Chapter 4

The p-spin interaction model

The Hamiltonian of this model replaces the 2-spin interactions of the SK model by p-spin interactions. The Hamiltonian is given by

H_{N}(σ) = −(\frac{p!}{2N^{p−1}} )^{\frac{1}{2}} ∑ g_{i_{1},...,i_{p}}σ_{i_{1}} … σ_{i_{p}}. (4.1)

The summation is over all possible choices of indices i_{1} < ⋯ < i_{p}. The g_{i_{1},...,i_{p}} are i.i.d. N (0, 1). We will soon explain why the case p ≥ 3 is fundamentally different from the case p = 2. To understand the normalization factor in (4.1) we observe that for two configurations σ, σ′, we have

E(H_{N}(σ)H_{N}(σ′)) = \frac{p!}{2N^{p−1}} ∑_{i_{1}<···<i_{p}} σ_{i_{1}} … σ_{i_{p}}σ′_{i_{1}} … σ′_{i_{p}} (4.2)

= \frac{1}{2N^{p−1}} ∑_{d} σ_{i_{1}}σ′_{i_{1}}σ_{i_{2}}σ′_{i_{2}} … σ_{i_{p}}σ′_{i_{p}}

= \frac{N}{2} (\frac{σ · σ′}{N} )^{p} − \frac{1}{2N^{p−1}} ∑_{nd} σ_{i_{1}}σ′_{i_{1}}σ_{i_{2}}σ′_{i_{2}} … σ_{i_{p}}σ′_{i_{p}}.

Here, ∑_{d} denotes the sum over all possible choices of indexes i_{1}, …, i_{p} that are all distinct, while ∑_{nd} denotes the sum over all choices of indexes such that at least two of them coincide. In particular we have

\frac{N}{2} − K ≤ E(H_{N}(σ)^{2}) ≤ \frac{N}{2} . (4.3)

(Following our convention, K = K(p) depends upon p only).

32 4. The p-spin interaction model

If we view (4.2) as

E(H_{N}(σ)H_{N}(σ′)) ≃ \frac{N}{2} (\frac{σ · σ′}{N} )^{p},

and we observe that since |σ·σ′| ≤ N, as p increases, there is less correlation between H_{N}(σ) and H_{N}(σ′) if σ ≠ ±σ′. Thus, as p → ∞ the model should resemble more and more the REM. This intuition is correct, but we must keep in mind that we cannot exchange the limits p → ∞ and N → ∞.

Let us note another consequence of (4.2).

Lemma 4.1.

E((H_{N}(σ) + H_{N}(σ′))^{2}) ≤ N(1 + R(σ, σ′)^{p}).

Let us define the number β_{p} by

(4.4) β_{p} = sup\{β | lim_{N→∞} p_{N}(β) = \frac{β^{2}}{4} + log 2\},

where of course p_{N}(β) corresponds to the Hamiltonian (4.1). It follows from Theorems 3.3 and 3.13 that β_{2} = 1. For p ≥ 3, the value of β_{p} is not rigorously known, but we will approximate it. To state our result, we need the function

φ(t) = \frac{1}{2}((1 + t) log (1 + t) + (1 − t) log (1 − t)). (4.5)

This function is probably best understood by noting that φ(0) = φ′(0) = 0, φ′′(t) = 1/(1 − t^{2}). Thus φ(t) ≥ \frac{t^{2}}{2} , and φ(t) ≃ \frac{t^{2}}{2} for small t. This function arises through the following result, a special case of the Chernov bounds for the binomial law.

Lemma 4.2. If t ≥ 0, we have

card{σ | ∑_{i≤N} σ_{i} ≥ Nt} ≤ 2^{N} exp(−Nφ(t)). (4.6)

Proof. We observe that, for each λ

∑ exp λ ∑_{i≤N} σ_{i} = 2^{N}(ch λ)^{N} = 2^{N} exp N log ch λ,

so that the left-hand side of (4.6) is bounded by

2^{N} inf_{λ≥0} exp N(log ch λ − λt) = 2^{N} exp(−Nφ(t)).  □

Theorem 4.3. For p ≥ 2 we have

inf_{0<t<1} 2(1 + t^{−p})φ(t) ≤ β_{p}^{2} ≤ 4 log 2. (4.7)

The right-hand side inequality follows from (2.18).

Challenge problem 4.4. Show that in fact

β_{p}^{2} < 4 log 2.

Since φ(t) ≥ \frac{t^{2}}{2} , for p = 2, the left-hand side of (4.7) is 1, the correct value. Thus Theorem 4.3 recovers Theorem 3.3. This is not a surprise: the proofs are very similar.

Proof of Theorem 4.3. It will closely follow that of (2.22). Consider

β^{2} ≤ inf_{0<t<1} 2(1 + t^{−p})φ(t). (4.8)

We define Y as in (2.29) and using (4.3), (2.34), we have

E Y ≥ \frac{2^{N−1}}{K\sqrt{N}} exp(−\frac{N^{2}β^{2}}{8(\frac{N}{2} − K)}).

Now,

\frac{N^{2}β^{2}}{8(\frac{N}{2} − K)} ≤ \frac{Nβ^{2}}{4(1 − \frac{2K}{N} )} ≤ \frac{Nβ^{2}}{4} + K,

so that

E Y ≥ \frac{1}{K\sqrt{N}} 2^{N} exp(−\frac{Nβ^{2}}{4} ). (4.9)

Now, we have

E Y^{2} = ∑_{σ,σ′} A(σ, σ′), (4.10)

where

A(σ, σ′) = P(\{−H_{N}(σ) ≥ \frac{Nβ}{2} , −H_{N}(σ′) ≥ \frac{Nβ}{2} \})

≤ P({−(H_{N}(σ) + H_{N}(σ′)) ≥ Nβ})

≤ exp(−\frac{Nβ^{2}}{2(1 + R(σ, σ′)^{p})})

≤ exp(−\frac{Nβ^{2}}{2(1 + |R(σ, σ′)|^{p})}),

using successively (2.34) and Lemma 4.1. We observe that |σ · σ′| is an integer. Thus if in (4.10) we regroup the pairs σ, σ′ for which |σ · σ′| = tN, we get

(4.11) E Y^{2} ≤ ∑_{\begin{matrix}0≤t≤1 \\ Nt integer \end{matrix}} exp(−\frac{Nβ^{2}}{2(1 + t^{p})}) card\{σ, σ′ | |σ · σ′| = tN\}.

Now, by (4.3)

(4.12)  card{σ, σ′ | |σ · σ′| = tN} = 2^{N} card{σ | ∑_{i≤N} σ_{i} = Nt}
≤ 2^{2N+1} exp(−Nφ(t)).

Moreover, (4.8) implies that

\frac{β^{2}}{2(1 + t^{p})} + φ(t) ≥ \frac{β^{2}}{2}

so that (4.9), (4.12) imply

E Y^{2} ≤ (N + 1)2^{2N} exp(−\frac{Nβ^{2}}{2} ) ≤ KN^{2}(E Y)^{2}. (4.13)

The proof is then completed as in Proposition 2.6. □ Proposition 4.5. If p is large enough we have 2\sqrt{log 2} (1 − 2^{−p}) ≤ β_{p} ≤ 2\sqrt{log 2}. (4.14)

This follows from Theorem 4.3 and (4.16) below. Lemma 4.6. If p is large enough, the following holds. If v ≥ 2^{− \frac{p}{2}} then

inf_{0≤t≤1−v} 2(1 + t^{−p})φ(t) ≥ 4 log 2 + \frac{p}{4}v. (4.15)

Moreover (4.16) inf_{0≤t≤1} 2(1 + t^{−p})φ(t) ≥ 4 log 2(1 − \frac{2^{−p}}{log 2}) + O(p^{2}2^{−2p}). where O(A) denotes a quantity such that |O(A)| ≤ LA. Proof. Since φ(t) ≥ \frac{t^{2}}{2} , we have, by convexity,

2(1 + t^{−p})φ(t) ≥ t^{2−p} ≥ 1 + (p − 2)(1 − t) ≥ 4 log 2 + \frac{p}{2}(1 − t),

if p is large and 1−t ≥ \frac{4}{p}. Thus we need only to consider values of t ≥ 1− \frac{4}{p}. Setting u = 1 − t, we observe that, by convexity,

1 + t^{−p} ≥ 2 + pu
(1 + t) log(1 + t) ≥ 2 log 2 − (1 + log 2)u,

and thus

2(1 + t^{−p})φ(t) ≥ ψ(u) := (2 + pu)(2 log 2 − (1 + log 2)u + u log u). Now,

(4.17)  ψ(u) = 4 log 2 + 2u(p − 1) log 2 − 2u
+ 2u log u − pu^{2}(1 + log 2) + pu^{2} log u.

For u ≤ \frac{4}{p}, we have

ψ(u) ≥ 4 log 2 + u(2(p − 1) log 2 − 2 + 2 log u − 4(1 + log 2) − 4 log \frac{p}{4}) (4.18)

≥ 4 log 2 + \frac{pu}{4},

provided log u ≥ −\frac{p}{2} log 2 and p is large enough. This proves (4.15). To prove (4.16), we can assume u ≤ 2^{− \frac{p}{2}} . In that case, rather than (4.18), we obtain

(4.19)  ψ(u) ≥ 4 log 2 + u[2(p − 1) log 2 − 2 + 2 log u
− p 2^{− \frac{p}{2}} (1 + log 2) − p 2^{− \frac{p}{2}} log(2^{\frac{p}{2}} )],

so that (for p large)

ψ(u) ≥ 4 log 2 if u ≥ 2^{−p+3}.

But, if u ≤ 2^{−p+3}, then (4.17) yields

ψ(u) ≥ 4 log 2 + 2u((p − 1) log 2 − 1) + 2u log u + O(p^{2}2^{−2p}),

and the right-hand side has its minimum at u = 2^{−p+1}, from which (4.16) follows. □

Our purpose is not to study in detail the p-spin interaction model for β ≤ β_{p}. Rather we will be interested in the case β > β_{p}.

Challenge problem 4.7. Find a convincing version of Theorem 3.13 for the present model.

Challenge problem 4.8. Find a convincing version of Proposition 2.9. What is the optimal size of I?

Challenge problem 4.9. More generally, which of properties proved in the setting of Chapter 3 for p = 2 can be generalized to the case p ≥ 3?

Challenge problem 4.10. Prove that given β > 0, there is p(β) > 0, θ = θ(β) such that if p ≥ p(β), the Gibbs’ measure of the p-spin interaction model is θ-shattered (as in Definition 3.9) at inverse temperature β with overwhelming probability.

Open Problem 4.11. Does there exist p > 0 such that the Gibbs’ measure of the p-spin interaction model is shattered for each β > 0?

We now turn to the main topic of this chapter, the search for estimates on the overlaps that will be valid not only for β ≤ β_{p} but also for β larger. To avoid having to distinguish cases we assume p even. In that case, Gibbs’ measure is invariant by global symmetry around zero.

Our estimates on the overlap will rule out the possibility that the overlaps belong to certain intervals. This justifies the following definition.

Definition 4.12. We say that an interval I is negligible (at a given β) if

E G_{N}^{⊗2}(\{σ, σ′ | |R(σ, σ′)| ∈ I\}) ≤ K exp(−\frac{N}{K} ).

We will prove that certain intervals are negligible. This is the key to the low temperature results of Chapter 8, but will not be needed before that.

Our goal is the following (I am grateful to A. Bovier for having pointed out to me that my argument gives the result without any restriction on β)

Theorem 4.13. There exists a number p_{0} such that if p ≥ p_{0} then for all values of β, the interval [2^{− \frac{p}{4}} , 1 − 2^{− \frac{p}{2}} ] is negligible.

We consider the function ξ(β) given by

(4.20) ξ(β) = \begin{cases}log 2 + \frac{β^{2}}{4} & if β ≤ 2\sqrt{log 2}, \\ β\sqrt{log 2} & if β ≥ 2\sqrt{log 2}. \end{cases}

This function occurs in Propositions 2.5 and 2.6.

Lemma 4.14. For all p sufficiently large, we have

lim inf_{N→+∞} p_{N}(β) ≥ ξ(β) − 2^{−p}β. (4.21)

Proof. The function β ↦ p_{N}(β) is convex, and by definition of β_{p}, for β ≤ β_{p} its limit is log 2 + β^{2}/4. It is elementary to conclude that

lim inf_{N→+∞} p′_{N}(β_{p}) ≥ β_{p}/2,

so that, using convexity of p_{N} again, for β ≥ β_{p} we have

lim inf_{N→+∞} p_{N}(β) ≥ log 2 + \frac{β_{p}^{2}}{4} + \frac{1}{2}β_{p}(β − β_{p}),

and thus, by an elementary computation

lim inf_{N→+∞} p_{N}(β) ≥ ξ(β) − \frac{1}{2}(2\sqrt{log 2} − β_{p})β.

But it follows from (4.14) that 2\sqrt{log 2} − β_{p} ≤ 2^{−p+1}. □

We now define, for |u| ≤ 1,

A_{N}(u) = ∑ exp(−β(H_{N}(σ) + H_{N}(σ′))), (4.22)

where the sum is over the pairs of configurations σ, σ′ that satisfy R(σ, σ′) = u.

Lemma 4.15. We have

\frac{1}{N} E log A_{N}(u) ≤ ξ_{u}(β) (4.23)

where

ξ_{u}(β) = \frac{β^{2}}{2} (1 + u^{p}) + 2 log 2 − φ(u)

if

\frac{β^{2}}{2} (1 + u^{p}) ≤ 2 log 2 − φ(u) (4.24)

and

ξ_{u}(β) = β\sqrt{2(1 + u^{p})(2 log 2 − φ(u))}

if (4.24) fails.

Proof. It is identical to that of Proposition 2.5 if one observes that when R(σ, σ′) = u,

E((H_{N}(σ) + H_{N}(σ′))^{2}) ≤ N(1 + u^{p})

by Lemma 4.1, and that the number of pairs in the summation (4.22) is at most 2^{2N} exp(−Nφ(u)). □

Using Proposition 2.2 (and its extension to the left-hand side of (4.23)), we see that to prove Theorem 4.13 it suffices to prove the following

Lemma 4.16. If |u| ∈ [2^{−p/4}, 1 − 2^{−p/2}] and if p is large enough, we have

ξ_{u}(β) ≤ 2ξ(β) − β2^{− \frac{p}{2} −4}. (4.25)

Proof. We can assume u ≥ 0 by symmetry. The basic fact is that if |u| ∈ [2^{−p/4}, 1 − 2^{−p/2}] (and p is large enough) then

2u^{p} log 2 − (1 + u^{p})φ(u) ≤ −2^{− \frac{p}{2} −2}. (4.26)

To prove this, we observe that if u^{p−2} ≤ 1/8, this is at most

\frac{u^{2}}{4} − \frac{u^{2}}{2} ≤ −\frac{u^{2}}{4} ≤ −2^{− \frac{p}{2} −2}.

If u^{p−2} ≥ 1/8, we appeal to (4.15) with v = 2^{−p/2} to see that

(1 + u^{p})φ(u) ≥ (2 log 2 + \frac{p 2^{−p/2}}{4} )u^{p}

so that the left-hand side of (4.26) is at most

−\frac{p 2^{−p/2}}{4} (\frac{1}{8})^{p/(p−2)} ≤ −2^{− \frac{p}{2} −2}

for p large enough. This proves (4.26).

To prove (4.25), we first assume that (4.24) fails. We observe that this implies

β^{2} ≥ log 2 ≥ \frac{1}{2}. (4.27)

We write

2(1 + u^{p})(2 log 2 − φ(u)) = 4 log 2 + 2(2u^{p} log 2 − (1 + u^{p})φ(u))
≤ 4 log 2 − 2^{− \frac{p}{2} −1}

by (4.26), and thus

\sqrt{2(1 + u^{p})(2 log 2 − φ(u))} ≤ 2\sqrt{log 2} − 2^{− \frac{p}{2} −3}.

Since 2ξ(β) ≥ 2β\sqrt{log 2}, (4.25) follows by (4.27).

Finally, if (4.24) holds,

ξ_{u}(β) = \frac{β^{2}}{2} + 2 log 2 + \frac{β^{2}}{2} u^{p} − φ(u). (4.28)

By (4.24),

\frac{β^{2}}{2} ≤ 2 log 2 − φ(u) (4.29)

so that, by (4.28),

ξ_{u}(β) ≤ \frac{β^{2}}{2} + 2 log 2 + 2u^{p} log 2 − (1 + u^{p})φ(u)

≤ \frac{β^{2}}{2} + 2 log 2 − 2^{− \frac{p}{2} −2}

by (4.26). Now, by (4.24), β^{2} < 4 log 2, and β^{2}/2 + 2 log 2 = 2ξ(β). □

Challenge problem 4.17. Show that if β < β_{p} and x > 0, the interval [x, 1] is negligible.