Theoretical and Mathematical Physics, 173(1): 1377–1386 (2012)

THE UNIQUENESS CONDITION FOR A WEAKLY PERIODIC GIBBS MEASURE FOR THE HARD-CORE MODEL c U. A. Rozikov∗ and R. M. Khakimov† 

We study the hard-core model on the Cayley tree. We show that this model admits only periodic Gibbs measures with the period two. We find sufficient conditions for all weakly periodic Gibbs measures to be translation invariant.

Keywords: Cayley tree, configuration, hard-core model, Gibbs measure, periodic measure, weakly periodic measure

1. Introduction The main problem associated with a Hamiltonian is to describe all Gibbs measures corresponding to it. The reader can find the definition of the Gibbs measure and of other subjects related to Gibbs measure theory, for example, in [1]–[3]. This problem was extensively studied for the Ising model on a Cayley tree. For instance, a noncountable set of boundary Gibbs measures was constructed in [4], and the necessary and sufficient condition for a nonordered phase of the Ising model on the Cayley tree to have a boundary was found in [5]. Periodic Gibbs measures on a Cayley tree have been studied for various models of statistical mechanics in numerous papers (see, e.g., [6]–[13]). These measures were mostly translation invariant or periodic with the period two. Moreover, it was proved that for many models on a Cayley tree, the set of periodic Gibbs measures is very poor, i.e., that only period-two periodic Gibbs measures exist (see, e.g., [8]–[12]). The notion of the Gibbs measure was extended to a more general notion of the weakly periodic Gibbs measure introduced in [14], [15], where the existence of such measures for the Ising model on the Cayley tree was also proved. The hard-core (HC) model on the Cayley tree was studied in [13], where it was proved that the translation-invariant Gibbs measure is unique for this model. Moreover, under certain conditions on the HC-model parameters, we have proved the nonuniqueness of periodic Gibbs measures with the period two. We devote the present paper to studying a weakly periodic measure for the HC model on a Cayley tree and to proving the uniqueness of such measures under certain conditions. This paper is organized as follows. In Sec. 2, we present the basic definitions and known facts. In Sec. 3, we describe the systems of functional equations and prove that the only periodic Gibbs measures admitted by the HC model are those with the period two. In Sec. 4, we find sufficient conditions under which all weakly periodic Gibbs measures are translation invariant. ∗

Institute of Mathematics at National University of Uzbekistan, Tashkent, Uzbekistan, e-mail: [email protected]. †

Namangan State University, Namangan, Uzbekistan, e-mail: [email protected].

Translated from Teoreticheskaya i Matematicheskaya Fizika, Vol. 173, No. 1, pp. 60–70, October, 2012. Original article submitted February 20, 2012. 0040-5779/12/1731-1377

1377

2. Definitions and known facts A Cayley tree τ k of order k ≥ 1 is an infinite tree, i.e., a graph without cycles and such that each of its vertices is incident to exactly k+1 edges. Let τ k = (V, L, i), where V is the set of vertices of τ k , L is the set of edges, and i is the incidence function setting its endpoints x, y ∈ V into correspondence with each edge l ∈ L. If i(l) = {x, y}, then x and y are called the nearest-neighbor vertices, and we use the notation l = x, y. The distance d(x, y), x, y ∈ V , in the Cayley tree is defined by the formula d(x, y) = min{d | ∃x = x0 , x1 , . . . , xd−1 , xd = y ∈ V such that x0 , x1 , . . . , xd−1 , xd }. For a fixed x0 ∈ V , we set Wn = {x ∈ V | d(x, x0 ) = n},

Vn = {x ∈ V | d(x, x0 ) ≤ n}.

(1)

For x ∈ Wn , we set S(x) = {y ∈ Wn+1 : d(x, y) = 1}. Let Φ = {0, 1} and σ ∈ ΦV be a configuration, i.e., σ = {σ(x) ∈ Φ : x ∈ V }, where we set σ(x) = 1 if the vertex x of the Cayley tree is occupied and σ(x) = 0 if it is vacant. A configuration σ is said to be admissible if σ(x)σ(y) = 0 for any pair of nearest neighbors x, y in V (in Vn or Wn correspondingly). We let Ω (ΩVn or ΩWn ) denote the set of such configurations. Obviously, Ω ⊂ ΦV . The HC-model Hamiltonian is defined as ⎧  ⎨J x∈V σ(x), σ ∈ Ω, (2) H(σ) = ⎩+∞, σ∈ / Ω, where J ∈ R. Let B be the σ-algebra generated by cylindric subsets of Ω. For an arbitrary n, we let BVn = {σ ∈ Ω : σ|Vn = σn } denote the subalgebra of B, where σ|Vn is the restriction of σ to Vn and σn : x ∈ Vn → σn (x) is an admissible configuration in Vn . Definition 1. For λ > 0, the HC-model Gibbs measure is a probabilistic measure μ on (Ω, B) such that for any n and σn ∈ ΩVn , we have  μ{σ ∈ Ω : σ|Vn = σn } =

μ(dω) Pn (σn |ωWn+1 ),

(3)

Ω

where Pn (σn | ωWn+1 ) =

e−H(σn ) 1(σn ∨ ω|Wn+1 ∈ ΩVn+1 ). Zn (λ; ω|Wn+1 )

Here, the symbol ∨ denotes the union of configurations, and Zn (λ; ω|Wn+1 ) is the normalization multiplier with the boundary condition ω|Wn : Zn (λ; ω|Wn+1 ) =



e−H(σn ) 1( σ ∨ ω|Wn+1 ∈ ΩVn+1 ).

(4)

σn ∈ΩVn

It is known that we can represent τ k as Gk , which is the free product of k+1 order-two cyclic groups k be a subgroup of Gk . If a Gibbs measure is invariant with the corresponding generators a1 , . . . , ak+1 . Let G k -periodic measure. with respect to some finite-index subgroup Gk ⊂ Gk , it is called a G 1378

It is known [13] that in correspondence with each Gibbs measure of the HC model on the Cayley tree, we can set the entity of quantities z = {zx , x ∈ Gk } satisfying the equality zx =



(1 + λzy )−1 ,

(5)

y∈S(x)

where λ = eJ > 0 is a parameter. We set f (z) ≡ f (z, λ) =

1 , (1 + λz)k

λcr =

kk . (k − 1)k+1

Proposition 1 [13]. If λ ≤ λcr , then functional equation (5) admits a unique solution zx = z ∗ , where z > 0 is a unique solution of the equation z = f (z). If λ > λcr , then any solution zx of functional equation (5) satisfies the inequality ∗

z− ≤ zx ≤ z+

∀x ∈ V,

where z− , z+ > 0 are uniquely defined by the equations z− = f (f (z− )) and z+ = f (z− ) or, equivalently, z− = lim f (2n−1) (1), n→∞

z+ = lim f (2n) (1). n→∞

Here, f (n+1) (z) = f (f (n) (z)), n = 0, 1, 2, . . . . In particular, we have (1 + λ)−k < zx < 1, x ∈ V . k -periodic if zyx = zx for each Definition 2. An entity of quantities z = {zx , x ∈ Gk } is said to be G x ∈ Gk , y ∈ Gk ; Gk -periodic entities are said to be translation invariant. For any x ∈ Gk , the set {y ∈ Gk : x, y}\S(x) consists of a single element, denoted by x↓ (see [14], [16]). k = {H1 , . . . , Hr } be the quotient group, where G k is the normal divisor of index r ≥ 1. Let Gk /G k -weakly periodic if zx = zij Definition 3. An entity of quantities z = {zx , x ∈ Gk } is said to be G for x ∈ Hi , x↓ ∈ Hj for each x ∈ Gk . We note that a weakly periodic entity z coincides with the standard periodic entity (see Definition 2) if the value of zx is independent of x↓ . k -(weakly) k -(weakly) periodic if it corresponds to the G Definition 4. A measure μ is said to be G periodic entity of quantities z.

3. Periodic measures and the system of equations (2)

(2)

The existence of Gk -periodic Gibbs measures, where Gk is a subgroup consisting of words of even length, was shown in [13] under certain conditions on λ and k. The following theorem shows that only (2) Gk -periodic Gibbs measures can exist for the HC model on the Cayley tree. Theorem 1. For any normal divisor G ⊂ Gk , any G-periodic Gibbs measure of the HC model is either (2) translation invariant or a Gk -periodic Gibbs measure. Proof. Let g(z, λ) = (1 + λz)−1 . Clearly g(y, λ) = g(z, λ) if and only if y = z. Therefore, the proof of the theorem is equivalent to the proof of Theorem 2 in [9]. 1379

The following theorem establishes a condition on a normal divisor G ⊂ Gk under which any G-periodic Gibbs measure of the HC model is translation invariant. Theorem 2. If G ∩ {a1 , . . . , ak+1 } = ∅ for a normal divisor G ⊂ Gk (i.e., G contains at least one generator), then any G-periodic Gibbs measure of the HC model is translation invariant. Proof. The proof is analogous to the proof of Theorem 3 in [9]. Here, we study G-weakly periodic Gibbs measures for any normal divisor G of index two. Let ∅ = A ⊂ Nk = {1, 2, . . . , k + 1}. It is known that every index-two subgroup of the group Gk has the structure   HA = x ∈ Gk : wx (ai ) is even , i∈A

where wx (ai ) is the number of letters ai in the word x ∈ Gk [6]. We note that an HA -weakly periodic entity z has the form ⎧ ⎪ z00 , x ∈ HA , x↓ ∈ HA , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨z01 , x ∈ HA , x↓ ∈ Gk \ HA , zx = (6) ⎪ ⎪ z , x ∈ G \ H , x ∈ H , 10 k A ↓ A ⎪ ⎪ ⎪ ⎪ ⎩ z11 , x ∈ Gk \ HA , x↓ ∈ Gk \ HA . For simplicity, we introduce the notation z1 = z00 , z2 = z01 , z3 = z10 , z4 = z11 , and |A| = i, where |A| is the cardinality of the set A. By equality (5), the quantities zi , i = 1, . . . , 4, then satisfy the system of equations 1 1 1 1 z1 = , z2 = , i k−i i−1 (1 + λz3 ) (1 + λz1 ) (1 + λz3 ) (1 + λz1 )k−i+1 (7) 1 1 1 1 z3 = , z4 = . (1 + λz2 )i−1 (1 + λz4 )k−i+1 (1 + λz2 )i (1 + λz4 )k−i (2)

Remark 1. For A = Nk , the HA -corresponding index-two normal divisor coincides with Gk . This (2) (2) case was studied in [9]. Moreover, we can easily see that Gk -weakly periodic Gibbs measures are Gk (2) periodic. We therefore do not consider the case HA = Gk . Remark 2. It is easy to verify that for solutions of system (7), we have z1 = z2 if and only if z3 = z4 . A solution (z1 , z2 , z3 , z4 ) of system of equations (7) is a translation-invariant Gibbs measure if (2) z1 = z2 = z3 = z4 . It is a Gk -(weakly) periodic Gibbs measure if (z1 = z2 , z3 = z4 ) z1 = z2 , z3 = z4 . Remark 3. Obviously, HA ∩ {a1 , . . . , ak+1 } = ∅ for any A = Nk . By Theorem 2, an HA -periodic Gibbs measure is therefore translation invariant. This means that system of equations (7) does not admit solutions (z1 , z2 , z3 , z4 ) such that z1 = z2 , z3 = z4 , and z1 = z3 . We rewrite system of equations (7) in the form  z1 =  z3 = 1380

1 + λz1 1 + λz3 1 + λz4 1 + λz2

i

1 , (1 + λz1 )k

i−1

1 , (1 + λz4 )k

i−1 1 1 + λz1 z2 = , 1 + λz3 (1 + λz1 )k i  1 1 + λz4 z4 = . 1 + λz2 (1 + λz4 )k 

(7 )

Dividing the first equation in this system by the second and the third by the fourth, we obtain the system of equations z1 1 + λz1 z3 1 + λz2 = , = . (8) z2 1 + λz3 z4 1 + λz4 Using this system, we rewrite system (7 ) in the form  z1 =  z3 =

z1 z2 z4 z3

i



1 , (1 + λz1 )k

i−1

z2 = 

1 , (1 + λz4 )k

z4 =

z1 z2 z4 z3

i−1 i

1 , (1 + λz1 )k (9)

1 . (1 + λz4 )k

We find z2 from the first equation in (9) and z3 from the fourth equation. Substituting the obtained quantities in the fourth and first equations in system (7), we then obtain z1 =

(1 + λz4 )k ((1 + λz4 )k/i +

1−1/i i λz4 )

1 , (1 + λz1 )k−i (10)

(1 + λz1 )k

1 z4 = 1−1/i i (1 + λz )k−i k/i 4 ((1 + λz1 ) + λz ) 1

or z1 = ϕ(z4 )ψ(z1 ), where ϕ(z) =

z4 = ϕ(z1 )ψ(z4 ),

(1 + λz)k , ((1 + λz)k/i + λz 1−1/i )i

ψ(z) =

(11)

1 . (1 + λz)k−i

4. The uniqueness conditions for the weakly periodic Gibbs measure The following lemma is useful for our studies. Lemma 1. 1. If we have the equalities z1 = z4 and z2 = z3 in system of equations (7), then z1 = z2 = z3 = z4 . 2. The equality z1 = z4 holds if and only if z2 = z3 . Proof. 1. We consider the difference z1 − z3 =

1 (1 + λz3

)i (1

+ λz1

)k−i

−

1 (1 + λz2

)i−1 (1

+ λz4 )k−i+1

=

=

(1 + λz2 )i−1 (1 + λz4 )k−i+1 − (1 + λz3 )i (1 + λz1 )k−i = (1 + λz3 )i (1 + λz1 )k−i (1 + λz2 )i−1 (1 + λz4 )k−i+1

=

(1 + λz2 )i−1 (1 + λz4 )k−i+1 − (1 + λz2 )i−1 (1 + λz1 )k−i+1 + (1 + λz3 )i (1 + λz1 )k−i (1 + λz2 )i−1 (1 + λz4 )k−i+1 +

(1 + λz2 )i−1 (1 + λz1 )k−i+1 − (1 + λz3 )i (1 + λz1 )k−i = (1 + λz3 )i (1 + λz1 )k−i (1 + λz2 )i−1 (1 + λz4 )k−i+1

= A1 (z4 − z1 ) + B1 (z2 − z3 ) + C1 (z1 − z3 ), 1381

i.e., z1 − z3 = A1 (z4 − z1 ) + B1 (z2 − z3 ) + C1 (z1 − z3 ),

(12)

where A1 =

λ[(1 + λz4 )k−i + (1 + λz4 )k−i−1 (1 + λz1 ) + · · · + (1 + λz1 )k−i ] , (1 + λz1 )k−i (1 + λz3 )i (1 + λz4 )k−i+1

B1 =

λ(1 + λz1 )[(1 + λz2 )i−2 + (1 + λz2 )i−3 (1 + λz3 ) + · · · + (1 + λz3 )i−2 ] , (1 + λz2 )i−1 (1 + λz3 )i (1 + λz4 )k−i+1

C1 =

(1 + λz2

)i−1 (1

λ . + λz3 )(1 + λz4 )k−i+1

The quantities A1 , B1 , and C1 are positive here. Moreover, from the third equation in system (7), we obtain C1 = λz3 /(1 + λz3 ). From the lemma condition z1 = z4 and z2 = z3 and from equality (12), we obtain z1 = z2 = z3 = z4 . 2. Let z1 = z4 . From system (8), we then have z3 /z2 = (1 + λz2 )/(1 + λz3 ), whence z2 = z3 . Let z2 = z3 . Then z1 /z4 = (1 + λz1 )/(1 + λz4 ), whence we have z1 = z4 . The lemma is proved. Theorem 3. Let one of the conditions 1. i = 1, 2. i = (k + 1)/2, 3. i = k, 4. i = k − 1, where i = |A|, be satisfied. Then all the HA -weakly periodic HC-model Gibbs measures are translation invariant. Proof. To prove the statement, it suffices to demonstrate that system of functional equations (11) has only roots of the form z1 = z2 = z3 = z4 . For this, we subtract the second equation in system (11) from the first, and using the Lagrange theorem on the mean for the functions ϕ(z) and ψ(z) on (z1 , z4 ), we obtain z1 − z4 = ϕ(z4 )ψ(z1 ) − ϕ(z1 )ψ(z4 ) = = ϕ(z4 )ψ(z1 ) − ϕ(z1 )ψ(z1 ) + ϕ(z1 )ψ(z1 ) − ϕ(z1 )ψ(z4 ) = = ψ(z1 )[ϕ(z4 ) − ϕ(z1 )] + ϕ(z1 )[ψ(z1 ) − ψ(z4 )] = = ψ(z1 )ϕ (ξ)(z4 − z1 ) + ϕ(z1 )ψ  (η)(z1 − z4 ), where ξ ∈ (z1 , z4 ) and η ∈ (z1 , z4 ). Hence, (z1 − z4 )[1 + ψ(z1 )ϕ (ξ) − ϕ(z1 )ψ  (η)] = 0.

(13)

This equation implies that z1 = z4 , or 1 + ψ(z1 )ϕ (ξ) − ϕ(z1 )ψ  (η) = 0. 1382

(14)

Here, ψ  (η) = −λ(k − i)/(1 + λη)k−i+1 ≤ 0, i.e., the function ψ(η) decreases because i ≤ k, λ > 0, η > 0, and the derivative of ϕ(z) at a point ξ ∈ (z1 , z4 ) is ϕ (ξ) =

λ(1 + λξ)k−1 [λξ(k − i + 1) − i + 1] . ξ 1/i [(1 + λξ)k/i + λξ 1−1/i ]i+1

1. In the case i = 1, the above formulas imply that ϕ (ξ) > 0. Because ψ  (η) ≤ 0, Eq. (14) has no solutions. Hence, z1 = z4 , and by Lemma 1, we have z1 = z2 = z3 = z4 . 2. In the case i = (k + 1)/2, we show that Eq. (14) has no solutions, i.e., ψ(z1 )ϕ (ξ) − ϕ(z1 )ψ  (η) > −1.

(15)

By Proposition 1, we obtain 1 < ξ < 1, (1 + λ)k

1 < z1 < 1. (1 + λ)k

For i = (k + 1)/2, we then have the estimate ψ(z1 )ϕ (ξ) − ϕ(z1 )ψ  (η) =

λ(1 + λξ)k−1 [λξ(k − (k + 1)/2 + 1) − (k + 1)/2 + 1] 1 + (1 + λz1 )k−(k+1)/2 ξ 2/(k+1) [(1 + λξ)2k/(k+1) + λξ 1−2/(k+1) ](k+1)/2+1 (1 + λz1 )k

+ [(1 + λz1 >

+

1−2/(k+1) (k+1)/2 λz1 ]

λ(k − (k + 1)/2) > (1 + λη)k−(k+1)/2+1

λ(1 + λ/(1 + λ)k )k−1 [(λ/(1 + λ)k )(k + 1)/2 − (k + 1)/2 + 1] + (1 + λ)(k−1)/2 [(1 + λ)2k/(k+1) + λ](k+3)/2 +

>

)2k/(k+1)

(1 + λ/(1 + λ)k )k λ(k − 1) > 2[(1 + λ)2k/(k+1) + λ](k+1)/2 (1 + λ)(k+1)/2

λ[(1 + λ)k + λ]k−1 [λ(k + 1) − (k − 1)(1 + λ)k ] + 2(1 + λ)k2 +k+1 [(1 + λ)2k + λ]k+3 +

[(1 + λ)k + λ]k λ(k − 1) > −1. 2(1 + λ)k2 +k+1 [(1 + λ)2k + λ]k+3

We hence obtain λ[(1 + λ)k + λ]k−1 [λ(k + 1) − (k − 1)(1 + λ)k ] + + λ(k − 1)[(1 + λ)k + λ]k + 2(1 + λ)k

2

+k+1

[(1 + λ)2k + λ]k+3 =

= λ[(1 + λ)k + λ]k−1 [λ(k + 1) − (k − 1)(1 + λ)k + (k − 1)[(1 + λ)k + λ]] + + 2(1 + λ)k

2

+k+1

[(1 + λ)2k + λ]k+3 > 0

because λ(k + 1) − (k − 1)(1 + λ)k + (k − 1)[(1 + λ)k + λ] = 2kλ > 0. System of equations (7) therefore only has solutions of the form z1 = z2 = z3 = z4 in this case. 1383

3. In the case i = k, by Proposition 1, we have 1 < ξ < 1. (1 + λ)k We then have the estimate λ(1 + λξ)k−1 [λξ(k − i + 1) − i + 1] > ξ 1/i [(1 + λξ)k/i + λξ 1−1/i ]i+1

ϕ (ξ) =

λ(1 + λ/(1 + λ)k )k−1 [λ(k − i + 1)/(1 + λ)k − i + 1] . [(1 + λ)k/i + λ]i+1

>

(16)

For i = k, ψ(z) = 1 and hence ψ  (z) = 0. Therefore, Eq. (14) becomes ϕ (ξ) + 1 = 0, and inequality (16) becomes ϕ (ξ) >

λ[(1 + λ)k + λ]k−1 [λ + (1 − k)(1 + λ)k ] . (1 + 2λ)k+1 (1 + λ)k2

Obviously, Eq. (14) has no solutions for ϕ (ξ) > −1. To verify this, it suffices to show that the right-hand side of the last inequality is greater than −1: λ[(1 + λ)k + λ]k−1 [λ + (1 − k)(1 + λ)k ] > −1. (1 + 2λ)k+1 (1 + λ)k2 Indeed, 2

λ[(1 + λ)k + λ]k−1 [λ + (1 − k)(1 + λ)k ] + (1 + 2λ)k+1 (1 + λ)k > 0 or λ2 [(1 + λ)k + λ]k−1 + (1 + λ)k [(1 + 2λ)k+1 (1 + λ)k

2

−k

− (k − 1)λ[(1 + λ)k + λ]k−1 ] > 0.

We show that the difference (1 + 2λ)k+1 (1 + λ)k

2

−k

− (k − 1)λ[(1 + λ)k + λ]k−1

is positive. For this, we use the Newton binomial formula to expand (1 + 2λ)k+1 and [(1 + λ)k + λ]k−1 and regroup the terms as (1 + 2λ)k+1 (1 + λ)k

2

−k

− (k − 1)λ[(1 + λ)k + λ]k−1 = 2

−k

2

−k

− k + 1)] +   2 k(k + 2) k−1 2 (1 + λ)k −2k − (k − 1)2 + + (1 + λ)k λk−1 2   k(k + 1) k−2 (k − 1)(k − 2) 2k k−2 k2 −3k 2 + (k − 1) (1 + λ) − + (1 + λ) λ 3 2   2 k(k + 1) 3 (k − 1)(k − 2) 2 (1 + λ)2k − + + · · · + (1 + λ)k −3k λ3 (k − 1) 3 2   k2 −2k 2 k(k + 1) 2 k 2 (1 + λ) − (k − 1) + + (1 + λ) λ 2

= (2λ)k+1 (1 + λ)k

+ (1 + λ)k 1384

2

−k

+ λk [(k + 1)2k (1 + λ)k

λ[(k + 1)2 − (k − 1)] + (1 + λ)k

2

−k

.

It is easy to verify that every difference in the square brackets in the right-hand side of this equality is positive for the corresponding values of k. Hence, ϕ (ξ) > −1, which means that Eq. (13) for i = k has roots only of the form z1 = z4 . Hence, by Lemma 1, system of equations (7) has a unique solution with z1 = z2 = z3 = z4 under this condition. 4. In the case i = k − 1, the proof is analogous to the preceding cases. Using Proposition 1, we prove inequality (15) for this case, i.e., for i = k − 1. Indeed, we have ψ(z1 )ϕ (ξ) − ϕ(z1 )ψ  (η) =

(1 + λz1 +

>

(k−2)/(k−1) k−1 ] (1

[(1 + λz1 )k/(k−1) + λz1

+ λη)2

>

λ((1 + λ)k + λ)k > [(1 + λ)k/(k−1) + λ]k−1 (1 + λ)k2 +2

λ((1 + λ)k + λ)k−1 (2λ + (2 − k)(1 + λ)k ) + (1 + λ)k2 +2 [(1 + λ)k + λ]k +

=

λ(1 + λz1 )k

λ((1 + λ)k + λ)k−1 (2λ + (2 − k)(1 + λ)k ) + (1 + λ)k2 +1 [(1 + λ)k/(k−1) + λ]k +

>

λ(1 + λξ)k−1 (2λξ − k + 2) + + λξ)k/(k−1) + λξ (k−2)/(k−1) ]k

)ξ 1/(k−1) [(1

λ((1 + λ)k + λ)k = [(1 + λ)k + λ]k−1 (1 + λ)k2 +2

λ[2λ + (2 − k)(1 + λ)k ] + λ((1 + λ)k + λ)2 > −1. (1 + λ)k2 +2 [(1 + λ)k + λ]

Hence, 2λ2 + λ((1 + λ)k + λ)2 + (1 + λ)k [(1 + λ)k

2

+2

+ λ(1 + λ)k

2

−k+2

− (k − 2)λ] > 0

because the expression in the square brackets is positive. The theorem is proved. For i ∈ {1, (k + 1)/2, k − 1, k}, this theorem implies that independently of the values of λ, all weakly periodic Gibbs measures are translation invariant. The following theorem provides conditions for λ at all values of i. Theorem 4. For λ ∈ (0, λcr ] ∪ [(i − 1)/(z− (k − i + 1)), +∞), all HA -weakly periodic Gibbs measures for the HC model are translation invariant. Here, z− and λcr are defined in Proposition 1. Proof. The case λ ∈ (0, λcr ] follows from Proposition 1. We take Eq. (14). It obviously admits no solutions if ϕ (ξ) ≥ 0, which means that λξ(k − i + 1) − i + 1 ≥ 0 or λ≥

i−1 . ξ(k − i + 1)

By Proposition 1, we have z− ≤ ξ. Replacing ξ with z− , we then obtain the statement of Theorem 4 in the right-hand side of the last inequality. 1385

Acknowledgments. One of the authors (U. A. R.) thanks the Institute des Hautes Etudes Scientifiques (Bures-sur-Yvette, France) for the financial support of his stay at the IHES in January–April 2012 and the Program IMU/CDC for the support.

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