Chapter 8

Emergence and separation of the lumps in the p-spin interaction model

We go back to the study of the p-spin interaction model with no external field, and our efforts will focus on the low temperature region. Theorem 8.1. There exists a number p_{0} such that, if p ≥ p_{0}, then for β ≤ 2^{\frac{p}{2} −6}, with overwhelming probability we have

G_{N}^{⊗2}(\{|R(σ, σ′)| ∈ [2^{− \frac{p}{4}} , 1 − 2^{− \frac{p}{2}} ]\}) ≤ exp(−\frac{N}{K} ). (8.1)

Proof. This follows from Theorem 4.13 and Fubini’s theorem. □

Thus, for large p, |R(σ, σ′)| is typically close to 0 or 1. This fact has strong consequences about Gibbs’ measure.

It is convenient and will somewhat clarify matters to consider a more general setting. Let

S_{N} = {σ ∈ ℝ^{N} | ∑_{i≤N} σ_{i}^{2} = N},

90 8. The lumps in the p-spin interaction model

be the sphere of radius \sqrt{N} (so that Σ_{N} ⊂ S_{N}), and for σ, σ′ in S_{N}, we write

R(σ, σ′) = \frac{σ · σ′}{N} = \frac{1}{N} ∑_{i≤N} σ_{i}σ′_{i}. (8.2)

Consider now a probability measure μ on S_{N}, and assume that for a certain (small) number a,

μ ⊗ μ(T) = ε (8.3)

is very small, where

T = {(σ, σ′) | |R(σ, σ′)| ∈ [a, 1 − a]}.

What does (8.3) tell us about μ? First it could happen that (8.3) is true simply because it is already true that

μ ⊗ μ({(σ, σ′) | |R(σ, σ′)| ≥ a}) (8.4)

is very small. This is for example the case if μ is uniform on Σ_{N}. It could also happen that μ is concentrated on a single point, or more generally on a set A such that R(σ, σ′) ≥ 1 − a if (σ, σ′) ∈ A. We will show that the general situation is a mixture of the previous two cases.

The construction will involve thinking of S_{N} as a metric space, equipped with the distance

d(σ, σ′) = arccos(R(σ, σ′)).

We consider δ given by

δ = arccos(1 − a) (8.5)

so that, if a is small enough, we have

a < cos 7δ, (8.6)

and for σ in S_{N}, we consider the sets

C_{1}(σ) = {σ′ ∈ S_{N} | |R(σ, σ′)| ≥ cos δ}
C(σ) = {σ′ ∈ S_{N} | |R(σ, σ′)| ≥ cos 3δ}
D(σ) = {σ′ ∈ S_{N} | cos 6δ ≤ |R(σ, σ′)| < cos 2δ}.

Thus C_{1}(σ) (resp. C(σ)) is the union of two balls of S_{N} for the metric d, of radius δ (resp. 3δ) that are centered at σ and −σ. One should keep in mind that μ might (and, in the case of G_{N}, will) be invariant by the symmetry σ ↦ −σ, so that all our constructions should respect this symmetry.

Definition 8.2. A set C(σ) such that

μ(C_{1}(σ)) ≥ ε^{\frac{1}{3}} (8.7)

is called a lump.

Thus, a lump is a small set (its diameter is ≤ 6δ) and it carries a significant mass. The motivation of D(σ) is that this is “nearly empty space” around a lump C(σ).

Lemma 8.3. For each lump C(σ) we have

μ(D(σ)) ≤ ε^{\frac{2}{3}} . (8.8)

Proof. By (8.3), (8.7) it suffices to prove that

C_{1}(σ) × D(σ) ⊂ T.

Consider σ′ ∈ C_{1}(σ), σ′′ ∈ D(σ). Without loss of generality we can assume R(σ, σ′) ≥ 0, R(σ, σ′′) ≥ 0. Thus d(σ′, σ) ≤ δ, 2δ ≤ d(σ, σ′′) ≤ 6δ, so that δ < d(σ′, σ′′) ≤ 7δ, i.e. cos 7δ ≤ R(σ′, σ′′) ≤ cos δ, and (σ′, σ′′) ∈ T using (8.6). □

Theorem 8.4. Under (8.3), (8.6), we can find a disjoint family (C_{α})_{α≤M} of lumps such that

μ ⊗ μ(\{(σ, σ′) | |R(σ′, σ′′)| ≥ a\}\ ⋃_{α≤M} C_{α}^{2}) ≤ 3ε^{\frac{1}{3}} . (8.9)

Comment. Here, and below, α is an integer. The content of the theorem is that essentially the only way one can have |R(σ, σ′)| ≥ a is that σ, σ′ belong to the same lump.

This of course contains information only if

μ ⊗ μ({(σ, σ′) | |R(σ, σ′)| ≥ a}) (8.10)

is not too small, or equivalently under (8.3), if

μ ⊗ μ({(σ, σ′) | |R(σ, σ′)| ≥ 1 − a})

is not too small.

Proof of Theorem 8.4. We consider a maximal disjoint family of lumps (C_{α})_{α≤M}. (It might very well happen that M = 0) and we prove (8.9). First, by (8.7) and since the lumps C_{α} are disjoint, we have Mε^{\frac{1}{3}} ≤ 1, so that M ≤ ε^{− \frac{1}{3}} . Next, if C_{α} = C(σ_{α}), by Lemma 8.3 we have μ(D(σ_{α})) ≤ ε^{\frac{2}{3}} , so that if D = ⋃_{α≤M} D(σ_{α}) we have μ(D) ≤ Mε^{\frac{2}{3}} ≤ ε^{\frac{1}{3}} . Thus to prove (8.9), it suffices to prove that

{(σ, σ′) ∈ S_{N}^{2} | |R(σ, σ′)| ≥ 1 − a}) ⊂ S′ ∪ (D × S_{N}) ∪ ( ⋃_{α≤M} C_{α}^{2}), (8.11)

where

S′ = \{(σ, σ′) | σ′ ∈ C_{1}(σ), μ(C_{1}(σ)) ≤ ε^{\frac{1}{3}} \}.

This implies (8.9) because μ ⊗ μ(S′) ≤ ε^{\frac{1}{3}} by Fubini’s theorem. To prove (8.11), consider (σ, σ′) ∈ S_{N}^{2}, with |R(σ, σ′)| ≥ 1 − a and assume that (σ, σ′) ∉ S′, that is, μ(C_{1}(σ)) ≥ ε^{\frac{1}{3}} . The maximality of the family (C_{α})_{α≤M} implies that for some α we have

C(σ) ∩ C(σ_{α}) ≠ ∅. (8.12)

Let us assume for clarity R(σ, σ_{α}) > 0, R(σ, σ′) > 0. Then by (8.12) we have d(σ, σ_{α}) ≤ 6δ. If d(σ, σ_{α}) ≥ 2δ, we are done because σ ∈ D(σ_{α}) ⊂ D. If d(σ, σ_{α}) ≤ 2δ, then, since R(σ, σ′) ≥ 1 − a, we have d(σ, σ′) ≤ δ, so that d(σ′, σ_{α}) ≤ 3δ and (σ, σ′) ∈ C_{α}^{2}. □

Challenge problem 8.5. Formulate and prove a statement showing that the lumps are essentially determined.

We now go back to the case of Gibbs’ measure. When (8.3) occurs for μ = G_{N}, under (8.6), we can construct the lumps (C_{α})_{α≤M} where ε is the left-hand side of (8.3). It is perfectly possible a priori that M = 0, that there exist no lumps, or that there exist few of them, and that they are far from exhausting all of G_{N}. To keep notations simple, it is better not to distinguish these various cases, and to adopt a procedure that exhausts all of the configuration space Σ_{N}. This will be done in a trivial way (this does not bring more information than Theorem 8.4). After having constructed the lumps (C_{α})_{α≤M} we continue the construction by enumerating all the remaining pairs {σ, −σ} as C_{M+1}, … until Σ_{N} is exhausted. This happens at a certain index α_{0}, and for α > α_{0} we set C_{α} = ∅; we thus define a sequence of lumps (C_{α})_{α≥1}. (Only the (C_{α})_{α≤M} actually deserve the name of lumps).

Theorem 8.6. There exists a number p_{0} such that if p ≥ p_{0}, then for β ≤ 2^{\frac{p}{2} −6}, we can find a decomposition (C_{α})_{α≥1} of Σ_{N} in disjoint sets with the following properties:

(8.13) Each set C_{α} is symmetric, i.e. σ ∈ C_{α} ⟹ −σ ∈ C_{α}.

σ, σ′ ∈ C_{α} ⟹ |R(σ, σ′)| ≥ 1 − 2^{− \frac{p}{4} +6}. (8.14)

E(G_{N}^{2}(\{(σ, σ′) | |R(σ, σ′)| ≥ 2^{− \frac{p}{4}} \} \ ⋃_{α≥1} C_{α}^{2})) ≤ exp(−\frac{N}{K} ). (8.15)

Proof. We apply the construction of Theorem 8.4 for each value of the disorder with a = 2^{− \frac{p}{4}} . To prove (8.14), we observe that σ, σ′ ∈ C_{α} ⟹ R(σ, σ′) ≥ cos 6δ where cos δ = 1 − a and that cos 6δ ≥ 1 − 2^{6}(1 − cos δ) for small δ. □

Of course one should repeat that all the information contained in this statement is contained in (8.15); so for example if it is already true that

E G_{N}^{2}(\{σ, σ′ | |R(σ, σ′)| ≥ 2^{− \frac{p}{4}} \}) ≤ exp(−\frac{N}{K} ) (8.16)

(as in the case where β < β_{p}) a decomposition such as that of Theorem 8.6 is uninteresting. On the other hand, things become interesting as soon a (8.16) fails. Indeed, in that case, since the lumps are disjoint, we have ∑_{α≥1} G_{N}(C_{α}) = 1, so that

E max_{α} G_{N}(C_{α}) ≥ E(∑_{α≥1} G_{N}(C_{α})^{2})

= E(G_{N} ⊗ G_{N}(⋃ C_{α}^{2}))

≥ E(G_{N} ⊗ G_{N}(\{(σ, σ′) | |R(σ, σ′)| ≥ 2^{− \frac{p}{4}} \})) − exp(−\frac{N}{K} ),

where we have used (8.15) in the last line. The ultimate goal (for which we must refer to [T7]) is to show that the “total mass of the macroscopic lumps is one”.

We can reformulate (8.15) by saying that if σ, σ′ do not belong to the same lump, then (generically) we have |R(σ, σ′)| ≤ 2^{− \frac{p}{4}} . The rest of the chapter is devoted to show that in fact not only do we have |R(σ, σ′)| ≤ 2^{− \frac{p}{4}} , but in fact R(σ, σ′) ≃ 0. The proof will rely upon the cavity method, and our first task is to set up notation to do this. Given σ in Σ_{N}, we set

T (σ) = (σ_{i_{1}} … σ_{i_{p−1}}),

where the index ranges over all families i_{1} < i_{2} < ⋯ < i_{p−1}. We consider a family of independent N (0, 1) variables,

g = (g_{i_{1}...i_{p−1}}), (8.17)

with the same range of indexes as above, and we define

g · T (σ) = ∑ g_{i_{1}...i_{p−1}}σ_{i_{1}} … σ_{i_{p−1}}

where the range of the index is as above.

We recall the Hamiltonian H_{N}(σ) of (4.1). Consider σ_{N+1} ∈ {−1, 1}. Then we have

(8.18)  −βH_{N}(σ) + σ_{N+1}β(\frac{p!}{2N^{p−1}} )^{\frac{1}{2}} T (g) · σ

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

where the sum is over 1 ≤ i_{1} < ⋯ < i_{p} ≤ N + 1, and where

g_{i_{1}...i_{p−1}i_{p}} = g_{i_{1}...i_{p−1}}

when i_{p} = N + 1. If we recall the notation

ϱ = (σ_{i})_{i≤N+1} ∈ Σ_{N+1},

we see that the right-hand side of (8.18) is

−β′H_{N+1}(ϱ),

where

β′ = (\frac{N + 1}{N})^{\frac{p−1}{2}} . (8.19)

That is, (8.18) is the Hamiltonian of an (N + 1)-spin system at a slightly lower temperature. We denote by G_{N+1} = G_{N+1}(β′) the corresponding Gibbs’ measure and by ⟨ · ⟩′ average with respect to this measure: we will use the same conventions as for the SK model. Then (5.19) holds, except that now

E = E(σ, σ_{N+1}) = exp σ_{N+1}β(\frac{p!}{2N^{p−1}} )^{\frac{1}{2}} g · T (σ) (8.20)

and in (5.22) we now have

E_{k} = exp(∑_{l≤k} σ_{N+1}^{l}β(\frac{p!}{2N^{p−1}} )^{\frac{1}{2}} g · T (σ^{l})). (8.21)

Even though everything looks pretty much like in the case p = 2, it is good to observe that

(8.22) E(((\frac{p!}{2N^{p−1}} )^{\frac{1}{2}} g · T (σ))^{2}) = \frac{p!}{2N^{p−1}} \binom{N}{p − 1}

= \frac{p!}{2N^{p−1}} \frac{N(N − 1) … (N − p + 1)}{(p − 1)!}

≃ \frac{p}{2}.

Thus as p increases, the exponent in E_{k} gets larger, and is harder to control. This is unavoidable: as p increases, there is less relationship between the structure of the (N +1)-spin system and the structure of the N-spin system. The good news is that, on the other hand, the correlation between the variables g · T (σ) decreases.

Lemma 8.7. We have

|T (σ) · T (σ′) − \frac{(σ · σ′)^{p−1}}{(p − 1)!} | ≤ K(p)N^{p−2}. (8.23)

Proof. We have, setting a_{i} = σ_{i}σ′_{i}

T (σ) · T (σ′) = ∑_{i_{1}<···<i_{p−1}} σ_{i_{1}} … σ_{i_{p−1}}σ′_{i_{1}} … σ′_{i_{p−1}}

= ∑_{i_{1}<···<i_{p−1}} a_{i_{1}} … a_{i_{p−1}}

= \frac{1}{(p − 1)!} ∑_{d} a_{i_{1}} … a_{i_{p−1}},

where ∑_{d} means that the summation is over indices i_{1}, … i_{p−1} all different (but not necessarily ordered). Now

(σ · σ′)^{p−1} = (∑_{i≤N} a_{i})^{p−1} = ∑ a_{i_{1}} … a_{i_{p−1}},

where the summation is over all choices of i_{1} ≤ N, …, i_{p} ≤ N. For each i, we have |a_{i}| ≤ 1 and there are at most K(p)N^{p−2} choices of i_{1}, …, i_{p−1} that are not all different. □ Theorem 8.8. There exists a number p_{0} such that if p ≥ p_{0} and β ≤ 2^{\frac{p}{2} −6}, we have

E⟨R(σ, σ′)^{2}1_{\{|R(σ,σ′)|≤1/2\}}⟩ ≤ \frac{K(p)}{N}. (8.24)

We will show that

β ≤ 2^{\frac{p}{2} −6} ⟹ A_{N}(β) ≤ \frac{K(p)}{N} (8.25)

where A_{N}(β) = E⟨R(σ^{1}, σ^{2})^{2}1_{{|R(σ^{1},σ^{2})|≤c}}⟩, (8.26) and where c = 2^{− \frac{p}{4} +1}. This implies (8.24) by Theorem 4.13.

We use symmetry between sites to see that A_{N}(β) = E⟨σ_{N}^{1}σ_{N}^{2}R(σ^{1}, σ^{2})1_{{|R(σ^{1},σ^{2})|≤c}}⟩, (8.27) and, changing N into N + 1 and β into β′, we get A_{N+1}(β′) = E⟨σ_{N+1}^{1}σ_{N+1}^{2}R(ϱ^{1}, ϱ^{2})1_{{|R(ϱ^{1},ϱ^{2})|≤c}}⟩. (8.28) We have set ϱ^{l} = (σ_{1}^{l}, …, σ_{N+1}^{l}) (8.29) so that

R(ϱ^{1}, ϱ^{2}) = \frac{N}{N + 1}R(σ^{1}, σ^{2}) + \frac{σ_{N+1}^{1}σ_{N+1}^{2}}{N + 1} . (8.30)

This implies that

A_{N+1}(β′) ≤ \frac{1}{N + 1} + \frac{N}{N + 1} E⟨σ_{N+1}^{1}σ_{N+1}^{2}R(σ^{1}, σ^{2})1_{\{|R(ϱ^{1},ϱ^{2})|≤c\}}⟩′. (8.31)

Now (8.30) implies

|R(ϱ^{1}, ϱ^{2}) − R(σ^{1}, σ^{2})| ≤ \frac{2}{N + 1}.

Thus

\frac{2}{N + 1} ≤ \frac{c}{2} ⟹ |1_{\{|R(ϱ^{1},ϱ^{2})|≤c\}} − 1_{\{|R(σ^{1},σ^{2})|≤c\}}| ≤ 1_{\{ \frac{c}{2} ≤|R(ϱ^{1},ϱ^{2})|≤ \frac{3c}{2} \}},

and thus, if β′ ≤ 2^{\frac{p}{2} −6}, we deduce from Theorem 4.13 that

(8.32)  A_{N+1}(β′) ≤ \frac{1}{N + 1} + exp(−\frac{N}{K} )

+ \frac{N}{N + 1} E⟨σ_{N+1}^{1}σ_{N+2}^{2}R(σ^{1}, σ^{2})1_{\{|R(σ^{1},σ^{2})|≤c\}}⟩.

Setting

f = f(σ^{1}, σ^{2}) = R(σ^{1}, σ^{2})1_{{|R(σ^{1},σ^{2})|≤c}},

we are now in a position to use the cavity method to write

E⟨σ_{N+1}^{1}σ_{N+1}^{2}f⟩′ = E \frac{⟨f Av ε_{1}ε_{2}E_{2}⟩}{Z^{2}}, (8.33)

where, as usual, ε_{l} = σ_{N+1}^{l}. The family g of (8.17) is independent of the disorder involved in G_{N} (the variables g_{i_{1},...,i_{p}}) so we will try to evaluate

E_{g} \frac{⟨f Av ε_{1}ε_{2}E_{2}⟩}{Z^{2}}. (8.34)

This will be done using Kahane’s principle. The proof will resemble that of Proposition 6.6. Let us write

μ(σ) = G_{N}({σ}).

For x = (x(σ)) ∈ ℝ^{Σ_{N}} we write

F_{1}(x) = ∑ μ(σ^{1})μ(σ^{2})f(σ^{1}, σ^{2}) Av ε_{1}ε_{2} exp ∑_{l≤2} ε_{l}x(σ^{l})
= ∑ μ(σ^{1})μ(σ^{2})f(σ^{1}, σ^{2}) sh x(σ^{1}) sh x(σ^{2})
F_{2}(x) = ∑ μ(σ^{1})μ(σ^{2}) Av exp ∑_{l≤2} ε_{l}x(σ^{l})

= (∑ μ(σ) ch x(σ))^{2}

where the summations are over all values of σ^{1}, σ^{2}. We consider

F (x) = \frac{F_{1}(x)}{F_{2}(x)}.

Consider the Gaussian family v = (v(σ)) where

v(σ) = β(\frac{p!}{2N^{p−1}} )^{\frac{1}{2}} g · T (σ), (8.35)

so that the quantity (8.34) is E_{g} F (v).

We consider a centered Gaussian family u such that

(8.36)  (σ, σ′) ∈ ⋃_{α} C_{α}^{2} ⟹ E u(σ)u(σ′) = E v(σ)v(σ′)
(8.37)  (σ, σ′) ∉ ⋃_{α} C_{α}^{2} ⟹ E u(σ)u(σ′) = 0.

The first condition means that u and v have the same covariance inside a lump, while the second condition means that different lumps are independent for u.

It should be obvious that such a family exists. The idea of (8.36) is that we do not disturb anything inside a lump, while making the lumps independent. In this manner, it is not an obstacle to the proof that we do not know what happens inside the lumps.

Lemma 8.9. E_{g} F (u) = 0.

Proof. We observe that we have

(8.38)  F_{1}(x) = ∑ f(σ^{1}, σ^{2})μ(σ^{1})μ(σ^{2}) sh x(σ^{1}) sh x(σ^{2})
(8.39)  F_{2}(x) = ∑ μ(σ^{1})μ(σ^{2}) ch x(σ^{1}) ch x(σ^{2}).

We will prove that, given σ^{1}, σ^{2} we have E_{g} A(u, σ^{1}, σ^{2}) = 0 where

A(u, σ^{1}, σ^{2}) = \frac{f(σ^{1}, σ^{2}) sh u(σ^{1}) sh u(σ^{2})}{F_{2}(u)}.

Certainly we can assume |R(σ^{1}, σ^{2})| ≤ \frac{1}{2}, for otherwise f(σ^{1}, σ^{2}) = 0. Consider α such that σ^{1} ∈ C_{α}. Then σ^{2} ∉ C_{α}, for otherwise |R(σ^{1}, σ^{2})| > 1/2. Consider the Gaussian family u_{α} given by

u_{α}(σ) = u(σ) if σ ∉ C_{α}
u_{α}(σ) = −u(σ) if σ ∈ C_{α}.

Then u_{α} and u have the same distribution by (8.36), (8.37) so that

E_{g} A(u, σ^{1}, σ^{2}) = E_{g} A(u_{α}, σ^{1}, σ^{2}). (8.40)

But we have u_{α}(σ^{1}) = −u(σ^{1}), u_{α}(σ^{2}) = u(σ^{2}), F_{2}(u_{α}) = F_{2}(u) so that

A(u, σ^{1}, σ^{2}) = −A(u_{α}, σ^{1}, σ^{2}),

and combining with (8.40) this proves the lemma. □

To estimate E_{g} F (v) − E_{g} F (u), we will use (6.38). That is, we set

u_{t} = \sqrt{1 − t} u + \sqrt{t} v
θ(t) = E_{g} F (u_{t})

so that

E_{g} F (v) = E_{g} F (v) − E_{g} F (u) = ∫_{0}^{1} θ′(t) dt. (8.41)

To compute θ′(t), it will be convenient to introduce the quantity B(σ^{1}, σ^{2}) given by

B(σ^{1}, σ^{2}) = \frac{β^{2}p!}{2N^{p−1}} T (σ^{1}) · T (σ^{2})

when σ^{1}, σ^{2} do not belong to the same lump, and B(σ^{1}, σ^{2}) = 0 otherwise.

Lemma 8.10. We have

θ′(t) = ∑_{r=3}^{6} E_{g} F_{r}(u_{t}) (8.42)

where

F_{3}(x) = \frac{1}{F_{2}(x)} ∑ f(σ^{1}, σ^{2})μ(σ^{1})μ(σ^{2})B(σ^{1}, σ^{2}) ch x(σ^{1}) ch x(σ^{2}) (8.43)

(the summation is over σ^{1}, σ^{2} in Σ_{N}),

(8.44)  F_{4}(x) = −\frac{F_{1}(x)}{F_{2}^{2}(x)} ∑ μ(σ^{1})μ(σ^{2})B(σ^{1}, σ^{2}) sh x(σ^{1}) sh x(σ^{2})
(8.45)  F_{5}(x) = −\frac{4}{F_{2}^{2}(x)} ∑ μ(σ^{1})μ(σ^{2})μ(σ^{3})μ(σ^{4})f(σ^{1}, σ^{2})
× B(σ^{2}, σ^{3}) sh x(σ^{1}) ch x(σ^{2}) sh x(σ^{3}) ch x(σ^{4})
(8.46)  F_{6}(x) = \frac{4F_{1}(x)}{F_{2}^{3}(x)} ∑ μ(σ^{1})μ(σ^{2})B(σ^{1}, σ^{2}) sh x(σ^{1}) sh x(σ^{2}).

Proof. We use (6.46). Now the indexes a, b belong to Σ_{N}. If a = η^{1}, b = η^{2}, we have

E(v_{a}v_{b} − u_{a}u_{b}) = B(η^{1}, η^{2}). (8.47)

Since F is a ratio, the quantity

\frac{1}{2} ∑ E(v_{a}v_{b} − u_{a}u_{b})\frac{∂^{2}F}{∂x_{a}∂x_{b}}

is the sum of four terms, as in the formula

\frac{∂^{2}}{∂x_{a}∂x_{b}} (\frac{F_{1}}{F_{2}} ) = \frac{∂^{2}F_{1}}{∂x_{a}∂x_{b}} \frac{1}{F_{2}} − F_{1} \frac{∂^{2}F_{2}}{∂x_{a}∂x_{b}} \frac{1}{F_{2}^{2}}

− (\frac{∂F_{1}}{∂x_{a}} \frac{∂F_{2}}{∂x_{b}} + \frac{∂F_{1}}{∂x_{b}} \frac{∂F_{2}}{∂x_{a}} ) \frac{1}{F_{2}^{2}} + 2 F_{1} (\frac{∂F_{2}}{∂x_{a}} \frac{∂F_{2}}{∂x_{b}} ) \frac{1}{F_{2}^{3}} .

These terms are F_{3} to F_{6} respectively. Let us see how (8.45) arises. For a = η_{1}, we have

\frac{∂F_{1}}{∂x_{a}} = \begin{matrix}∑ \\ σ & ^{2} \end{matrix} μ(η^{1})μ(σ^{2})f(η^{1}, σ^{2}) ch x(η^{1}) sh x(σ^{2})

+ \begin{matrix}∑ \\ σ & ^{1} \end{matrix} μ(σ^{1})μ(η^{1})f(σ^{1}, η^{1}) sh x(σ^{1}) ch x(η^{1})

= 2 \begin{matrix}∑ \\ σ & ^{2} \end{matrix} μ(η^{1})μ(σ^{2})f(η^{1}, σ^{2}) ch x(η^{1}) sh x(σ^{2})

by symmetry. Similarly, if b = η^{2},

\frac{∂F_{2}}{∂x_{b}} = 2 \begin{matrix}∑ \\ σ & ^{1} \end{matrix} μ(σ^{1})μ(η^{2}) ch x(σ^{1}) sh x(η^{2}).

Writing \frac{∂F_{1}}{∂x_{a}} \frac{∂F_{2}}{∂x_{b}} as a double sum, summing over η^{1}, η^{2} we get (8.45). □ Corollary 8.11. We have

|θ′(t)| ≤ 10β^{2}p2^{−(p−3)( \frac{p}{4} −1)} E_{g} \frac{F_{7}(u_{t})}{F_{2}(u_{t})} + S (8.48)

where

F_{7}(x) = ∑ μ(σ^{1})μ(σ^{2})f^{2}(σ^{1}, σ^{2}) ch x(σ^{1}) ch x(σ^{2})

and

E S ≤ \frac{K(p)}{N}.

Proof. We use the inequalities |sh x| ≤ ch x and |f| ≤ 1 to see that

∑_{r=3}^{6} |F_{r}(x)| ≤ 10\frac{F_{8}(x)}{F_{2}(x)}, (8.49)

with

F_{8}(x) = ∑ μ(σ^{1})μ(σ^{2})B^{2}(σ^{1}, σ^{2}) ch x(σ^{1}) ch x(σ^{2}).

Now, from (8.23) we see that

|B(σ^{1}, σ^{2})| ≤ \frac{β^{2}p}{2}|R^{p−1}(σ^{1}, σ^{2})| + \frac{K(p)}{N}.

Since B(σ^{1}, σ^{2}) = 0 when (σ^{1}, σ^{2}) ∈ ∪C_{α}^{2} we have

|B(σ^{1}, σ^{2})| ≤ \frac{β^{2}p}{2}|R(σ^{1}, σ^{2})|^{p−1}1_{\{|R|≤c\}} + \frac{β^{2}p}{2}1_{C} + \frac{K(p)}{N} (8.50)

where

C = {(σ^{1}, σ^{2}) | |R(σ^{1}, σ^{2})| ≥ c; (σ^{1}, σ^{2}) ∉ ∪C_{α}^{2}}.

Thus,

|B(σ^{1}, σ^{2})| ≤ \frac{β^{2}p}{2}c^{p−3}R(σ^{1}, σ^{2})^{2}1_{\{|R|≤c\}} + \frac{β^{2}p}{2}1_{C} + \frac{K(p)}{N}. (8.51)

Using elementary inequalities such as (a+b)^{2} ≤ 2(a^{2}+b^{2}), \sqrt{a + b} ≤ \sqrt{a}+\sqrt{b}, etc., from (8.49), (8.51) we see that (since F_{2} ≥ 1)

(8.52) ∑_{r=3}^{6} |F_{r}(x)| ≤ 10β^{2}pc^{p−2} \frac{F_{7}(x)}{F_{2}(x)} + \frac{K(p)}{N}

+ β^{2}p ∑ μ(σ^{1})μ(σ^{2})1_{C}(σ^{1}, σ^{2}) ch x(σ^{1}) ch x(σ^{2}).

We now appeal to (8.39) to obtain (8.48) with

S = \frac{K(p)}{N} + β^{2}p E_{g}(∑ μ(σ^{1})μ(σ^{2})1_{C}(σ^{1}, σ^{2}) ch u_{t}(σ^{1}) ch u_{t}(σ^{2})).

It follows from (8.15) that the expected value of S (in the disorder) is ≤ K(p)/N. □ Proposition 8.12. If p ≥ p_{0}, β ≤ 2^{\frac{p}{2} −6} we have θ′_{1}(t) ≥ −θ_{1}(t) − S_{1}, (8.53) where

θ_{1}(t) = E_{g} \frac{F_{7}(u_{t})}{F_{2}(u_{t})} (8.54)

and

E S_{1} ≤ \frac{K(p)}{N}.

Proof. It is similar to the proof of Corollary 8.11. □ Proof of Theorem 8.8. First, we claim that (8.53) implies 0 ≤ t ≤ 1 ⟹ θ_{1}(t) ≤ e^{1−t}θ_{1}(t) + S_{1}(1 − t). (8.55) Indeed, if u = 1 − t

ξ(u) = θ_{1}(1 − u) − e^{u}θ_{1}(1) − S_{1}u

satisfies ξ(0) = 0, ξ′(u) ≤ ξ(u), so that ξ(u) ≤ 0.

If we combine (8.55) with (8.54), (8.48), we see that for p ≥ p_{0}, β ≤ 2^{\frac{p}{4} −6} we have

|θ′(t)| ≤ \frac{1}{2}θ_{1}(1) + S + S_{1}, (8.56)

and, combining with (8.38), we have

|E_{g} \frac{⟨f Av ε_{1}ε_{2}E_{2}⟩}{Z^{2}} | = |E_{g} F (v)| ≤ \frac{1}{2} E_{g} \frac{F_{7}(v)}{F_{1}(v)} + S + S_{1}.

Recalling (8.33), and taking expectation in the disorder, we have

(8.57)  |E⟨σ_{N+1}^{1}σ_{N+1}^{2}f⟩′| ≤ \frac{1}{2} E \frac{F_{7}(v)}{F_{1}(v)} + \frac{K(p)}{N}

= \frac{1}{2} E⟨R(σ^{1}, σ^{2})^{2}1_{\{R≤c\}}⟩′ + \frac{K(p)}{N}.

By an argument similar to that in the proof of (8.33), we see that E⟨σ_{N+1}^{1}σ_{N+1}^{2}f⟩ and E⟨R(σ^{1}, σ^{2})^{2} 1_{{R≤c}}⟩′ differ from A_{N+1}(β′) by at most K(p)/N, so that (8.57) yields

A_{N+1}(β′) ≤ \frac{K(p)}{N}.  □

Challenge problem 8.13. Improve Theorem 8.8 to an exponential inequality.

If we combine Theorems 4.13 and 8.8, we see that, if

B = ⋃_{α} C_{α}^{2},

then

|E⟨R(σ^{1}, σ^{2})^{p}1_{B}⟩ − E⟨R(σ^{1}, σ^{2})^{p}⟩| ≤ \frac{K(p)}{N}. (8.58)

Also, since (1 − 2^{− \frac{p}{4}} )^{p} ≥ \frac{1}{2} for large p, we have

σ, σ′ ∈ C_{α} ⟹ \frac{1}{2} ≤ R(σ^{1}, σ^{2})^{p} ≤ 1

so that, if

w_{α} = G_{N}(C_{α}), (8.59)

\frac{1}{2} ∑ w_{α}^{2} ≤ ⟨R(σ^{1}, σ^{2})^{p}⟩ ≤ ∑ w_{α}^{2},

and, combining with (8.58),

\frac{1}{2} E ∑ w_{α}^{2} − \frac{K(p)}{N} ≤ E⟨R(σ^{1}, σ^{2})^{p}⟩ ≤ E ∑ w_{α}^{2} + \frac{K(p)}{N}. (8.60)

This allows us to obtain information about the weights (8.59). The following is a generalization of Lemma 5.13.

Lemma 8.14. We have

| \frac{d}{dβ} p_{N}(β) − \frac{β}{2} (1 − E⟨R(σ, σ′)^{p}⟩)| ≤ \frac{K}{N} . (8.61)

Proof. The same computation as in Lemma 5.13 gives

\frac{dp_{N}}{dβ} = \frac{βp!}{2N^{p}} ∑_{i_{1}<···<i_{p}}(1 − E⟨σ_{i_{1}}σ′_{i_{1}} … σ_{i_{p}}σ′_{i_{p}}⟩).

Changing p into p + 1, (8.23) shows that

|\frac{p!}{N^{p}} ∑_{i_{1}<···<i_{p}} σ_{i_{1}}σ′_{i_{1}} … σ_{i_{p}}σ′_{i_{p}} − R(σ, σ′)^{p}| ≤ \frac{K}{N} .  □

Corollary 8.15. If β > 2\sqrt{log 2}, then

lim inf_{N→∞} E⟨R(σ, σ′)^{p}⟩ > 0, (8.62)

and consequently, if β ≤ 2^{\frac{p}{4} −6},

lim inf_{N→∞} E ∑_{α≥1} w_{α}^{2} > 0. (8.63)

Proof. (8.62) follows from (8.61) and Lemma 4.14 and this implies (8.63) by (8.60). □

Comment. Observe that (8.62) and Theorem 8.6 prove (7.43).

Since ∑ w_{α} = 1, a consequence of (8.63) is that lim inf_{N→∞} E max w_{α} > 0: For large N, with positive probability, there exists at least one lump of mass of order 1. Our theory of lumps is not empty. (It is in fact shown in [T7] that the lumps do carry all the mass).

Proof of Proposition 7.11. We write

A_{N,k}(β) = E⟨R(σ, σ′)^{k}⟩.

Proceeding as in (8.37), we obtain

A_{N+1,2}(β′) ≤ \frac{1}{N + 1} + \frac{N}{N + 1} E \frac{1}{Z^{2}} ⟨R(σ^{1}, σ^{2}) Av ε_{1}ε_{2}E_{2}⟩. (8.64)

We note that ⟨R(σ^{1}, σ^{2}) Av ε_{1}ε_{2}E_{2}⟩ is a square (as in the proof of Proposition 5.2) so that, since Z ≥ 0,

A_{N+1,2}(β′) ≤ \frac{1}{N + 1} + E⟨R(σ^{1}, σ^{2}) Av ε_{1}ε_{2}E_{2}⟩. (8.65)

We recall that

E_{2} = exp β(\frac{4!}{2N^{3}} )^{\frac{1}{2}} ∑_{l≤2} ε_{l}g · T (σ^{l}),

so that

E_{g} Av ε_{1}ε_{2}E_{2} = B sh(\frac{β^{2}4!}{2N^{3}}T (σ^{1}) · T (σ^{2}))

for

B = exp \frac{β^{2}4!}{2N^{3}} \frac{N(N − 1)(N − 2)}{3!} ≤ exp 2β^{2}.

Assuming β ≤ 3, and using Lemma 8.7, we then have

A_{N+1,2}(β′) ≤ \frac{L_{1}}{N + 1} + L_{1}A_{N,4}(β). (8.66)

To prove Proposition 7.11, we now argue by contradiction. If this proposition was wrong, we could say that

∃N_{0}, ∀N ≥ N_{0}, ∀β < 3,

A_{N,2}(β) ≤ \frac{2L_{1}}{N} ⟹ A_{N,4}(β) ≤ \frac{1}{4}A_{N,2}(β). (8.67)

Let us define

B(N) = sup\{β > 0 | A_{N,2}(β) ≤ \frac{2L_{1}}{N} \}.

The function A_{N,2}( ·) is continuous, and A_{N,2}(0) = 1/N, so that

A_{N,2}(B(N)) = \frac{2L_{1}}{N} .

Thus, if β ≤ B(N), β ≤ 3 by (8.67) we have

A_{N,4}(β) ≤ \frac{1}{4}A_{N,2}(β) ≤ \frac{1}{2} \frac{L_{1}}{N} ,

so that by (8.66)

A_{N+1,2}(β′) ≤ \frac{L_{1}}{N + 1} + \frac{1}{2} \frac{L_{1}}{N} ≤ \frac{2L_{1}}{N + 1},

and this means that β′ ≤ B(N + 1). Thus we have proved that

B(N) ≤ 3 ⟹ B(N + 1) ≥ (\frac{N + 1}{N})^{\frac{3}{2}} B(N).

This implies that there are arbitrarily large values of N for which B(N) ≥ 3. This however contradicts (8.57). □