Uniqueness of the Gibbs Measure for the Anti-ferromagnetic Potts Model on the Inﬁnite 1 -Regular Tree for Large 1

In this paper we prove that for any integer q ≥ 5, the anti-ferromagnetic q -state Potts model on the inﬁnite (cid:2) -regular tree has a unique Gibbs measure for all edge interaction parameters w ∈ [ 1 − q /(cid:2), 1 ) , provided (cid:2) is large enough. This conﬁrms a longstanding folklore conjecture


Introduction
The Potts model is a statistical model, originally invented to study ferromagnetism [Pot52]; it also plays a central role in probability theory, combinatorics and computer science, see e.g.[Sok05] for background.
Let G = (V, E) be a finite graph.The anti-ferromagnetic Potts model on the graph G has two parameters, a number of states, or colors, q ∈ Z ≥2 and an edge interaction parameter w = e kJ/T , with J < 0 being a coupling constant, k the Boltzmann constant and T the temperature.The case q = 2 is also known as the zero-field Ising model.A configuration is a map σ : V → [q] := {1, . . ., q}.Associated with such a configuration is the weight w m (σ) , where m(σ) is the number of edges e = {u, v} ∈ E for which σ(u) = σ(v).There is a natural probability measure, the Gibbs measure Pr G;q,w [•], on the collection of configurations Ω = {σ : V → [q]} in which a configuration is sampled proportionally to its weight.Formally, for a given configuration φ : V → [q] the on a finite graph is unique, this is no longer the case for all infinite lattices.The transition from having a unique Gibbs measure to multiple Gibbs measures in terms of the temperature is referred to as a phase transition in statistical physics [Geo88,FV17] and it is an important problem to determine the exact temperature, the critical temperature, T c , at which this happens.There exist predictions for the critical temperature on several lattices in the physics literature by Baxter [Bax82,Bax86] (see also [SS97] for more details and further references), but it turns out to be hard to prove these rigorously cf.[SS97].
In the present paper we consider the anti-ferromagnetic Potts model on the infinite ∆-regular tree, T ∆ = (V, E), also known as the Bethe lattice, or Cayley tree.We briefly recall the formal definition of a Gibbs measure in this situation following [BW99,BW02].See [Roz21] for a survey on this topic in general.
The sigma algebra is generated by sets of the form U σ := {φ : V → [q] | φ ↾ U = σ}, where U ⊂ V is a finite set and σ : U → [q].Let w ∈ (0, 1).A probability measure µ on this sigma algebra is then called a Gibbs measure for the q-state anti-ferromagnetic Potts model on T ∆ at w, if for all finite U ⊂ V and µ-a.e.φ : V → [q] the following holds where ∂U denotes the collection of vertices in U that have a neighbor in V \ U and U • := U \ ∂U .We note that the probability in the right-hand side of this equation is determined in the finite graph induced by U , T ∆ [U ].Moreover, we note that for any w ∈ (0, 1) there exists at least one such Gibbs measure.
For a number of states q ≥ 2 define It is a longstanding folklore conjecture (cf.[BGG + 20, page 746]) that the Gibbs measure is unique if and only if w ≥ w c (where the inequality should be read as strict if q = ∆.)We note that using the well known Dobrushin uniqueness theorem, one obtains uniqueness of the Gibbs measure provided w > 1 − q 2∆ cf.[BCKL13,SS97], which is still far way from the conjectured threshold.The conjecture was confirmed by Jonasson for the case w = 0 [Jon02], by Srivastava, Sinclair and Thurley [SST14] for q = 2 (see also [Geo88]; in this case one can map the model to a ferromagnetic model since the tree is bipartite, which is much better understood), by Galanis, Goldberg and Yang for q = 3 [GGY18] and by three of the authors of the present paper for q = 4 and ∆ ≥ 5 [dBBR23].Our main result is a confirmation of this conjecture for all q ≥ 5 provided the degree of the tree is large enough.
Main Theorem.For each integer q ≥ 5 there exists ∆ 0 ∈ N such that for each ∆ ≥ ∆ 0 and each w ∈ [w c , 1) the q-state anti-ferromagnetic Potts model with edge interaction parameter w has a unique Gibbs measure on the infinite ∆-regular tree T ∆ .
It has long been known that there are multiple Gibbs measures when w < w c [PdLM83,PdLM87], see also [GŠV15]) and [BR19, KR17, GRR17, KRK14].We will briefly indicate below Lemma 2.2 how one can deduce this.Our main results therefore pinpoints the critical temperature for the anti-ferromagnetic Potts model on the infinite regular tree for large enough degree.For later reference we will refer to w c as the uniqueness threshold.
In Theorem 2.1 below, we will reformulate our main theorem in terms of the conditional distribution of the color of the root vertex of T ∆ conditioned on a fixed coloring of the vertices at a certain distance from the root, showing that this distribution converges to the uniform distribution as the distance tends to infinity.We in fact show that this convergence is exponentially fast for subcritical w (i.e.w > w c ).

Motivation from computer science
There is a surprising connection between phase transitions on the infinite regular tree and transitions in the computational complexity of approximately computing partition function of 2-state models (not necessarily the Potts model) on bounded degree graphs.For parameters inside the uniqueness region there is an efficient algorithm for this task [Wei06, LLY13, SST14], while for parameters for which there are multiple Gibbs measures on the infinite regular tree, the problem is NP-hard [SS14,GŠV16].It is conjectured that a similar phenomenon holds for a larger number of states.
While the picture for q-state models for q ≥ 3 is far from clear, some progress has been made on this problem for the anti-ferromagnetic Potts model.On the hardness side, Galanis, Štefankvovič and Vigoda [GŠV15] showed that for even numbers ∆ ≥ 4 and any integer q ≥ 3, approximating the partition function of the Potts model Z(G; q, w) is NP-hard on the family of graphs of maximum degree ∆ for any 0 ≤ w < 1 − q/∆ = w c , which we now know to be the uniqueness threshold (for ∆ large enough).On the other side, much less is known about the existence of efficient algorithms for approximating Z(G; q, w) or sampling from the measure Pr G;q,w for the class of bounded degree graphs when w > w c .Implicit in [BDPR21] there is an efficient algorithm for this problem whenever 1 − αq/∆ < w ≤ 1, with α = 1/e, which has been improved to α = 1/2 in [LSS0].
For random regular graphs of large enough degree, our main result implies an efficient randomized algorithm to approximately sample from the Gibbs measure Pr G;q,w for any w c < w ≤ 1 by a result of Blanca, Galanis, Goldberg, Štefankovič, Vigoda and Yang [BGG + 20, Theorem 2.7].See also [CLMM23] for a very recent improvement.In [Eft22], Efthymiou proved a similar result for Erdős-Rényi random graphs without the assumption that w c is equal to the uniqueness threshold on the tree.At the very least this indicates that the uniqueness threshold on the infinite regular tree plays an important role in the study of the complexity of approximating the partition function of and sampling from the Potts model on bounded degree graphs.

Approach
Our approach to prove the main theorem is based on the approach from [dBBR23] for the cases q = 3, 4. As is well known, to prove uniqueness it suffices to show that for a given root vertex, say v, the probability that v receives a color i ∈ [q], conditioned on the event that the vertices at distance n from v receive a fixed coloring, converges to 1/q as n → ∞ regardless of the fixed coloring of the vertices at distance n.Instead of looking at these probabilities, we look at ratios of these probabilities.It then suffices to show that these converge to 1.The ratios at the root vertex v can be expressed as a rational function of the ratios at the neighbors of v. See Lemma 2.2 below.This function is rather difficult to analyze directly and as in [dBBR23] we analyze a simpler function coupled with a geometric approach.A key new ingredient of our approach is to take the limit of ∆, the degree of the tree, to infinity and analyze the resulting function.This function turns out be even simpler and behaves much better in a geometric sense.With some work we translate the results for the limit case back to the finite case and therefore obtain results for ∆ large enough.This is inspired by a recent paper [BBP21] in which this idea was used to give a precise description of the location of the zeros of the independence polynomial for bounded degree graphs of large degree.

Organization
In the next section we give a more technical overview of our approach.In particular we recall some results from [dBBR23] that we will use and set up some terminology.We also gather two results that will be used to prove our main theorem, leaving the proofs of these results to Section 3 and Section 4 respectively.Assuming these results, the main theorem will be proved in Subsection 2.4.
2 Preliminaries, setup and proof outline

Reformulation of the main result
We will reformulate our main theorem here in terms of the conditional distribution of the color of the root vertex of T ∆ conditioned on a fixed coloring of the vertices at a certain distance from the root.
Let ∆ ≥ 2 be an integer.In what follows it will be convenient to write d = ∆ − 1.For a positive integer n we denote by T n d+1 the finite tree obtained from T d+1 by fixing a root vertex r, deleting all vertices at distance more than n from the root, deleting one of the neighbors of r and keeping the connected component containing r.We denote the set of leaves of T n d+1 by Λ n , except when n = 0, in which case we let Λ 0 = {r}.For a positive integer q we call a map τ : Λ n → [q] a boundary condition at level n.
The following theorem may be seen as a more precise form of our main result.
Theorem 2.1.Let q ≥ 3 be a positive integer.There exist constants C > 0 and d 0 > 0 such that for all integers d ≥ d 0 and all α ∈ (0, 1) the following holds for any i ∈ {1, . . ., q}: for any boundary condition τ at level n and edge interaction w(α) = 1 − αq d+1 , Remark 1.We can in fact strengthen (3) in two ways.First of all, for any α < α < 1 there exists a constant C α > 0 such that the right-hand side of (3) can be replaced by C α αn .Secondly, for any fixed d ≥ d 0 there exist a constant C d > 0 such that the right-hand side of (3) can be replaced by As is well known (see e.g.[dBBR23, Lemma 1.3]2 ) Theorem 2.1 directly implies our main theorem.Therefore the remainder of the paper is devoted to proving Theorem 2.1.
We now outline how we do this.

Log-ratios of probabilities
Theorem 2.1 is formulated in terms of certain conditional probabilities.For our purposes it turns out to be convenient to reformulate this into log-ratios of these probabilities.To introduce these, we recall some relevant definitions from [dBBR23].Throughout we fix an integer q ≥ 3. Given a (finite) graph G = (V, E) and a subset U ⊆ V of vertices, we call τ : U → [q] a boundary condition on G.We say vertices in U are fixed and vertices in V \ U are free.The partition function restricted to τ is defined as We just write Z(G) if U, τ and q, w are clear from the context.Given a boundary condition τ : U → [q], a free vertex v ∈ V \ U and a state i ∈ [q] we define τ v,i as the unique boundary condition on U ∪ {v} that extends τ and associates i to v. When U and τ are clear from the context, we will denote Z U∪{v},τv,i (G) as Z v i (G).Let τ : U → [q] be a boundary condition and v ∈ V be a free vertex.For any i ∈ [q] we define the log-ratio RG,v,i as where log denotes the natural logarithm.Note that RG,v,q = 0. We moreover remark that RG,v,i can be interpreted as the logarithm of the ratio of the probabilities that the root gets color i (resp.q) conditioned on the event that U is colored according to τ .For trees the log-ratios at the root vertex can be recursively computed from the log-ratios of its neighbors.To describe this compactly we introduce some notation that will be used extensively throughout the paper.Fix d ∈ R >1 and let α ∈ (0, 1].Define the maps G d,α;i , F d,α;i : R q−1 → R for i ∈ {1, . . ., q − 1} as and Define the map F d,α : R q−1 → R q−1 whose ith coordinate function is given by F d,α;i (x 1 , . . ., x q−1 ) and define G d,α similarly.To suppress notation we write . We also define exp(x 1 , . . ., x q−1 ) = (exp(x 1 ), . . ., exp(x q−1 )) and log(x 1 , . . ., x q−1 ) = (log(x 1 ), . . ., log(x q−1 )).
We note that G d,α and F d,α are analytic in 1/d near 0 when viewing d as a variable.We will now use the map F d,α to give a compact description of the tree recurrence for log-ratios.
Lemma 2.2.Let T = (V, E) be a tree, τ : U → [q] a boundary condition on U V .Let v be a free vertex of degree d ≥ 1 with neighbors v 1 , . . ., v d .Denote T i for the tree that is the connected component of T − v containing v i .Restrict τ to each T i in the natural way.Write Ri,j for the log-ratio RTi,vi,j .Then for α such that w = 1 a convex combination of the images of the map F d,α .
Proof.By focusing on the jth entry of the left-hand side and substituting R T,v,j := exp( RT,v,j ), we see that (6) follows from the well known recursion for ratios See e.g.[dBBR23] for a proof of this.
We note that if the boundary condition τ is constant on the leaves of the tree T n d+1 , then the log-ratios at the root can be obtained by iterating the univariate function f given by f (x) = F d,α (x, . . ., x) at w = w(α).The point x = 0 is a fixed point of f ; it satisfies |f ′ (0)| ≤ 1 if and only if w ≥ w c .From this it is not difficult to extract that there exist multiple Gibbs measures when w < w c .
Denote 0 for the zero vector in R q−1 .(Throughout we will denote vectors in boldface.)We define for any n ≥ 1 the set of possible log-ratio vectors Here the ratios RT n d+1 ,r,1 depend on τ but this is not visible in the notation.The following lemma shows how the recursion from Lemma 2.2 will be used.
Lemma 2.3.Let q ≥ 3 and d ≥ 2 be integers.If there exists a sequence {T n } n≥1 of convex subsets of R q−1 with the following properties: Proof.The proof is straightforward and analogous to the proof of Lemma 2.3 in [dBBR23] and we therefore omit it.
We note that the lemma is only stated for α = 1.An analogues statement for α ∈ (0, 1) and F d replaced by F d,α with a more accurate dependence of N on ε follows from a certain monotonicity of F d,α , as will be explained in the proof of Theorem 2.1 below.
In the next section we construct a family of convex sets that allows us to form a sequence {T n } n≥1 with the properties required by the lemma.

Construction of suitable convex sets
We need the standard q − 2-simplex, which we denote as The symmetric group S q acts on R q by permuting entries of vectors.Consider R q−1 ⊂ R q as the subspace spanned by {e 1 − e q , . . ., e q−1 − e q }, where e i denotes the ith standard base vector in R q .This induces a linear action of S q on R q−1 , also known as the the standard representation of S q and denoted by x → π • x for x ∈ R q−1 and π ∈ S q .The following lemma shows that the map F d,α is S q -equivariant for any α ∈ (0, 1], essentially because the action permutes the q colors of the Potts model and no color plays a special role.
The following two propositions capture the image of P c under applications of the map F d .
Proposition 2.5.Let q ≥ 3 be an integer.Then there exists Proposition 2.6.Let q ≥ 3 be an integer.There exists d 2 > 0 such that for all d ≥ d 2 the following holds: for any c ∈ (0, q + 1] there exists 0 < c ′ < c such that An intuitive explanation for why we need F •2 d and cannot work with F d directly is that the derivative of F d at 0 is equal to −Id, which reflects the fact that we are dealing with an antiferromagnetic model, while the derivative of F •2 d at 0 is equal to Id.We postpone the proofs of the two results above to the subsequent sections.A crucial ingredient in both proofs will be to analyze the limit lim d→∞ F d .We first utilize the two propositions to give a proof of Theorem 2.1.

A proof of Theorem 2.1
Fix an integer q ≥ 3. Let d 1 , d 2 be the constants from Proposition 2.5 and 2.6 respectively.Let d 0 ≥ max{d 1 , d 2 } large enough to be determined below.Note that the log-ratios at depth 0 are of the form ∞ • e i and −∞ • 1, where 1 denotes the all ones vector.This comes from the fact that the probabilities at level 0 are either 1 or 0 and so the ratios are of the form 1 + ∞e i or 0. This implies that the log-ratios at depth 1 are convex combinations of So for d ≥ d 0 and d 0 large enough they are certainly contained in P q+1 .
We start with the proof of (2).We construct a decreasing sequence {c n } n∈N and let T 2n−1 = P cn .For even n > 0 we set T n = F d (P cn−1 ), which is convex by Proposition 2.5.We set c 1 = q + 1 and for n ≥ 1, given c n , we can choose, by Proposition 2.6, c n+1 < c n so that F •2 d (P cn ) ⊆ P cn+1 .Choose such a c n+1 as small as possible.We claim that the sequence {c n } n∈N converges to 0. Suppose not then it must have a limit c > 0. Choose c ′ < c such that F •2 d (P c ) ⊆ P c ′ .Then for n large enough we must have F •2 d (P cn ) ⊆ P c/2+c ′ /2 , contradicting the choice of c n+1 .Since {c n } n∈N converges to 0, it follows that the sequence T n converges to {0}.With Lemma 2.3 this implies (2).
Lemma 2.7.For any α ∈ (0, 1], any x ∈ R q−1 and any integer d there is . By basic properties of the logarithm, (3) now quickly follows.This finishes the proof of Theorem 2.1.
The strengthening mentioned in Remark 1 can be derived from the fact that the derivative of F d,α at 0 is equal to −αd d+1−α Id.Note that αd d+1−α < α for all α ∈ (0, 1) and d.Therefore on a small enough open ball B around 0 the operator norm of the derivative of F d,α can be bounded by α for all d ≥ d 0 (and by α for fixed d ≥ d 0 ).Then for any integer n ≥ 0, F •n d,α (B) ⊂ αn B (α n B respectively).For n 0 large enough P cn 0 is contained in this ball B. For n > 2n 0 we then set T n = αn−2n0 B (α n−2n0 B respectively).The statements in the remark now follow quickly.

The d → ∞ limit map
As mentioned above, an important tool in our approach is to analyze the maps with coordinate functions We write G ∞;i (x 1 , . . ., x q−1 ) = q 1−xi By Lemma 2.4 for any π ∈ S q , any x ∈ R q−1 and any d we have π As the action of π on R q−1 does not depend on d, we immediately see π In the next two sections we will prove Propositions 2.5 and 2.6.The idea is to first prove a variant of these propositions for the map F ∞ and then use that F d → F ∞ uniformly to finally prove the actual statements.We use the description of P c as intersection of half spaces π • H ≥−c in Section 3 and the description as the union of the π • D c in Section 4.

Convexity of the forward image of P c
This section is dedicated to proving Proposition 2.5.
Fix an integer q ≥ 3.For µ ∈ R we define the half space H ≥µ as in (9).The half space H ≤µ is defined similarly.We denote by H µ the affine space which is the boundary of H ≤µ .
In what follows we will often use that the map G ∞ is a fractional linear transformation and thus preserves lines and hence maps convex sets to convex sets, see e.g.[BV04, Section 2.3].
Lemma 3.1.For all c > 0, the set exp( is strictly convex. Proof.Since G ∞ is a fractional linear transformation, it preserves convex sets.It therefore suffices to show that exp(H ≥−c ) is strictly convex.
To this end take any x, y ∈ exp(H ≥−c ) and let λ ∈ (0, 1).We need to show that λx+(1−λ)y ∈ exp(H ≥−c ).By strict concavity of the logarithm we have In what follows we need the angle between the tangent space of G ∞ (exp(H −c )) for c > 0 at G ∞ (x) for any x ∈ exp(H −c ) and the space H 0 .This angle is defined as the angle of a normal vector of the tangent space pointing towards the interior of G ∞ (exp(H ≥−c )) and the vector −1 (which is a normal vector of H 0 ).Lemma 3.2.For any c ∈ [0, q + 1] and any x ∈ exp(H −c ) the angle between the tangent space of G ∞ (exp(H −c )) at G ∞ (x) and H 0 is strictly less than π/2.
Proof.We will first show that the tangent space cannot be orthogonal to H 0 .
The map G ∞ is invertible (when restricted to R q−1 >0 ) with inverse G −1 ∞ whose coordinate functions are given by is given by the gradient of the function g.Thus to show that this tangent space is not orthogonal to H 0 , we need to show that x 1 x 2 Figure 1: Depicting the situation in Lemma 3.3, for q = 3, c = 2 and y = 1 20 .The domain Dom c of the function h y,c which we define in the proof of Lemma 3.3 is made by choosing a ′ = −3.

We have
• −q( q−1 i=1 y i + q) + q(q − 1)y j ( y) > 0 for each k, all terms in the final sum are nonzero and have the same sign.This proves (14).
Since the angle between the tangent space of G ∞ (exp(H −c )) at G ∞ (x) and H 0 depends continuously on x this angle should either be always less than π/2 or always be bigger.Since by the previous lemma the set G ∞ (exp(H ≥−c )) is convex, it is the former.
We next continue with the finite case.We will need the following definition.The hypograph of a function f : D → R is the region {(x, y) | x ∈ D, y ≤ f (x)}.Below we will consider a hypersurface contained in R q−1 that we view as the graph of a function with domain contained in H 0 .In this context the hypograph of such a function is again contained in R q−1 , but the 'positive y-axis' points in the direction of 1 as seen from 0 ∈ H 0 .Lemma 3.3.There exists y 1 > 0 such that for all y ∈ [0, y 1 ) and c ∈ [0, q + 1] the set F y (P c ) is contained in the hypograph of a concave function, h y,c , with a convex compact domain in H 0 .
Proof.We first prove that for any x ∈ H 0 and c ∈ [0, q + 1] there exists an open neighborhood such that the following holds for any (y ′ , c ′ , x ′ ) ∈ W c,x : the angle between the tangent space of where we denote x c := x − c q−1 1 ∈ H −c .To see this note that by the previous lemma we have that the tangent space of F ∞ (H −c ) at F ∞ (x c ) is not orthogonal to H 0 and in fact makes an angle of less than π/2 with H 0 .Say it has angle π/2 − γ.Since (y, c, x) → F 1/y (x c ) is analytic, there exists an open neighborhood W 0 of (0, c, x) such that for any (y ′ , x ′ , c ′ ) ∈ W 0 the angle between the tangent space of F 1/y ′ (H −c ′ ) at F 1/y ′ (x ′ c ′ ) and H 0 is at most π/2 − γ/2.Clearly, W 0 contains an open neighborhood of (0, c, x) of the form Y × C × X proving (15).
Next fix c ∈ [0, q + 1] and x ∈ H 0 and write W c,x = Y × C × X. Together with the implicit function theorem, (15) now implies that for each y ′ ∈ Y and any c ′ ∈ C, that locally at x c ′ , F 1/y ′ (H −c ) is the graph of an analytic function f y ′ ,c ′ ,x on an open domain contained in H 0 .Here we use that F 1/y is invertible with analytic inverse.By choosing Y and C small enough, we may by continuity assume that we have a common open domain, D c,x , for these functions for all c ′ ∈ C and y ′ ∈ Y , where we may moreover assume that these functions are all defined on the closure of D c,x .
We next claim, provided the neighbourhood W = Y c,x × C c,x is chosen small enough, that for each y ′ ∈ Y and c ′ ∈ C, the largest eigenvalue of the Hessian f y ′ ,c ′ ,x on D c,x is strictly less than 0. (16) To see this we note that by the previous lemma we know that F ∞ (H ≥−c ) is strictly convex.Therefore the Hessian 3 of f 0,c,x on D c,x is negative definite, say its largest eigenvalue is δ < 0. Similarly as before, there exists an open neighborhood W ′ ⊆ W of (0, c) of the form W ′ = Y ′ × C ′ such that for each y ′ ∈ Y ′ and c ′ ∈ C ′ , the function f y ′ ,c ′ ,x has a negative definite Hessian with largest eigenvalue at most δ/2 < 0 for each z ∈ D c,x (by compactness of the closure of D c,x ).We now want to patch all these function to form a global function on a compact and convex domain.
We first collect some properties of F 1/y that will allow us to define the domain.
First of all note that by compactness there exists a > 0 such that for each c ∈ [0, q + 1], exp(P c ) ⊂ H ≤a (where the inclusion is strict).We now fix such a value of a. Since G ∞ is S qequivariant, we know that G ∞ (H ≤a ) = H ≥a ′ for some a ′ ∈ R. We now choose y * > 0 small enough such that the following two inclusions hold for all y ∈ [0, y * ] and c ∈ [0, q + 1] where proj H0 denotes the orthogonal projection onto the space H 0 .The first inclusion holds since F 1/y converges uniformly to F ∞ as y → 0. For the second inclusion note that Because exp(H −c ) ∩ H ≤a is compact, the desired conclusion follows since F 1/y → F ∞ uniformly as y → 0.
Let us now consider for c ∈ [0, q + 1] the projection Indeed, it is the continuous image of the compact set exp(H ≥−q−1 ) ∩ H ≤a under the map 3 Recall that the Hessian of a function f : . When these partial derivatives are continuous and the domain U is convex, f is concave if and only if its Hessian is negative definite at each point of the domain U [BV04].
By (18) Dom c is contained in proj H0 (F 1/y (H −c )) for all y ∈ [0, y * ] and c ∈ [0, q + 1].It follows that the sets Y c,x × C c,x × D c,x , where x ranges over H 0 and c over [0, q + 1], form an open cover of {0} × ∪ c∈[0,q+1] ({c} × Dom c ).Since the latter set is compact by (19), we can take a finite sub cover.Therefore there exists y 1 > 0 such that for each y ∈ [0, y 1 ) and each c ∈ [0, q + 1] we obtain a unique global function h y,c on the union of these finitely many domains, which by ( 16) has a strictly negative definite Hessian.By construction the union of these domains contains Dom c for each c ∈ [0, q + 1].Consequently, restricted to Dom c , h y,c is a concave function for each y ∈ [0, y 1 ) and c ∈ [0, q + 1].By (17), it follows that F 1/y (P c ) is contained in the hypograph of h y,c , as desired.
We can now finally prove Proposition 2.5, which we restate here for convenience.
Proposition 2.5.Let q ≥ 3 be an integer.Then there exists d 1 > 0 such that for all d ≥ d 1 and c ∈ [0, q + 1], F d (P c ) is convex.
Proof.By the previous lemma we conclude that for d larger than 1/y 1 , F d (P c ) is contained in the hypograph of the function h 1/d,c , denoted by hypo(h c,1/d ) and moreover that this hypograph is convex, as the function h 1/d,c is concave on a convex domain.
Since P c is invariant under the S q -action, it follows that and therefore by Lemma 2.4, We now claim that the final inclusion in ( 20) is in fact an equality.To see the other inclusion, take some z ∈ ∩ π∈Sq π • hypo(h 1/d,c ).By symmetry, we may assume that z is contained in R q−1 ≥0 .Then z is equal to F d (x) for some x ∈ H ≥−c ∩ R q−1 ≤0 , implying that z is indeed contained in F d (P c ).This then implies that F d (P c ) is indeed convex being equal to the intersection of the convex sets π • hypo(h 1/d,c ).

Forward invariance of P c in two iterations
This section is dedicated to proving Proposition 2.6.We start with a version of the proposition for d = ∞ and after that consider finite d.

Two iterations of F ∞
Let Φ : R q−1 → R q−1 be defined by where we use •, • to denote the standard inner product on R q−1 .
This subsection is devoted to proving the following result.
Proposition 4.1.For any c ≥ 0 we have By the definition of P c in terms of D c , (11), and the S q -equivariance of the map F ∞ and hence of the map Φ, it suffices to prove this for P c replaced by D c .This can be derived from the following two statements: (i) For any c ≥ 0 the minimum of Φ(x), 1 on −c∆ is attained at −c/(q − 1) • 1.
(ii) For any c > 0 we have φ(c) < c.Indeed, these statements imply that for any c > 0 we have that Φ(−c∆) ⊆ D φ(c) D c .Clearly this is sufficient, since D c = ∪ 0≤c ′ ≤c − c ′ ∆ and therefore We next prove both statements, starting with the first one.

Statement (i)
Proposition 4.2.Let c ≥ 0. Then for any x ∈ −c∆ we have that Moreover, equality happens only at x = −c q−1 1.Before giving a proof, let us fix some further notation.By definition we have where we recall that F ∞;j denotes the jth coordinate function of F ∞ .Thus the ith coordinate of the gradient of Φ(x), 1 is given by q 3 e xi e F∞;i(x) (1 + q−1 j=1 e xj ) + q−1 j=1 e F∞;j(x) (1 − e xj ) q−1 j=1 e F∞;j(x) + 1 Let us define the following functions v i : R q−1 → R for i = 1, . . ., q − 1 as where we write Then we see that . ., e xq−1 ), Proof of Proposition 4.2.First of all observe that the function Φ(x), 1 is invariant under the permutation of the coordinates of x.Thus we can assume that and not all the coordinates of x are equal.Now it is enough to show that there exists a vector 0 = w ∈ R q−1 such that in the direction of w the function is (strictly) decreasing, w, 1 = 0 and x + t 0 w ∈ U for some small t 0 > 0. Let which is finite, since not all of the coordinates of x are equal.
Using the notation defined above, we obtain where C > 0 and y = exp(x).In particular, So to conclude that g ′ (0) < 0 and finish the proof, we need to show that Lemma 4.3 shows that we may assume y satisfies 1 ≥ y 1 = y 2 = . . .= y ℓ > y ℓ+1 ≥ y ℓ+2 = . . .= y q−1 ≥ 0. Lemma 4.4 below shows that for those vectors y (21) is indeed true.So by combining Lemma 4.3 and Lemma 4.4 below we obtain (21) and finish the proof.
where x ∈ R q−1 is defined as Proof.By continuity, it suffices to show where x ∈ R q−1 is defined as for 1 ≤ j ≤ q − 1 and any i ≥ ℓ + 2.
For t ∈ R we define y(t) by for j = 1, . . ., q − 1.Note that y(0) = y and y(y i /2 − y i+1 /2) = x.We further define After a straightforward calculation we can express ∆(t) as where we write and we write G ℓ (t) = G ∞;ℓ (y(t)) when ℓ ∈ {i, i + 1}.This notation indicates that G ℓ is a constant function of t when ℓ ∈ {i, i + 1}.Now observe that the function appearing in the last row, since its second derivative is given by As g(t) = g(y i − y i+1 − t), we obtain that g(t) has a unique minimizer in [0, y i − y i+1 ] exactly at t such that = y i − y i+1 − t.In other words, t = y i − x i+1 2 is the unique minimizer of g(t) on this interval and thus for ∆(t).This implies (22) and hence the lemma.
Lemma 4.4.Let 1 ≥ x 1 > x 2 ≥ x 3 ≥ 0 and q − 2 ≥ l ≥ 1.Then Proof.The algebraic manipulations that are done in this proof, while elementary, involve quite large expressions.Therefore we have supplied additional Mathematica code in Appendix A that can be used to verify the computations.We define ∆(y 1 , y 2 , y 3 ; t) := (y 1 y 3 (t − l − 1) + (l + 1)y 1 + (l + 1)y 1 y 2 − ly 2 ) e A1(y1,y2,y3;t) + (−y 2 y 3 (t − l − 1) − (l + 1)y 1 y 2 + y 1 − 2y 2 ) e A2(y1,y2,y3;t) + (y 1 − y 2 ) (1 − y 3 ) (t − l − 1)e A3(y1,y2,y3;t) , where A i (y 1 , y 2 , y 3 ; t) := (t + 1)(1 − y i ) 1 + ly 1 + y 2 + (t − (l + 1))y 3 for i = 1, 2, 3 (see Listing 1).One can check that We will treat t as a variable and vary it while keeping the values that appear in the exponents constant.To that effect let C i = A i (x 1 , x 2 , x 3 ; q − 1) and define These values are chosen such that for t 0 = q − 1 we have y i (t 0 ) = x i and A i (y 1 (t), y 2 (t), y 3 (t); t) = C i independently of t for i = 1, 2, 3 (see Listings 2 and 3).Therefore ∆(y 1 (t), y 2 (t), y 3 (t); t) is a rational function of t and we want to show that it is positive at t = q − 1.We can explicitly calculate that where r is a linear function (see Listing 4).It is thus enough to show that r(q − 1) > 0. We will do this by showing that r(l + 1) > 0 and that the slope of r is positive.We find that r(l + 1) is equal to where This is part of the output of Listing 5. Note that by construction, since Therefore the sum of the coefficients of e C1 and e C2 satisfies Now we will separate two cases depending on the sign of the coefficient of u 2 .If u 2 is non-negative, then If u 2 is negative, then In particular 2 + (1 The slope of r is given by This is part of the output of Listing 5. To show that this is positive we show that s • e −C2 is positive.Because both 1 + C 3 − C 1 and C 2 − C 1 are positive we find which is positive because 0 ≤ C 1 < C 2 ≤ C 3 .This concludes the proof.
We now continue with the second statement.

Statement (ii)
Proposition 4.5.For any x > 0 we have that Proof.The statement is equivalent to φ(x) < x.
Thus their composition has the following Taylor expansion around 0: This implies that there exists c 0 > 0 and A > 0, such that for any c 0 ≥ x ≥ 0 we have The next proposition implies forward invariance of P c under F •2 d for c small enough and d large enough.
Putting this together and making use of the triangle inequality, we obtain that for any 0 for some constant K > 0 (using that the 2-norm and the 1-norm are equivalent on R q−1 .)Now let us fix 0 < c 0 ≤ min{c 1 , c ′ 0 } small enough such that K2C 1 c 0 < A/4 and fix a y 0 > 0 such that for any any 0 ≤ y ≤ y 0 we have KA 3 (y) ≤ A/4.

Proof of Proposition 2.6
We are now ready to prove Proposition 2.6, which we restate here for convenience.
Proposition 2.6.Let q ≥ 3 be an integer.There exists d 2 > 0 such that for all integers d ≥ d 2 the following holds: for any c ∈ (0, q + 1] there exists 0 < c ′ < c such that Proof.We know by Proposition 4.7 there is a d 0 > 0 and a c 0 > 0 such that for d ≥ d

Concluding remarks
Although we have only proved uniqueness of the Gibbs measure on the infinite regular tree for a sufficiently large degree d, our method could conceivably be extended to smaller values of d.With the aid of a computer we managed to check that for q = 3 and q = 4 and all d ≥ 2 the map F •2 d maps P c into P φ d (−c) , where φ d is the restriction of −F •2 d to the line R • 1.It seems reasonable to expect that for other small values of q a similar statement could be proved.A general approach is elusive so far.It is moreover also not clear that F d (P c ) is convex, not even for q = 3.In fact, for q = 3 and c large enough F 3 (P c ) is not convex.But for reasonable values of c it does appear to be convex.For larger values of q this is even less clear.
Knowing that there is a unique Gibbs measure on the infinite regular tree is by itself not sufficient to design efficient algorithms to approximately compute the partition function/sample from the associated distribution on all bounded degree graphs.One needs a stronger notion of decay of correlations, often called strong spatial mixing [Wei06, GK12, GKM15, LY13] or absence of complex zeros for the partition function near the real interval [w, 1] [Bar16, PR17, BDPR21, LSS0].It is not clear whether our current approach is capable of proving such statements (these certainly do not follow automatically), but we hope that it may serve as a building block in determining the threshold(s) for strong spatial mixing and absence of complex zeros.We note that even for the case w = 0, corresponding to proper colorings, the best known bounds for strong spatial mixing on the infinite tree [EGH + 19] are still far from the uniqueness threshold.Very recently (after the current article was posted to the arXiv) these bounds have been significantly improved [CLMM23].
The functions y i (t) are defined as follows.Listing 3: Verification that y i (q − 1) = x i .This expression yields {x 1 , x 2 , x 3 } The function r(t) can subsequently be found with the following code.
Listing 4: The function r It can be observed that r is indeed linear in t.To calculate r(l + 1) and the slope of r we use the following piece of code.
0 and c ∈ (0, c 0 ) there exist c ′ < c such that F •2 d (D c ) ⊂ D c ′ .As P c = ∪ π∈Sq π • D c , we see by Lemma 2.4 that for d ≥ d 0 and c ∈ (0, c 0 ) we have F •2 d (P c ) ⊂ P c ′ .Next we consider c ∈ [c 0 , q + 1].By Proposition 4.1 we know F •2 ∞ (P c ) ⊂ P φ(c) and φ(c) < c for any c > 0. As F d converges to F ∞ uniformly, we see for each c ∈ [c 0 , q + 1] there is a d c > 0 large enough such that for d ≥ d c and c ′ = c/2 + φ(c)/2 we have F •2 d (P ĉ) P c ′ for all ĉ sufficiently close to c.By compactness of [c 0 , q + 1], we obtain that there is a d max > 0 such that for any d > d max and any c ∈ [c 0 , q + 1] there exists c ′ < c such that F •2 d (P c ) P c ′ .The proposition now follows by taking d 2 = max(d 0 , d max ).