The Curie–Weiss model with complex temperature: phase transitions Mira Shamis1, Ofer Zeitouni2 January 11, 2017 7 1 0 2 Abstract n We study the partition function and free energy of the Curie–Weiss model with a J complex temperature, and partially describe its phase transitions. As a consequence, 9 we obtain information on the locations of zeros of the partition function. ] R 1 Introduction P . h t An important component of large deviations theory is Varadhan’s lemma, which states that a m if a sequence of probability measures µ satisfies the large deviations principle in a (Polish) N [ space X with speed N and rate function I, then for any bounded continuous function f : X → R, 1 (cid:90) v 1 lim log eNf(x)µ (dx) = sup(f(x)−I(x)). (1.1) 5 N 7 N→∞ N x∈X 3 See [2] for a precise statement, relaxed assumptions, and applications. 2 0 In many applications, considering real-valued f is too restrictive, and one may be inter- 1. ested in relaxing it to allow for complex-valued f. Statistical mechanics provides for a rich 0 class of examples; we mention in particular the Yang–Lee theory, where the complex pertur- 7 bation is in form of a magnetic field, or the quantum spin chain models [5], where quantitites 1 : of interest such as emptiness formation can be formulated as exponential asymptotics of the v i type (1.1) with complex integrand. Note that in such examples, because f is multiplied by X N, relatively small changes in phase may lead to sign changes of the integrand in (1.1) and r a therefore to cancelations. It seems maybe naive at this point to hope for a general theory, which would consist of an analogueof (1.1). Ourgoalinthispaperismoremodest: weconsideronesimpleexample,the Curie–Weiss model with complex temperature, and partially develop the asymptotic theory concerning its partition function. While we are not able to give a complete description of 1Department of Mathematics, Weizmann Institute of Science, Rehovot 7610001, Israel and School of Mathematical Sciences, Queen Mary University of London, Mile End Road, London E1 4NS, England. E-mail: m.shamis@qmul.ac.uk. Supported in part by ISF grant 147/15. 2Department of Mathematics, Weizmann Institute of Science, Rehovot 7610001, Israel. E-mail: ofer.zeitouni@weizmann.ac.il. Supported in part by ISF grant 147/15. 1 the associated phase diagram, we will be able to show that the phase diagram is not trivial. As a consequence of our analysis, we will also obtain information on the (complex) zeros of the partition function, of importance in the Yang–Lee theory of phase transitions, see [8, Theorem 2]. WebeginbyintroducingtheCurie–Weissmodelthatwewillconsider. Letσ = (σ ,...,σ ) ∈ 1 N {−1,+1}N. Define the Hamiltonian N 1 (cid:88) N H (σ) = − σ σ = − (m (σ))2, (1.2) N i j N 2N 2 i,j=1 where the magnetization is m (σ) = 1 (cid:80)N σ . For β ∈ C, let Z denote the free energy, N N i=1 i β,N i.e. (cid:90) (cid:90) N 1 (cid:88) (cid:89) Z = exp(−βH (σ)) = ··· exp(−βH (σ)) µ(dσ ), (1.3) β,N 2N N N i σ∈{−1,+1}N i=1 where µ(dσ) = 1(δ +δ ). 2 1 −1 When β is real, it is an easy exercise to apply Varadhan’s lemma (1.1) and Cramer’s theorem concerning the large deviations of m in order to conclude that N (cid:40) 1 0, β ∈ (−∞,1] F = lim log|Z | = β β,N N→∞ N > 0, β ∈ (1,∞). (F is referred to as the Free Energy.) More refined analysis (see e.g. [3]) yields that for β β ∈ R\{1}, Z = A eNFβ(1+o(1)), (1.4) β,N β where A > 0 is some constant that depends only on β; this is due to the Gaussian nature of β √ the fluctuations of N(m −m∗(β)), where m∗(β) is the asymptotic magnetization, under N the measure exp(−βH (σ))(cid:81)N µ(dσ )/Z . Also, m∗(β) = 0 for β ≤ 1. N i=1 i β,N When β = (1+(cid:15)+iR) ∈ C, one expects to similarly have a separation between a region where F = 0 and F (cid:54)= 0. In particular, one predicts the existence of a critical curve C in β β the complex plane, passing through 1, that divides the complex plane into a region where F = 0 and its complement where F (cid:54)= 0. β β For symmetry reasons, it is enough to consider R ≥ 0. Our first result describes a region where F vanishes. β Theorem 1.1. There exist constants c,c(cid:48),(cid:15) > 0 so that, with β = 1 + (cid:15) + iR, if either √ 0 0 < (cid:15) ≤ (cid:15) and c (cid:15) ≤ R ≤ √c(cid:48) or (cid:15) < 0, then 0 (cid:15) (cid:115) β 1 Z = (1+o (1)), lim log|Z | = 0. β,N β −β2 (cid:15),R N→∞ N β,N Remark 1.1. One can make the constants c,c(cid:48),(cid:15) explicit. Our proof gives (cid:15) = 1/9,c = √ √ 0 0 20,c(cid:48) = π/ 32, but these are certainly not optimal constants. 2 Remark 1.2. It is possible to also treat the case of (cid:15) = 0, where one may observe a transition as function of R: for R = 0, it is standard, see [7, Theorem 2], that Z is asymptotic to a β,N constant multiple of N1/4, while a local analysis near the saddle point 0 reveals that if R > 0 is small then Z is asymptotic to an (R-dependent) constant, see Theorem 1.3 below. β,N Our next result shows that along a particular curve that is asymptotic to 1+i∞ and to ∞+πi, indeed F > 0. β Theorem 1.2. For β = 1+(cid:15)+iR on the curve R (cid:18)1+ π (cid:19) 1+(cid:15) = log R , π < R < ∞, (1.5) 2π 1− π R we have, for some constant A(cid:101) , that β 1 |Z | = Z A(cid:101) (1+o(1)), lim log|Z | > 0. β,N Reβ,N β β,N N→∞ N √ Remark 1.3. The curve in Theorem 1.2 is asymptotic to c(cid:48)(cid:48)/ (cid:15) as (cid:15) → 0; compare with Theorem 1.1, noting that c(cid:48)(cid:48) (cid:54)= c(cid:48). In a neighborhood of β = 1, we actually can give a complete description of the transition away from F = 0. Define the even function h (u) = u2/2β−logcoshu for u ∈ C, |u| < π/2. β β With this definition we will see in Proposition 2.1 that (cid:115) N (cid:90) ∞ Z = e−Nhβ(u)du. (1.6) β,N 2πβ −∞ In Claim 5.1 below we show that for some c > 0 small and 0 < |β −1| ≤ c, h(cid:48) (u) has three β zeros in a neighborhood of 0: 0,±u . β Theorem 1.3. There exists c(cid:48) ≤ c such that for 0 < |β −1| ≤ c(cid:48) (cid:0) (cid:0) (cid:1)(cid:1) 1. Z = √1 1+O 1 when Reβ ≤ 1, β,N 1−β N 2. Z = √1 (cid:0)1+O(cid:0)1(cid:1)(cid:1)+2(cid:113) β e−Nhβ(uβ)(cid:0)1+O(cid:0)1(cid:1)(cid:1) when Reβ ≥ 1, β,N 1−β N β−β2+u2 N β and for any δ > 0 the implicit constants are uniform in δ ≤ |β −1| ≤ c(cid:48). See Figure 1 for a schematic illustration of our theorems. We remark that on the line Reβ = 1 we will see (as a consequence of Claim 1.4 below) that Reh (u ) > 0 (except for β β β = 1). In particular, the two statements in Theorem 1.3 coincide on that line. In Theorem 1.3, an important role is played by those β with Reh (±u ) = 0. These are β β characterized by the following claim. Claim 1.4. There exist c,C > 0 and a smooth function (cid:15) (cid:55)→ R ((cid:15)) on [−c,c] such that 0 |R ((cid:15))−(cid:15)| ≤ C(cid:15)2 and the following holds for β = 1+(cid:15)+iR: 0 • If |R| < |R ((cid:15))|, then Reh (±u ) < 0, 0 β β 3 0.2 6 5 0.1 4 0.0 3 2 (cid:45)0.1 1 (cid:45)0.2 0 1.0 1.1 1.2 1.3 1.4 1.5 0.8 0.9 1.0 1.1 1.2 Figure1: Schematicillustrationofourresults. Left: F = 0totheleftofthepurplecurveon β the left (Theorem 1.1); F < 0 on the red curve on the right (Theorem 1.2). We conjecture β that the two phases are separated by a curve similar to the one schematically depicted in black. The three curves are asymptotic at infinity to the line Reβ = 1. Right: The vicinity of the critical point β = 1: here F > 0 in the purple region on the right, and F = 0 in the β β blue region on the left. • If |R| = |R ((cid:15))|, then Reh (±u ) = 0, 0 β β • If |R| > |R ((cid:15))|, then Reh (±u ) > 0. 0 β β For c as above, define the critical curve Γ = {β = 1+(cid:15)+iR| 0 ≤ (cid:15) ≤ c, R = ±R ((cid:15))}. 0 Theorem 1.3 allows us to describe the location of zeros of Z , and show that in a neigh- β,N borhood of β = 1, they are close to the critical curve Γ. Define (cid:115) 1 β Ψ (β) = √ +2 e−Nhβ(uβ), Reβ ≥ 1. N 1−β β −β2 +u2 β The zeros of Ψ (β) near β = 1 lie near the critical curve Γ; we will show that the zeros of N Z are close to those zeros. β,N Corollary 1.5. For any δ > 0 the following holds for N ≥ N (δ). The zeros of Z in 0 β,N δ < |β −1| < c(cid:48) lie in Reβ > 1, and for any zero β of Z there exists a unique zero β(cid:48) of β,N Ψ (β) such that |β −β(cid:48)| < Cδ. Vice versa, for any zero β(cid:48) of Ψ (β) with δ < |β(cid:48) −1| < c(cid:48) N N2 N there exists a unique zero β of Z with |β −β(cid:48)| < Cδ. In particular, the zeros of Z lie β,N N2 β,N within C N−1 from the critical curve. δ We also obtain information on the empirical measure of zeros of Z , in a neighborhood N of the critical point β = 1. For c(cid:48) > 0 small, introduce the scaled zero-counting measure 1 (cid:88) µ = δ . (1.7) N β N β:|β−1|≤c(cid:48),Z =0 β,N 4 Define a positive measure µ on C, supported on Γ := Γ∩{|β −1| ≤ c(cid:48)} as follows: For a c(cid:48) segment I of Γ connecting a = 1+(cid:15) +iR ((cid:15) ) ∈ Γ , j = 1,2, with (cid:15) > (cid:15) ≥ 0, set j j 0 j c(cid:48) 2 1 1 µ(I) = (Imh (u )−Imh (u )), (1.8) 2π a2 a2 a1 a1 and extend µ by symmetry to the lower half plane. One checks that µ is a finite positive measure on C. Corollary 1.6. µ → µ in the weak topology for positive measures on C. N N→∞ The results above do not completely characterize the phase diagram of the Curie–Weiss model. In Section 6, we discuss this point and present a conjecture for the critical curve separating the region where the free energy vanishes asymptotically from that where it is strictly positive. 2 Integral representation and preliminaries The proofs of all of the theorems are based on the saddle-point analysis of the following integral representation. Proposition 2.1. If Reβ > 0, then (cid:18)βN(cid:19)1/2(cid:90) ∞ (cid:18) N (cid:19)1/2(cid:90) ∞ Z = exp(−Nf (u))du = exp(−Nh (u))du, (2.1) β,N β β 2π 2πβ −∞ −∞ where f (u) = βu2 −log(cosh(βu)), h (u) = u2/2β−logcoshu, and the branch of the square β 2 √ β root is choosen so that 1 = 1. Proof. Let X ,X ,...,X be independent identically distributed Bernoulli random vari- 1 2 N ables: P{X = 1} = P{X = −1} = 1, and let β ∈ C. Then, using E to denote expectation j j 2 with respect to these random variables, we have   (cid:32) (cid:33)2 Zβ,N = EexpβN 1 (cid:88)N Xj  = E(cid:90) exp(cid:18)−u2 +u(cid:112)βN(cid:80)Xj(cid:19) √du 2 N 2 N 2π j=1 (cid:114) (cid:90) (cid:18) (cid:19) (cid:32) N (cid:33) βN βN (cid:88) = du˜exp − u˜2 Eexp βu˜ X , j 2π 2 j=1 where the second equality uses the Hubbard-Stratonovich transformation and the last uses the change of variables u˜ = √u . Since for any a we have Eexp(aX) = cosha and using the βN assumption that X are i.i.d random variables, we obtain j (cid:32) (cid:33) N N (cid:88) (cid:89) Eexp βu˜ X = Eexp(βu˜X ) = (Eexp(βu˜X ))N = (cosh(βu˜))N . j j j j=1 j=1 Combining the last two displays gives (cid:114) (cid:90) (cid:18) (cid:19) (cid:114) (cid:90) βN βN βN Z = du˜exp − u˜2 +N logcosh(βu˜) = exp(−Nf (u))du. β,N β 2π 2 2π 5 3 Proof of Theorem 1.1 The proof of Theorem 1.1 for (cid:15) < 0 follows from known asymptotics for the Curie–Weiss model. Indeed, for such (cid:15) and with R = 0 and σ2 = 1/(1−(cid:15)), we have by [7, Theorem 2] (cid:15) √ that Z (cid:16) σ and, under the measure e−(1+(cid:15))HN(σ)/Z , we have by [3] that Nm 1+(cid:15),N (cid:15) 1+(cid:15),N N converges in distribution to a centered Gaussian random variable of variance σ2 := −1/(cid:15). (cid:15) We then obtain that with (cid:15) < 0, (cid:90) 1 Z (cid:16) √ e−(1−(cid:15))u2/2−iRu2/2du, β,N 2π which gives the claim. The proof in case (cid:15) > 0 follows a saddle-point analysis of the integral representation from Proposition 2.1. Throughout, C denote constants that may depend on (cid:15) and R but not on i anything else. The following preliminary claims play an important role in the analysis. √ √ Claim 3.1. For any (cid:15) ≤ 1, u ≤ 8(cid:15), 24(cid:15) ≤ R ≤ √π , 9 128(cid:15) u2 1 u2 Ref (u) = (1+(cid:15)) − log(cosh2(u(1+(cid:15)))−sin2(uR)) ≥ (cid:15) . β 2 2 2 Proof. By Taylor expansion, t2 t4 t2 t4 cosht ≤ 1+ +cosht ≤ 1+ + 2 24 2 12 where the last inequality used that t ≤ 3/2 and therefore cosht ≤ 2.4. Using again t < 3/2 we obtain 3 cosh2t ≤ 1+t2 + t4. 4 √ From the assumptions we get that (1 + (cid:15))u ≤ (1 + (cid:15)) 8(cid:15) < 3/2. Therefore, using again (cid:15) < 1/9, (cid:18) 3u2(1+(cid:15))4(cid:19) cosh2((1+(cid:15))u) ≤ 1+u2 +(cid:15)u2 2+(cid:15)+ ≤ 1+u2 +12(cid:15)u2. 4(cid:15) For Ru ≤ π we have sin2(Ru) ≥ R2u2/2, hence 4 R2u2 cosh2((1+(cid:15))u)−sin2(Ru) ≤ 1+u2 +12(cid:15)u2 − . 2 √ Since R ≥ 24(cid:15), we have R2u2/2 ≥ 12(cid:15)u2 and therefore cosh2((1+(cid:15))u)−sin2(Ru) ≤ 1+u2 ≤ eu2, and therefore u2 1 u2 Ref (u) = (1+(cid:15)) − log(cosh2(u(1+(cid:15)))−sin2(uR)) ≥ (cid:15) . β 2 2 2 6 The next claim handles larger values of the argument u. √ Claim 3.2. Let (cid:15) < 1/9. For any t ≥ (1+(cid:15)) 8(cid:15), (cid:18) t2(cid:19) cosht ≤ exp (1−(cid:15)) . (3.1) 2 √ In particular, for u ≥ 8(cid:15), u2 (cid:15)2u2 Ref (u) ≥ (1+(cid:15)) −log(cosh(u(1+(cid:15))) ≥ . (3.2) β 2 2 Proof. By Taylor expansion we have (cid:88)∞ t2k t2 (cid:20) t2 t4 (cid:21) (cid:88)∞ t2k cosht = = 1+ (1−(cid:15))+ (cid:15) + + , (2k)! 2 2 24 (2k)! k=0 k≥3 and (cid:18) t2(cid:19) t2 t4 (cid:88)∞ t2k(1−(cid:15))k exp (1−(cid:15)) = 1+ (1−(cid:15))+ (1−(cid:15))2 + . 2 2 8 2kk! k≥3 For k ≥ 3 we have, if (cid:15) ≤ 1/2, (2k)! (k +1)···(2k) 1 = ≥ 2k ≥ . 2kk! 2k (1−(cid:15))k Therefore (cid:88)∞ t2k (cid:88)∞ t2k(1−(cid:15))k ≤ , (2k)! 2kk! k≥3 k≥3 √ and, since (cid:15) ≤ 1 and t ≥ (1+(cid:15)) 8(cid:15), 9 t4 t4 (cid:15)t2 t4 (cid:18) 1 1 (cid:19) (1−(cid:15))2 − − ≥ (1−(cid:15))2 − − ≥ 0. 8 24 2 8 3 2(1+(cid:15))2 This completes the proof of (3.1). To see (3.2), take t = (1+(cid:15))u, which satisfies the assumptions leading to (3.1). Then, (cid:18) u2(cid:19) (cid:18)(1+(cid:15))(1−(cid:15)2)u2(cid:19) cosh((1+(cid:15))u) ≤ exp (1−(cid:15))(1+(cid:15))2 ≤ exp . (3.3) 2 2 Hence, we have, using the monotonicity of the logarithm and (3.3) u2 u2 (1+(cid:15))(1−(cid:15)2)u2 (cid:15)2u2 (1+(cid:15)) −log(cosh(u(1+(cid:15)))) ≥ (1+(cid:15)) − ≥ . 2 2 2 2 7 We continue with the proof of Theorem 1.1, considering the regime (cid:15) > 0, and R as in the statement of Claim 3.1. In view of Proposition 2.1, we write (cid:90) ∞ (cid:20)(cid:90) −δ (cid:90) ∞ (cid:90) δ(cid:21) e−Nfβ(u)du = + + e−Nfβ(u)du = I(cid:48) +I +I = 2I +I , (3.4) 1 1 2 1 2 −∞ −∞ δ −δ where the last equality follows from the symmetry. Note that f (0) = f(cid:48)(0) = 0. Hence, β β u = 0 is a saddle point. We will show below that the main contribution to the integral comes from a neighborhood of this saddle point. We will choose δ = N−2/5 so that Nδ3 → 0 and √ δ N → ∞ as N → ∞. We begin by estimating I . 1 (cid:34) √ (cid:35) (cid:90) ∞ (cid:90) 8(cid:15) (cid:90) ∞ I = e−Nfβ(u)du = + e−Nfβ(u)du = W +W . 1 √ 1 2 δ δ 8(cid:15) Using Claim 3.1, we have √ √ (cid:90) 8(cid:15) (cid:90) 8(cid:15) √ |W1| ≤ e−NRefβ(u)du ≤ e−N(cid:15)22u2du ≤ e−N(cid:15)22δ2( 8(cid:15)−δ) ≤ e−cN1/5, (3.5) δ δ for some constant c > 0. To estimate W , we use (3.1) of Claim 3.2 and obtain 2 (cid:90) ∞ (cid:90) ∞ |W2| ≤ √ e−NRefβ(u)du ≤ √ e−N(cid:15)u22du ≤ e−CN, (3.6) 8(cid:15) 8(cid:15) where C = C((cid:15)) > 0 is some constant. Combining (3.5) and (3.6) we get |I | ≤ e−CN1/5 +e−CN ≤ e−CÑ1/5. (3.7) 1 We turn to estimating I . Denote by 2 u2 P (u) = (β −β2) 2 2 the Taylor approximation of f (u) to second order. Note that, by the assumptions on R β Re(β −β2) = −(cid:15)−(cid:15)2 +R2 ≥ −2(cid:15)+R2 > 0. For |u| < δ for our choice of δ we have O(u3) = O(N−6/5). We get (cid:90) δ (cid:90) δ e−Nfβ(u)du = e−NP2(u)e−N(fβ(u)−P2(u))du −δ −δ (3.8) (cid:90) δ (cid:90) δ (cid:0) (cid:1) = e−NP2(u)du+ e−NP2(u) e−N(fβ(u)−P2(u)) −1 du. −δ −δ Since for any |x| < 1/2: |ex −1| < 2x, we obtain (cid:12) (cid:12) (cid:12)e−N(fβ(u)−P2(u)) −1(cid:12) ≤ C N|f (u)−P (u)| ≤ C N−1/5, (3.9) 1 β 2 2 8 where the constants depend only on β. Combining (3.8) and (3.9) we obtain (cid:12)(cid:12)(cid:90) δ (cid:90) δ (cid:12)(cid:12) (cid:90) δ 1 (cid:12) e−Nfβ(u)du− e−NP2(u)du(cid:12) ≤ C N−1/5 e−NReP2(u)du ≤ C N−1/5√ , (3.10) 2 3 (cid:12) (cid:12) N −δ −δ −δ Now we have the following inequality (cid:12)(cid:90) ∞ (cid:90) δ (cid:12) (cid:90) ∞ (cid:90) ∞ (cid:12)(cid:12) e−NP2(u)du− e−NP2(u)du(cid:12)(cid:12) ≤ 2 e−NReP2(u)du ≤ 2 e−NRe(β−β2)u2δdu (cid:12) (cid:12) −∞ −δ δ δ (3.11) 1 = e−C5N1/5. C N3/5 4 Since (cid:115) (cid:90) ∞ 2π e−NP2(u)du = , N(β −β2) −∞ we obtain combining (3.10) and (3.11) that (cid:12) (cid:115) (cid:12) (cid:12) 2π (cid:12) C (cid:12)I − (cid:12) ≤ 6 . (cid:12) 2 N(β −β2)(cid:12) N7/10 (cid:12) (cid:12) Using the estimate (3.7) on |I | we obtain 1 (cid:12) (cid:115) (cid:12) (cid:12)(cid:90) ∞ 2π (cid:12) C (cid:12) e−Nfβ(u)du− (cid:12) ≤ 7 . (cid:12) N(β −β2)(cid:12) N7/10 (cid:12) −∞ (cid:12) This concludes the proof of the theorem. 4 Proof of Theorem 1.2 Proof. Observe that u2 f (u) = (1+(cid:15)) −log(cosh((1+(cid:15))u)), f(cid:48) (u) = (1+(cid:15))(u−tanh((1+(cid:15))u))). 1+(cid:15) 2 1+(cid:15) By the assumption (1.5) we get (cid:16) π(cid:17) (cid:18) π (cid:18) π R (cid:18)1+ π (cid:19)(cid:19)(cid:19) f(cid:48) ± = (1+(cid:15)) ± −tanh ± log R = 0, and f(cid:48) (0) = 0. 1+(cid:15) R R R2π 1− π 1+(cid:15) R (cid:0) (cid:1) Thesearetheonlyrealzerosoff(cid:48) andf ±π < 0, inparticulartheminimumoff (u) 1+(cid:15) 1+(cid:15) R 1+(cid:15) over R is achieved at u = ±π/R. We claim that the same is true of u2 1 Ref (u) = (1+(cid:15)) − log(cosh2((1+(cid:15))u)−sin2(Ru)). β 2 2 Indeed, by the monotonicity of the logarithm, for any u ∈ R we get Ref (u) ≥ f (u). On β 1+(cid:15) the other hand, Ref (±π) = f (±π) < 0. This yields that the minimum of Ref (u) over β R 1+(cid:15) R β 9 R is achieved at ±π/R, as claimed. We note in passing that f(cid:48)(±π/R) = 0, i.e. the points β ±π/R, which minimize Ref (u) over u ∈ R, are in fact saddle points. β Now we estimate the integral as before. Let δ = N−2/5 as in the proof of Theorem 1.1. SimilarlytotheproofofTheorem1.1wedividetheintegralintopiecesandusethesymmetry to obtain (cid:34) (cid:35) (cid:90) ∞ (cid:90) π−δ (cid:90) π+δ (cid:90) ∞ R R e−Nfβ(u)du = 2 +2 +2 e−Nfβ(u)du = 2I +2I +2I . 1 2 3 −∞ 0 π−δ π+δ R R Also let (cid:34) (cid:35) (cid:90) ∞ (cid:90) π−δ (cid:90) π+δ (cid:90) ∞ R R e−Nf1+(cid:15)(u)du = 2 +2 +2 e−Nf1+(cid:15)(u)du = 2I(cid:98) +2I(cid:98) +2I(cid:98). 1 2 3 −∞ 0 π−δ π+δ R R Then, for any j = 1,2,3 we get |I | ≤ I(cid:98). j j We will show that I(cid:98)1,I(cid:98)3 ≤ Ce−CN1/5e−Nf1+(cid:15)(Rπ), (4.1) (cid:114) (cid:113) (cid:113) 2π |I2| |fβ(cid:48)(cid:48)(π/R)|(1+o(1)) = I(cid:98)2 f1(cid:48)(cid:48)+(cid:15)(π/R)(1+o(1)) = N e−Nf1+(cid:15)(Rπ), (4.2) and this, together with the fact that Ref (π/R) = f (π/R) < 0, will prove the theorem β 1+(cid:15) and also (1.4). We start from the estimate of I(cid:98). We write 3 (cid:34) (cid:35) (cid:90) ∞ (cid:90) 3 (cid:90) ∞ I(cid:98) = e−Nf1+(cid:15)(u)du = + e−Nf1+(cid:15)(u)du =: W +W . 3 1 2 π+δ π+δ 3 R R To estimate W , note that since cosh(x) ≤ ex for x real, we have that for u ≥ 3, 2 f (u) = (1+(cid:15))u2/2−logcosh((1+(cid:15))u) ≥ (1+(cid:15))u(u/2−1) ≥ (1+(cid:15))u/2. 1+(cid:15) Thus, W ≤ e−3N/2. (4.3) 2 To estimate W , note that for any 3 ≥ u > π/R+δ the function u (cid:55)→ f (u) is increasing 1 1+(cid:15) and therefore f (π/R+δ) ≥ f (π/R)+cδ2. Thus, 1+(cid:15) 1+(cid:15) W1 ≤ Ce−N(f1+(cid:15)(Rπ)+cδ2) = Ce−Nf1+(cid:15)(Rπ)e−CN1/5. (4.4) Combining (4.3) and (4.4) yields (4.1) for I(cid:98). On the other hand, since f (u) is decreasing 3 1+(cid:15) for any 0 ≤ u ≤ π/R−δ with the minimum at π/R−δ, in the same way we obtain I(cid:98)1 ≤ Ce−Nf1+(cid:15)(Rπ)e−CN1/5, (4.5) which proves (4.1) for I(cid:98). 1 10

The Curie-Weiss model with complex temperature: phase transitions

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