On the Almeida-Thouless transition line in the Sherrington-Kirkpatrick model with centered Gaussian external field

We study the phase transition of the free energy in the Sherrington-Kirkpatrick mean-field spin glass model with centered Gaussian external field. We show that the corresponding Almeida-Thouless line is the correct transition curve that distinguishes between the replica symmetric and replica symmetry breaking solutions in the Parisi formula.


Introduction and main results
The famous Sherrington-Kirkpatrick (SK) mean-field spin glass model was introduced in [12] aiming to explain some unusual magnetic behavior of certain alloys. By means of the replica method, it was also proposed in [12] that the thermodynamic limit of the free energy in the SK model can be solved by the replica symmetric ansatz at very high temperature. A complete picture was later settled in the seminal works [10,11] of Parisi, in which he adapted an ultrametric ansatz and deduced a variational formula for the limiting free energy at all temperature, known as the Parisi formula. This formula was rigorously established by Talagrand [13] utilizing the replica symmetry breaking bound discovered by Guerra [5]. See [8] for physicists' studies of the SK model as well as [9,14,15] for the recent mathematical progress.
For any N ≥ 1, the Hamiltonian of the SK model is defined as where g ij 's are i.i.d. standard Gaussian. The parameters β > 0 and h ∈ R are the (inverse) temperature and external field, respectively. Define the free energy as The famous Parisi formula [9,15] asserts that almost surely, Here, Pr([0, 1]) is the collection of all probability distribution functions on [0, 1] equipped with the L 1 (dx) distance and P is a functional on Pr([0, 1]) defined as where Φ α is the weak solution [6] to with boundary condition Φ(1, x) = log cosh x. It is known [2] that the Parisi formula has a unique minimizer denoted by α P . We say that the Parisi formula is solved by the replica symmetric solution if α P = 1 [q,1] for some q ∈ [0, 1] and is solved by the replica symmetry breaking solution if otherwise. For any β, h > 0, let q = q(β, h) be the unique solution (see [4] and [14, Proposition A.14.1]) to where z is standard Gaussian. In [1], it was conjectured that for β, h > 0, the SK model is solved by the replica symmetric solution if and only if (β, h) satisfies In other words, for β, h > 0, the following equation, known as the Almeida-Thouless line, characterizes the transition between the replica symmetric and replica symmetry breaking solutions. Toninelli [16] proved that above the AT line, i.e, β 2 Ecosh −4 (h + βz √ q) > 1, the solution to the Parisi formula is replica symmetry breaking. Later Talagrand [15] and Jagannath-Tobasco [7] showed that inside the AT line, β 2 Ecosh −4 (h + βz √ q) ≤ 1, there exist fairly large regimes in which the Parisi formula is solved by replica symmetric solution.
As parts of their regimes are not up to the AT line, verifying the exactness of the AT line remains open.
In this short note, we investigate the SK model with centered Gaussian external field and show that the corresponding AT line is indeed the transition line distinguishing between the replica symmetric and replica symmetry breaking solutions in the Parisi formula. For β, h > 0, the Hamiltonian of the SK model with centered Gaussian external field is defined as where (g ij ) 1≤i<j≤N and (ξ i ) 1≤i≤N are i.i.d. standard normal and are independent of each other. The free energy associated to this Hamiltonian is given by In a similar manner, the limiting free energy can be expressed by the Parisi formula as in (3) with a replacement of h by hξ for ξ a standard Gaussian random variable, more precisely, we have that almost surely, where and Φ α is defined as (3). Note that the proof of this formula is identically the same as those in [9,15] with no essential modifications. Moreover, as in [2], the Parisi formula here also has a unique minimizer denoted by α P . We again say that the Parisi formula is solved by the replica symmetric solution if α P = 1 [q,1] for some q ∈ [0, 1] and is solved by the replica symmetry breaking solution if otherwise. The AT line corresponds to our model is formulated analogously. Let z, ξ be i.i.d. standard Gaussian random variables. For β, h > 0, let q = q(β, h) be the unique fixed point of where the existence and uniqueness of q are guaranteed thanks to the Latala-Guerra lemma. 1 The AT line associated to the SK model with centered Gaussian external field is the collection 1 In [4] and [14, Proposition A.14.1], it is known that for any r ∈ R, is also strictly decreasing on (0, ∞). Since f (∞) = 0 and f (0) = ∞, there exists a unique x 0 such that f (x 0 ) = 1. This ensures that (6) has a unique solution.  (5).
The following is our main result.
The rest of the paper is organized as follows. In Section 2, we gather some results regarding the directional derivative of the functional P and some consequences following the first order optimality of the Parisi variational formula. The proof of Theorem 1 is presented in Section 3. Acknowledgements. The author thanks the anonymous referees for some helpful comments regarding the presentation of this paper. Special thanks are due to Si Tang for running the simulations in Figures 1 and 2.

Proposition 2.
If α 0 is the minimizer of P, then every point in the support of α 0 must satisfy φ α 0 (s) = s.
In the case of the SK model with deterministic external field (1), the two propositions above were established in Theorem 2 and Proposition 1 in [3], respectively, where the statements are essentially the same as Propositions 1 and 2 with a replacement of hξ by h. As the proofs in [3] can be directly applied to Propositions 1 and 2 with no major changes, we do not reproduce the details here. The following lemma provides a simpler expression for φ α . Proof. Conditionally on ξ, let E ξ be the expectation associated to the probability measure From the Girsanov theorem, under E ξ , is a standard Brownian motion for which we denote by (W ξ (s)) 0≤s≤1 . Set M ξ (s) = hξ + βW ξ (s). Under E ξ , write 1 0 βα(r)∂ x Φ(r, X ξ α (r))dW (r) + Consequently, Next, note that from Itô's formula and (3), for 0 ≤ t < s ≤ 1, and then from the Fubini theorem, Plugging this into (9) yields that where we used that (M (s)) 0≤s≤1 d = (M ξ (s)) 0≤s≤1 conditionally on ξ. Finally, taking expectation in ξ completes our proof.
While Lemma 1 holds for any α, the important case that we shall use in our main proof is when α = 1 [q,1] for some q ∈ [0, 1]. In this case, one can compute φ α more explicitly. Let z, z be i.i.d. standard Gaussian independent of ξ. Denote by E the expectation with respect to z only.
, if s ∈ [q, 1], and , if s ∈ [q, 1], and that Plugging these along with the assumption α = 1 [q,1] into Lemma 1 establishes (10). As for (11), note that for any twice differentiable function f with f ∞ < ∞, we can compute by using the Gaussian integration by parts 2 to obtain Applying this equation and cosh(x) = cosh(x) 2 Let Z = (z 1 , . . . , z n ) be a centered Gaussian random vector and let F be a differentiable function on R n with to the E -expectations in the first equation of (10), the first equation of (11) follows by a straightforward computation. To obtain the second equation in (11), write for z 1 , z 2 i.i.d. standard Gaussian independent of z. From the last equation and (tanh(x)) = cosh −2 (x), we have Using Gaussian integration by parts yields that

Proof of Theorem 1
Throughout the entire proof, we let q be the unique solution to (6). First, we assume that (β, h) lies above the AT line, i.e., We show that α P can not be replica symmetric. If not, then we must have that α P = 1 [q ,1] for some q ∈ [0, 1] and from Proposition 2 and Lemma 2, q must satisfy E tanh 2 (hξ + βz q ) = φ α P (q ) = q .

Remark 1.
In the case of the SK model with non-random external field (1), the corresponding φ α (s) for α = 1 [q,1] and q satisfying (4) is the same as in Lemma 2 except that hξ is replaced by h. The same argument in Theorem 1 enables us to show that the Parisi formula can not be solved by the replica symmetric solution if (β, h) lies outside the AT line. However, if (β, h) lies inside the AT line, while it can still be shown that φ α (s) ≥ s for all s ∈ [0, q), it is unclear why φ α (s) ≤ s for all s ∈ [q, 1]. In this case, one can still use the FKG inequality to obtain that for q ≤ s ≤ 1, φ α (s) ≤ β 2 E cosh −4 (h + βz √ s), but a numerical simulation, see Figure 2, suggests that the following does not always hold: for all q ≤ s ≤ 1.