INTRODUCTION AND MAIN RESULTS

The Sherrington-Kirkpatrick (SK) model is an important mean-field spin glass that was introduced to explain some unusual magnetic behaviors of certain alloys. For a given disorder (random variable) h with finite second moment, a given (inverse) temperature 𝛽 > 0, and any integer N ≥ 1, its Hamiltonian is defined as

N , Random Struct Alg. 2024;64:267-286. wileyonlinelibrary.com/journal/rsa © 2023 Wiley Periodicals LLC 267

where (h ij ) 1≤i,j≤N are i.i.d. copies of h. One of the main objectives in the study of the SK model is to understand the limit of the free energy, (𝜎) , which has attracted a lot of attention in physics as well as in mathematics communities, see, for instance, [17,18,[21][22][23]. This paper is concerned with the order of fluctuation for the free energy. When h is standard Gaussian, it can be checked directly from the Gaussian-Poincaré inequality that Var(F N (𝛽)) = O(N). It is natural to ask whether one can improve this bound as Var(F N (𝛽)) = o(N), a phenomenon called superconcentration, introduced in the pioneering work of Chatterjee [10,11]. In light of this notion, superconcentration was established with the bound that for any 𝛽 > 0, there exists a constant C = C(𝛽) > 0 such that

When 𝛽 ≤ 1∕ √ 2, much sharper bounds were also obtained in the literature. In the case of 𝛽 < 1∕ √ 2, [2] showed that F N (𝛽) satisfies a central limiting theorem, and their result implies that Var(F N (𝛽)) = Θ(1) 1 . At 𝛽 = 1∕ √ 2, it was predicted by [4,19] that a sharp phase transition should occur, namely, Var(F N (𝛽)) = Θ(log N). Along this conjecture, a partial result Var(F N (𝛽)) = O((log N) 2 ) was known by the authors, see [14]. Interestingly, if one now considers the SK model in the presence of an external field, that is, replacing -H N (𝜎) with -H N (𝜎) + r ∑ N i=1 𝜎 i for some r > 0 in the free energy, then it was known in [13] that the corresponding free energy obeys a central limit theorem and Var(F N (𝛽)) = Θ(N), agreeing with the rate obtained from the Gaussian-Poincaré inequality instead of exhibiting superconcentration.

While the aforementioned results addressed superconcentration assuming that the disorder is Gaussian, we aim to investigate this phenomenon for more general choice of disorders. Note that for an arbitrary h with finite second moment, the Efron-Stein inequality readily implies Var(F N (𝛽)) = O(N). In contrast to this bound, we say that the free energy is superconcentrated if Var(F N (𝛽)) = o(N).

To state our main results, we assume that Eh = 0 and Eh 2 = 1. We express h = f (g) for some nondecreasing function f and a standard Gaussian random variable g. Let g 1 , g 2 be independent copies of g. For 0 ≤ t ≤ 1, define

where

Note that w(t) → 1 as t ↑ 1 if we further assume E|h| 3 < ∞. Indeed, since f is nondecreasing, f (g 1 t )f (g 2 t ) converges to f (g) 2 almost surely as t ↑ 1 and for any M > 0, denoting

On the other hand, from the Hölder inequality and using the union bound,

These together yield the desired limit. With this, our first main result shows that superconcentration for the free energy holds for any h with a finite third moment, where the upper bound for the variance is related to the rate of convergence of w(t) at 1.

Theorem 1.

There exist positive constants c, K > 0 depending only on 𝛽 such that whenever h satisfies Eh = 0, Eh 2 = 1, and E|h| 3 < ∞, we have

In the next main result, we let f be an arbitrary absolutely continuous function and take h = f (g). Note that now f is not necessarily nondecreasing. If we make the assumption that Eh = 0, Eh 2 = 1, and E|f ′ (g)| 3 < ∞, then we can obtain superconcentration for the free energy in the same rate as (1). Theorem 2. There exists a constant K > 0 depending only on 𝛽 such that whenever h satisfies that Eh = 0, Eh 2 = 1, and f is absolutely continuous with E|f ′ (g)| 3 < ∞, we have

A few remarks are in order.

Remark 1. We do not expect to obtain the bound N∕ log N in Theorem 2 directly from Theorem 1. Nevertheless, if in Theorem 1 we assume additionally that f is differentiable and

, which follows from the mean value theorem and the fact that w ′ is nondecreasing, see (8) below. As a result,

and this implies that Var(F N (𝛽)) = O(N log log N∕ log N).

Remark 2. Under the assumption that the first four moments of h agree with those of g and h has a finite fifth moment, one can manage to match the first and second moments of the free energies associated to h and g asymptotically by using the approximate Gaussian integration by parts, which will lead to Var(F N ) = O(N∕ log N) as ( 1) and ( 3), see, for example, [9]. Our main results address superconcentration by reducing the moment assumption as much as possible.

Remark 3. In a more general framework, one can consider the mixed even p-spin model, whose Hamiltonian is defined as

where h i 1 , … ,i 2p for all 1 ≤ i 1 , … , i 2p ≤ N and all p ≥ 1 are i.i.d. copies of h and (𝛽 p ) p≥1 is a real sequence with ∑ ∞ p=1 2 p 𝛽 2 p < ∞. In [10], Chatterjee showed that the corresponding free energy is again superconcentrated as long as h is standard normal. We point out that the same results in Theorems 1 and 2 also hold in this setting by the same proofs.

We now discuss three applications of Theorems 1 and 2. ) .

To show this, we use Theorem 1 by expressing h = f (g) for

where Φ is the CDF of g. Note that f is nondecreasing. Denote 𝛾 = Φ -1 (p). A direct computation shows that

) .

To compute these probabilities, we write

On the other hand,

From these and using a 2 p + b 2 (1p) = 1, we arrive at

Thus, we can find a constant 𝑑 > 0 such that if t is sufficiently close to 1,

Theorem 1 then implies that for some constants C, c > 0 and for N large,

and our claim follows by noting that

Proof Sketch. Our proof is based on Chatterjee's interpolation argument (see [10,11]) in proving superconcentration for the free energy in the SK model with Gaussian disorder h = g. The argument starts by noting that F N (𝛽) is a function of i.i.d. standard Gaussian g = (g ij ) 1≤i,j≤N and writing Var(F N (𝛽)) = 𝜙(1) -𝜙(0), where for independent copies g 1 and g 2 of g,

and F = F N (𝛽). The first key step uses Gaussian integration by parts inductively to show that 𝜓(a) = 𝜙 ′ (e -a ) for a ≥ 0 is a completely monotone function, and from the Bernstein theorem, this function can be represented as 𝜓(a) = ∫ [0,∞) e -as 𝜇(𝑑s) for some positive measure 𝜇. Consequently, from Hölder's inequality, for all 0 < s ≤ t < 1,

In the second step, one relates 𝜙 ′ (t) to the second moment of the cross overlap associated to the interpolated spin system. In particular, by employing the so-called Latala argument (see [11], Lemma 10.4), it can be shown that 𝜙 ′ (s) = O(1) as long as s is small enough, where the fact that g is symmetric was heavily used. On the other hand, the above inequality makes it possible to get 𝜙 ′ (t) = O(N 1-log t∕ log s ) whenever t ≥ s. With these, one readily obtains the desired bound O(N∕ log N) utilizing the relation Var(F N (𝛽)) = ∫ 1 0 𝜙 ′ (t) 𝑑t. In our argument, we adapt the following interpolation

Now the terms in 𝜙 ′ (t) involve f ′ (see (12) below). While (5) remains valid, the main difficulty arises in obtaining an useful bound for 𝜙 ′ (s) with small s. To this end, for technical purposes, we adapt the convexity argument in [14] by considering the coupled free energy (15) instead of using the Latala approach. Our control in some sense relies on an approximate Gaussian integration by parts argument throughout.

Universality of superconcentration in other models. Superconcentration does not only exhibit in mean-field spin glass models, but also in random growth models on the integer lattice such as first-passage percolation [15], directed polymers [3], frog model [8].

For first-passage percolation, after a series of work [6,7,15], it is shown that under a 2 + log moment assumption, the model exhibits superconcentration, and it does not depend on the distribution of the disorder. Similar results hold for many related models. In [12], Chatterjee shows that superconcentration holds in a certain type of "surface growth models", which includes directed last-passage percolation and directed polymers, under the assumption that the disorder is a Lipschitz function of a Gaussian random variable.

The approach to superconcentration for growth models relies on the idea in [7], which consists of two components: one is the L 1 -L 2 bound by Talagrand (or its variants), and the other is the translation invariance of the model. Even though the superconcentration results look very similar in both mean-field spin glasses and random growth models (the upper bounds for the variances are also of order N∕ log N), this approach does not seem to work in mean-field spin glass models in any obvious way, due to the fact that spin glass models and growth models are very different in nature.

2

PROOF OF THEOREM 1

Some auxiliary lemmas

In this subsection, we shall gather three elementary lemmas that will be used in our main controls later.

t and g 2 t be defined as in (4). The first lemma controls the derivative of the expectation associated to this interpolation. Although the proof has appeared in [10], we still provide a proof for completeness. We say that

smooth and all of its partial derivatives are of moderate growth. Define

Then for any 0 ≤ t ≤ 1,

and for any 0

Proof. By symmetry,

Using Gaussian integration by part yields

and this gives (6). As for the second assertion, note that each term in 𝜙 ′ (t) is of the same form as that of 𝜙(t). Hence, we can apply induction to show that

Here, if we set 𝜓(a) = 𝜙 ′ (e -a ) for 0 ≤ a < ∞, then 𝜓 is completely monotone, that is, (-1) n 𝜓 (n) (a) ≥ 0 for all 0 < a < ∞ and n ≥ 0. From this and the Bernstein theorem (see for instance [16], Sec. XIII.4), one can express 𝜓(a) = ∫ e -ax 𝜇(𝑑x) for all a > 0 by some finite positive measure 𝜇 on [0, ∞). From the Hölder inequality, for any 0 < a < b,

and this is equivalent to (7). ▪ Lemma 2. Assume that Y, X 1 , X 2 are random variables with finite second moment and

From this and EY = 0, putting x = X, multiplying by Y and taking expectation on both sides complete our proof. ▪ Lemma 3. Let 𝜓 be a differentiable function on R and be of moderate growth. Then

Proof.

is continuous, Γ is closed and we can write Γ c as a disjoint union of open intervals ⋃ l∈I (a l , b l ), where I ⊆ N is some index set. Here, on each (a l , b l ), 𝜓 takes a fixed sign, and on {a 1 , b 1 , a 2 , b 2 , … }, 𝜓 = 0. From this, we can rewrite

for some sequence {w l } l∈I with w l = 1 or -1. Now, we compute directly

From ( 9), the assertion follows. ▪

Main controls

Throughout this entire subsection, we assume that f satisfies the following:

(A) ∶ f is nondecreasing and smooth with Ef (g) = 0, E f (g) 2 = 1, and E|f (g)| 3 < ∞, and its derivatives of all orders are of moderate growth.

Let h = f (g). Let g = (g ij ) 1≤i,j≤N , g 1 = (g 1 ij ) 1≤i,j≤N , and g 2 = (g 2 ij ) 1≤i,j≤N be i.i.d. standard Gaussian. For any 0 ≤ t ≤ 1, set

Define

where for 𝓁 = 1, 2,

and g 𝓁 t,ij is the (i, j)-th entry of g 𝓁 t . From (6),

Here, ⟨⋅⟩ t is the Gibbs expectation associated to the following Gibbs measure

Recall that w(t) = Ef (g 1 t )f (g 2 t ). From Lemma 1, we also have w

Under the assumption (A), we have for 0 < t < 1 that

where

]

] ,

where the inequality used Lemma 3 for

] and the last equality used Gaussian integration by parts with respect to g 2 . From the Hölder inequality and independence, it follows that

The same bound is also valid for E|X 2 |U. From these, putting Y = U -EU, and using the fact that f ′ ≥ 0, we have

Using Gaussian integration by parts and the Cauchy-Schwarz inequality,

It follows that max

From these and Lemma 2, for X ∶= (X 1 , X 2 ) and twice differentiable L with max 𝓁=1,2 ||𝜕 x 𝓁 L|| ∞ ≤ 𝛾, we arrive at

Now for fixed i, j, conditionally on g 1 t and g 2 t except g 1 t,ij and g 2 t,ij , we express ⟨𝜎 i 𝜎 j 𝜏 i 𝜏 j ⟩ t as L(X) in distribution. A direct computation gives

where (𝜎 1 , 𝜏 1 ) and (𝜎 2 , 𝜏 2 ) are i.i.d. samples from the Gibbs measure associated to ⟨⋅⟩ t . From these, max 𝓁=1,2 ||𝜕 x 𝓁 L|| ∞ ≤ 2𝛽N -1∕2 . Consequently, from ( 14) and using conditional expectation,

Summing these up over all i, j completes our proof. ▪ Lemma 5. Assume that (A) holds. There exists a constant K depending only on 𝛽 such that whenever t ∈ [0, 1) satisfies

we have

2024, 2, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/rsa.21183 by UNIVERSITY OF MINNESOTA 170 WILSON LIBRARY, Wiley Online Library on [01/07/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License where R 1

Proof. For t ∈ [0, 1] and 𝜆 ≥ 0, consider

Denote by ⟨⋅⟩ t,𝜆 the Gibbs average with respect to the i.i.d. samples (𝜎 𝓁 , 𝜏 𝓁 ) 𝓁≥1 from the Gibbs measure associated to this free energy Q. A direct differentiation and Gaussian integration by parts yield that

In the same manner as the proof of Lemma 4, if we let

Therefore, from ( 14), for D(t) ∶= 16𝛽 3 R 0 (t), we have

From this, whenever 0 ≤ w(t) ≤ 𝜆,

which implies that

Let 𝜆 0 > 0 such that 2𝛽 2 𝜆 0 < 1. For any t ≥ 0 satisfying 2𝛽 2 (𝜆 0 + w(t)) < 1, if we plug 𝜆 = 𝜆 0 + w(t) into the above inequality, then

Finally, since

where the third inequality used ( 16) and the fact that Q(t, 0) = Q(0, 0). Here, Q(t, 0) = Q(0, 0) holds because, recalling the definition Z 𝓁 t from (11),

In order to bound the right-hand side of the last inequality, our next step is to show that we can essentially replace f (g 1 ij ) and f

in Q(0, ⋅) by i.i.d. standard normal random variables by using approximate Gaussian integration by parts. Let

1≤i,j≤N and

) . This is essentially the same as Q(0, 𝜆) with the replacement of

and

. For any 0 ≤ s ≤ 1, set the interpolated free energy

) .

Similar to the Gibbs expectation ⟨⋅⟩ t,𝜆 , we let ⟨⋅⟩ ′ s be the Gibbs expectation with respect to the i.i.d. (𝜎 𝓁 , 𝜏 𝓁 ) 𝓁≥1 sampled from the Gibbs measure associated to the free energy 𝜌(s, 𝜆). It follows that

Here, the second term can be computed by the usual Gaussian integration by parts,

As for the first term, note that Eh = 0, Eh 2 = 1, and E|h| 3 < ∞, we can use approximate Gaussian integration by parts (see, e.g., [5], Lemma 2.2) to obtain

Consequently, we arrive at

This together with (17) implies that

) .

In the last step, note that

Observe that due to the symmetry of (z 1 ij ) and (z 2 ij ), under the expectation E⟨⋅⟩ ′ 0 , NR(𝜎, 𝜏)

we have

where recalling (13), the last inequality used the bound

Recalling that

and taking 𝜆 0 = 1∕(4𝛽 2 ), whenever t satisfies

we have

This completes our proof. ▪

Proof of Theorem 1

Smooth case: First, we show that Theorem 1 holds under the assumption (A). Recall 𝜙 from (10).

It suffices to bound 𝜙(1) -𝜙(0). Denote 𝜂(t) = E⟨R(𝜎, 𝜏) 2 ⟩ t . For 0 < r < 1, write

By Lemma 4 and integration by parts, the first term is bounded above by

where we dropped w(0)𝜂(0) and ∫ r 0 w(t)𝜂 ′ (t) 𝑑t since they are both nonnegative, which follow by noting that the summand of the Gibbs expectation in 𝜂(t) is of the form F(g 1 t )F(g 2 t ) for some bounded smooth function F.

On the other hand, observe that 𝜙 ′ (t) ≤ 𝛽 2 Nw ′ (t). It follows that

10982418, 2024, 2, Downloaded from https://onlinelibrary.wiley.com/doi/10.1002/rsa.21183 by UNIVERSITY OF MINNESOTA 170 WILSON LIBRARY, Wiley Online Library on [01/07/2024]. See the Terms and Conditions (https://onlinelibrary.wiley.com/terms-and-conditions) on Wiley Online Library for rules of use; OA articles are governed by the applicable Creative Commons License

Combining these yields that

) .

Next, from Lemma 5, we fix 0 < s < 1 such that

.

By using (7), for any r satisfying s ≤ r < 1,

Consequently, there exists some K ′ > 0 depending only on 𝛽 such that for any r satisfying s ≤ r < 1,

To control the right-hand side, let N ≥ 2 and take r = (log N)

Note that s ≤ r ≤ 1 and that if a = 1r, then 1 -= (log N) 2 log s∕ log N = (log N) -2 log(s -1 )∕ log N . Using the bound 1cx ≤ (1x) c for all x ∈ [0, 1] and c ≥ 1 implies that

and thus, there exists some C depending only on s such that

On the other hand, from our choice of r,

Putting ( 21) and ( 22) back into (20) yields that

where K ′′ is a constant depending only on 𝛽. This proves Theorem 1 under assumption (A). General case: Assume that h satisfies Eh = 0, Eh 2 = 1 and E|h| 3 < ∞ and h can be written as h = f (g) for some nondecreasing f , where g is a standard normal random variable. For any integer n ≥ 1, set f n (x) = max(min(f (x), n), -n). Let h n = f n (g) for

Note that Eh n = 0 and Eh 2 n = 1. Also, we have |f n (g)| ≤ |f (g)| for all n, and hence by the dominated convergence theorem, E|f n (g)-f (g)| 3 → 0 as n → ∞. Thus, E|h n -h| 3 → 0 and Ef n (g 1 t )f n (g 2 t ) → w(t) as n → ∞. From these, if we can show that h n enjoys the inequality in Theorem 1, then so does h. To this end, for any fixed since f n is bounded and nondecreasing, we can construct a sequence of smooth and nondecreasing functions (f n,k ) k≥1 of moderate growth (for instance, take

satisfies the upper bound in Theorem 1 for any k ≥ 1, we can pass to the limit k → ∞ to obtain the same bound for h n , completing our proof.

PROOF OF THEOREM 2

Smooth case: Assume that f satisfies the extra assumption that f is smooth and its derivatives of all orders are of moderate growth. Recall 𝜙(t) from (10). Note that in the proof of Lemma 4, we can bound

by the Hölder inequality. Thus, the statement of Lemma 4 is valid with the replacement of R 0 (t) by the constant

that is,

where we recall that 𝜂(t) = E⟨R(𝜎, 𝜏) 2 ⟩ t . From this bound, it can also be checked directly that (18) holds with D(t) being replaced by D 0 = 16𝛽 3 C 0 . Moreover, as long as 2𝛽 2

we have

Letting 𝜆 0 = 1∕(4𝛽 2 ), this inequality then implies that whenever

we have

Now, by using ( 23), for any 0 < s < 1,

where the last inequality used monotonicity of 𝜂. Here, we can select and fix s satisfying that 4𝛽 2 log (1s) -1 < 1 so that we can apply (24) to bound 𝜂(s) ≤ C 1 (s)N -1∕2 . In a similar manner as that of (19), we can bound that for any s ≤ r ≤ 1,

where the second inequality used the bound that for any x > 1,

Putting these together, we arrive at

for some universal constant K depending only on 𝛽. Note that the following Gaussian-Poincaré inequality holds (see, e.g., [20], Eq. (2.5)),

where C is a universal constant independent of f . We can bound each E|h| 3 in our main control above by E|f ′ (g)| 3 . Together with the trivial bound Ef ′ (g) 2 ≤ 1 + E|f ′ (g)| 3 , we obtain the desired inequality (3).

General case:

We continue to handle the general case in Theorem 2. First of all, we argue that without loss of generality, we can assume that f is uniformly bounded on R. Indeed, consider the absolutely continuous function f M = max(-M, min(M, f )) for M ≥ 1. We see that |f In addition, because f n is compactly supported, |f n | 3 is integrable on R with respect to the Lebesgue measure. Since a.e.

we also have

With these, for any n ≥ 1, there exists a sequence of smooth functions (𝜙 n,k ) k≥1 with compact support such that 𝜙 n,k → f n and 𝜙 ′ n,k → f ′ n as k → ∞ under the L 3 -norm with respect to the Lebesgue measure on R (see, for instance, [1], Corollary 3.23). This readily implies that Here, the first limit readily implies that the variance of the free energy associated to f n,k (g) converges to that associated to h in the limit k → ∞ and then n → ∞, while the second limit leads to Since h n,k ∶= f n,k (g) satisfies all the assumptions in Theorem 2 and the derivatives of f n,k of all orders are of moderate growth by the compact supportiveness of 𝜙 n,k , from the smooth case above, the inequality (3) holds for h n,k , from which sending the limit in the order k → ∞ and then n → ∞ completes our proof.