equal to zero, and let μ̂(n)

1 , μ̂(n)2 , . . . , μ̂

(n)

(72)∈ RER∗[5] be

the corresponding solutions if we replace λ(n) with λ̂(n) in (16). Then, one verifies thatthere is an absolute constant C > 0 such that for all n ≥ 5 and any i ∈ {1, 2, . . . , (72

)}there is π ∈ B5 such that μ̂

(n)i (π) < −Cn−1/2 + o(n−1/2). Next, note that for any

i ∈ {1, 2, . . . , (72)} and S ⊆ [5], we have

A(5)μ(n)i (S) − A(5)μ̂

(n)i (S) = λ(n)(S) − λ̂(n)(S) = o(n−1/2),

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Journal of Theoretical Probability

Since A(5) does not depend on n, it follows that μi (S) − μ̂i (S) = o(n−1/2) for allS ⊆ [5], implying, in particular, thatμ(n)

i has the same sign as the corresponding entryof μ̂i at at least one entry which is negative in μ̂i . This gives the desired conclusion. ��

5.2 The Random Cluster Model is Almost Never a Limiting Color Representation

In this section, we give a proof of Theorem 1.5. A main tool in the proof of this resultwill be the next lemma. The main idea of the proof of Theorem 1.5 will be to showthat the necessary condition given by this lemma does in fact not hold for the randomcluster model.

Throughout this section, we will use the following notation. For a fixed n ∈ N, aset S ⊆ [n] and a partition π ∈ Bn , define

A(S, π) := 1(π |S has at most one odd-sized partition element).

If |S| is odd and A(S, π) = 1, let TS,π denote the unique odd-sized partition elementof π |S .Lemma 5.3 Let n ∈ N and let G be a graph with n vertices. Let β > 0 and assume thatμ ∈ �−1

1/2(νG,β,0) arises as a subsequential limit of color representations of XG,β,h

as h → 0. Finally, let S ⊆ [n] be such that |S| is odd. Then,

n

[ ∑π∈Bn

A(S, π)|TS,π | μ(π)

]=[ ∑i∈[n]

∑π∈Bn

|T{i},π | μ(π)

]·[ ∑

π∈Bn

A(S, π) μ(π)

].

(17)

Proof Notefirst that by combiningTheorem5.1,Remark 5.2 andLemma7.3, it followsthat for any T ⊆ [n], we have

∑π∈Bn

A(T , π) μ(π) =⎧⎨⎩

n∑

σ∈{0,1}n (2‖σ‖−n)χT (σ ) νG,β,0(σ )∑i∈[n]

∑σ∈{0,1}n (2‖σ‖−n)(2σi−1) νG,β,0(σ )

if |T | is odd∑

σ∈{0,1}n χT (σ ) νG,β,0(σ ) if |T | is even.(18)

Next, note that for any such set T ⊆ [n], we have∑

σ∈{0,1}n(2‖σ‖ − n)χT (σ ) νG,β,0(σ ) =

∑σ∈{0,1}n

χT (σ )∑i∈[n]

(2σi − 1) νG,β,0(σ )

=∑i∈[n]

∑σ∈{0,1}n

χT�{i}(σ ) νG,β,0(σ ).

Since |S| is odd, |S�{i}| is even for each i ∈ [n] , and hence, by (18), we have

∑i∈[n]

∑σ∈{0,1}n

χS�{i}(σ ) νG,β,0(σ ) =∑i∈[n]

∑π∈Bn

A(S�{i}, π)μ(π)

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If A(S, π) = 1 for some π ∈ Bn , then since |S| is odd, π has exactly one odd-sizedpartition element, which is equal to the restriction of the partition element TS,π of π

to S. Further, since |S�{ j}| is even for each j ∈ [n], A(S�{ j}, π) = 1 if and only ifj ∈ TS,π , and hence

∑π∈Bn

∑j∈[n]

A(S�{ j}, π) μ(π) =∑

π∈Bn

A(S, π)|TS,π | μ(π)

Combining the three previous inline equations, we obtain

∑σ∈{0,1}n

(2‖σ‖ − n)χS(σ ) νG,β,0(σ ) =∑

π∈Bn

A(S, π)|TS,π | μ(π). (19)

Plugging this into (18), and recalling that |S| is odd by assumption, we get

∑π∈Bn

A(S, π) μ(π) = n∑

σ∈{0,1}n (2‖σ‖ − n)χS(σ ) νG,β,0(σ )∑i∈[n]

∑σ∈{0,1}n (2‖σ‖ − n)(2σi − 1) νG,β,0(σ )

= n∑

π∈BnA(S, π)|TS,π | μ(π)∑

i∈[n]∑

π∈Bn|T{i},π | μ(π)

.

Rearranging this equation, we obtain (17), which is the desired conclusion. ��The purpose of the next lemma is to provide a simpler way to calculate the sums

in (17) when μ is a random cluster model on a finite graph G by translating such sumsinto corresponding sums for Bernoulli percolation.

Lemma 5.4 Let G = (V , E) be a connected and vertex transitive graph. Fix somer ∈ (0, 1) and define r̂ = r

2−r . Further, let m be the length of the shortest cycle in G.Then, there is a constant Z ′′

G,r̂ such that for any function f : Bn → R, we have that

Z ′G,r

Z ′′G,r̂

∑π∈BV

f (π)μG,r (π)

=∑

w∈{0,1}Er̂‖w‖(1 − r̂)|E |−‖w‖ f (π [w]) (1 + 1((V , Ew) contains a cycle))

+ O(r̂ xm+1).

Proof First recall the definition of the random cluster model corresponding to a param-eter q ≥ 0;

μG,r ,q(π) = 1

Z ′G,r ,q

∑

w∈{0,1}E :π [w]=π

r‖w‖(1 − r)|E |−‖w‖q‖π [w]‖, π ∈ BV .

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where Z ′G,r ,q is a normalizing constant ensuring that this is a probability measure.

This implies, in particular, that for any π ∈ BV , we have

Z ′G,r ,q μG,r ,q(π) =

∑

w∈{0,1}E :π [w]=π

r‖w‖(1 − r)|E |−‖w‖q‖π [w]‖

=∑

w∈{0,1}E :π [w]=π

r‖w‖(1 − r)|E |−‖w‖q |V |−‖w‖

· (1 + (q − 1)1((V , Ew) contains a cycle)) + O(rm+1)

= (1 − r)|E |q |V | ∑

w∈{0,1}E :π [w]=π

(r

q(1 − r)

)‖w‖

· (1 + (q − 1)1((V , Ew) contains a cycle)

) + O(rm+1).

If we define r̂ = rr+q(1−r) , then we obtain

Z ′G,r ,q μG,r ,q(π) = (1 − r)|E |q |V | ∑

w∈{0,1}E :π [w]=π

(r̂

1 − r̂

)‖w‖

· (1 + (q − 1)1((V , Ew) contains a cycle)) + O(rm+1).

= (1 − r)|E |q |V |

(1 − r̂)|E |∑

w∈{0,1}E :π [w]=π

r̂‖w‖(1 − r̂)|E |−‖w‖

· (1 + (q − 1)1((V , Ew) contains a cycle)) + O(rm+1).

In particular, this implies that for ant function f : Bn → R, we have

Z ′G,r ,q

∑π∈BV

f (π)μG,r ,q(π) = (1 − r)|E |q |V |

(1 − r̂)|E |∑

w∈{0,1}Ef (π [w]) r̂‖w‖(1 − r̂)|E |−‖w‖

· (1+(q−1)1((V , Ew) contains a cycle))+O(rm+1).

If we let q = 2 the desired conclusion now follows. ��We now use the previous lemma, Lemma 5.4, to give a version of Lemma 5.3 for

Bernoulli percolation. To this end, for r ∈ (0, 1) and q ≥ 0, let μ̂G,r ,q be the measureon {0, 1}E defined by

μ̂G,r̂ ,0(w) := r̂‖w‖(1 − r̂)|E |−‖w‖q‖π [w]‖, w ∈ {0, 1}E .

We then have the following lemma.

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Lemma 5.5 Let n ∈ N and let G = (V , E) be a connected graph on n vertices. Let mbe the length of the shortest cycle in G. Further, let S ⊆ V satisfy |S| = 3 and let (m)

Sbe the number of length m cycles in G which contain all the vertices in S. Assumethat β > 0 is such that μG,1−e−2β is a subsequential limit of color representations ofνG,β,h as h → 0. Set r̂ = r

2−r . Then,

n[ ∑

w∈{0,1}EA(S, π [w])(|TS,π [w]| − 1

)μ̂G,r̂ ,0(w)

]+ n

[

(m)S r̂m · (m − 1)

]

=[ ∑i∈[n]

∑

w∈{0,1}E

(|T{i},π [w]| − 1)μ̂G,r̂ ,0(w)

]·[ ∑

w∈{0,1}EA(S, π [w]) μ̂G,r̂ ,0(w)

]

+ O(r̂m+1).

(20)

Proof Note first that by Lemma 5.3, we have that

n ·[∑π∈Bn

A(S, π)|TS,π | μG,r (π)]

=[∑i∈[n]

∑π∈Bn

|T{i},π | μG,r (π)]

·[∑π∈Bn

A(S, π) μG,r (π)]

or equivalently,

[ ∑w∈{0,1}n

n · μ̂G,r ,2(w)]

·[ ∑w∈{0,1}n

A(S, π [w])(|TS,π [w]| − 1)μ̂G,r ,2(w)

]

=[∑i∈[n]

∑w∈{0,1}n

(|T{i},π [w]| − 1)μ̂G,r ,2(w)

]·[ ∑w∈{0,1}n

A(S, π [w]) μ̂G,r ,2(w)].

By Lemma 5.4, this implies that

[ ∑

w∈{0,1}En(1 + 1(π [w] contains a cycle)) · μ̂G,r̂ ,0(w)

]

·[ ∑

w∈{0,1}EA(S, π [w])(|TS,π [w]| − 1

)(1 + 1(π [w] contains a cycle)) μ̂G,r̂ ,0(w)

]

=[∑i∈[n]

∑

w∈{0,1}E

(|T{i},π [w]| − 1)(1 + 1(π [w] contains a cycle)) μ̂G,r̂ ,0(w)

]

·[ ∑

w∈{0,1}EA(S, π [w])(1 + 1(π [w] contains a cycle)) μ̂G,r̂ ,0(w)

]+ O(r̂m+1).

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Noting that, by assumption, one needs at leastm edges to make a cycle inG, it followsthat the previous equation is equivalent to that

n[ ∑

w∈{0,1}EA(S, π [w])(|TS,π [w]| − 1

)μ̂G,r̂ ,0(w)

]+ n

[

(m)S r̂m · (m − 1)

]

=[∑i∈[n]

∑

w∈{0,1}E

(|T{i},π [w]| − 1)μ̂G,r̂ ,0(w)

]·[ ∑

w∈{0,1}EA(S, π [w]) μ̂G,r̂ ,0(w)

]

+ O(r̂m+1)

which is the desired conclusion. ��We are now ready to give a proof of Theorem 1.5.

Proof Fix any S ⊆ V with |S| = 3. Note that with the notation of Lemma 5.5, wehave that m = 3 and that (3)

S = 1. Moreover, one easily verifies that

∑

w∈{0,1}EA(S, π [w]) μ̂G,r̂ ,0(w) = (1 − (1 − r̂)3) + (1 − r̂)3 · 3(n − 3)r̂2 + O(r̂3).

and that

∑i∈[n]

∑

w∈{0,1}E

(|T{i},π [w]| − 1)μ̂G,r̂ ,0(w) = (2 − 1) · (n − 1) r̂(1 − r̂)2(n−2)

+ (3 − 1) ·(n − 1

2

)(3

1

)r̂2 + O(r̂3).

Moreover, using Table 1, one sees that

∑

w∈{0,1}EA(S, π [w])(|TS,π [w]|−1

)μ̂G,r̂ ,0(w)=3(n−1)r̂2+2(7−12n+3n2)r̂3+O(r̂4)

(21)

This implies, in particular, that

[ ∑

w∈{0,1}EA(S, π [w])(|TS,π [w]| − 1

)μ̂G,r̂ ,0(w)

]+[

(m)S r̂m · (m − 1)

]

= 3(n − 1)r̂2 + 2(8 − 12n + 3n2)r̂3 + O(r̂4)

and that[ ∑

w∈{0,1}E

(|T{1},π [w]| − 1)μ̂G,r̂ ,0(w)

]·[ ∑

w∈{0,1}EA(S, π [w]) μ̂G,r̂ ,0(w)

]+ O(r̂m+1)

= 3(n − 1)r̂2 + 6(n − 1)(n − 3)r̂3 + O(r̂4)

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Journal of Theoretical Probability

Table 1 The local edge patternswe need to consider (up topermutations) to obtain theexpression in (21).

Pattern |TS,π [w]| Probability up to O(r̂4)

(i) 2 3(n − 3)(n − 4)r̂3

(ii) 4 (n − 3)r̂3

(iii) 2 3(n − 3)r̂2(1 − r̂)2(n−2)

(iv) 3 3(n − 3)(n − 4) r̂3

(v) 3 3(n−3

2)r̂3

(vi) 4 3 · 2(n − 3)r̂3

(vii) 3 3r̂2(1 − r̂)1+3(n−3)

(viii) 4 3 · 2(n − 3)r̂3

(ix) 4 3(n − 3)r̂3

(x) 3 r̂3

and hence (20) does not hold. By Lemma 5.5, this implies the desired conclusion. ��

6 Color Representations for large h

In this section we show that for any finite graph G and any β > 0, νG,β,h has a colorrepresentation for all sufficiently large h. The main idea of the proof is to show that avery specific formal solution to (2), which was first used in [8], is in fact nonnegative,and hence a color representation of νG,β,h , whenever h is large enough.

Proof of Theorem 1.7 Fix n ≥ 1 and β > 0.For T ⊆ [n]with |T | ≥ 2, recall thatwe letπ [T ]denote the unique partitionπ ∈ Bn

which is such that T is a partition element of π , and all other partition elements hassize one. Let π [∅] denote the unique partition π ∈ Bn in which all partition elementsare singletons. Let ph := νKn ,β,h(1{1}). Define

μ(π) :=

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Journal of Theoretical Probability

⎧⎪⎪⎨⎪⎪⎩

∑S⊆[n] : T⊆S

(−1)|S|−|T | ∑S′ : S′⊆S(−(1−ph))|S|−|S′|νKn ,β,h(0S

′)

ph(−(1−ph))|S|+p|S|h (1−ph)

if π = π [T ], |T | ≥ 2

1 − ∑T⊆[n] : |T |≥2 μ(π [T ]) if π = π [∅]

0 else.

(22)

By the proof of Theorem 1.6 in [8], we have �∗p(μ) = νKn ,β,h . We will show that,

given the assumptions, μ(π) ≥ 0 for all π ∈ Bn , and hence μ ∈ �−1p (νKn ,β,h). To

this end, note first that for σ ∈ {0, 1}n , we have

νKn ,β,h(σ ) = Z−1Kn ,β,h exp

((√β∑i∈[n]

(2σi − 1) + h/√

β)2

/2),

where ZKn ,β,h is the corresponding normalizing constant. This implies, in particular,that for any S ⊆ [n],

limh→∞

νKn ,β,h(0S1[n]\S)νKn ,β,h(1[n])

= limh→∞ e−2|S|(h+β(n−|S|)) =

{1 if S = ∅0 else.

(23)

This implies, in particular, that limh→∞ ph = 1 and that as h → ∞, we have

ZKn ,β,h ∼ νKn ,β,h(0∅1[n]) → 1.

Similarly, one shows that for any S ⊆ [n], we have

νKn ,β,h(0S) ∼

{νKn ,β,h(0S1[n]\S) if β|S| ≤ h,

νKn ,β,h(0[n]1∅) if β|S| ≥ h.(24)

Since β(n−1) ≤ h by assumption, it follows that νKn ,β,h(0S) ∼ νKn ,β,h(0S1[n]\S)for all S ⊆ [n], and hence, in particular, that ph ∼ νKn ,β,h(0[1]1[n]\{1}).

Combining these observations, it follows that for any S ⊆ [n] and any k ∈ S, wehave

(1 − ph) νKn ,β,h(0S\{k})νKn ,β,h(0S)

∼ νKn ,β,h(0{1}1[n]\{1}) νKn ,β,h(0S\{k}1[n]\(S\{k}))νKn ,β,h(0∅1[n]) νKn ,β,h(0S1[n]\S)

= e−4β(|S|−1)

(25)

and hence when h is sufficiently large,

νKn ,β,h(0S) ∼

[ |S|∏i=1

pe4β(|S|−i)]νKn ,β,h(1

∅) = (1 − ph)|S|e2β|S|(|S|−1). (26)

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Journal of Theoretical Probability

We will now use the equations above to describe the behavior of (22). To this end,fix some S ⊆ [n] and note that by combining (26) and (25), we obtain

∑S′ : S′⊆S

(−1)|S|−|S′|(1 − ph)|S|−|S′|νKn ,β,h(0

S′)

= (−1)|S|(1 − ph)|S| ∑

S′ : S′⊆S

(−1)|S′|e−2β|S′|(|S′|−1) + o(νKn ,β,h(0S)).

Using (22), it follows that if T ⊆ [n] and |T | ≥ 2, then

μ(π [T ]) =∑

S⊆[n] : T⊆S

(−1)|S|−|T | ∑S′ : S′⊆S(−(1 − ph))|S|−|S′|νKn ,β,h(0S

′)

ph(−(1 − ph))|S| + p|S|h (1 − ph)

=∑

S⊆[n] : T⊆S

(−1)|S|−|T |[(−(1 − ph))|S| ∑

S′ : S′⊆S(−1)|S′|e2β|S′|(|S′|−1) + o(νKn ,β,h(0S))

]

1 − ph + o(1 − ph)

= 1

1 − ph

∑S⊆[n] : T⊆S

(−1)|T |(1 − ph)|S| ∑

S′ : S′⊆S

(−1)|S′|e2β|S′|(|S′|−1)

+ o

(νKn ,β,h(0T )

1 − ph

).

We now rewrite the previous equation. To this end, note first that

∑S⊆[n] : T⊆S

(1 − ph)|S| ∑

S′ : S′⊆S

(−1)|S′|e2β|S′|(|S′|−1)

=∑S′⊆[n]

(−1)|S′|e2β|S′|(|S′|−1)∑

S⊆[n] : T∪S′⊆S

(1 − ph)|S|

=∑S′⊆[n]

(−1)|S′|e2β|S′|(|S′|−1)(1 − ph)|T∪S′|(2 − ph)

n−|T∪S′|.

Again using (26), it follows that the largest terms in this sum is of order νKn ,β,h(0T ),and hence, using symmetry, it follows that the previous equation is equal to

∑S′⊆[n] : |S′|≤|T |

(−1)|S′|e2β|S′|(|S′|−1) · (1 − ph)|T∪S′|(2 − ph)

n−|T∪S′|

+ o(νKn ,β,h(0T )).

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Journal of Theoretical Probability

Again using that νKn ,β,h is invariant under permutations, it follows that this is equalto

|T |∑i=0

(−1)i e2βi(i−1)i∑

j=0

(|T |j

)(n − |T |i − j

)(1 − ph)

|T |+(i− j)(1 + p)n−(i− j)

+ o(νKn ,β,h(0T ))

= (1 − ph)|T |

|T |∑i=0

(−1)i e2βi(i−1)(|T | + n(1 − ph))i (2 − ph)

n−i + o(νKn ,β,h(0T )).

Summing up, we have showed that for any T ⊆ [n] with |T | ≥ 2, we have that

μ(π [T ]) = (1 − ph)

|T |−1|T |∑i=0

(−1)|T |−i e2βi(i−1)(|T | + n(1 − ph))i (2 − ph)

n−i

+ o

(νKn ,β,h(0T )

1 − ph

). (27)

Since for any positive and strictly increasing function f : N → R, we have that

|T |∑i=0

(−1)|T |−i ei f (i) >

|T |∑i=|T |−1

(−1)|T |−i ei f (i) > 0,

it follows that

|T |∑i=0

(−1)|T |−i e2βi(i−1)(|T | + n(1 − ph))i (2 − ph)

n−i

≥|T |∑

i=|T |−1

(−1)|T |−i e2βi(i−1)(|T | + n(1 − ph))i (2 − ph)

n−i

= (|T | + n(1 − ph))|T |−1(2 − ph)

n−|T |e2β|T |(|T |−1)

·[

(|T | + n(1 − ph)) − e−4β(|T |−1)(2 − ph)

].

This is clearly larger than zero, and in fact, by (26), as h tends to infinity, it is asymptoticto

(1 − ph)−|T |νKn ,β,h(0

T ) |T ||T |−1[

|T | − e−4β(|T |−1)].

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Journal of Theoretical Probability

This implies that the error term in (27) is much smaller than the rest of the expression,which is strictly positive, i.e.,

μ(π [T ]) ∼ (1 − ph)

|T |−1|T |∑i=0

(−1)|T |−i e2βi(i−1)(|T | + n(1 − ph))i (2 − ph)

n−i > 0.

It now remains to show only that μ(π [∅]) > 0. To this end, note first that for any

T ⊆ [n] with |T | ≥ 2, we have that

(1 − ph)|T |−1

|T |∑i=0

(−1)|T |−i e2βi(i−1)(|T | + n(1 − ph))i (2 − ph)

n−i

≤ (1 − ph)−1(2n)n

|T |∑i=0

(1 − ph)|T |−i (1 − ph)

i e2βi(i−1).

By (26), (1− ph)i e2βi(i−1) ∼ νKn ,β,h(0S). Since νKn ,β,h(0S) tends to zero as h → ∞for any S ⊆ [n] with |S| ≥ 1, it follows that limh→∞ μ

(π [T ]) = 0 provided that

limh→∞ νKn ,β,h(0T )/(1 − ph) = 0. To see that this holds, note simply that by (26),

νKn ,β,h(0S)/(1 − ph) ∼ νKn ,β,h(0T 1[n]\T )

νKn ,β,h(0[1]1[n]\[1])= e−2(|T |−1)(|h|+β(n−|T |−1))

which clearly tends to zero as h → ∞. This concludes the proof. ��Remark 6.1 The previous proof shows that the formal solution to (2) given by (22)is a color representation of νKn ,β,h whenever β is fixed and h is sufficiently large.One might hence ask if (22) is also nonnegative for fixed h �= 0 as β → 0. This is,however, not the case, and one can if fact show that for any fixed h �= 0, when β > 0is sufficiently small, μ(π [T ]) < 0 for any T ⊆ [n] which is such that |T | is odd.Remark 6.2 Given that the relationship we get between β and h in Theorem 1.7 is quitefar from the conjectured result, one might try to get a stronger result by optimizing theabove proof. However, using Mathematica, one can check that the particular formalsolution to (2) given by (22) in fact not nonnegative when (n − 1)β > h whenn = 3, 4, 5.

7 Technical Lemmas

In this section we collect and prove the technical lemmas which have been usedthroughout the paper.

Lemma 7.1 Let n ∈ N and let G = (V , E) be a graph with n vertices. Further, letβ > 0 be fixed and, for h ≥ 0, define pG,β(h) := νG,β,h(1{1}). Finally, let σ ∈ {0, 1}n.

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Journal of Theoretical Probability

Then,

dνG,β,h(σ )

dpG,β

|h=0 = 2(2‖σ‖ − n) νG,β,0(σ )∑σ̂∈{0,1}n (2σ̂1 − 1)(2‖σ̂‖ − n) νG,β,0(σ̂ )

= 2n(2‖σ‖ − n) νG,β,0(σ )∑σ̂∈{−1,1}n (2‖σ̂‖ − n)2 νG,β,0(σ̂ )

.

Proof By definition, for any h ≥ 0 we have

νG,β,h(σ ) = Z−1G,β,h exp(β

∑{i, j}∈E

(2σi − 1)(2σ j − 1) + h(2‖σ‖ − n)),

where

ZG,β,h =∑

σ̂∈{0,1}nexp(β

∑{i, j}∈E

(2σ̂i − 1)(2σ̂ j − 1) + h(2‖σ̂‖ − n)).

If we differentiate ZG,β,h νG,β,h(σ ) with respect to h, we get

d(ZG,β,h νG,β,h(σ )

)

dh|h=0 = (2‖σ‖ − n) ZG,β,0 νG,β,0(σ ).

From this it follows that

dZG,β,h

dh|h=0 =

∑σ̂∈{0,1}n

d(ZG,β,h νG,β,h(σ̂ )

)

dh|h=0

=∑

σ̂∈{0,1}n

(2‖σ̂‖ − n

)ZG,β,0 νG,β,0(σ̂ )

= ZG,β,0 E

[2∥∥XG,β,0

∥∥ − n)]

= ZG,β,0 · 0 = 0.

As

d(ZG,β,h νG,β,h(σ )

)

dh= dZG,β,h

dhνG,β,h(σ ) + ZG,β,h

dνG,β,h(σ )

dh

it follows that

dνG,β,h(σ )

dh|h=0 = (

2‖σ‖ − n)νG,β,0(σ )

and hence

dpG,β

dh|h=0 =

∑σ̂∈{0,1}n :

σ̂1=1

(2‖σ̂‖ − n

)νG,β,0(σ̂ ).

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Journal of Theoretical Probability

Combining the two previous equations, we obtain

dνG,β,h(σ )

dpG,β|h=0=

dνG,β,h(σ )

dh

(dpG,β

dh

)−1|h=0=

(2‖σ‖ − n

)νG,β,0(σ )∑

σ̂∈{0,1}n : σ̂1=1(2‖σ̂‖ − n

)νG,β,0(σ̂ )

.

The desired conclusion now follows by noting that

∑σ̂∈{0,1}n : σ̂1=1

(2‖σ̂‖ − n

)νG,β,0(σ̂ ) = 1

2

∑σ̂∈{0,1}n

(2σ̂1 − 1)(2‖σ̂‖ − n

)νG,β,0(σ̂ )

= 1

2n

∑σ̂∈{0,1}n

(2‖σ̂‖ − n

)2νG,β,0(σ̂ ).

��Lemma 7.2 Let n ∈ N. Further, let ν ∈ P({0, 1}n) and let S ⊆ [n]. Then,

∑T : T⊆S

ν(1T )(−2)|T | =∑

T : T⊆S

(−1)|T |ν(1T 0S\T ).

Proof

∑S′ : S′⊆S

ν(1S′)(−2)|S′| =

∑S′ : S′⊆S

∑T : T⊆S′

ν(1S′)(−1)|S′|

=∑

T : T⊆S

∑S′ : T⊆S′⊆S

ν(1S′)(−1)|S′|

=∑

T : T⊆S

∑S′′ : S′′⊆S\T

ν(1T∪S′′)(−1)|T∪S′′|

=∑

T : T⊆S

(−1)|T | ∑S′′ : S′′⊆S\T

ν(1T∪S′′)(−1)|S′′|

=∑

T : T⊆S

(−1)|T |ν(1T 0S\T ).

��Lemma 7.3 Let n ∈ N and let G be a graph with n vertices. Further, let β > 0 befixed and, for h ≥ 0, define pG,β(h) := νG,β,h(1{1}). Finally, let S ⊆ [n]. Then, thefollowing two equations hold.

(i)

∑S′⊆S

(−2)|S| νG,β,0(1S′

) = (−1)|S| ∑σ∈{0,1}n

χS(σ ) νG,β,0(σ )

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Journal of Theoretical Probability

(ii)

∑S′⊆S

(−2)|S|−1 dνG,β,h(1S′)

dpG,β

|h=0

= n(−1)|S|+1

∑σ∈{0,1}n

(2‖σ‖ − n

)χS(σ ) νG,β,0(σ )

∑σ∈{0,1}n

(2‖σ‖ − n

)2νG,β,0(σ )

.

Proof By Lemma 7.2, we have that

∑T : T⊆S

(−2)|T |νG,β,0(1T ) =

∑T : T⊆S

(−1)|T |νG,β,0(1T 0S\T )

=∑

T : T⊆[n](−1)|T∩S| νG,β,0(1

T 0[n]\T ) =∑

σ∈{0,1}n

∏i∈S

(−(2σi − 1)) νG,β,0(σ )

= (−1)|S| ∑σ∈{0,1}n

∏i∈S

(2σi − 1) νG,β,0(σ ) = (−1)|S| ∑σ∈{0,1}n

χS(σ ) νG,β,0(σ )

and hence (i) holds. To see that (ii) holds, note first that by the same argument asabove, it follows that

∑T : T⊆S

(−2)|T |−1 dνG,β,h(1T )

dpG,β

|h=0= 2−1(−1)|S|+1∑

σ∈{0,1}nχS(σ )

dνG,β,h(σ )

dpG,β

|h=0 .

Applying Lemma 7.1, we obtain (ii). ��

Lemma 7.4 Let n ∈ N, β > 0 and h ≥ 0. Then,

∑σ∈{0,1}n

χ∅(σ ) νKn ,β,h(σ ) = 1

n∑

σ∈{0,1}n

(2‖σ‖ − 1

)χ{1}(σ ) νKn ,β,h(σ ) =

∑σ∈{0,1}n

(2‖σ‖ − n

)2νKn ,β,h(σ )

(n)2∑

σ∈{0,1}nχ[2](σ ) νKn ,β,h(σ ) =

∑σ∈{0,1}n

(2‖σ‖ − n

)2νKn ,β,h(σ ) − n

(n)3∑

σ∈{0,1}n

(2‖σ‖ − n

)χ[3](σ ) νKn ,β,h(σ )

=∑

σ∈{0,1}n

(2‖σ‖ − n

)4νKn ,β,h(σ ) − (3n − 2)

∑σ∈{0,1}n

(2‖σ‖ − n

)2νKn ,β,h(σ )

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Journal of Theoretical Probability

(n)4∑

σ∈{0,1}nχ[4](σ ) νKn ,β,h(σ )

=∑

σ∈{0,1}n

(2‖σ‖ − n

)4νKn ,β,h(σ ) − 2(3n − 4)

∑σ∈{0,1}n

(2‖σ‖ − n

)2νKn ,β,h(σ )

+ 3(n2 − 2n)

and

(n)5∑

σ∈{0,1}n

(2‖σ‖ − n

)χ[5](σ ) νKn ,β,h(σ )

=∑

σ∈{0,1}n

(2‖σ‖ − n

)6νKn ,β,h(σ ) − 10(n − 2)

∑σ∈{0,1}n

(2‖σ‖ − n

)4νKn ,β,h(σ )

+ (15n2 − 50n + 24)∑

σ∈{0,1}n

(2‖σ‖ − n

)2νKn ,β,h(σ ).

Proof By symmetry, for each m ∈ [n] we have that(n

m

) ∑σ∈{0,1}n

χ[m](σ ) νKn ,β,h(σ ) =∑

S⊆[n] :|S|=m

∑σ∈{0,1}n

χS(σ ) νKn ,β,h(σ )

=∑

σ∈{0,1}nνKn ,β,h(σ )

∑S⊆[n] :|S|=m

χS(σ )

(28)

and, completely analogously,

(n

m

) ∑

σ∈{0,1}n(2‖σ‖ − n

)χ[m](σ ) νKn ,β,h(σ ) =

∑

σ∈{0,1}n(2‖σ‖ − n

)νKn ,β,h(σ )

∑S⊆[n] :|S|=m

χS(σ ).

(29)

Now fix some σ ∈ {0, 1}n . Then, we have that∑i∈[n]

1σi=1 = ‖σ‖

and hence

∑S⊆[n] : |S|=m

χS(σ ) =n−‖σ‖∑i=0

(n − ‖σ‖

i

)( ‖σ‖m − i

)(−1)i .

When m ∈ [5], it follows that

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Journal of Theoretical Probability

∑S⊆[n] : |S|=m

χS(σ )

=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

‖σ‖ if m = 1(2‖σ‖−n

)22! − n

2 if m = 2(2‖σ‖−n

)33! − (3n−2)·

(2‖σ‖−n

)6 if m = 3(

2‖σ‖−n)4

4! − (3n−4)·(2‖σ‖−n

)212 + n2−2n

8 if m = 4(2‖σ‖−n

)55! − (n−2)·

(2‖σ‖−n

)312 + (15n2−50n+24)·

(2‖σ‖−n

)120 if m = 5.

By plugging these expressions into Eqs. (28) and (29), the desired conclusion follows.��

Acknowledgements The author acknowledges support from the European Research Council, Grant No.682537, from Stiftelsen G S Magnusons fond and from Knut and Alice Wallenbergs stiftelse. The authorwould like to thank Jeffrey E. Steif for many interesting conversation, and in particular for mentioning theexample given in Proposition 1.3. The author would also like to thank the anonymous referee for makingseveral useful comments.

Funding Open access funding provided by Royal Institute of Technology.

OpenAccess This article is licensedunder aCreativeCommonsAttribution 4.0 InternationalLicense,whichpermits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you giveappropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence,and indicate if changes were made. The images or other third party material in this article are includedin the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. Ifmaterial is not included in the article’s Creative Commons licence and your intended use is not permittedby statutory regulation or exceeds the permitted use, you will need to obtain permission directly from thecopyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

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