Color representations of Ising models

In \cite{st2017}, the authors defined so-called \emph{general divide and color models}. One of the most well-known examples of such a model is the Ising model with external field $h = 0$, which has a color representation given by the random cluster model. In this paper, we give necessary and sufficient conditions for this color representation to be unique. We also show that if one considers the Ising model 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 which provides color representations of Ising models with external fields cannot, in general, be a color representation. Furthermore, we show that there can be at most finitely many $\beta>0$ at which the random cluster model can be continuously extended to a color representation for $h \not = 0$.


Introduction
A simple mechanism for constructing random variables with a positive dependency structure is the so called divide and color model. This was first introduced in [8] but had already arisen in many different contexts. Definition 1.1. A {0, 1}-valued process X := (X i ) i∈S is a 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. Assign a "color" in {0, 1} to each partition element of π as follows. Independently of the chosen partition π, and independently for different partition elements in π, assign, with probability p, all the variables in a partition element the value 1 and with probability 1 − p assign all the variables the value 0.
This final {0, 1}-valued process X is called the divide and color model associated to (µ, p), and we say that (µ, p) is a color representation of X.
As detailed in [8], many processes in probability theory are divide and color models; one of the most prominent examples being the Ising model with zero external field. To define this model, let G = (V, E) be a finite connected graph with vertex set V and edge set E. We say that a random vector X = X β,h = (σ i ) i∈V ∈ {−1, 1} V is an Ising model with interaction parameter β > 0 and external field h ∈ R if X has probability density function ν G,β,h proportional to exp β {i,j}∈E It is well known that the Ising model has a color representation when h = 0 given by the random cluster model (using the set {−1, 1} as colors). To define the random cluster model µ G,p we first, for w ∈ {0, 1} E and r ∈ (0, 1), define the probability measureμ =μ G,r on {0, 1} E bŷ where C(w) is the number of connected components of the subgraph (V, E w ) of G with edge set E w := {e ∈ E : w(e) = 1}, and Z G,r is a normalizing constant ensuring that this is indeed a probability measure. Next, given an element w ∈ {0, 1} E , let π(w) denote the partition of V whose partition elements correspond to the connected subgraphs of (V, E w ). The corresponding measure on the set of partitions of V , denoted by B V , will be denoted by µ G,r . In other words, for π ∈ B V we have that µ G,r (π) =μ G,r w ∈ {0, 1} E : π(w) = π .
The main goal of this paper is to investigate how divide and color representations relate to Ising models, both in the presence and absence of an external field. Our first result is the following theorem, which states that X β,0 in fact has at least two distinct color representations. Theorem 1.2. Let X β,0 be the Ising model on a connected graph G with at least three vertices. Then X β,0 has at least two distinct color representations. Furthermore, if G is not a tree, then there are at least two distinct color representations of X β,0 which has support only at partitions with partition elements inducing connected subgraphs of G.
We remark that if G has only one or two vertices, then it is known from Theorem 2.1 in [8] (see also Theorems 1.2 and 1.4 in [4]) that the Ising model X β,0 on G has a unique color representation, given by the random cluster model.
To get an intuition for what should happen when h > 0, we first look at a few toy examples. Proposition 1.3. Let X β,h be the Ising model on a complete graph with three vertices. Then (i) X β,h has a unique color representation for any β > 0 and h > 0 (ii) the (unique) color representation of X β,0 obtained by letting h → 0 in the unique color representation of X β,h is not the random cluster model.
Interestingly, if we increase the number of vertices in the underlying graph by one, the picture becomes more complicated. Proposition 1.4. Let X h,β be the Ising model on a complete graph with four vertices. To simplify notation, set x = e 2β and y = e 2h . Then X β,h has a color representation if and only if h = 0 or In particular, if we let β 0 := log(2 + √ 3)/2, then the following holds.
(i) If β < β 0 , then X β,h has a color representation for all sufficiently small h, (ii) If β > β 0 , then X β,h has no color representation for any sufficiently small h.
Moreover, the random cluster model does not arise as a limit of color representations of X β,h as h → 0, for any β > 0.
Interestingly, this already shows that there are Ising models X β,h which does not have color representations.
The proof of Proposition 1.4 can in principle be extended directly to any n ≥ 4, but it quickly becomes computationally very heavy and the analogues of (1) become quite involved (see Remark 4.1 for the analogue expression when n = 5). In Figure 1, we draw the set of all pairs (β, h) which satisfies the inequality in (1) together with the corresponding set when n = 5.
This figure, together with the previous two propositions, suggests that the following conjectures should hold for Ising models X β,h on complete graphs.
I. If β > 0 is sufficiently small, then X β,h has a color representation for all h ∈ R.
II. For each β > 0, X β,h has a color representations for all sufficiently large h ∈ R.
III. If β is sufficiently large, then X β,h has no color representation for any sufficiently small h > 0.
IV. If β, h > 0 and X β,h has a color representation, then so does X β ,h and X β,h for all h > h and β ∈ (0, β).
V. The random cluster model corresponding to X β,0 does not arise as a subsequential limit of color representations of X β,h as h → 0.
The next result concerns the last of these conjectures.
Theorem 1.5. Let X β,h be the Ising model on a connected transitive graph G on n vertices. Then the random cluster model on G can arise as a subsequential limit of color representations of X β,h when h → 0 for at most n(n − 1) different values of β. shows the sets of all (β, h) ∈ R 2 + which are such that the Ising model X β,h on K 4 (green) and K 5 (magenta) has at least one color representation (see Proposition 1.4 and Remark 4.1).
Interestingly, Theorem 1.5 does not require n to be large. The set of exceptional values of β where the random cluster model could arise as a limit is a consequence of the proof strategy used, and could most likely be shown to be empty by using a different proof.
Our next result shows that the third of the conjectures above is true when n is sufficiently large. Theorem 1.6. Let n ∈ N,β > 0 and β :=β/n. Further, let X β,h be the Ising model on a complete graph with n vertices. Ifβ ≥β c = 1 and n is sufficiently large, then X has no color representation for any sufficiently small h.
As a consequence of this theorem, the fifth conjecture is true when β > 1/n and n is sufficiently large.
Finally, our last result gives a partial answer to the second conjecture.
Then the Ising model X β(h),h on K n has a color representation for all sufficiently large |h|.
Simulations suggests that the previous result should be possible to extend to β(h) < |h|. This is a much stronger statement, especially for large n, and would require a different proof strategy. The assumption that (n − 1)β(h) ≤ h, made in Theorem 1.7, is however a quite natural condition, since this exactly corresponds to that ν G,β, The rest of this paper will be structured as follows. In Section 2, we give the background and definitions needed for the rest of the paper. In Section 3, we give a proof of Theorem 1.2. In Section 4, we give proofs of Propositions 1.3 and 1.4, and also discuss the case G = K 5 . Next, in Section 5, we prove Theorem 1.5 and Theorem 1.6 and in Section 6, we give a proof of Theorem 1.7. Finally, in Section 7, we state and prove a few technical lemmas which are used throughout the paper.

Background and notation
The main purpose of this section is to give definitions of the notation used throughout the paper.
We will use both {0, 1} n and Σ n := {−1, 1} n to denote binary strings of length n, and will use the natural identification of 0 and −1. For disjoint sets S, T ⊆ [n], we will let 1 S 0 T denote the unique binary string σ ∈ {0, 1} S∪T with σ i = 1 for all i ∈ S and σ j = 0 for all j ∈ T . Whenever ν is a signed measure on {0, 1} n we will use this notation to write ν(1 S ) := T ⊆[n]\S ν(1 S∪T 0 [n]\(S∪T ) ). Next, whenever σ is a binary string in {−1, 1} n or {0, 1} n , we will write σ := i∈[n] σ i . Finally, when S ⊆ [n] and σ ∈ {−1, 1}, we will let σ S : We now give some notation for working with set partitions. Firstly, the set of partitions of [n] will be denoted by B n . If π ∈ B n has partition elements π 1 , π 2 , . . . , π m , we will sometimes write π = (π 1 , . . . , π m ). Give a partition π ∈ B n and a binary string σ, we write π σ if σ is constant on the partition elements of π. If π σ, π i is a partition element of π and j ∈ π i , we write σ(π i ) := σ j . Note that this function is well defined exactly when π σ. Next, we let π denote the number of partition elements of π. Combining these notations, if σ ∈ {0, 1} n (or σ ∈ {−1, 1} n ) is constant on the partition elements of π (i.e. if π σ, then we let σ π := π i=1 σ(π i ). Finally, if S ⊆ [n],we write π S to denote the restriction of the π to the set S (so that π S ∈ B S ).

The linear operator
Let X = (X i ) i∈[n] be a {0, 1}-valued vector and let ν = ν X denote the probability density function of X. It was observed in [8] that if X has a color representation (µ, p), then µ and ν satisfy the following set of linear equations.
Moreover, whenever µ = (µ(π)) π∈Bn is non-negative and satisfies these equations, (µ, p) is a color representation of X. A signed measure µ on B N which satisfies (2), but which is not necessarily non-negative, will be called a formal solution to (2). The matrix corresponding to the linear system of equations given in (2) is for σ ∈ {0, 1} n and π ∈ B n given by It was shown in [4] that A n,1/2 has rank 2 n−1 and that when p ∈ (0, 1)\{1/2}, A n,p has rank 2 n − n.
To simplify notation, when µ is a signed measure on B n and π ∈ B n , we sometimes write µ π := µ(π). Also, when thinking about (2) as a system of linear equations, we will abuse notation slightly and let µ denote both the signed measure and the corresponding vector (µ(π)) π∈Bn , given some unspecified but arbitrary ordering of B n .

Other generalizations of the random cluster model
When h = 0, there are generalizations of the random cluster model (see e.g. [1]), from which the Ising model is obtained by independently assigning colors to different partition elements. These have been shown to have properties which can be used in similar ways as analogue properties of the random cluster model. However, these models use different color probabilities for different partition elements, and are hence not divide and color models. On the other hand, Proposition 1.4 shows that there are graphs G and parameters β, h > 0 such that X β,h ∼ ν G,β,h has no color representation. This motivates considering less restrictive generalizations of the random cluster model, such as the one given in [1].

Color representations of X β,0
In this section we give a proof of Theorem 1.2.
Proof of Theorem 1.2. Let V be the set of vertices of G, and let n := |V |. Fix any labelling of the vertices of G which uses the elements of [n] as labels and in this way, identify binary strings σ ∈ {0, 1} V with the corresponding binary strings in {0, 1} n .
Next, note that when p = 1/2, σ ∈ {0, 1} n and π ∈ B n , If we for S ⊆ [n] and π ∈ B n define then a standard inclusion-exclusion argument shows 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 is row equivalent to A , and hence also to A. By Theorem 1.2 in [4], A has rank 2 n−1 , and hence the same is true for A . Now note that if S ⊆ [n], π ∈ B n , and we let T 1 , T 2 , . . . , T π S denote the partition elements of π S , then we have that = I(π S has only even sized partition elements).

and hence
A (S, π) = I(π S has only even sized partition elements).
Let T be a spanning tree of G. Let B T n ⊆ B n denote the partitions of [n] whose partition elements induce connected subgraphs of T . Note that the number of such partitions is equal to 2 n−1 . For S ⊆ [n] with |S| even and σ ∈ B T n , define A T (S, π) := I(π S has only even sized partition elements) Then A T is a submatrix of A . We will show that A T has full rank. Since A T is a 2 n−1 by 2 n−1 matrix, this is equivalent to having non-zero determinant. To see that det A T = 0, note first that if |S| is even and σ ∈ B T n , then π has only even sized partition elements and any finer partition of S has at least one odd sized partition element Since all partition elements of π ∈ B T n induce connected subgraphs of G, B is a permutation matrix. Since permutation matrices have non-zero determinant, this implies that B, and hence also A T , has full rank. Since A T has 2 n−1 rows and columns, this implies in particular that A T has rank 2 n−1 . On the other hand, A T is a submatrix of A , and A is row equivalent to A which also has rank 2 n−1 . This implies in particular that when we solve (2), we can use the columns corresponding to partitions in B T n as dependent variables. Now recall that since X = X β,0 is the Ising model on some graph G, X has at least one color representation (µ, 2 −1 ), with µ being a random cluster model. The random cluster model gives strictly positive mass to all partitions π whose partition elements induce connected subgraphs of G. In particular, it gives strictly positive mass to all partitions in B T n . If we use the columns corresponding to partitions in B T n as dependent variables, then all dependent variables are given positive mass by µ. Since n ≥ 3, there is at least one free variable. By continuity, it follows that we can find another color representation 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 ∈ B n \B T , whose partition elements induce connected subgraphs of G (but not T ). From this the desired conclusion follows.
Proof of Proposition 1.4. Fix β > 0 and h > 0 and let p := P (X β,h 1 = 1). Since ν K4,β,h is permutation invariant, it follows that if X β,h has a color representation, then it has at least one color representation which is invariant under the action of S 4 . It is easy to check that there are exactly five different partitions in B 4 /S 4 , namely (1234), (123, 4), (12, 34), (1,2,34) and (1,2,3,4). By Theorem 1.5(i) in [4], the linear subspace spanned by all symmetric formal solutions of A 4,p µ = ν K4,β,h has dimension one. By linearity, this implies in particular that if X β,h has a color representation, then it has at least one color representation which is S 4 -invariant and gives zero mass to at least one of the partitions in B 4 /S 4 . This gives us one unique solution for each choice of partition in B 4 /S 4 .
Solving the corresponding linear equation systems in Mathematica and studying the solutions, after some work, one obtains the desired necessary and sufficient condition.

Existence
The goal of this subsection is to provide a proof of Theorem 1.6. A main tool in the proof of this theorem will be the following result from [4].
x (9) Remark 5.2. By applying elementary row operations to the system of linear equations in (9), one obtains the following equivalent system of linear equations.
π∈Bn I(π S has at most one odd sized partition element) µ(π) (See the last equation in the proof of Theorem 1.7 in [4].) The proof of Theorem 1.6 will be divided into two parts, corresponding to the two casesβ > 1 andβ = 1.
Proof of Theorem 1.6 whenβ > 1 (the supercritical regime). Assume that n ∈ N is large and let X Kn,β,0 be the Ising model on K n with parameter β > 0 and external field h = 0. Let zβ be the unique positive root to the equation z = tanh(βz). Then it is well known (see e.g. [3], p. 181) that and hence, by Lemma 7.4, as n → ∞ we have that By Remark 5.2, any color representation µ of ν Kn,β,0 which arise as a subsequential limit of color representations of ν Kn,β,h as h → 0 must satisfy (10). By Lemma 7.3, for any S ⊆ [n] we have that S ⊆S (−2) |S| ν Kn,β,0 (1 S ) = (−1) |S| · σ∈Σn σ S · ν Kn,β,0 (σ) and S ⊆S (−2) |S|+1 ν Kn,β,0 (1 S ) = (−1) |S|+1 · σ∈Σn σ σ S · ν Kn,β,0 (σ) σ∈Σn σ σ {1} · ν Kn,β,0 (σ) and hence (10) is equivalent to that π∈Bn I(π S has at most one odd sized partition element) µ(π) Now note that ν Kn,β,0 is permutation invariant. This implies in particular that if (12) has a non-negative solution, then it also has a S n -invariant nonnegative solution µ (n) . Next, observe that the restriction µ (n, [4]) of any such solution to {1, 2, 3, 4} is a solution to the rows of the original equation system corresponding to sets S ⊆ {1, 2, 3, 4}. If one only considers symmetric solutions to this restricted system, then, by Theorem 1.5(i) in [4], the corresponding null space has dimension one. Since this is a linear system if equations, if a nonnegative solution exists, then there will also exist a non-negative solution with at least one (out of the five) entries being equal to zero. Since there are one free variable and five non-free variables, at most five such solutions can exist.
Let the right hand side of (12) be denoted by λ (n) = (λ Using this, one verifies that the five symmetric solutions to with at least one zero entry are given by . Above, all the negative entries are displayed in red. Since all of these solutions have at least one strictly negative entry, it follows that all solutions µ (∞) to (13) must have at least one strictly negative entry. Now let µ (n) be a solution to A (n,symm) µ (n) = λ (n) . Then the restriction µ (n, [4]) of µ (n) to [4] must satisfy A (4,symm) µ (n, [4] But this implies that for all S ⊆ [n] we have that which implies that q (n, [4]) must have negative entries for all sufficiently large n, and hence the desired conclusion follows.
We now give a proof of Theorem 1.6 in the caseβ = 1. This proof is very similar to the previous proof, but requires that we look at the distribution of first five coordinates of X Kn,β,0 and are more careful with the asymptotics. Since we need to consider many more solutions in this case, we only provide a sketch here.
Let λ (n) = (λ (n) (S)) S⊆[n] be the right hand side of (12) and letλ (n) be the vector containing only the largest order term of each entry of the same vector, i.e. let If we repeat the procedure we used in the super-critical case, we get only nonnegative solutions if we project onto any set of size at most four. If we project onto the set {1, 2, 3, 4, 5} however, one can verify that all extremal solutions have at least one strictly negative component of the form Cn −1/2 + o(n −1/2 ) where C < 0. Let the extremal solutions to A (5) µ =λ (n) be labelled byμ 1 ,μ 2 , etc. and let µ 1 , µ 2 , etc. be the corresponding solutions to A (5) µ = λ (n) . Then for any i and S ⊆ [5], and hence µ i (S) −μ i (S) = o(n −1/2 ), implying in particular that each entry of µ i has the same signs as the corresponding entry ofμ i at at least one entry which is negative inμ i . This gives the desired conclusion.

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 result will be the next lemma. The main idea of the proof of Theorem 1.5 will then be to show that the necessary condition given by this lemma does in fact not hold for the random cluster model. Throughout this section, we will use the following notation. For a fixed n ∈ N, a set S ⊆ [n] and a partition π ∈ B n , define A(S, π) := I(π S has at most one odd sized partition element).
If |S| is odd and A(S, π) = 1, let T S,π denote the unique odd sized partition element of π.
Rearranging this equation, we obtain (14), and hence the desired conclusion follows.
The purpose of the next lemma is to provide a simpler way to calculate the sums in (14) when µ is the random cluster model on G, by translating such sums into corresponding sums for Bernoulli percolation.
Lemma 5.4. Let G = (V, E) be a connected and vertex transitive graph. Fix some r ∈ (0, 1) and definer = r 2−r . Further, let m be the length of the shortest loop in G. Then there is a constant Z G,r such that for any function f : B n → R, we have that Proof. First recall the definition of the random cluster measure µ G,r on {−1, 1} V .
This implies in particular that and hence for any function f : B n → R we have that Since 2 |V | 2−r 2 |E| = O(1), the desired conclusion follows.
We now use the previous lemma, Lemma 5.4, to give a version of Lemma 5.3 for Bernoulli percolation. To this end, forr ∈ (0, 1), letμ P G,r be the measure on {0, 1} E defined bŷ We then have the following lemma. be the number of length m cycles in G which contain all the vertices in S. Let X β,h ∼ ν G,β,h be the Ising model on G and assume further that the corresponding random cluster model µ G,r arises as a limit of color representations of ν β,h as h → 0. Setr = r 2−r . Then Proof. Fix any set S ⊆ [n] with |S| = 3. Note first that by Lemma 5.3, we have that n · π∈Bn A(S, π)|T S,π | µ G,r (π) = i∈[n] π∈Bn |T {i},π | µ G,r (π) · π∈Bn A(S, π) µ G,r (π) or equivalently, w∈{0,1} n A(S, π(w))μ G,r (w) .
We are now ready to give a proof of Theorem 1.5.
Remark 5.6. It is a relatively simple exercise to show that (17) in fact holds for all unrooted infinite regular trees (letting m = ∞). Using this fact, one can subtract the "tree-part" from (17) to get a slightly simpler expression, which can be used to reach the same conclusion as above for any finite vertex transitive graph.

Color representations for large |h|
In this section we show that for any β > 0, ν G,β,h has a color representation for all sufficiently large h. The main idea of the proof is to show that a very specific formal solution to (2), which was first used in [4], is in fact non-negative, and hence a color representation of ν G,β,h , whenever h is large enough.
Proof of Theorem 1.7. For T ⊆ [n] with |T | = 1, let π T denote the unique partition π ∈ B n which is such that T is a partition element of π, and all other partition elements has size one. Let p = p(K n , β, h) := ν Kn,β,h (1 {1} ). By the proof of Theorem 1.6 in [4], for any p ∈ (0, 1)\{1/2}, one formal solution of (2) is given by We will show that, given the assumptions, this is a non-negative solution to (2), and hence a color representation of X β,h ∼ ν Kn,β,h . To this end, note first that for σ ∈ Σ n , we have that This implies in particular that p(K n , β, h) → 0 and that Z β,h ∼ ν Kn,β,h (1 ∅ 0 [n] ) → 1 as h → ∞. Similarly, one shows that Since β(n − 1) ≤ h by assumption, it follows that ν Kn,β, , and hence in particular that p(K n , β, h) ∼ ν Kn,β,h (1 [1] 0 [n]\ [1] ). Combining these observations, it follows that for any S ⊆ [n] and any k ∈ S, we have that and hence when h is sufficiently large, We will now use the equations above to describe the behavior of (19). To this end, note first that combining (23) We will now rewrite the previous equation. To this end, note first that Again using that ν Kn,β,h is invariant under permutations, it follows that this is equal to Summing up, we have showed that for any T ⊆ [n] with |T | ≥ 2, we have that Since for any positive and strictly increasing function f : N → R, we have that This is clearly larger than zero, and in fact, by (23), as p tends to zero, it is asymptotic to This implies that the error term in (24) is much smaller than the rest of the expression, which is strictly positive, i.e.
Remark 6.1. The previous proof shows that the formal solution to (2) given by (19) is a color representation of ν β,h whenever β is fixed and h is sufficiently large. One might hence ask if (19) is also non-negative 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 > 0 is sufficiently small, µ(σ 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 quite far from the conjectured result, one might try to get a stronger result by optimizing the proof above. However, using Mathematica, one can check that the particular formal solution to (2) given by (19) in fact not non-negative when (n − 1)β > h when n = 3, 4, 5.
Proof. Let S ⊆ [n]. By Lemma 7.2, we have that if m = 5.
By plugging these expressions into equations (25) and (26), the desired conclusion follows.