A CLT for the total energy of the two-dimensional critical Ising model

TL;DR

This paper establishes a central limit theorem for the total energy of the two-dimensional critical Ising model with specific boundary conditions, showing it converges to a Gaussian distribution under certain growth conditions for the lattice dimensions.

Contribution

It proves a new CLT for the total energy of the 2D critical Ising model with mixed boundary conditions, extending understanding of its fluctuation behavior.

Findings

Total energy fluctuations follow a Gaussian distribution asymptotically.

The result holds for lattice dimensions satisfying a growth condition involving logarithms.

Provides a quantitative description of energy distribution at criticality.

Abstract

Consider the Ising model on $([1,2N]\times[1,2M])\cap\mathbb{Z}^2$ at critical temperature with periodic boundary condition in the horizontal direction and free boundary condition in the vertical direction. Let $E_{M,N}$ be its total energy (or Hamiltonian). Suppose $M$ is a function of $N$ satisfying $M\geq N/(\ln N)^{\alpha}$ for some $\alpha\in[0,1)$. In particular, one may take $M=N$. We prove that \begin{equation*} \frac{E_{M,N}+4\sqrt{2}M N-(4/\pi)N\ln N}{\sqrt{(32/\pi)MN\ln N}} \end{equation*} converges weakly to a standard Gaussian distribution as $N\rightarrow\infty$.

Equations

$$\frac{E_{M,N}+4\sqrt{2}MN-(4/\pi )N\ln N}{\sqrt{(32/\pi )MN\ln N}}$$

$${\mathbb{P}}_{\Lambda }(\sigma ):=\frac{1}{Z_{\Lambda }(\beta )}{e}^{\beta {\sum }_{\{i,j\}}{\sigma }_i{\sigma }_j},$$

$$M_{\Lambda }:=\sum _{i\in \Lambda }{\sigma }_i,$$

$$E_{\Lambda }:=-\sum _{\{i,j\}}{\sigma }_i{\sigma }_j.$$

$$\chi :=\sum _{j\in {\mathbb{Z}}^d}\text{Cov}({\sigma }_0,{\sigma }_j)<\infty .$$

$$E_{M,N}:=-\sum _{j=1}^{2M}\sum _{k=1}^{2N}{\sigma }_{j,k}{\sigma }_{j,k+1}-\sum _{j=1}^{2M-1}\sum _{k=1}^{2N}{\sigma }_{j,k}{\sigma }_{j+1,k},$$

$${\mathbb{P}}_{M,N}^{\beta }(\sigma ):=\frac{1}{Z_{M,N}(\beta )}{e}^{-\beta E_{M,N}},$$

$$Z_{M,N}(\beta ):=\sum _{\sigma \in \{-1,+1{\}}^{{\Lambda }_{M,N}}}{e}^{-\beta E_{M,N}}$$

$$\underset{N\to \infty }{\lim }\frac{N(\ln \ln N{)}^2}{M\ln N}=0.$$

$${\hat{E}}_{M,N}:=\frac{E_{M,N}+4\sqrt{2}MN-(4/\pi )N\ln N}{\sqrt{4MN\ln N}}.$$

$$\underset{N\to \infty }{\lim }⟨e^{t{\hat{E}}_{M,N}}{⟩}_{M,N}^{{\beta }_c}=e^{4t^2/\pi },$$

$$a(-\ln a{)}^{-1/2}\sum _{\{x,y\}}[{\sigma }_x{\sigma }_y-\sqrt{2}/2]{\delta }_{(x+y)/2}⟹\text{Gaussian white noise as }a\downarrow 0,$$

$$a^{-1}[⟨{\sigma }_{x_a(z)}{\sigma }_{y_a(z)}{⟩}_{{\Lambda }_a}^{{\beta }_c}-\sqrt{2}/2]$$

$$⟨{ϵ}_{z_1}{ϵ}_{z_2}⟩\approx \frac{C}{|z_1-z_2{|}^2},$$

$$⟨{ϵ}_{z_1}{ϵ}_{z_2}\dots {ϵ}_{z_n}⟩=\left|\frac{1}{n!2^n}\sum _{\tau \in S^{2n}}\prod _{j=1}^n{A}_{\tau (2j-1)\tau (2j)}\right|,$$

$$Z_{M,N}(\beta )=(2\sinh (2\beta ){)}^{2MN}(\cosh (\beta ){)}^{-2N}\prod _{\theta }\left[\frac{e^{2M{\gamma }_{\theta }}+e^{-2M{\gamma }_{\theta }}}{2}+\frac{e^{2M{\gamma }_{\theta }}-e^{-2M{\gamma }_{\theta }}}{2}{g}_{\theta }\right],$$

$$\cosh ({\gamma }_{\theta })=\coth (2\beta )\cosh (2\beta )-\cos (\theta )\text{ with }{\gamma }_{\theta }\ge 0,$$

$$g_{\theta }=\frac{\coth (2\beta )-\cosh (2\beta )\cos (\theta )}{\sinh ({\gamma }_{\theta })}.$$

$$\sinh (2{\beta }_c)=1,\cosh (2{\beta }_c)=\sqrt{2},\cosh ({\gamma }_{\theta }){|}_{\beta ={\beta }_c}=2-\cos \theta ,$$

$$\sinh ({\gamma }_{\theta }){|}_{\beta ={\beta }_c}=\sqrt{3-4\cos \theta +{\cos }^2\theta }.$$

$$⟨e^{s{E}_{M,N}}{⟩}_{M,N}^{\beta }=\frac{Z_{M,N}(\beta -s)}{Z_{M,N}(\beta )}.$$

$$⟨e^{s{E}_{M,N}}{⟩}_{M,N}^{\beta }=\frac{{\sum }_{\sigma }{e}^{s{E}_{M,N}}{e}^{-\beta E_{M,N}}}{Z_{M,N}(\beta )}=\frac{Z_{M,N}(\beta -s)}{Z_{M,N}(\beta )}.$$

$$\ln Z_{M,N}(\beta )=2MN\ln (2\sinh (2\beta ))-2N\ln (\cosh (\beta ))+2M\sum _{\theta }{\gamma }_{\theta }+\sum _{\theta }{f}_{\theta },$$

$$f_{\theta }:=\ln [1+e^{-4M{\gamma }_{\theta }}+(1-e^{-4M{\gamma }_{\theta }})g_{\theta }]-\ln 2.$$

$$L_1(\beta ):=2MN\ln (2\sinh (2\beta )),L_2(\beta ):=2N\ln (\cosh (\beta )),$$

$$L_3(\beta ):=2M\sum _{\theta }{\gamma }_{\theta },L_4(\beta ):=\sum _{\theta }{f}_{\theta }.$$

$$\underset{N\to \infty }{\lim }\left[L_1\left({\beta }_c-t/\sqrt{4MN\ln N}\right)-L_1({\beta }_c)+\frac{t}{\sqrt{4MN\ln N}}4\sqrt{2}MN\right]=0,$$

$$\underset{N\to \infty }{\lim }\left[L_2\left({\beta }_c-t/\sqrt{4MN\ln N}\right)-L_2({\beta }_c)\right]=0,$$

$$\underset{N\to \infty }{\lim }\left[L_3\left({\beta }_c-t/\sqrt{4MN\ln N}\right)-L_3({\beta }_c)\right]=\frac{4t^2}{\pi },$$

$$\underset{N\to \infty }{\lim }\left[L_4\left({\beta }_c-t/\sqrt{4MN\ln N}\right)-L_4({\beta }_c)-\frac{t}{\sqrt{4MN\ln N}}\frac{4}{\pi }N\ln N\right]=0.$$

$$\ln {⟨e^{t{\hat{E}}_{M,N}}⟩}_{M,N}^{{\beta }_c}=\ln {⟨e^{t{E}_{M,N}/\sqrt{4MN\ln N}}⟩}_{M,N}^{{\beta }_c}+\frac{t}{\sqrt{4MN\ln N}}\left[4\sqrt{2}MN-\frac{4}{\pi }N\ln N\right]$$

$$=\ln Z_{M,N}\left({\beta }_c-t/\sqrt{4MN\ln N}\right)-\ln Z_{M,N}({\beta }_c)+\frac{t}{\sqrt{4MN\ln N}}\left[4\sqrt{2}MN-\frac{4}{\pi }N\ln N\right].$$

$$\ln {⟨e^{t{\hat{E}}_{M,N}}⟩}_{M,N}^{{\beta }_c}=\left[L_1\left({\beta }_c-t/\sqrt{4MN\ln N}\right)-L_1({\beta }_c)+\frac{t}{\sqrt{4MN\ln N}}4\sqrt{2}MN\right]$$

$$-\left[L_2\left({\beta }_c-t/\sqrt{4MN\ln N}\right)-L_2({\beta }_c)\right]+\left[L_3\left({\beta }_c-t/\sqrt{4MN\ln N}\right)-L_3({\beta }_c)\right]$$

Full text

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converges weakly to a standard Gaussian distribution as $N\rightarrow\infty$.

1. Introduction

Recall that the classical Ising model at inverse temperature $\beta$ on $\Lambda\subset\mathbb{Z}^{d}$ with free boundary condition is defined by the probability measure $\mathbb{P}_{\Lambda}$ on $\{-1,+1\}^{\Lambda}$ such that for each $\sigma\in\{-1,+1\}^{\Lambda}$,

[TABLE]

where the sum is over all nearest neighbor pairs in $\Lambda$, and $Z_{\Lambda}(\beta)$ is the partition function (which is the normalization constant needed to make this a probability measure). The total magnetization and total energy (or Hamiltonian) are

[TABLE]

It was proved in [12] that under the full-plane Ising measure $\mathbb{P}$ (with the corresponding expectation $\mathbb{E}$), $((M_{\Lambda}-\mathbb{E}M_{\Lambda})/\text{Var}{M_{\Lambda}},(E_{\Lambda}-\mathbb{E}E_{\Lambda})/\text{Var}{E_{\Lambda}})$ converges weakly to a standard bivariate Gaussian distribution (i.e., the two components are independent and each is a mean [math] variance $1$ Gaussian random variable) if the susceptibility

[TABLE]

In particular, this implies that such a convergence holds when $d=2$ and $\beta\neq\beta_{c}$ where $\beta_{c}$ is the critical inverse temperature. A similar Gaussian limit was obtained for the total magnetization and total energy on one side of a rectangle when $d=2$ and $\beta=\beta_{c}$ in [5] (see also [1] for the total magnetization only), and for $(M_{\Lambda},E_{\Lambda})$ when $d>4$ and any $\beta\in[0,\beta_{c}]$ in [6]. When $d=2$ and $\beta=\beta_{c}$, it was proved in [3] that $(M_{\Lambda}-\mathbb{E}M_{\Lambda})/\text{Var}{M_{\Lambda}}$ converges weakly to a non-Gaussian limit; and for the Ising model on a $(2N)\times(2M)$ rectangle with periodic boundary condition in the horizontal direction and free boundary condition in the vertical direction (with the total energy denoted by $E_{M,N}$), it was proved in [4] that $(E_{M,N}-\mathbb{E}E_{M,N})/\sqrt{2M2N\ln N}$ converges to a Gaussian distribution by first taking $M\rightarrow\infty$ and then $N\rightarrow\infty$. One disadvantage of this iterated limit is that one does not see the effect from the boundary condition. In this paper, for the same Ising model as considered in [4], we prove a central limit theorem (CLT) for $E_{M,N}$ when both $M$ and $N\rightarrow\infty$ simultaneously. More precisely, we consider the Ising model on $\Lambda_{M,N}:=([1,2N]\times[1,2M])\cap\mathbb{Z}^{2}$ with the total energy

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where $k=2N+1$ is identified with $k=1$. For each $\sigma\in\{-1,+1\}^{\Lambda_{M,N}}$, we have

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where

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is the partition function.

Our main result is

Theorem 1**.**

Consider the Ising model on $\Lambda_{M,N}$ at critical temperature with periodic boundary condition in the horizontal direction and free boundary condition in the vertical direction (i.e., with the Hamiltonian given by (5)). Suppose that $M\in(0,\infty)$ is a function of $N$ satisfying

[TABLE]

Let $\hat{E}_{M,N}$ be the normalized random variable

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Then for each $t\geq 0$,

[TABLE]

where $\langle\cdot\rangle_{M,N}^{\beta_{c}}$ denotes the expectation with respect to $\mathbb{P}_{M,N}^{\beta_{c}}$. In particular, this implies that $\hat{E}_{M,N}$ converges weakly to a Gaussian distribution with mean [math] and variance $8/\pi$ as $N\rightarrow\infty$.

Remark 1**.**

We believe a similar CLT holds for the critical Ising model with other boundary conditions (e.g., free, all $+$, all $-$). Furthermore, for the critical Ising model on the rescaled lattice $a\mathbb{Z}^{2}$, we expect that the renormalized energy field

[TABLE]

where the sum is over all nearest neighbor pairs in $a\mathbb{Z}^{2}$ and $\delta_{(x+y)/2}$ is a unit Dirac point measure at $(x+y)/2$.

Remark 2**.**

Let $\Lambda_{a}:=a\mathbb{Z}^{2}\cap\Lambda$ be the $a$-approximation of $\Lambda$. For any $z\in V$, let $x_{a}(z)y_{a}(z)$ be the edge which is closest to $z$. It was proved in [10] that under free or all $+$ boundary condition,

[TABLE]

has a conformally covariant limit as $a\downarrow 0$. See also [9] for a generalization of this result to $n$-point energy correlation functions. Even though the results of [10, 9] do not apply directly to the boundary condition considered in Theorem 1, they suggest the $N\ln N$ behavior (resulted from the free boundary condition) in the expectation of $E_{M,N}$ since the limit of (12) has an order of $[\text{dist}(z,\partial\Lambda)]^{-1}$ where $\text{dist}(z,\partial\Lambda)$ denotes the Euclidean distance between $z$ and the boundary of $D$.

Remark 3**.**

For the full-plane critical Ising model, Hecht [8] showed that the truncated two-point energy correlation function has the following behavior

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where $\epsilon_{z_{i}}:=\langle\sigma_{x(z_{i})}\sigma_{y(z_{i})}-\sqrt{2}/2\rangle$ with $\{x(z_{i}),y(z_{i})\}$ the closest edge to $z_{i}$. In [7], it was shown that (see (2.3) and (2.13) there)

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where $S^{2n}$ denotes the set of $(2n)!$ permutations of $\{1,2,\dots,2n\}$ and $A_{ij}:=\langle\epsilon_{z_{i}}\epsilon_{z_{j}}\rangle$. Equation (14) without the modulus is Isserlis’s formula (or Wick’s formula) for the multivariate Gaussian distribution. This is one of the motivations of the current paper: the critical scaling limit of the magnetization field was established in [3] and it is natural to ask if an analogous result holds for the energy field. Theorem 1 suggests that a scaling limit of the energy field (with correlations behaving like (13) and (14)) may not exist in the usual probabilistic sense (i.e., pairing the limiting field against some nice test functions to get random variables).

We prove Theorem 1 in the next section, our method is similar to that of [1, 4, 5]. Namely, we first write the moment generating function of $E_{M,N}$ as a ratio of two partition functions (at different temperatures), and then use the explicit formula for the partition function to derive the asymptotic behavior of this moment generating function.

2. Proof of the main theorem

The following lemma about the partition function from [11] is essential to the proof of Theorem 1.

Lemma 1**.**

The partition function defined in (7) is

[TABLE]

where the product is over $\theta=\pi(2n-1)/(2N)$ with $n=1,2,\dots,N$, and

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[TABLE]

Proof.

See Sections 2 and 3 of Chapter VI in [11]. ∎

Remark 4**.**

The partition function (15) differs from (7) in [4] by a factor of $2^{2N}$. By checking the particular case $\beta=0$, one can see that (15) is the correct one. But such a difference does not affect the computation of $\langle e^{tE_{M,N}}\rangle_{\Lambda_{M,N}}$ since the latter is the ratio of two partition functions (see Lemma 3).

It is well-known that the critical inverse temperature for the two-dimensional Ising model is $\beta_{c}=\ln(1+\sqrt{2})/2$. We will use the following computations many times in the paper.

Lemma 2**.**

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Proof.

The lemma follows from trivial computations. ∎

Lemma 3**.**

For any $s\in\mathbb{R}$ and $\beta\geq 0$,

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Proof.

[TABLE]

∎

By Lemma 1, we have

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where

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We define $L_{i}$ for $i=1,2,3,4$ by

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In the rest of this paper, we always assume $M$ is a function of $N$ satisfying (8). Theorem 1 will follow from the following estimates about $L_{i}$’s.

Proposition 1**.**

Suppose $M$ is a function of $N$ satisfying (8). Then for each $t\geq 0$, we have

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Let us prove Theorem 1 under the assumption of Proposition 1.

Proof of Theorem 1 (modulo proving Proposition 1).

For $\hat{E}_{M,N}$ defined in (9), we have by Lemma 3 that

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Using (22)-(25), we can write

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This completes the proof of the first part of Theorem 1 (i.e., (10)) by applying Proposition 1. The second part of Theorem 1 follows from a standard probability argument (see, e.g., Problem 30.4 of [2]). ∎

The first two limits in Proposition 1 are easy to prove.

Proof of (26) and (27) in Proposition 1.

Note that $L_{1}^{\prime}(\beta)=4MN\coth(2\beta)$ and $L_{1}^{\prime\prime}(\beta)=-8MN\operatorname{csch}^{2}(2\beta)$. So by the Taylor expansion of $L_{1}$ around $\beta_{c}$ and Lemma 2, we have

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where $\tilde{\beta}\in\left(\beta_{c}-t/\sqrt{4MN\ln N},\beta_{c}\right)$. By Lemma 2, $|L_{1}^{\prime\prime}(\tilde{\beta})|=8MN\operatorname{csch}^{2}(2\tilde{\beta})\leq 16MN$ for any $\tilde{\beta}\in\left(\beta_{c}-t/\sqrt{4MN\ln N},\beta_{c}\right)$ if $N$ is large. This completes the proof of (26). Similarly, the Taylor expansion of $L_{2}$ around $\beta_{c}$ gives

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where $\tilde{\beta}\in\left(\beta_{c}-t/\sqrt{4MN\ln N},\beta_{c}\right)$. It is clear that $|\tanh(\tilde{\beta})|\leq 1$ for any such $\tilde{\beta}$ whenever $N$ is large. Combining this and our assumption on $M$ (i.e., (8)) completes the proof of (27). ∎

The following three lemmas will be very useful when we deal with the Taylor expansions of $L_{3}(\beta)$ and $L_{4}(\beta)$.

Lemma 4**.**

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For each large $N$, each $\beta\in\left(\beta_{c}-1/\sqrt{4MN\ln N},\beta_{c}+1/\sqrt{4MN\ln N}\right)$ and each $\theta\in(0,\pi]$, we have

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Proof.

The proof of (32) is trivial. The inequality (33) follows from the monotonicity of $\operatorname{csch}$, Lemma 2 and the mean value theorem. The inequality (34) follows from

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and

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The inequality (35) follows from

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and (36). ∎

Lemma 5**.**

There exist constants $C_{1},C_{2}\in(0,\infty)$ such that for all large $N\in\mathbb{N}$,

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Proof.

Let $H_{n}:=\sum_{k=1}^{n}1/k$ be the $n$-th harmonic number. It is well-known that

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where $\hat{\gamma}$ is the Euler-Mascheroni constant. The lemma follows by the following observation:

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∎

Lemma 6**.**

There exist constants $C_{3},C_{4}\in(0,\infty)$ such that for all large $N$ and all $\beta>0$,

[TABLE]

Proof.

Recall that

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By (32) in Lemma 4,

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where the last inequality follows since

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Therefore, by Lemma 5 and $\sum_{n=1}^{\infty}1/(2n-1)^{2}<\infty$, for all large $N$,

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∎

From (16), we can compute (all derivatives are respect to $\beta$)

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[TABLE]

[TABLE]

By Lemma 2, we have

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We are ready to prove (28) in Proposition 1.

Proof of (28) in Proposition 1.

The Taylor expansion of $L_{3}$ (see (25)) around $\beta=\beta_{c}$ implies that there exists $\tilde{\beta}\in\left(\beta_{c}-t/\sqrt{4MN\ln N},\beta_{c}\right)$ such that

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where we have used (45) in the last equality.

Note that $1/\sqrt{3-4\cos\theta-\cos^{2}\theta}-1/\theta$ is a continuous function of $\theta$ on $[0,\pi]$ if we define its value at $\theta=0$ being [math] since

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Therefore, there exists a constant $C_{5}\in(0,\infty)$ such that for all $N\in\mathbb{N}$,

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Now we have

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where we have used (47) and $\theta=(2n-1)\pi/(2N)$ in the second equality and Lemma 5 in the last equality.

Next, we prove that the remainder in (2) vanishes as $N\rightarrow\infty$. By (2), Lemmas 2 and 4, there exist constants $C_{6},C_{7},C_{8}\in(0,\infty)$ such that for all large $N$, each $\tilde{\beta}\in\left(\beta_{c}-t/\sqrt{4MN\ln N},\beta_{c}\right)$ and each $\theta\in(0,\pi]$,

[TABLE]

Therefore, by Lemma 6, we have for all large $N$,

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This and (8) imply

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Combining (2), (2) and (49), we finish the proof of (28). ∎

The last and more difficult function we need to deal with is $L_{4}(\beta)=\sum_{\theta}f_{\theta}$. We first compute the derivatives of $f_{\theta}$ (with respect to $\beta$). By (23), (16), (17) and (42), we have

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where

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By Lemma 2, (17), (45), (50) and (52), we have

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where

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We need the following lemma to analyze the Taylor expansion of $L^{4}(\beta)$ around $\beta_{c}$. Let us emphasize again that $M$ is a function of $N$ satisfying (8).

Lemma 7**.**

There exist constants $C_{9},C_{10}\in(0,\infty)$ such that for each $\beta>0$ and each large $N$,

[TABLE]

Proof.

By (16), and (32) in Lemma 4, we have

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where the last inequality follows since $(\sqrt{3-\cos\theta}+\sqrt{1-\cos\theta})/\sqrt{2}\geq 1$ for any $\theta\in[0,\pi]$. It is easy to see that (see, e.g., the proof of Lemma 5)

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For the other sum in the RHS of (2), we have (using (41) in the second inequality)

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where the the last inequality follows from Lemma 5 and

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From (8), we have

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Combining (2)-(61), we get (56). The inequality (57) follows from (40) and (56). ∎

The Taylor expansion of $L_{4}$ (see (25)) around $\beta=\beta_{c}$ gives

[TABLE]

where $\tilde{\beta}\in\left(\beta_{c}-t/\sqrt{4MN\ln N},\beta_{c}\right)$. The following lemma is about the asymptotic behavior of the first term on the RHS of (62).

Lemma 8**.**

[TABLE]

Proof.

From (54), we have (recall that $\eta_{\theta}\geq 1$ from (55))

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Since $\lim_{\theta\downarrow 0}\left[(1-\cos\theta)^{-1/2}-\sqrt{2}\theta^{-1}\right]=0$, there exists a constant $C_{11}\in(0,\infty)$ such that for each $N\in\mathbb{N}$,

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The last sum in the RHS of (2) contains

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where we have used (55), and (56) from Lemma 7 in the last equality. The remaining sum that we have not analyzed is

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By applying (56) from Lemma 7 , we get (noting that $\sqrt{2(1-\cos x)/(3-\cos x)}\in[0,1]$)

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The second sum on the RHS of (2) is

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Since the limit of the function in the brackets as $\theta\downarrow 0$ is $-2\sqrt{2}$, there exists a constant $C_{12}\in(0,\infty)$ such that for all $N\in\mathbb{N}$,

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Combining (2)-(70), we get

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By using (8) and noting that $\theta=(2n-1)\pi/(2N)$, we have

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This completes the proof of the lemma by applying Lemma 5. ∎

Our last lemma is about the asymptotic behavior of the second term on the RHS of (62).

Lemma 9**.**

For any $\tilde{\beta}\in\left(\beta_{c}-t/\sqrt{4MN\ln N},\beta_{c}\right)$, we have

[TABLE]

Proof.

Since $g_{\theta}\geq 0$ for each $\beta\in(0,\beta_{c}]$, we have for each $\beta\in(0,\beta_{c}]$ and $\theta\in(0,\pi]$,

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Applying this to (51), we obtain for each $\beta\in(0,\beta_{c}]$ and $\theta\in(0,\pi]$,

[TABLE]

where $g_{\theta}^{\prime}$ and $g_{\theta}^{\prime\prime}$ are defined in (52) and (53). By (17), (42), and Lemmas 2 and 4, one has for all large $N$, all $\beta\in\left(\beta_{c}-1/\sqrt{4MN\ln N},\beta_{c}+1/\sqrt{4MN\ln N}\right)$ and all $\theta\in(0,\pi]$,

[TABLE]

By (42), (2), (52), (53), and Lemmas 2 and 4, there exist constants $C_{13},\dots,C_{18}\in(0,\infty)$ such that for all large $N$, all $\beta\in\left(\beta_{c}-1/\sqrt{4MN\ln N},\beta_{c}+1/\sqrt{4MN\ln N}\right)$ and all $\theta\in(0,\pi]$,

[TABLE]

Combining (2)-(77), we get that there exist constants $C_{19},C_{20},C_{21}\in(0,\infty)$ such that

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for all large $N$, all $\beta\in\left(\beta_{c}-1/\sqrt{4MN\ln N},\beta_{c}\right)$ and all $\theta\in(0,\pi]$. This, (8), and Lemmas 6 and 7 complete the proof of the lemma. ∎

Remark 5**.**

The only place where we actually use $t\geq 0$ in the proof of Proposition 1 (and thus Theorem 1) is (72). It seems possible to generalize this proof to $t\in\mathbb{R}$ by a more careful analysis of (51) or using $f^{\prime\prime\prime}_{\theta}$.

We have all the ingredients to prove (29) in Proposition 1.

Proof of (29) in Proposition 1.

This follows from (62), and Lemmas 8 and 9. ∎

Acknowledgements

This research was partially supported by STCSM grant 17YF1413300. The author thanks Chuck Newman for many useful discussions related to this work.