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Existence of two-step replica symmetry breaking for the spherical mixed p-spin glass at zero temperature

We provide the first examples of two-step replica symmetry breaking (2-RSB) models for the spherical mixed p-spin glass at zero temperature. Precisely, we show that for a certain class of mixtures, the Parisi measure at zero temperature is purely atomic and has exactly three distinct points in its support. We then derive a few consequences for the topology of the random landscape in these cases. Our main result also provides a negative answer to a question raised in 2011 by Auffinger and Ben Arous about the classification of pure-like and full mixture models.

The measure ν P = ν P (ξ) on [0, 1] is called the Parisi measure at zero temperature. Its importance lies primarily on the fact that it describes the energy landscape of H N near the global minimum [5]; for instance lim The aim of this article is to study the structure of the support of the Parisi measure at zero temperature as a function of the model ξ. Our main result shows that it is possible to find functions ξ such that the Parisi measure of the model is atomic with exactly three atoms in its support. 1 This is referred as two levels of replica symmetry breaking (2-RSB) in mathematical physics nomenclature (k-RSB refers to atomic measures with k + 1 atoms). For more information on the physics literature, we refer the reader to [18]. Precisely, we prove the following. Theorem 1. There exist models ξ such that for some positive constants m 1 , m 2 , ∆ P and q ∈ (0, 1) one has ν P (ξ) = m 1 δ 0 + m 2 δ q + ∆ P δ 1 .
The classification of levels of replica symmetry breaking (RSB) has a long history in the physics community, see [18,Chapter 3]. Although one can artificially cook up stochastic processes with arbitrary levels of finite RSB (for instance the GREM [8,14]), it was believed that only replica symmetric (RS), 1-RSB and FRSB models would appear "naturally". 2 For instance, all models defined on the discrete hypercube would exhibit infinite levels of replica symmetry breaking at low enough temperature (FRSB), that is, the Parisi measure will have infinite many points in its support 3 . Moreover, it is known that the spherical pure p-spin model is 1-RSB at low temperature (see [20,Section 4]).
More recently, existence of 2-RSB spherical models were suggested at positive temperature in the physics literature in an insightful paper by Crisanti-Leuzzi [11] (see also [10,12,17] for a long debate on these cases). Several rigorous results on properties of Parisi measures appear in the works of Talagrand [21], Panchenko-Talagrand [20], and more recently in [1-3, 5, 6, 15, 19]. We refer the readers to Talagrand [21] and Auffinger-Chen [4] for an introduction on Parisi measures.
As far as we know, Theorem 1 is the first rigorous result to provide "natural" examples of spin glass models beyond RS, 1-RSB and FRSB phases. This result is somewhat surprising. Indeed, prior to the results of this paper, it was perceived that the spherical model had only one of two possible behaviors at zero temperature: the Parisi measure would either be 1-RSB (as in the pure p-spin) or FRSB. This distinction was the motivation behind the classification of pure-like and full mixture models introduced in [1]. It turns out that this is a tepid classification: we find pure-like 2-RSB models and full mixture 2-RSB models. Furthermore, the examples that we construct provide a negative answer to Question [1,Question 4.1] that does not involve using the spherical SK model. We refer the reader to Remark 1 in Section 4 for more on these examples.
A word of comment is needed. The structure of the Parisi measure at zero temperature in the case of Ising spins (when S N is replaced by {±1} N ) was recently determined in [6]. The result there is strikingly different: on {±1} N , all models have infinite levels of replica symmetry breaking, as predicted by physicists. Comparing with Theorem 1, this shows that the sphere has a richer collection of models than the hypercube (at least in terms of different levels of RSB).
We also study the landscape of H N when the model is 2-RSB. We provide information on the topology of level sets near the global minimum of H N . In order to state our results, set, for any
Let us now describe a major consequence of Theorem 2. For any ε, η, K > 0, denote by (ii) O N contains at least Ke N/K many elements.
It is known [5, Proposition 2] that for any ε, η > 0, there exists K > 0 such that for any N ≥ 1, If we combine (1) with Theorem 2 we obtain the following orthogonal decomposition of local minima of H N . Let Corollary 3. There exist η ′ > 0, ǫ ′ > 0, K > 0 so that for all N ≥ 1 the event defined by (ii) Any continuous path connecting two points in Crt(η ′ ) must leave the level set L(η ′ + ǫ ′ ), A major feature of Theorem 2 and Corollary 3 is that we can always find exponentially many local minima of H N around the ground state energy. Furthermore, if we think H N as a random landscape on the sphere, in order to go from one deep local minimum to another one we must climb a diverging energy barrier (in N ). This fact was predicted to hold in 1-RSB models and known for the pure p-spin [2,5]. Now that such phenomenon occurs in the 2-RSB phase, it is natural to ask the following question: Question 1. In the family of spherical mixed p-spin models, do there exist k-RSB models (at zero or positive temperature) for any k ≥ 3, k ∈ N?
We anticipate the answer to be positive. In fact, our results for 2-RSB suggest that the energy landscape of k-RSB models for k ≥ 3 should satisfy Theorem 2 and Corollary 3 provided such models exist. However, since we have discovered 2-RSB examples, the next desirable step is not only to answer the question above but to give a complete characterization of the Parisi measure as a function of ξ. This seems beyond reach at this moment.
The organization of the paper is the following. In the next section, we establish sufficient and necessary conditions for the model to be 2-RSB. These conditions follow from a careful analysis of a criterion developed by Chen and Sen in [9]. This criterion was also explored in the 1-RSB case by Auffinger and Chen in [5]. Compared to the 1-RSB case, the complexity of the analysis of the 2-RSB is acutely more demanding. In Section 3, we provide sufficient conditions for s + p models (i.e. ξ(x) = (1 − λ)x s + λx p ) to be 2-RSB. These conditions are easier to check in practice, and are indeed verified in Section 4 for some models, proving Theorem 1. We study the energy landscape of 2-RSB models in Section 5.
We start this section by recalling the following criterion derived in Chen-Sen [9]. Similar criteria also appeared in [5,21]. Recall from Section 1 that ν ∈ K is a measure of the form Theorem 4 (Chen-Sen [9]). For ν ∈ K, let Here the notation ν is different from the one used in [9, Theorem 2], but it coincides with [5,15].
Note that and for q < u ≤ 1, By Theorem 4, ν is the Parisi measure for ξ if and only if g(0) = 0, g(q) = 0, g ′ (q) = 0, The condition g ′ (q) = 0 follows from other conditions. We include it here because it simplifies a lot of computations in the following. Unraveling the first four conditions by using the expressions of g(u) and ν((r, 1]), we find Now, let z 1 = A 1 q/∆, z 2 = A 2 (1 − q)/∆. From the third and fourth equations above, we have It follows that From (3) and (4), we observe that z 1 and ∆ are determined by q and z 2 . Using the variables z 1 , z 2 , we can rewrite the first, second, and third equations as Equations (3) and (7) are not independent and one of them together with (5) and (6) determines q, z 1 , z 2 which are the unknowns for the problem. Consider This was used to determine the 1-RSB phase in [5]. Note that (5) degenerates to (8) as q → 1 and (6) degenerates to (8) as q → 0. The condition g(u) ≥ 0 can be rewritten as for 0 ≤ u ≤ q and for q ≤ u ≤ 1, In what follows, we always assume z 1 is determined by q and z 2 via (4) while sometimes we still write z 1 for convenience. Plugging (3) into (5) and (6), and eliminating z 1 using (4), we find Despite their complicated appearance, these equations are our starting point to find 2-RSB models. First, we show that the zero sets given in (11) and (12) have some nice properties. Note that both functions f 1 and f 2 are C ∞ in their domains.
Proof. We start by proving (i). Let w = 1 + z 2 and set ϕ(w) = f 1 (q, w − 1). Taking derivative in w, we have Note that qξ ′ (q) > 2ξ(q) for q > 0. Solving the quadratic equation given by ϕ ′ (w) = 0, we find the two roots One can directly check that On the other hand, we note that Putting what we have shown together, the function f 1 (q, ·) is strictly decreasing on (−1, z s 2 ) and on (z b 2 , +∞), and is strictly increasing on (z s 2 , z b 2 ); moreover, z s 2 is a local minimum and z b 2 is a local maximum for f 1 (q, ·). It follows that f 1 (q, ·) has exactly two zeros in (−1, +∞), one is strictly less than z s 2 and the other is z b 2 . We now proceed to prove (ii). Note that the function z 2 → 1+z 2 z 2 2 log(1 + z 2 ) − 1 z 2 is a strictly decreasing function from 1 2 to 0 as z 2 goes from 0 to +∞. By continuity, it suffices to show that for any q, f 2 (q, 0+) and f 2 (q, +∞) have different signs. Since ξ ′ (q) = ∞ p=2 c 2 p pq p−1 and But for x ∈ (0, 1) The following Lemma provides one extra condition.
Proof. Since u = q is a local minimum of g(u), we have g ′ (q) = 0 and g ′′ (q) ≥ 0. Recall that Then The assertion follows by plugging ∆ 2 as given in (3) into this equation.
These functions come from substituting the term ∆ in (9) and (10) using (3). Last, define We summarize the above calculations in the following Proposition.

Proposition 7.
A measure ν P defined as in (2) is the Parisi measure for the model ξ if and only if (3) holds, z 1 > 0 and qz 2 > (1 − q)z 1 , and Furthermore, condition (19) holds if and h 2 (u) < 0 in a neighborhood (q, q + δ) of q and h 2 (u) has only one zero in (q, 1).
3 Spherical s + p models: a simpler criterion Although Proposition 7 characterizes a possible 2-RSB phase, it is hard to solve the nonlinear system given by (18). Here we try to reduce the difficulty by focusing on s + p models, i.e. we will take for some λ ∈ (0, 1), p > s ≥ 3. Since p + (p + 1) models are known to be in the 1-RSB phase [5], we assume p ≥ s + 2.
Proof. Note that for the s + p model we consider here, h 1 (u) u | u=0+ < 0. It follows that h 1 (u) < 0 in a neighborhood of 0. Since g 1 (0) = g 1 (q) = 0, g ′ 1 (u) and h 1 (u) have at least one zero in (0, q). By Proposition 7, it remains to show that h 1 (u) has only one zero in (0, q). Note that There are four sign changes for the coefficients of h 1 (u). By Descartes' rule of signs, h 1 (u) has at most four strictly positive roots counting multiplicity. A calculation yields The condition (15) implies that h ′ 1 (q) ≤ 0. We also know h 1 (+∞) < 0 and h 1 (1) > 0. Then there is at least one zero greater than 1. Since has a zero in (q, 1) and at least three zeros in [q, +∞). Therefore h 1 has exactly one zero in (0, q). If h ′ 1 (q) = 0, then h 1 (u) has a root at q with multiplicity at least two. It follows that h 1 (u) has at least three roots in [q, +∞) and thus has at most one root in (0, q).
Proof. By Lemmas 8 and 9, conditions (9) and (10) are verified. Note that (3) and (4) are equivalent. Now the assertion follows from the criterion given in Proposition 7.
Let us now explain how we will check the three conditions in Theorem 10. Here for convenience of writing, we use q, z 2 both for a solution of the system (18) and for the set of solutions of f 2 (q, z 2 ) = 0. It should be clear from the context what we mean. The general principle we follow here is to show that the implicit function z 2 = φ(q) defined by (12) is strictly decreasing under some condition on λ, and then use intermediate value property of continuous functions to estimate q, z 2 in a small interval.
First, we compute dz 2 dq with implicit differentiation: Solving for 1+z 2 (12) and plugging into the above equation, we find ) .
With the elementary inequality log(1 + z) ≥ 2z 2+z for z > 0, we immediately see that the denominator is strictly positive. The numerator is simplified as It follows that dz 2 dq < 0 if and only if Proposition 11. We have dz 2 dq < 0 provided λ = 0 or We remark that it is not difficult to see that we need to impose some restriction in λ in order to obtain dz 2 dq < 0. The rest of the section is devoted to the proof of the above proposition.
Proof. The argument is a little technical but still elementary. The difficulty here is that we don't have efficient methods to determine the sign of the polynomial in (22). Although not obvious, we will factor out a factor (q − 1) 3 and then show that the remaining factor has all coefficients positive, which is guaranteed by condition (23).
We will use the elementary identity 1 − q n = (1 − q) n−1 k=0 q k repeatedly. Since ξ(1) = 1, we have Let us start with analyzing III. For simplicity, we denote the term in the parenthesis by III ′ =

III
(1−λ)λs . By change of variable k = p − 1 − ℓ in the second summand, we have Now let III ′′ = III ′ /(q − 1) and we will find a closed formula for III ′′ . By expanding all the sums and recombining, we find Using the elementary summation formulae for m 1 j and m 1 j 2 , we have q n 1 6 (n + 1)(n + 2)(n + 3), where Note that the coefficient of n is negative, which implies that III ′′ 2 is an increasing function of n for n ∈ N. We see also that the coefficients of q n for n ≤ s − 3 are all positive.
Let us now turn to the quantity IV and define IV ′ = IV (1−λ)λp . By the same argument as for III ′ , we find a factor q − 1 in IV ′ and define IV ′′ = IV ′ /(q − 1). Then we have q n 1 6 (n + 1)(n + 2)(n + 3) where Similarly, we define I ′′ = I (1−λ) 2 s(q−1) and II ′′ = Both are positive because all their coefficients are positive. We will drop the positive term I ′′ and determine the sign of 1 q−1 (II + III + IV ). Note that for n ≤ s − 3 or n ≥ p + s − 4, the coefficients of 1 q−1 (II + III + IV ) are all positive. For p − 2 ≤ n ≤ p + s − 5, we have as p + s − 5 − n ≥ 0 and p − s > 0. So the coefficients are strictly positive for p − 2 ≤ n ≤ p + s − 5. It remains to determine the coefficients for s − 2 ≤ n ≤ p − 3. Let II ′′ 2 = 1 6 (n + 1)(n + 2)(n + 3) be the coefficient of q n in II ′′ . Note that II ′′ 2 , III ′′ 2 and IV ′′ 2 are all increasing as n increases. It suffices to consider the coefficient of q s−2 in 1 q−1 (II + III + IV ), which is This quantity is no less than zero exactly as (23) holds. Thus dz 2 dq is strictly negative because we have strict positivity for the coefficients when p−2 ≤ n ≤ p+s−5. This ends the proof of Proposition 11.

Examples of 2-RSB models and the proof of Theorem 1
We will use Theorem 10 and Proposition 11 to determine models that are in the 2-RSB phase. The method is as follows: we first use the intermediate value property for the continuous functions f 1 and f 2 defined in (11) and (12), and condition (23) to estimate q, z 2 in a small interval, and then check the conditions in Theorem 10. The zero set of f 1 is more involved, but to guarantee a solution for (11) and (12) we only need a mild property of this zero set, which we now explain.
Suppose we know a solution of (11) and (12) satisfies q ∈ [q − , q + ], z 2 ∈ [z − 2 , z + 2 ] for some 0 < q − < q + < 1 and z − 2 , z + 2 > 0. Then from (4) we know ensures z 1 > 0, and Since (4) is used to compute z 1 , both equations of (15) will be fulfilled provided Substituting z 1 by (4), we know Clearly, as q increases, the first term is decreasing while the second is increasing. So in order to check h 1 (1) > 0 it suffices to check Since it follows that h 2 (0) < 0. Before going to the examples, we define Our examples below show that among the s + p, s ≥ 3, 2-RSB models, there exist both pure-like and full mixture models. Therefore, this provides a strong negative answer to the question of Auffinger and Ben Arous, and shows that in general the level of RSB cannot be classified by the critical points consideration.
The next examples provide the proof of Theorem 1.

Example 1 (A pure-like 2-RSB model). Consider
Since G > 0, this model is pure-like. By Proposition 11, a calculation shows that dz 2 dq < 0 provided λ > 7 39 ; and here we have λ = 2 7 > 7 39 . By numerical calculations, we obtain the following: 2). Since z 2 is strictly decreasing as q increases by Proposition 11, the zero set of f 2 (q, z 2 ) satisfies that if q ∈ (0.743, 0.747), z 2 ∈ (3.17, 3.25). To see there is a solution for (11) and (12), the first line shows that f 1 (0.743, z 2 ) has a zero z 1 2 in (3.2, 3.22) and that f 1 (0.747, z 2 ) has a zero z 2 2 in (3.2, 3.22). Since (24) holds, there is a continuous curve connecting the two points (0.743, z 1 2 ) and (0.747, z 2 2 ) as explained above. By continuity there is at least one common element in (q, Remark 2. The 3 + 16 model was predicted by physicists to be 2-RSB for some choice of coefficients and some positive temperature; see [11, Figure 2]. However, it is not clear from the prediction in [11] if the zero temperature case had the same behavior. For instance, by [6] all mixed p-spin (Ising) models are FRSB at zero temperature, while this is in general not expected to be true throughout the low temperature regime.
Example 2 (A full mixture 2-RSB model). For our second example we take Since G < 0, this is a full mixture model. By numerical calculations, we have The following examples show that 2-RSB exists even for convex functions ξ. Convexity of ξ was used before to simplify many questions in spin glasses. Example 3. ξ(x) = 5 6 x 4 + 1 6 x 40 . Since G < 0, this model is full mixture. By proposition 11, dz 2 dq < 0 if λ > 7/107, which is satisfied here. Numerical calculations yield that f 1 (0.89, 3.6) > 0, f 1 (0.89, 3.8) < 0, f 1 (0.9, 3.7) > 0, f 1 (0.9, 3.9) < 0, f 2 (0.89, 3.9) > 0, f 2 (0.89, 4.1) < 0, f 2 (0.9, 3.5) > 0, f 2 (0.9, 3.7) < 0, and condition (24) holds. By the same analysis as above, there is a solution to (11) and (12)   Remark 4. Let us stress that our argument here is very specific to the 2-RSB phase and to s + p models in many aspects. The difficulty goes beyond perseverance. For instance, the four sign change phenomenon for h 1 and h 2 observed in Lemmas 8 and 9 does not happen if we consider a model with three mixed spins. Proposition 11 is tricky to observe, but it is essential in order to bypass the properties of the zero set of f 1 as defined in (11), which is hard to study analytically. In summary, our argument is a combination of various non-trivial, specific observations for the s + p models and for the 2-RSB phase. In fact, many miraculous coincidences occur so that our argument works.
Let U = {(B, ν) ∈ R × K :ν(0) < B} and define the Parisi functional U by Then the Parisi formula [5,Theorem 10] states that and the the infimum is achieved by a unique (B P , ν P ) ∈ U . In this case Let γ be a 2-RSB Parisi measure given as in (2). Using (31) and dropping the subscript, we have Proof of Theorem 2. Let ξ be a 2-RSB convex model with Parisi measure given by γ as above.
Using the change of variables (3) and comparing with (33), we can rewrite The rest of the proof follows the same argument as for [5,Theorem 6]. We reproduce it here for completeness and clarity for the general audience.
From (36), this means that P N (η/2, S) ≤ Ke −N/K for all N ≥ N 0 and this clearly implies (34), ending the proof of the theorem.
Remark 5. Note that Examples 3 and 4 considered in the previous section satisfy the assumptions (and conclusion) of Theorem 2.
Comparing ψ ′ 1 (a) with the function h 1 (u) as in (16), we see that ψ ′ 1 (a) and h 1 (a) have the same sign for 0 ≤ a ≤ q. Similarly, note that ψ 2 (q) = 0 and ψ 2 (1) = 0 by (3) and (6). Moreover, a calculation gives Comparing with h 2 (u) as in (17), we see that ψ ′ 2 (b) and h 2 (b) have the same sign for q ≤ b ≤ 1. This means that ψ 1 (a) < 0, a ∈ (0, q) and ψ 2 (b) < 0, b ∈ (q, 1) will hold once the conditions (20) and (21)  Let x, y ∈ L(η) so that |R 12 (x, y)| ≤ ǫ. By choosing ǫ small enough, any path between x and y must contain two points that have overlap in (S ε ∪ −S ε ) c . From Theorem 2 this implies that with high probability C x ∩ C y = ∅, where η is taken from Theorem 2. Since H N is a smooth function almost surely, each of this disjoint connected components must contain at least a local minimum. Now choose η ′ and ǫ ′ so that 0 < η ′ + ǫ ′ < η. Note that L(η ′ ) ⊆ L(η). The Corollary now follows directly by equation (1), since the random set O N guarantees the existence of exponentially many of such points in L(η ′ ).
We end the paper with one extra remark about the computations above. This may be of independent interest.
Remark 7. The functions h 1 and h 2 are essential for the 2-RSB phase. Indeed, we have seen in Remark 6 that the derivatives of ψ 1 and g 1 (resp. ψ 2 and g 2 ) have the same sign as h 1 (resp. h 2 ) in their domain. In fact, h 1 and h 2 also appear in another situation.