# A spectral gap throughout the high-temperature Sherrington-Kirkpatrick phase

*An expository walkthrough, written to be readable with high-school mathematics.*

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## 0. Prologue: the paper in one sentence

If you know what a magnet is, and you know that flipping one little magnet influences its neighbours, here is the entire paper in one sentence:

> **In a special random magnet called the Sherrington-Kirkpatrick (SK) model, for every fixed temperature above a threshold (inverse temperature beta < 1), and for a random choice of the couplings, re-rolling individual spins at unit rate pulls every observable back to its average at a rate bounded below by a constant that does not depend on the number of spins n -- and this holds for every coupling matrix except a vanishing fraction of them.**

The phrase "rate bounded below by a constant" is a *spectral gap*. Making that precise, proving it for all functions at once, on a single random matrix, is the entire job. The rest of this exposition is a map of how the proof works. The paper is technically long; it is compressed here heavily. Every logical step of the proof appears; only the routine chasing of constants is summarized.

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## 1. The cast of characters

### 1.1 Spins

A *spin* is a little arrow pointing up or down. Take n of them, so a *spin configuration* is a point

$$ x = (x_1, \dots, x_n) \in \{-1, +1\}^n. $$

We may flip a single spin i, writing x^(i) for the configuration with x_i replaced by -x_i.

### 1.2 The SK interaction matrix

The magnet is defined by a symmetric n x n matrix J of *couplings*, with all diagonal entries zero and, for i < j, independent Gaussian entries

$$ J_{ij} \sim N(0, j/n), \qquad j := \beta^2. $$

Each pair of spins interacts weakly (standard deviation of order j/n), but there are about n^2/2 pairs, so the total interaction felt by one spin is of order one. The entries are random and signed: some pairs want to align, some want to anti-align. This is what makes the model a *spin glass* rather than an ordinary magnet. The parameter beta is the *inverse temperature*; j = beta^2 is its square.

The diagonal of J is zero, and the diagonal never matters: since x_i^2 = 1, adding any diagonal matrix to J adds only a constant to the energy below, leaving the Gibbs measure unchanged. We will sometimes temporarily put a Gaussian diagonal on J for matrix arguments, and remove it again.

### 1.3 The Gibbs measure and temperature

Temperature enters through a probability distribution on configurations. Fix an *external field* h in R^n (a small push on each spin). The *Gibbs measure* (or *posterior*) is

$$ \mu_h(x) = \frac{1}{Z_h} \exp\Big( \tfrac12 x^{\mathsf T} J x + h^{\mathsf T} x \Big), \qquad Z_h = \sum_{x} \exp\Big( \tfrac12 x^{\mathsf T} J x + h^{\mathsf T} x \Big). $$

The theorem concerns the *zero-field* measure mu_0. Smaller beta means higher temperature: the energy ½ x^T J x matters less and the measure becomes more spread out. The special temperature is beta = 1; we work throughout at fixed beta < 1.

### 1.4 The heat-bath update

The dynamics we care about is *heat-bath* (Glauber) updating. Look at spin i and freeze all other spins. Under mu_h the spin i is independent of the rest in a simple way, because the energy splits:

$$ \tfrac12 x^{\mathsf T} J x = x_i \cdot (Jx)_i + (\text{terms not involving } x_i), $$

using that J has zero diagonal (so (Jx)_i = sum_{k != i} J_{ik} x_k). Hence, conditionally on all other spins,

$$ \text{Prob}(x_i = +1) = \frac{e^{h_i + (Jx)_i}}{e^{h_i + (Jx)_i} + e^{-h_i - (Jx)_i}}, $$

a logistic (heat-bath) rule. The conditional *mean* and *variance* of x_i are therefore

$$ m_i := \tanh\big( h_i + (Jx)_i \big), \qquad v_i := \sech^2\big( h_i + (Jx)_i \big) = 1 - m_i^2. $$

The heat-bath update redraws x_i from this two-point distribution, leaving the others fixed.

### 1.5 The conditional projection, the Dirichlet form, and the gap

Write P_i^h for "conditional average over spin i given the others":

$$ (P_i^h f)(x) = \textstyle\frac{f(x) + f(x^{(i)})}{2} + x_i m_i \cdot \frac{f(x) - f(x^{(i)})}{2x_i}. $$

This is the orthogonal projection onto functions that do not depend on x_i, and it is symmetric in L^2(mu_h). The natural measure of "how much does resampling site i change f" is the squared change E_{mu_h}(f - P_i^h f)^2. Summing over all sites defines the *unscaled Dirichlet form*

$$ \mathcal D_h(f) = \sum_{i=1}^n \mathbb E_{\mu_h} (f - P_i^h f)^2. $$

The *generator* of the continuous-time process in which each site rings at rate one is

$$ \mathcal L_h = \sum_i (P_i^h - I), $$

self-adjoint and negative semidefinite in L^2(mu_h), satisfying the integration-by-parts identity

$$ \mathbb E_{\mu_h}\big[ f\, (-\mathcal L_h f) \big] = \mathcal D_h(f). $$

The *spectral gap* of this generator is

$$ \text{gap}_h = \inf_{\substack{f \text{ nonconstant}}} \frac{\mathcal D_h(f)}{\operatorname{Var}_{\mu_h}(f)}. $$

This gap is the reciprocal of the slowest relaxation time of the process: if gap_h >= gamma, every observable relaxes to its mean at rate at least gamma. A gap bounded below by a constant (independent of n) means the system forgets its past in O(1) time.

The unscaled Dirichlet form is convenient because it *does not* grow with n; a function that depends on one spin has Theta(1) Dirichlet form and Theta(1) variance. Bounding the ratio by a constant is exactly the goal.

### 1.6 The discrete-time chain

A discrete attempt chooses one site uniformly and applies P_i^h. Its transition operator is P_h = n^{-1} sum_i P_i^h, so

$$ \mathcal L_h = n(P_h - I). $$

Thus the discrete chain's gap is the continuous gap divided by n. A constant unscaled gap therefore gives a discrete gap of order 1/n, i.e. mixing in order n attempts. All the work is in the unscaled gap; the discrete chain is a one-line conversion.

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## 2. The question, and why it is hard

### 2.1 Where a gap comes from

Resampling one site changes the law only where that site is not already pinned down by the rest. The local quantity controlling this is the *conditional variance* v_i = 1 - m_i^2: if the neighbours of spin i overwhelmingly force it up or down, then v_i is small and updating site i barely matters; if the neighbours leave it nearly free, v_i is close to 1 and updating site i matters a lot.

The unscaled Dirichlet form literally averages these local changes, and the elementary identity

$$ \mathcal D_h(f) = \mathbb E_{\mu_h}\Big[ \sum_i v_i\,(d_i f)^2 \Big], \qquad d_i f(x) = \frac{f(x) - f(x^{(i)})}{2 x_i}, $$

shows this explicitly: d_i f is the discrete derivative of f in direction i, and it is weighted by the local variance v_i. So the gap question becomes: **is the weighted discrete gradient always large compared with the variance?**

### 2.2 The naive route and where it stalls

A first instinct is Dobrushin's uniqueness argument. The conditional means m_i depend on the other spins, and the *influence* of spin j on the mean of spin i is (up to sign) J_{ij} v_j, which is at most |J_{ij}|. Assembling these into an influence matrix and demanding that its spectral radius be less than 1 gives a Poincare inequality (this is Wu's theorem, 2006). For SK that spectral radius is essentially the operator norm of J, which is about 2 sqrt j = 2 beta. So this route covers only beta < 1/2 at best. It is a real theorem, but it stops well short of beta = 1.

The obstruction is real: at low temperature the couplings form a dense, signed mesh, individual influences are not small, and the "one site at a time" picture genuinely struggles. What rescues the argument between beta = 1/2 and beta = 1 is the *sign structure* of J together with a completely different idea, namely observing the spins through noise and controlling the *posterior* measures. That is the engine of this paper.

### 2.3 What "for all functions at once" demands

The theorem must hold for every real function f on the cube simultaneously, for a *single random matrix* J, with probability tending to one over J. It is not enough to prove an estimate on average over J, nor to prove it for linear functions a^T x alone. The proof must pin down the gap on one matrix event of probability tending to one, uniformly over an n-dependent, huge family of functions. This is why essentially every estimate below is stated as "on a typical J, simultaneously for all ...".

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## 3. The main theorem

**Theorem (spectral gap throughout the high-temperature phase).** Fix 0 < beta < 1 and put j = beta^2. Let J be a symmetric n x n matrix with zero diagonal and, for i < j, independent J_{ij} ~ N(0, j/n). Let mu_0 be the zero-field Gibbs measure. Then there is a finite constant C_beta, depending only on beta, such that

$$ \mathbb P_J\Big\{ \operatorname{Var}_{\mu_0}(f) \le C_\beta\, \mathcal D_0(f) \ \text{ for every } f : \{-1,1\}^n \to \mathbb R \Big\} \longrightarrow 1 \qquad (n \to \infty). $$

Equivalently, the unscaled spectral gap of the heat-bath generator is at least 1/C_beta with probability tending to one. Consequently the discrete-time heat-bath chain that picks a uniform site at each step has spectral gap at least 1/(C_beta n) with probability tending to one.

Three remarks on the statement:

1. *The constant depends on beta, not on n.* It may blow up as beta approaches 1, and it is not optimized. The point is that it is finite for each fixed beta < 1.
2. *beta is fixed before n grows.* The theorem is a statement about one fixed temperature and a large magnet.
3. *The theorem concerns mu_0*, but the proof runs through Gibbs measures at *external fields* h = (external field), because those are exactly the posterior measures produced by observing spins through noise.

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## 4. The high-temperature phase, and what was known

### 4.1 Why beta = 1

At beta < 1 the model is in its *high-temperature (paramagnetic) phase*: the free energy is replica-symmetric, and the Gibbs measure has a bounded spin covariance. At beta = 1 a transition of spin-glass type begins (the AT line, in the physicists' language), and the picture changes. This paper lives strictly in the high-temperature phase and does not claim anything at or beyond beta = 1. The entire difficulty is to reach *all the way up to* 1, not merely a small neighbourhood of 0.

### 4.2 What the naive arguments reach, and what recent work achieves

Here is the state of play, informally, because the reader deserves to see that beta < 1 is not a small improvement over the obvious bounds.

- *Free energy.* Aizenman, Lebowitz and Ruelle (1987) proved the annealed free energy formula throughout beta < 1. This fixes the equilibrium thermodynamics but says nothing about dynamics.
- *Influence methods.* Wu (2006) gives a Poincare inequality under a Dobrushin-type spectral-radius condition, which for SK reaches only a threshold strictly below 1.
- *Log-Sobolev via a field decomposition.* Bauerschmidt and Bodineau (2019) prove a high-temperature LSI, but for the *full spin-flip* energy, which is a different (and in some sense harder, but not the same) object from the heat-bath Dirichlet form.
- *Spectral condition for the heat-bath gap.* Eldan, Koehler and Zeitouni (2022) give a general criterion that reaches beta < 1/4 for SK. Adhikari, Brennecke, Xu and Yau (2024) get constant unscaled gaps for mixed p-spin models under a small weighted condition on interaction strengths.
- *Mixing of the discrete chain.* Anari, Jain, Koehler, Pham and Vuong (2022) used entropic independence to get O(n log n) attempts for beta < 1/4. Anari, Koehler and Vuong (2024) extended the threshold to about 0.295. Wang (2026) proved worst-start mixing in O_beta(n log(n/eps)) attempts for every fixed beta < 1/2, uniformly over external fields. Boban, Li and Oveis Gharan (2026) got a constant unscaled gap and O(n^2) zero-field mixing for beta < 1/2 + eps_0, with a tiny absolute eps_0 >= 5 x 10^-5.
- *Equilibrium covariance.* El Alaoui and Gaitonde (2024) proved the expected operator norm of the zero-field spin covariance stays bounded for every beta < 1. Brennecke, Schertzer, Xu and Yau (2024) proved the operator-norm approximation Cov_{mu_0}(x) ~ ((1+beta^2) I - J)^{-1} in probability.

So static covariance bounds already cover the full range beta < 1, but they control only *linear* observables a^T x, i.e. variances of the form a^T Cov a. Turning that into an inequality for *all* functions is precisely the gap in the argument, and it is where the posterior machinery enters.

### 4.3 What is genuinely new here

The theorem reaches every fixed beta < 1 in the zero field. The mechanism is not a sharper influence estimate; it is a way to *transfer* functional inequalities from random posterior Gibbs measures mu_h (at the noisy observation fields h) back to the unobserved mu_0. The posterior measures are easy to control because an observation in Gaussian noise partially pins down each spin, dropping the effective local variances. The transfer, and the careful bookkeeping of the small probability that the posterior is bad, is the technical heart, taken up in Section 8 onward.

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## 5. The engine: observing spins through noise

The whole proof rests on one construction, from the stochastic-localization method of Eldan (2013) in the Gaussian form of El Alaoui and Montanari (2021). Instead of attacking mu_0 directly, we imagine *watching the system through a blurry instrument* and controlling the posterior law of the spins.

### 5.1 The observation channel

Draw a spin sample X from mu_0. Run, in continuous "observation time" t, the channel

$$ Y(t) = t X + B(t), \qquad 0 \le t \le T, $$

where B is a standard n-dimensional Brownian motion independent of X. Think of Y(t) as n independent noisy measurements, one per spin, with signal-to-noise ratio growing like t. Writing x^T Y(t) - t n / 2 for the log-likelihood (using that (Y(t) - t x)^2 / (2t) = ... and x_i^2 = 1), Bayes' rule gives

$$ \mathrm{law}(X \mid Y(t)) = \mu_{Y(t)}, $$

the Gibbs measure at the *random external field* h = Y(t). So the posterior measures that were introduced abstractly in Section 1.3 are exactly the fields the observation process visits. The zero field is the starting point t = 0; the observed fields drift away from zero as t grows.

### 5.2 The filtering identity

Let F_t be the observation history, E_t the average under mu_{Y(t)}, and Cov_t the covariance under it. The key fact of the channel is that for every function g,

$$ \mathrm d(\mathbb E_t g) = \operatorname{Cov}_t(g, x)^{\mathsf T} \mathrm d I(t), \qquad I(t) := Y(t) - \int_0^t \mathbb E_s x \, \mathrm ds. $$

Here I is the *innovation*, a Brownian motion in the observation filtration. In words: the posterior mean of g moves exactly proportionally to the posterior covariance of g with the spins, and the "noise" driving it is the part of the observation not yet explained by the current posterior. This is a martingale identity; one proves it by differentiating the Bayes formula with respect to Y and subtracting the predictable drift.

### 5.3 Three consequences that run the proof

The transfer argument uses nothing from the channel beyond the following three facts.

1. *Covariance becomes the rate of variance decay.* For f, the posterior variance V_t = Var_t(f) and second moment N_t = E_t(f^2) satisfy, by the filtering identity,

$$ \mathrm d V_t = \text{(martingale part)} \;-\; \|\operatorname{Cov}_t(f,x)\|^2\,\mathrm dt, $$

so the posterior variance decays at a rate equal to the squared covariance of f with the spins; the martingale part has zero mean, and the second moment N_t = E_t(f^2) is itself a martingale (a conditional expectation). So *the covariance controls the loss of posterior variance*.

2. *Dirichlet monotonicity.* At every deterministic time,

$$ \mathbb E\, \mathcal D_{Y(t)}(f) \le \mathcal D_0(f). $$

Resampling after observing cannot create more conditional variation than resampling before, by conditional-variance contraction. This is the statement that observation can only *help* the dynamics.

3. *The drift-energy (entropy) identity.* Change the initial spin law from mu_0 to the square reweighting f^2 mu_0 / (E f^2), and let P^f be the resulting observation law. If a_t = E_nu_t x - E_t x is the drift of the posterior mean under P^f, then the relative entropy of the stopped observation law has the exact form

$$ D\big(P^f \,\Vert\, P\big) = \frac12 \mathbb E^f \int_0^\tau \|a_t\|^2 \, \mathrm dt. $$

This is the entropy identity; it converts a bound on \|a_t\| (which is a *local mean* bound, Input 2b) into control of the weight that a rare, high-variance set can place on the observation law. It is the technical reason the exceptional paths can be dominated.

### 5.4 Why observing "helps"

Heuristically, after time t each spin is observed with signal-to-noise t, so each spin becomes better and better pinned down, and the effective local conditional variances v_i shrink. At large enough t the posterior Gibbs measure is provably close to a product measure, and a constant gap holds trivially (Input 3). The hard part is to not lose the variance on the way there and to not let a thin set of bad observations carry all the variance. That is precisely what facts (1)-(3) control.

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## 6. The strategy, and the three inputs

### 6.1 The plan in one paragraph

We will choose a large but fixed observation time T and a stopping time tau that freezes the process just before the posterior could become "bad". On the good event, a *terminal gap* (Input 3) gives the Poincare inequality at the observed field. On the way there, two *posterior estimates* (Inputs 2a, 2b) guarantee that variance cannot be lost too fast (Input 2a) and that exceptional paths cannot hold the variance (Input 2b). The probability that a path is bad is exponentially small, because the fields are *stable* along the path (Input 1). Combining these gives Var_{mu_0}(f) <= C_beta D_0(f) for every f at once, on a typical J.

Everything below reduces to establishing the three inputs on a typical matrix, and then assembling them. The assembly (Section 7) is short and honest; the inputs (Sections 11-14) are the technical mass.

### 6.2 Input 1: stable fields along observations

A field h is *good* if the linearized field equation at h is strictly positive, for every approximate solution and every admissible nearby diagonal. Precisely, with the Hessian

$$ \mathsf H_J(y,A) = I + A^{1/2}\Big( B(y) I - J - \tfrac{2j}{n} m(y) m(y)^{\mathsf T} \Big) A^{1/2}, \qquad B(y) = j\big(1 - \langle m(y)^2 \rangle\big), $$

where m(y) = tanh y, we call h good if the implication "small residual and nearby diagonal => positive Hessian" holds throughout the field.

**Input 1 (stability along observations).** Fix 0 < j < 1 and T < infinity. On an event of matrices J with probability tending to one, the following holds. If X ~ mu_0 and B_0 is independent Brownian motion, then with conditional probability at least 1 - e^{-a n}, simultaneously for all 0 <= t <= T and all approximate solutions y of the field equation with residual below 2 rho_0 sqrt(n), and all diagonal 0 <= A <= (1+eta) I with ||A - V_y||_{Frobenius} <= eps_A sqrt(n), one has the positive Hessian bound H_J(y,A) >= c I. Here a, c, rho_0, eps_A, eta > 0 are fixed (depending only on j, T), and V_y = diag(1 - m(y)^2).

There is a matching version with a stopping time tau_0, so that on [0, tau_0) the *smaller* residual threshold rho_0 sqrt(n) suffices. This is exactly what the transfer needs.

*Why it is true.* If a witness (y, A) with small residual and non-positive Hessian existed, a local-volume calculation would force the Gaussian observation integral to be exponentially large on a small ball around y, contradicting a scalar rate bound coming from an entropy inequality. The entropy input pins down the law of a typical observation, the local volume forces any unstable zero to leave a detectable footprint, and the two cannot coexist. This is Section 11.

### 6.3 Input 2a: the directional covariance bound

**Input 2a (directional covariance).** There is a constant C such that, at every good field h, for every function G,

$$ \big\| \operatorname{Cov}_{\mu_h}(G, x) \big\| \le C\,\big( \operatorname{Var}_{\mu_h}(G) + \mathcal D_h(G) \big)^{1/2}. $$

Equivalently, ||Cov(G,x)||^2 <= C'(Var(G)+D_h(G)). The constant is uniform in n, h and G. This is the strongest single estimate in the paper. It says the covariance of an *arbitrary* function with the spin vector is controlled by the function's own variance plus Dirichlet form. For a linear function it is a covariance-matrix statement, but for a general function it is far stronger, and it is what makes fact (1) of Section 5.3 produce a usable variance-retention inequality.

*Why it can hold.* The proof tests the residual identities of a finite TAP-type recursion against a cleverly chosen *implicit* test vector, a solution of a linear field equation with a strict inverse margin. Two scalar identities, one testing a direction and one testing an averaged coefficient, together with the weak residual rule, give the covariance bound. The subtlety is that the recursion need not converge; only two consecutive residuals are made small, by a telescoping identity. This is Sections 12 (and the residual machinery in Section 12's prerequisites).

### 6.4 Input 2b: the mean under square reweighting

**Input 2b (weighted mean).** For every sufficiently small fixed delta > 0 there is a constant C_delta such that, for N = E_{mu_h} f^2 > 0 and nu = f^2 mu_h / N,

$$ \frac{\big\| \mathbb E_\nu x - t_*(h) \big\|}{\sqrt n} \le C\,\delta + \frac{C_\delta}{\sqrt n}\Big(1 + \frac{\mathcal D_h(f)}{N}\Big). $$

Here t_*(h) = tanh r(h), and r(h) is the unique solution of the TAP fixed-point equation

$$ r - h + \big( j(1 - n^{-1}\|\tanh r\|^2) I - J \big)\tanh r = 0. $$

Existence and uniqueness of r at good fields is part of the stability analysis. Taking f = 1 recovers the same comparison for the posterior mean E_{mu_h} x itself.

The two constants play different roles and the separation is essential: C is independent of delta (it comes from root stability), while C_delta may grow as delta shrinks (it comes from the finite diagnostic length used to build the test functions). Fact (3) of Section 5.3 then bounds the bad-path entropy by 2 C T delta plus a term of order 1/sqrt(n). This is the crux of Section 13.

### 6.5 Input 3: the terminal gap

**Input 3 (terminal gap).** Fix 0 < j < 1. There are a finite time T and a rate c > 0, and matrix events of probability tending to one, such that for every matrix in the event, every initial spin law, and every pair (X, Z) with Z standard Gaussian,

$$ \operatorname{Var}_{\mu_{Y}}(f) \le 2\, \mathcal D_Y(f) \quad \text{for every } f, \qquad Y = T X + \sqrt T\, Z, $$

except on an event of observation probability at most e^{-c n}.

*Why it is true.* At a large fixed observation time, most spins are observed with very strong signal, so their local conditional variances v_i are exponentially small. The only spins with appreciable variance form a set of size at most alpha n for a small alpha. A *signed two-spin curvature inequality* (after Wang, 2026) reduces the Bochner-type identity for the generator to a quadratic form whose only dangerous piece on the small set is bounded using small principal submatrices of U = tanh J. This is Section 14.

### 6.6 The bookkeeping of the constants

The ordering of choices is the whole game in the assembly. One fixes, in order: T (from Input 3); then the stability constants, the good-field failure rate a, and the finite diagnostic length for Input 2a (these determine the variance-retention rate C_d and v = e^{-C_d T}); then the leading constant C of Input 2b (which depends only on root stability, not on delta); then delta (so that 2 C T delta <= v/8); then its diagnostic length and C_delta; and finally n. Every depth and tolerance is fixed *before* n grows. Nothing below depends on a quantity that is chosen after its use.

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## 7. Assembly: the three inputs imply the theorem

This section is the only place the main theorem is actually proved. It uses exactly Inputs 1, 2a, 2b, 3 and nothing else. Everything after it is the proof of those inputs. Read this section first even if you skip the technical parts: it is where the whole scheme snaps shut.

Throughout, fix a matrix J for which Inputs 1 and 3 hold and on which the diagnostic events of Inputs 2a and 2b hold; all of these hold with probability tending to one, and we intersect finitely many such events. Fix f; by homogeneity it suffices to handle the normalization

$$ \mathbb E_{\mu_0} f = 0, \qquad \mathbb E_{\mu_0} f^2 = 1, \qquad d_0 := \mathcal D_0(f) \le 1. $$

(The case d_0 > 1 is automatic: Var_{mu_0} f = 1 <= d_0 <= C d_0 for any C >= 1, choosing C_beta >= 4/v later.) Running the observation channel from Section 5, write

$$ M_t = \mathbb E_t f, \qquad N_t = \mathbb E_t f^2, \qquad V_t = N_t - M_t^2 = \operatorname{Var}_{\mu_{Y(t)}} f. $$

### 7.1 The stopping rule and the bad event

Let tau_0 be the stopping time from Input 1, and set tau = min(T, tau_0). This rule depends only on J and the observations, never on the test function f, which is essential: the same tau must serve all f at once.

Call a stopped path *bad* if tau_0 <= T, or if tau_0 > T and the terminal field Y(T) fails the terminal gap. On the good matrix event,

$$ \mathbb P(\text{bad}) \le 2 e^{-a n}, $$

combining the stability failure bound of Input 1 and the observation failure bound of Input 3, after decreasing the rate a if needed. The event is measurable with respect to the stopped observation filtration.

### 7.2 Variance cannot be lost too fast (Input 2a)

Let u(t) = E V_{t ∧ tau}. By the filtering identity (Section 5.3, fact 1) and optional stopping at bounded times,

$$ u'(t) = -\,\mathbb E\Big[ \mathbf 1_{\{t < \tau\}} \big\| \operatorname{Cov}_t(f, x) \big\|^2 \Big] \quad \text{for a.e. } t. $$

Before tau the field is good, so Input 2a applies, and because Cov_t(f,x) = Cov_t(f - E f, x) and Var_t(f) + D_t(f) is the natural quantity,

$$ \big\| \operatorname{Cov}_t(f, x) \big\|^2 \le C_{\mathrm d}\,\big( V_t + \mathcal D_{Y(t)}(f) \big) $$

with C_d fixed. Taking expectations and using Dirichlet monotonicity E D_{Y(t)}(f) <= D_0(f) = d_0 gives the scalar differential inequality

$$ u'(t) \ge -\,C_{\mathrm d}\,\big( u(t) + d_0 \big). $$

Solving this from u(0) = Var_{mu_0} f = 1 (set w = u + d_0, so w' >= -C_d w and w(0) = 1 + d_0) yields, with v := e^{-C_d T},

$$ \boxed{\ \mathbb E V_\tau \ \ge\ v - (1 - v)\, d_0.\ } \tag{7.1} $$

The mechanism: the covariance governs how fast posterior variance can drop, and the covariance is itself controlled by the variance plus the Dirichlet form. This is why a *directional* covariance estimate is exactly the right input.

### 7.3 Exceptional paths cannot hold the variance (Input 2b)

The inequality (7.1) alone does not rule out the case where nearly all the variance sits on the thin bad set. To exclude it, change the initial spin law from mu_0 to f^2 mu_0, a probability law by normalization, and let P^f be the corresponding observation law. Its density on the stopped filtration is N_tau, and N_tau is exactly the posterior second moment. Therefore

$$ \mathbb E\big[ \mathbf 1_{\mathrm{bad}} V_\tau \big] \le \mathbb E\big[ \mathbf 1_{\mathrm{bad}} N_\tau \big] = \mathbb P^f(\mathrm{bad}), $$

because V_tau <= N_tau always, and because integrating the density N_tau against the indicator of bad gives its P^f-probability. So bounding the variance on bad paths is the same as bounding the probability that the *reweighted* observation law is bad.

That is exactly what Input 2b targets. Let a_t = E_{nu_t} x - E_t x be the drift of the posterior mean under P^f, where nu_t = f^2 mu_{Y(t)} / N_t. By the entropy identity (Section 5.3, fact 3), with H_tau := D(P^f \Vert P) on the stopped filtration,

$$ H_\tau = \frac12 \mathbb E^f \int_0^\tau \| a_t \|^2 \, \mathrm dt. $$

Input 2b bounds b_t := \|a_t\| / sqrt(n) by 2 C delta + (C_delta / sqrt n)(2 + D_t(f)/N_t) before tau. Since b_t <= 2 (as ||a_t|| <= 2 sqrt n), we may use b_t^2 <= 2 b_t. At a deterministic time, the density identity and Dirichlet monotonicity give E^f [D_t(f)/N_t] = E D_t(f) <= d_0. Combining, and using tau <= T,

$$ \frac{H_\tau}{n} \le 2 C T \delta + \frac{C_\delta T}{\sqrt n}(2 + d_0) \le 2 C T \delta + \frac{3 C_\delta T}{\sqrt n}, \tag{7.2} $$

using d_0 <= 1 in the last step. The separation of C (independent of delta) and C_delta (allowed to blow up as delta shrinks) is what makes the next step work: we can first shrink delta to control the *leading* term, with C_delta growing but harmless.

Now convert the entropy bound into a bad-path probability bound. If an event has probability q under P and p under P^f, then the log-sum inequality gives D(P^f \Vert P) >= p log(1/q) - log 2. With q <= 2 e^{-a n}, so log(1/q) >= a n - log 2, and for large n,

$$ p = \mathbb P^f(\mathrm{bad}) \le \frac{H_\tau + \log 2}{a n - \log 2}. $$

Insert (7.2). For d_0 <= 1 and large n, this is at most

$$ \mathbb P^f(\mathrm{bad}) \le \frac{4 C T}{a}\,\delta + \frac{6 C_\delta T}{a \sqrt n} + \frac{2 \log 2}{a n}. $$

Choose delta > 0 once, so small that the first term is at most v/8 (this is possible because C is independent of delta). With that delta fixed, enlarge n until the last two terms sum to at most v/8. Then, simultaneously for all normalized f with d_0 <= 1,

$$ \boxed{\ \mathbb P^f(\mathrm{bad}) \le v/4, \ \ \text{hence}\ \ \mathbb E\big[ \mathbf 1_{\mathrm{bad}} V_\tau \big] \le v/4.\ } \tag{7.3} $$

### 7.4 The good paths and the final inequality (Input 3)

On a good path, tau = T and the terminal field Y(T) is a terminal good field, so Input 3 gives V_T = Var_{mu_{Y(T)}} f <= 2 D_{Y(T)}(f). Taking expectations and using Dirichlet monotonicity at the deterministic time T,

$$ \mathbb E\big[ \mathbf 1_{\mathrm{good}} V_\tau \big] \le 2\, \mathbb E\, \mathcal D_{Y(T)}(f) \le 2 d_0. \tag{7.4} $$

Now add (7.1), (7.3), (7.4) in the obvious way. On one hand E V_tau >= v - (1-v) d_0. On the other, E V_tau = E[1_bad V_tau] + E[1_good V_tau] <= v/4 + 2 d_0. Hence

$$ v - (1 - v) d_0 \le 2 d_0 + v/4, $$

which rearranges to (3v/4) <= (3 - v) d_0, i.e.

$$ d_0 \ge \frac{3 v}{4 (3 - v)} \ge \frac{v}{4}. $$

Therefore Var_{mu_0} f = 1 <= (4/v) D_0(f). Undoing the normalization (D_0 is quadratic and Var is quadratic in f, so both center and scale cleanly), and taking C_beta = 4/v = 4 e^{C_d T}, finishes the proof of the main theorem. The discrete chain follows from the one-line conversion of Section 1.6.

### 7.5 The picture, in one box

The logic of the assembly is the following loop of inequalities, for a fixed typical J:

- Good paths lose at most the terminal variance: E[1_good V_tau] <= 2 d_0. (Input 3, plus monotonicity.)
- Bad paths carry probability at most v/4 even under the square-reweighted law: E[1_bad V_tau] <= v/4. (Inputs 1, 2b, plus the entropy identity.)
- The total posterior variance cannot have dropped below v - (1-v) d_0. (Input 2a, plus the filtering identity and monotonicity.)

Feeding the upper bound into the lower bound forces d_0 >= v/4, which is the Poincare inequality. The three inputs are exactly the three transfers of information in this loop: covariance controls the drift, reweighting controls the exception, and the curvature at a large time controls the endpoint.

---

## 8. Toolbox A: Gaussian facts and discrete calculus

Before the technical inputs, collect the two elementary toolkits they use: facts about Gaussian randomness, and the discrete calculus on the cube.

### 8.1 Gaussian concentration and comparison

*Concentration.* If F is a real L-Lipschitz function of a standard Gaussian vector, then

$$ \mathbb P\{ |F - \mathbb E F| > u \} \le 2 \exp\!\big( -u^2 / (2 L^2) \big). $$

This is used everywhere to turn a bounded expectation into a simultaneous bound over finitely many objects, and to turn "typical" into "with probability at least 1 - C e^{-c n}".

*Sudakov-Fernique comparison.* If two centered Gaussian processes have ordered squared increments, then the one with larger increments has the larger expected supremum. It remains true after adding the same deterministic function to both. This comparison is how the largest eigenvalue of a random matrix is controlled near 2 sqrt(j), and how the quadratic-process margin in the matrix analysis is proved.

*Gaussian integration by parts.* Let W be a centered GOE matrix of scale j: W_{ab} ~ N(0, j/n) for a < b and W_{aa} ~ N(0, 2j/n). For a scalar function H of W,

$$ \mathbb E[ W_{ab} H(W) ] = \frac{j}{n}\, \mathbb E\big[ \mathrm d H(W)[E_{ab} + E_{ba}] \big], $$

where E_{ab} is the elementary matrix and dH[...] is the derivative in that symmetric direction. When a = b the direction is doubled, matching the doubled diagonal variance. This identity is the engine that computes every expectation of a product word in Section 9.

### 8.2 Discrete calculus on the cube

For a function f on {-1,1}^n define its *discrete derivative*

$$ d_i f(x) = \frac{f(x) - f(x^{(i)})}{2 x_i}, $$

where x^(i) is x with the i-th spin flipped. Because both numerator and denominator change sign under flip i, d_i f does *not* depend on x_i. These derivatives satisfy the discrete product rule

$$ d_i(fg) = f\, d_i g + g\, d_i f - 2 x_i (d_i f)(d_i g), $$

the extra term being the price of the second-order remainder. The Dirichlet form is exactly a weighted sum of their squares:

$$ \mathcal D_h(f) = \mathbb E_{\mu_h}\Big[ \sum_i v_i \, (d_i f)^2 \Big], \qquad v_i = 1 - m_i^2, $$

as noted in Section 2.1. Finally, define the *test norm*

$$ \| G \|_*^2 = \mathbb E_{\mu_h} G^2 + \mathcal D_h(G). $$

The residual estimates will all be stated as "at most C ||G||_*"; the point of the norm is that it measures both the size of G and its fluctuation under flipping one spin.

### 8.3 The level-1 conditional identity

The seed of all residual identities is the following exact identity, obtained by conditioning on all spins but the i-th and using that x_i - m_i is centred:

$$ \mathbb E_{\mu_h}\sum_i (x_i - m_i^1) G U_i = \mathbb E_{\mu_h}\sum_i v_i^1 \, d_i(G U_i), \qquad m^1 = \tanh(h + Jx), \ v^1 = \sech^2(h + Jx). $$

By the product rule, the right side is at most C ||G||_* whenever U has bounded norm and bounded sum of diagonal derivatives sum_i |d_i U_i|. This is why the whole method works with test vectors U that carry a bounded "diagonal derivative" cost; the recursion machinery of Section 12 constructs exactly such U.

---

## 9. Toolbox B: random-matrix estimates (words)

The matrix estimates are the technical foundation under Inputs 1 and 2a/2b. The reader who only wants the outline can read the statements and the one-paragraph "why"; the proofs are a finite amount of Gaussian bookkeeping.

### 9.1 Words, pairings, and their predictions

Couple J to a GOE W by J = W - diag(W), so J has the right zero-diagonal law. Fix eta > 0 with j (1+eta)^2 < 1, and let a_+ = 1 + eta. For a nonnegative diagonal A <= a_+ I, put

$$ \chi_A = \frac{1}{n}\operatorname{Tr} A, \qquad B_A = j \chi_A, \qquad K_A(z;M) = \big( I - z A M + z^2 B_A A \big)^{-1}. $$

*Words.* A *word* is an ordered product of diagonal matrices, copies of a symmetric matrix M, and at most one factor K_A(z;M). Its length is the number of factors. To a word one assigns a diagonal *prediction* P(Q) by the noncrossing-pairing rule: contract innermost pairs MDM across copies of M by

$$ M D M \;\longmapsto\; j\, \frac{\operatorname{Tr} D}{n} I, $$

combine adjacent diagonals, and sum over all complete noncrossing pairings. This is the operator-valued semicircular moment rule of Speicher (1998), with covariance D -> j (Tr D / n) I. A word with a factor K_A is handled by expanding K_A as the formal series sum_{r>=0} (zAM - z^2 B_A A)^r and applying the rule coefficient by coefficient.

*The inverse is well behaved even when A is small.* The associated symmetric matrix, S_A(z;M) = I + z^2 B_A A - z A^{1/2} M A^{1/2}, controls the inverse without needing a lower bound on the individual diagonal entries:

$$ K_A(z;M) = I + A^{1/2} S_A(z;M)^{-1} A^{1/2}\big( z M - z^2 B_A I \big), \qquad K_A(z;M) A = A^{1/2} S_A(z;M)^{-1} A^{1/2}. $$

These identities are just the resolvent identity (I - UV)^{-1} = I + U(I - VU)^{-1} V.

### 9.2 The simultaneous word estimate

**Lemma (simultaneous words).** Fix a length L, a diagonal bound M_0, and c > 0. There are constants C, c_1 > 0 and n_0 such that, on an event of probability at least 1 - C e^{-c_1 n} for n >= n_0, simultaneously for every word Q of length at most L, every real diagonal factor of norm at most M_0, and every 0 <= A <= a_+ I, one has

$$ \| Q \| \le C, \qquad \Big( \sum_{a \ne b} |Q_{ab}|^4 \Big)^{1/4} \le C, \qquad \| \operatorname{diag}(Q - \mathsf P(Q)) \|_2 \le C. $$

For a word containing K_A(J), the assertion is restricted to parameters with S_A(1;J) >= c I. The diagonal parameters A may be chosen *after* the random matrix is observed; this adaptive freedom is the whole reason the estimate is stated simultaneously.

*Why it is true, in one paragraph.* For a fixed deterministic diagonal, a Gaussian comparison proves S_A(z;W) stays uniformly positive along 0 <= z <= 1 with exponentially small failure. The diagonal expectation of a truncated word is then computed by Gaussian integration by parts; a finite loop recursion in the number of ordinary M letters shows that this expectation equals the pairing prediction up to O(1/n). Concentration plus a finite mesh in the diagonal parameters makes the three bounds hold simultaneously for all allowed A, using sign-conjugation symmetry to kill off-diagonal means. Truncating the inverse on the operator-norm ball makes every word globally Lipschitz, which is what allows the simultaneous mesh argument. Finally J = W - diag(W) is handled by bounding the removed diagonal in Frobenius norm (its squared norm is a chi-square, exponentially unlikely to be large).

### 9.3 Two corollaries used later

*Bilinear deterministic equivalents.* For a fixed deterministic diagonal A, fixed deterministic vectors v_1,...,v_m of bounded norm, and any tolerance eps, with G_A = A^{1/2} S_A(1;W)^{-1} A^{1/2} = K_A(W) A, all the bilinear forms

$$ v_r^{\mathsf T} (G_A - A) v_s, \qquad v_r^{\mathsf T}(G_A W - B_A A) v_s, \qquad v_r^{\mathsf T}(W G_A W - B_A I - B_A^2 A) v_s, $$

are within eps of their predicted values (A, B_A A, B_A I + B_A^2 A) on an event of probability at least 1 - C e^{-c_3 n}, uniformly in A and the vectors. This is the form actually consumed by the Hessian computation in Section 11.

*A regularized log-determinant bound.* Fix eps, an integer r_0, and M_2 < infinity. For every fixed small gamma > 0 and all large n, on an event of probability at least 1 - C e^{-c_4 n}, simultaneously for every diagonal 0 <= V <= I and every matrix E of rank at most r_0 and norm at most M_2, setting b = (1/n) Tr V and L = I + (jb I - W) V + E,

$$ \frac{1}{n} \log \det(L^{\mathsf T} L + \gamma I) \le j b^2 + \varepsilon. $$

Here V and E may depend on W. This is the determinant input to the local-volume lemma of Section 11. Its proof differentiates (1/n) log det L_z along z, uses the loop-equation identities of Section 9.2, adds a regularization gamma to give a global Lipschitz bound on the normalized log-determinant, and then meshes the cube of diagonal V, absorbing the finite-rank E by the rank inequality.

These are all the matrix facts the argument needs. Every use below cites one of the three lemmas of this section.

---

## 10. Toolbox C: the scalar entropy inequality

The stability input (Section 13) is proved by a Gaussian-integral calculation, and a *scalar* entropy inequality is what makes that integral decay away from an explicit equality curve. This section states and explains it.

### 10.1 The entropy inequality

Throughout, N(d,s) is the Gaussian law with mean d and *variance* s.

**Lemma (scalar entropy).** Let 0 < j < 1, d in R, s > 0, and let P be a probability law on R with finite second moment. Set

$$ m(y) = \tanh y, \quad q = \mathbb E_P m^2, \quad b = 1 - q, \quad a = \mathbb E_P[(y - d)m], \quad S = s + j q. $$

If s >= j q, then

$$ D(P \,\Vert\, N(d,s)) \ge \frac{j\,(a - s b)^2}{2 s S}, \qquad \text{equality iff } P = N(d,s). $$

*Reading it.* The inequality compares a general law P with a Gaussian N(d,s). The "mismatch" is measured by the scalar a - s b. For a Gaussian one has a = s b exactly (Gaussian integration by parts, see below), and then the inequality is trivial. The content is that *any* non-Gaussian law is forced to have relative entropy bounded below by a fixed multiple of the squared mismatch. This is exactly what one needs: away from the Gaussian curve, the entropy is strictly positive, so the associated Gaussian integral decays exponentially.

*How it is proved.* Two estimates are combined. First, a monotone change of variables T_u(y) = y - u tanh y (a diffeomorphism for u < 1) lets one write D as a Gaussian-variance term plus an average of log(1 - u(1 - m^2)); concavity of the logarithm gives a quadratic lower bound, and completing the square isolates the deficit h(u) = u^2/2 - B ell(u) (with ell(u) = -log(1-u) - u). For u in a range controlled by a_0 = 1 - B this deficit is *strictly positive*. Second, for the remaining range of the mismatch a - s b, one compares instead with a Gaussian of a different variance sv, using the exact second-moment identity and the Cauchy-Schwarz bound v >= C^2/(jk). A little one-variable calculus (the function f(C) = C^2/k - 1 - log(C^2/k) - (C-B)^2/(1+k)) then closes the remaining range. The equality case is read off: both bounds are strict unless P is exactly the Gaussian.

### 10.2 The consistent Gaussian law

In the stability calculation the Gaussian parameters are *not* free: they are tied to the law P through consistency equations. For t >= 0 and sigma >= 0, set

$$ M = \mathbb E_P \tanh y, \quad q = \mathbb E_P \tanh^2 y, \quad d = t + j M, \quad s = t + \sigma^2 + j q. $$

**Lemma (consistent Gaussian).** Fix 0 < j < 1. For each t >= 0, the scalar equation

$$ s_*(t) = t + j\, \mathbb E_{N(s_*(t), s_*(t))} \tanh y $$

has a unique nonnegative solution (with N(0,0) the point mass at 0). Its Gaussian P_*(t) = N(s_*(t), s_*(t)) is the unique Gaussian law satisfying the consistency equations with sigma = 0. Thus, among consistent laws with s > 0, equality in the scalar entropy inequality holds precisely at P_*(t). Moreover s_*(t) is Lipschitz with constant (1-j)^{-1}, and

$$ t \le s_*(t) \le \frac{t}{1-j}, \qquad q_*(t) := \mathbb E_{P_*(t)} \tanh^2 y = \frac{s_*(t) - t}{j} \le \frac{t}{1-j}. $$

*How it is proved.* First, symmetry. The density of N(s,s) satisfies rho_s(y)/rho_s(-y) = e^{2y}, and pairing y with -y gives the "Nishimori" identity E_{N(s,s)} tanh y = E_{N(s,s)} tanh^2 y. Together with a sign identity E_{N(d,s)}[tanh y sech^2 y] >= 0 for d >= 0, this forces any consistent Gaussian to have mean equal to variance, d = s. Second, the fixed point. Put F(s) = E_{N(s,s)} tanh y. Differentiating in both the mean and the variance gives F'(s) = E_{N(s,s)}[(1-m^2)(1-m)] in [0,1], so F is nonnegative and 1-Lipschitz with F(s) <= s. Then s -> t + j F(s) is a contraction of [0, infinity) with a unique fixed point s_*(t), and the bounds follow from 0 <= F(s) <= s.

### 10.3 Why a scalar inequality suffices for a vector problem

The Gaussian integral in Section 13 involves n i.i.d. coordinates, and the empirical law P_y = (1/n) sum_i delta_{y_i} is a general probability measure on R. The relative entropy D(P_y || N(d,s)) controls exactly the exponential rate of the integral, via the variational description of entropy worked into the compactness argument. So a scalar entropy inequality, plus its equality curve, is exactly the input needed. This is one of the places where the random, high-dimensional problem collapses to a one-dimensional statement about probability laws.

---

## 11. The TAP residual machinery

This section builds the test vectors that Inputs 2a and 2b need. It is deterministic: fix a field h_0 and a matrix J on the word events of Section 9, and take all expectations under mu_{h_0}. The goal is a family of vectors U, built by finitely many operations, for which the *residual* R_p of an iterated TAP recursion is provably small in the averaged sense. The recursion is *not* assumed to converge; the point is that a fixed finite depth already suffices.

### 11.1 The primary recursion

Start from a spin sample and iterate, with the Onsager correction of the TAP equations (Bolthausen 2014):

$$ m^0 = x, \quad b_0 = 0, \qquad m^l = \tanh h^l, \quad b_l = \langle 1 - (m^l)^2 \rangle, \quad R_l = m^{l-1} - m^l, $$

$$ h^{l+1} = h_0 + J m^l - j\, b_l\, m^{l-1} \qquad (l \ge 1). $$

Here <...> is the empirical average (1/n) sum_i. The vector R_l = m^{l-1} - m^l is the *residual* (the increment). It measures how far one iterate is from the next.

### 11.2 The telescoping identity

A one-line expansion of the squared increment gives

$$ \frac{\| R_l \|^2}{n} = b_l - b_{l-1} - \frac{2 S_l}{\sqrt n}, \qquad S_l := \frac{R_l^{\mathsf T} m^l}{\sqrt n}. $$

Since 0 <= b_l <= 1, the sum of the squared increments is at most b_d + (2/sqrt n) sum_l |S_l|. So if the scalar pairings S_l/sqrt(n) are small through a depth d large enough, then the sum of the ||R_l||^2/n is O(1), and therefore *some pair of consecutive residuals must be small*. This is the key finite-depth fact: convergence of the recursion is replaced by the existence of two consecutive small residuals, at a depth chosen in advance.

The mechanism also produces the moment bounds (proved by the weak residual rule below, using the test vector U = m^l / sqrt n):

$$ \mathbb E S_l^2 \le C_l, \qquad \mathbb E_\nu \Big( \frac{S_l}{\sqrt n} \Big)^2 \le \frac{C_l}{\sqrt n}\Big( 1 + \frac{\mathcal D_{h_0}(f)}{N} \Big), $$

where N = E f^2 and nu = f^2 mu_{h_0} / N. The first is an ordinary second moment; the second is the same after square reweighting, with the extra 1/sqrt(n) factor that the entropy argument of Section 7.3 requires.

### 11.3 Selecting a small-residual region smoothly

The moment bounds control S_l but not R_l itself. To get a *smooth* region where two consecutive residuals are genuinely small (so that the implicit construction of Section 11.6 lives there), introduce cutoffs.

**Lemma (residual selection).** For each sufficiently small fixed rho > 0, there is a fixed depth d and functions C_k, 2 <= k <= d, with: (i) 0 <= C_k <= 1, C_k is supported where max(||R_k||,||R_{k-1}||) <= rho sqrt n, equals 1 where both are at most rho sqrt n/2, and ||dC_k|| <= C_rho / sqrt n pointwise; (ii) with C_hat_k = C_k prod_{i<k}(1-C_i) and D = 1 - sum_k C_hat_k, one has 0 <= D <= 1 and

$$ \mathbb P_{\mu_{h_0}}(D > 0) \le \frac{C_\rho}{n}, \qquad \mathbb P_\nu(D > 0) \le \frac{C_\rho}{\sqrt n}\Big( 1 + \frac{\mathcal D_{h_0}(f)}{N} \Big). $$

*Why.* Choose a depth d with floor(d/2) rho^2/4 > 2 and then eps > 0 with 2 d eps < 1. Wherever max_l |S_l|/sqrt n <= eps, telescoping forces sum_l ||R_l||^2/n < 2; the disjoint pairs (1,2),(3,4),... then cannot all contain a residual above rho sqrt n/2, so at least one pair is small. The cutoffs turn this into a smooth support function, and the deficit D>0 can only happen on the large-deviation event max_l |S_l|/sqrt n > eps, whose probability is controlled by the two moment bounds and a finite union bound.

From here on, work on the region Omega_rho = {max(||R_k||,||R_{k-1}||) <= rho sqrt n} for the selected k. On Omega_rho, the root comparison (Section 13) gives ||h^k - r|| + ||m^{k-1} - t_*|| <= C rho sqrt n, where r = r(h_0) is the unique zero of the field equation and t_* = tanh r.

### 11.4 Admissible recipes and the weak residual rule

The residuals will be tested against vectors built by finitely many operations from the primary fields. Call such a vector *admissible from level p* if its formula only uses the primary fields h^l with l >= p; multiplication by bounded smooth coefficients of those fields is allowed, as is any linear combination of previously built auxiliary fields. A finite *recipe* is a finite such construction. Every vector in a recipe has bounded norm and bounded diagonal-derivative cost sum_i |d_i U_i|; the Onsager corrections are exactly what make this diagonal cost stay bounded (the "trace cancellation").

**Lemma (weak residual rule).** Let U be a terminal vector in a fixed recipe admissible from level p >= 1. For every scalar G,

$$ \big| \mathbb E_{\mu_{h_0}} G\, R_p^{\mathsf T} U \big| \le C\,\| G \|_*. $$

If moreover sup_x (|H_0| + ||dH_0||) <= C_0, then for every f,

$$ \big| \mathbb E_{\mu_{h_0}} f^2 H_0\, R_p^{\mathsf T} U \big| \le C\, \big( \mathbb E f^2 + \mathcal D_{h_0}(f) \big). $$

*Why.* The base case p = 1 is the exact level-1 conditional identity of Section 8.3: E sum_i (x_i - m_i^1) G U_i = E sum_i v_i^1 d_i(G U_i), split by the product rule. Its first piece is bounded by the diagonal-derivative cost of U, its second by the weighted Cauchy-Schwarz control of D_{h_0}(G). The induction step p -> p+1 uses the scalar kernel of Section 11.5 to turn R_{p+1} into earlier residuals plus one scalar correction that cancels exactly against the Onsager term, and applies the induction hypothesis to each remaining piece. The square-density version is the same argument with G replaced by f^2 and the identity d_i(f^2) = 2 f d_i f - 2 x_i (d_i f)^2.

### 11.5 The scalar kernel

To pass from R_{p+1} = tanh h^p - tanh h^{p+1} to earlier residuals, use the exact scalar identity, valid for all real z, r_0, a:

$$ \Phi(z;r_0,a) = \int_0^1 e^{a(1-u^2)/2}\frac{\cosh(u(z-r_0))}{\cosh z \cosh r_0}\,\mathrm du, $$

$$ \big[ (z - r_0 - a\tanh z) - a\,\partial_z \big]\Phi(z;r_0,a) = \tanh z - \tanh r_0. $$

This is a discrete "Stein identity" for the function tanh. It holds for either sign of a; at a = 0, Phi is the secant slope of tanh. The boundedness of Phi and of every fixed-order derivative of its ratios (uniformly in z, r_0 and in a from a bounded interval) is what keeps the recipes admissible when Phi is inserted as a coefficient.

### 11.6 The implicit auxiliary field

Input 2a needs one vector that is defined *implicitly*, as a solution of a linear equation, rather than by a finite recipe. On Omega_rho, set

$$ a = j(1 - b_{k-1} - q_*), \qquad A_i = \Phi(h_i^k; r_i, a), \qquad \chi = \langle A_i \rangle, $$

and introduce the coupled source and field

$$ w = w_0 + A(y^{\mathrm{imp}} - j c_1 t_*), \qquad y^{\mathrm{imp}} = J w - j\chi w - j c_1 m^{k-1}, \qquad c_1 = \langle \partial_k w \rangle. $$

Here w_0 is one of the two seeds A e (||e|| <= 1) or n^{-1/2} ∂_{r_0} Phi(h_i^k; r_i, a).

**Lemma (implicit construction).** For sufficiently small rho > 0, the equation above has a unique solution on Omega_{2rho}. There are constants independent of the primary depth k such that, throughout this region,

$$ \| w \| + \| y^{\mathrm{imp}} \| + \sqrt n\,|c_1| \le C, \qquad I + j\chi A - A^{1/2} J A^{1/2} \succeq c I. $$

The second display is the *inverse margin*. It says the linear operator defining the implicit field is positive definite by a fixed amount, uniformly in k and in the directional seed e.

*Why.* Eliminating y^imp and c_1 leaves a linear equation for w with matrix I + A L_2, whose symmetric normalization differs from I + j b_k A - A^{1/2} J A^{1/2} - (2j/n) A^{1/2} m^k (m^k)^T A^{1/2} by O(rho). The last symmetric matrix is positive by the good-field condition of Input 1 (Section 13), so for rho small the inverse exists and is bounded. The derivatives are then bounded by comparing the equation at x and at x^(i) and using the size bounds; the buffers are fixed before the depth, so nothing here depends on e or on k in a dangerous way.

### 11.7 The terminal residual identity

Finally, the implicit source and the weak residual rule combine into the identity that Input 2a actually tests:

**Lemma (terminal residual identity).** For the implicit construction at index k and every G supported on Omega_rho,

$$ \Big| \mathbb E G \Big[ \{ h^k - h_0 - (J - j b_{k-1} I) m^k \}^{\mathsf T} w - j (b_k - b_{k-1})\, n\, c_1 \Big] \Big| \le C\,\| G \|_*. $$

*Why.* The braced vector is JR_k - j b_{k-1}(R_{k-1}+R_k). Using symmetry of J, the field equation Jw = y^imp + j chi w + j c_1 m^{k-1}, and the inner-product identity R_p^T m^{p-1} = n(b_p - b_{p-1}) - R_p^T m^p, it becomes R_k^T y^imp + j(chi - b_{k-1}) R_k^T w - j b_{k-1} R_{k-1}^T w - j c_1 R_k^T m^k. The first three terms are weak residual rules at levels k and k-1, and the last uses the source m^k/sqrt n with the bounded multiplier sqrt n c_1 from the implicit construction. This is the identity that turns the *residual* R_k into something testable.

This is all the residual structure Inputs 2a and 2b require. Section 12 assembles the two posterior estimates from it.

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## 12. The two posterior estimates (Inputs 2a and 2b)

This section proves Inputs 2a and 2b at a fixed good field h_0, using the residual machinery. It is the technical center of the paper: the two estimates are what the assembly of Section 7 consumes. We keep the two proofs separate because they use *different* recipes: 2a needs the implicit test vector; 2b uses only ordinary recipes.

### 12.1 The comparison identity

Recall the setup of Section 11.6: on Omega_rho, with m = m^k, p = m - t_*, r the unique root of the field equation, a = j(1 - b_{k-1} - q_*), A_i = Phi(h_i^k; r_i, a), and L_* := sqrt n ( chi - (1 - q_*) ). The implicit source w satisfies w = w_0 + A(y^imp - j c_1 t_*). A direct expansion in which the root equation is subtracted, then the field equation for w is substituted, then the Stein identity of Section 11.5 is applied coordinatewise, gives the central comparison identity, valid on the buffered region for either seed w_0:

$$ \big( h^k - h_0 - (J - j b_{k-1} I) m \big)^{\mathsf T} w - j(b_k - b_{k-1}) n c_1 $$

$$ = \sum_i \big[ (h_i^k - r_i - a m_i) - a \partial_k \big] w_{0,i} + j(1 - q_* - \chi)\, p^{\mathsf T} w - j c_1 R_k^{\mathsf T} p. $$

Each piece now has a job. The left side is controlled by the terminal residual identity (Section 11.7). The last term is controlled by the weak residual rule with target p/sqrt n and multiplier sqrt n c_1. The middle term is where the good-field condition enters, and it is small on Omega_rho. The first term depends on the choice of seed w_0, and the two seeds test two different things.

### 12.2 Input 2a: the directional covariance bound

**Proposition (directional covariance).** There is a constant C such that, at every field satisfying the stable-field condition, for every G,

$$ \big\| \operatorname{Cov}_{\mu_{h_0}}(G, x) \big\| \le C\, \big( \operatorname{Var}_{\mu_{h_0}}(G) + \mathcal D_{h_0}(G) \big)^{1/2}. $$

The constant is uniform in n, h_0, G; only a fixed finite diagnostic length is needed.

*Proof mechanism.* The two seeds of Section 11.6 are used in order.

Step 1: control the averaged coefficient. Take the second seed, w_{0,i} = n^{-1/2} ∂_{r_0} Phi(h_i^k; r_i, a). By the second scalar identity of Section 11.6, the first sum in the comparison identity equals L_*. Test with G = C_k^2 L_*. The terminal residual identity and the weak residual rule bound the left side and the last term by C||C_k^2 L_*||_*; the cutoff derivative bounds give ||C_k^2 L_*||_* <= C + ||C_k L_*||_{L^2}. The middle term is at most C_0 rho |L_*| by the size estimates, and C_0 rho < 1/2 by choosing rho small *first* (before the depth). Absorbing gives

$$ \| C_k L_* \|_{L^2(\mu_{h_0})} \le C_k'. $$

Step 2: test a direction. Take the first seed, w_{0,i} = A_i e_i with any deterministic ||e|| <= 1. By the first scalar identity, the first sum equals p^T e = (m^k - t_*)^T e. Test with C_hat_k G. The terminal residual identity controls the left side; the last term is a weak residual rule; the middle term is bounded by the averaged coefficient bound of Step 1 together with Cauchy-Schwarz (using 0 <= C_hat_k <= C_k). Hence

$$ \big| \mathbb E\big[ \widehat C_k\, G\,(m^k - t_*)^{\mathsf T} e \big] \big| \le C\,\| G \|_*. $$

Adding the weak residual rule with the constant target e over levels 1..k replaces m^k - t_* by x - t_* at the same cost.

Step 3: sum over k, and treat the deficit. Summing over k and using sum_k C_hat_k = 1 - D, the deficit contributes at most 2 sqrt n ||G||_{L^2} Prob(D>0)^{1/2} <= C||G||_*, by the selection lemma. The result is |E[G (x - t_*)^T e]| <= C||G||_* uniformly over ||e|| <= 1. Centering G and taking the supremum over e gives the covariance bound. (Taking G = 1 additionally gives ||E_{\mu_{h_0}} x - t_*|| <= C.) Dimensions below the fixed construction threshold are covered by the elementary bounds ||Cov(G,x)||^2 <= n Var(G) and ||E x - t_*|| <= 2 sqrt n.

The essential subtlety: rho is chosen *before* the selection depth d, so the middle-term absorption C_0 rho < 1/2 does not carry a self-dependent choice of depth. This is why the ordering in Section 6.6 matters.

### 12.3 Input 2b: the mean under square reweighting

**Proposition (weighted mean).** For every sufficiently small fixed delta > 0 there is C_delta such that, for N = E f^2 > 0 and nu = f^2 mu_{h_0}/N,

$$ \frac{\big\| \mathbb E_\nu x - t_* \big\|}{\sqrt n} \le C\,\delta + \frac{C_\delta}{\sqrt n}\Big( 1 + \frac{\mathcal D_{h_0}(f)}{N} \Big). $$

The leading constant C is independent of delta and of the selection depth; only C_delta and the finite diagnostic length depend on delta. This estimate uses no implicit test vector.

*Proof mechanism.* Select the cutoffs with a fixed threshold rho <= delta, small enough for the root comparison. On each cutoff support, the residual-to-root comparison gives ||m^k - t_*|| <= C delta sqrt n, with the constant from the recurrence and the inverse bound of Section 13, independent of k. For any unit vector e, sum the *square-density* weak residual rule with target e through level k, using the multiplier C_hat_k. After dividing by N, this bounds |E_nu[C_hat_k (x - m^k)^T e]| <= C_delta (1 + D_{h_0}(f)/N). The cutoff weights sum to at most 1, so all root-comparison terms together cost at most C delta sqrt n. The leftover mass contributes at most 2 sqrt n Prob_nu(D>0), controlled by the weighted part of the selection lemma. Adding and dividing by sqrt n, then taking the supremum over e, proves the bound. Small dimensions are covered by ||E_nu x - t_*||/sqrt n <= 2 after enlarging C_delta, which leaves the leading constant C unchanged.

### 12.4 What the two constants are, and why the split matters

The constant C of Input 2b comes from root stability (Section 13) and is independent of delta; the constant C_delta comes from the finite diagnostic length of the ordinary recipe and blows up as delta -> 0. In the assembly of Section 7.3 this split is exactly what allows the leading term 2 C T delta to be made smaller than v/8 *first*, with C_delta's growth deferred to a term of order 1/sqrt(n) that is then killed by taking n large. Without the split, the choice of delta and the choice of n would be circular. This is the single most important bookkeeping point in the whole proof.

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## 13. Input 1: stable fields along observations

This is the longest technical input. It says that along a typical observation path, the field equation has a *strictly positive linearization* at every approximate solution. That positivity is what Section 11.6 used for its inverse margin, and what guarantees the unique root of Section 12.3.

### 13.1 The field equation and the Hessian

Fix a symmetric J and a field h_0. The *field map* is

$$ F_{J,h_0}(y) = y - h_0 + \big( B(y) I - J \big) m(y), \qquad m(y) = \tanh y, \quad q(y) = \langle m(y)^2 \rangle, \quad B(y) = j(1 - q(y)). $$

Its zero r is the TAP solution; t_*(h) = tanh r. Because m = tanh y has derivative V_y = diag(1 - m(y)^2), the derivative of F is I + (B I - J) V_y minus the rank-one variation of B, and its *symmetric* linearization, allowing a nearby diagonal A in place of V_y, is

$$ \mathsf H_J(y,A) = I + A^{1/2}\Big( B(y) I - J - \tfrac{2j}{n} m(y) m(y)^{\mathsf T} \Big) A^{1/2}. $$

The symmetric form is what we control; it avoids needing individual diagonal entries of A to be large.

### 13.2 The stability proposition (Input 1)

**Proposition (stability along observations).** Fix 0 < j < 1 and T < infinity. There are constants eta, c, eps_A, rho_0, a > 0, depending only on j, T, and events of matrices J with probability tending to one, such that for every J in the event: if X ~ mu_0 and B_0 is independent Brownian motion with Y(t) = tX + B_0(t), then with conditional probability at least 1 - e^{-an}, simultaneously for all 0 <= t <= T, all y, and all diagonal A,

$$ \Big( \| F_{J,Y(t)}(y) \| \le 2\rho_0\sqrt n, \ 0 \le A \le (1+\eta) I, \ \| A - V_y \|_{\mathrm{Fro}} \le \epsilon_A\sqrt n \Big) \ \Longrightarrow\ \mathsf H_J(y,A) \succeq c I. $$

We may require j(1+eta)^2 < 1 and ||J|| <= 2 sqrt(j) + theta on the event. The corollary with the stopping time tau_0 is the direct consequence: stop at the first time the *smaller* residual threshold rho_0 sqrt n is approached by an unstable witness, which happens with probability at most e^{-an}.

The proof has four parts: a scalar rate bound (13.3), a conditional Hessian positivity near the equality curve (13.4), a local-volume lower bound that turns any unstable zero into a detectable footprint (13.5), and a planting argument that removes the auxiliary matrix diagonal (13.6).

### 13.3 The Gaussian integral and its scalar rate

Work first under a *planted* Gaussian law,

$$ \widetilde J = W + \tfrac{j}{n} \mathbf 1 \mathbf 1^{\mathsf T}, \qquad h_0 = t \mathbf 1 + \sqrt t\, g, $$

with W a centered GOE of scale j, g standard Gaussian, and the diagonal of tilde J retained. For a set E of pairs (y, tilde J) define the Gaussian integral

$$ I_{t,\sigma}(E) = (2\pi\sigma^2)^{-n/2} \int \mathbf 1_E \exp\Big( -\frac{\| F_{\widetilde J,h_0}(y) \|^2}{2\sigma^2} + \frac{n j b(y)^2}{2} \Big)\,\mathrm dy. $$

The weight e^{n j b^2/2} compensates the exponential scale of the local determinant (it is exactly what the local-volume lemma compares to). The central computation is the *conditional reduction*: averaging the integral over the planted randomness gives

$$ \mathbb E\, I_{t,\sigma}(E) = \int \sqrt{\tfrac{s}{S}}\, G_{t,\sigma,n}(y)\, \mathbb P\Big( (y,\widetilde J)\in E \ \Big| \ W m + \sqrt{t+\sigma^2}\, g' = z \Big)\,\mathrm dy, $$

where G is an explicit product density times e^{n psi}, with psi = j(a_y - s b)^2/(2sS), d = t + jM, s = t + sigma^2 + jq, S = s + jq, and a_y = <(y - d 1)m>. The conditional probability is an ordinary Gaussian regression.

The scalar entropy inequality of Section 10 then gives psi - D(P_y || N(d,s)) <= 0, with equality precisely at the consistent Gaussian P_*(t) of Section 10.2. A compactness-and-cover argument (using the variational description of relative entropy) upgrades this pointwise rate to the *uniform scalar rate bound*: away from any open neighbourhood of the equality curve, the integral of G is at most e^{-a n}, uniformly in the time and in small sigma; and on every fixed second-moment ball the integral is at most e^{eps n} for any eps. This is where the scalar inequality pays off: it converts "entropy is positive off the curve" into "the Gaussian integral is exponentially small off the curve".

### 13.4 The conditional Hessian

Near the equality curve, the conditional law of the planted matrix given the observation is explicit: the GOE part perpendicular to the planted direction survives unchanged, while the parallel part and the rank-one correction are Gaussian regressions. Concretely, with u = m/sqrt(nq), Z_1 = (y - d 1)/sqrt(ns), and P_u = I - uu^T, one writes W = P_u W P_u + u w^T + w u^T + alpha uu^T, and the conditional means/variances of the three blocks give a fresh GOE W_0:

$$ P_u W P_u = P_u W_0 P_u, \qquad w = j\sqrt{q/s}\, Z_1 - \tfrac{ja_y}{s} u + \kappa P_u W_0 u, \qquad \alpha = \tfrac{2j(a_y + Bq)}{S} + \sqrt{\tfrac{t+\sigma^2}{S}}\, u^{\mathsf T} W_0 u. $$

Reducing to the *ideal* Gram comparison (replace empirical weighted averages by their equality-law values, sigma = 0, s = s_*(t)) leaves the identity minus a perturbation supported on three vectors X_1 = Y(1/sqrt n), M_1 = sqrt q Y(u), G_1 = Y(Z_1)/sqrt s, where Y(p) = S_A^{-1/2} A^{1/2} p. The coefficient matrix of this perturbation is

$$ C = j \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 - B/s & 1 \\ 0 & 1 & 0 \end{pmatrix}. $$

The three-vector Gram matrix H has entries built from b, e = E_{N(s,s)}[m^2(1-m^2)], and b/s - 2b + 6e, all obtained from two Gaussian symmetry identities (pairing y with -y and integrating by parts). The result is

$$ \det(I - HC) = (d_0 + 2 g_1)(d_0 + g_1)^2 > 0, \qquad d_0 = 1 - j b, \quad g_1 = j e. $$

The determinant is positive, and subtracting the rank-one piece j X_1 X_1^T is already positive definite (j||X_1||^2 = jb < 1) with the remaining rank-two piece having at most one non-positive direction. So the Schur complement, and therefore the conditional Hessian, is positive definite with a uniform margin. The passage from the ideal Gram computation to the actual one, and to all admissible diagonals A simultaneously, is by a finite mesh in A with a H\"older comparison; the omitted mass is absorbed by decreasing the Frobenius radius eps_*.

So: with conditional probability at least 1 - e^{-a_* n}, the Hessian H_{\widetilde J}(y,A) is positive definite near the equality curve, simultaneously for all admissible A near V_y and all y in the good scalar region.

### 13.5 Local volume: unstable zeros leave a footprint

The conditional Hessian controls typical y, but an *adversarial* unstable approximate zero must be ruled out at every y. The complementary half is a deterministic lower bound: a ball around any approximate zero contributes a definite exponential amount to the Gaussian integral.

**Lemma (local volume).** Fix j, K, r, eps. Suppose ||tilde J|| <= K and a set E contains the ball of radius r sqrt n around y. If the regularized determinant obeys (1/n) log det(L^T L + gamma I) <= j b(y)^2 + eps_1 and ||F(y)|| <= rho sqrt n, then

$$ (2\pi\sigma^2)^{-n/2} \int_E \exp\Big( -\frac{\| F(y') \|^2}{2\sigma^2} + \frac{n j b(y')^2}{2} \Big)\,\mathrm dy' \ge e^{-\epsilon n} $$

for all large n, provided eps_1 is small, gamma fixed, sigma small, rho small, in that order.

*Why.* Integrate over the Gaussian ellipsoid Delta = sigma Q^{-1/2} g'' with Q = L^T L + gamma I, restricted to ||g''|| <= 2 sqrt n. The choice of ellipsoid makes the linearized residual <= sigma ||g''||; the Taylor remainder and the variation of b are controlled by the Gaussian fourth moments; Jensen applied to the resulting conditional expectation, with the determinant hypothesis cancelling j b(y)^2/2, gives the e^{-eps n} lower bound. Regularizing by gamma removes the need for a sign assumption on det L (the parametrix has a sign, and this is what makes the lower bound valid rather than signed).

### 13.6 Excluding witnesses, meshing time, and undoing the planting

Let E_t(W) be the set of y for which *some* admissible A near V_y has a Hessian below c_E (small compared to c_*/2). The scalar rate bounds give E I_{t,sigma}(E_t) <= e^{-a_0 n} uniformly in t, sigma. On the other hand, the log-determinant lemma (Section 9.3) holds simultaneously at every y with rate eps_1, and an unstable approximate zero on that event forces I_{t,sigma}(E_t) >= e^{-eps n} by local volume. Choosing eps < a_0/2, Markov gives: the probability of any unstable witness at a fixed time is at most C e^{-a_2 n}.

To pass from a fixed time to the whole interval [0, T], take a deterministic mesh with step Delta. A Brownian increment over a step of length Delta is controlled by the reflection bound, and the deterministic drift contributes at most Delta sqrt n. Choosing Delta so the total normalized increment is at most rho/4 shows that a pathwise violation forces a violation at a mesh point; the Hessian and its witness do not see the field, and the residual increases by at most rho sqrt n/4. A union bound over the finite mesh gives the annealed path bound.

Finally, undo the planting. Let Z(J) = sum_x e^{x^T J x/2} and L(J) = Z(J)/E Z(J). Size-biasing the original law of (J, X) by L(J) makes X uniform and turns the off-diagonal entries of J into independent Gaussians with means j X_i X_j / n; adding a Gaussian diagonal and conjugating by D_X realizes the planted law exactly, and the witness conditions are invariant under this conjugation. To read the bound back, one needs a lower tail for L(J). Paley-Zygmund plus Gaussian concentration of log Z give, for every fixed delta > 0,

$$ \mathbb P\{ L(J) < e^{-\delta n} \} \longrightarrow 0. $$

Putting the planted path bound together with this lower tail, and choosing the rate a below a_4/2, yields the proposition. The measurability of the witness events is checked by restricting the parameter space to compacts and using that the witness inequalities are closed.

### 13.7 The root lemma

The stability proposition is a probabilistic statement. The clean deterministic consequence used by Sections 11 and 12 is:

**Lemma (root and inverse bound).** Fix j, K, c, r_0. Let J be symmetric with ||J|| <= K, and h_0 arbitrary. Suppose H_J(y,V_y) >= c I whenever ||F_{J,h_0}(y)|| <= r_0 sqrt n. Then F_{J,h_0} has exactly one zero r, and for ||F_{J,h_0}(y)|| < r_0 sqrt n,

$$ \| y - r \| \le C \| F_{J,h_0}(y) \|, \qquad C = 1 + (K + 3j)/c. $$

*Why.* The inverse F'(y)^{-1} has norm at most C on the region, by the resolvent identity and F'(y) = I + L_1 V_y with ||L_1|| <= K + 3j. For existence and uniqueness, work in m = tanh y and use the antiderivative potential Phi(m) = sum_i ∫_0^{m_i} atanh z dz - h_0^T m - ½ m^T J m + (j/2)||m||^2 - (j/4n)||m||^4. Its critical points are the zeros, and its Hessian at a zero is exactly H_J(y,V_y) >= c I, so critical points are strict local attractors. A box argument gives existence; the negative gradient flow and positivity of the Hessian make the basins open and disjoint, hence one zero. The distance bound comes from following the flow y'(s) = -F'(y)^{-1}F(y), on which F(y(s)) = (1-s)F(y); integrating the velocity bound gives ||y - r|| <= C||F(y)||.

This lemma is used twice: in Section 11.3 to compare m^{k-1} with t_* on Omega_rho, and in Section 12.3 for the root-comparison step. Its constant C is the origin of the delta-independent leading constant of Input 2b.

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## 14. Input 3: the terminal gap

The last input is a Poincare inequality valid after a long enough, but fixed, observation time T. It is the only place the *signed* structure of J is used directly in the terminal argument (through a two-spin curvature identity and small principal submatrices). The statement is about the observation law at time T, for *any* initial spin law.

### 14.1 The signed curvature inequality

Write the generator as H = sum_i (I - P_i). One checks Hf = sum_i (x_i - m_i^1) d_i f = sum_i (x_i - m_i^1) g_i, with g_i = d_i f and a_i = v_i^1 g_i. Each P_i is an orthogonal projection, so H is self-adjoint and positive, and E[f Hf] = D(f) = E sum_i a_i g_i.

Define the entrywise matrix quantities

$$ U_{ij} = \tanh J_{ij}, \qquad Q_{ij} = U_{ij}^2, \qquad r_4(U) = \max_i \sum_j U_{ij}^4, $$

with U_{ii} = Q_{ii} = 0.

**Lemma (signed curvature).** There are universal constants u_0 > 0 and C < infinity such that, for every n, every symmetric zero-diagonal J with max_{i,j}|U_{ij}| <= u_0, every h_0, and every f,

$$ \mathbb E_{\mu_{h_0}}[(Hf)^2] \ge (1 - C r_4(U))\, \mathcal D_{h_0}(f) - \mathbb E_{\mu_{h_0}}\big[ a^{\mathsf T} U a + C\, |a|^{\mathsf T} Q |a| \big]. $$

*Why.* Sum an *exact* two-spin calculation over pairs (i,j). Condition on all but spins i,j; their conditional density is proportional to exp(alpha x + gamma y + J_{ij} xy), with means a = tanh alpha, b = tanh gamma and mixing u = tanh J_{ij}. The density ratio against the product law has the exact form (1 + u x y)/(1 + u a b). Expand f on the four functions of (x,y); its two gradients have the form g_i = p + r(y-b), g_j = s + r(x-a), with a shared mixed coefficient r. Two one-spin identities reduce the cross term to four scalars; the mixed coefficients rp and rs *begin at order u^2* (their first-order parts cancel), and the r^2 coefficient is 1 + O(u). Adding u times the appropriate weight product and using Young's inequality to send the mixed terms into r^2/2 leaves a lower bound that is a negative signed pairing -u (a_i a_j) minus a quartic cost proportional to u^4 (v_i g_i^2 + v_j g_j^2). Summing over ordered pairs, the signed cost becomes a^T U a, the abs cost becomes |a|^T Q |a|, and the quartic cost is 2C r_4(U) D(f), giving the stated inequality. The cancellation keeping rp, rs at order u^2 is exactly why only a quartic row cost survives, not a quadratic one.

### 14.2 Small principal submatrices

The curvature inequality has a dangerous signed term a^T U a and an absolute term |a|^T Q |a|. If the coordinates with non-negligible a_i occupy a *small set* S, these terms are controlled by the principal submatrices U_{SS}, Q_{SS}.

**Lemma (small principal submatrices).** Fix j in (0,1). There is M = M(j) such that, for every eps > 0, there is alpha in (0, 1/2) with, with probability tending to one over J:

$$ \max_{i,j}|J_{ij}| \le n^{-1/4}, \quad \|J\|, \|U\|, \|Q\| \le M, \quad r_4(U) \le M n^{-1/2}, $$

and, for every S subset {1,...,n} with |S| <= alpha n, ||U_{SS}|| <= eps and ||Q_{SS}|| <= eps. The same M works for all eps.

*Why.* For a unit vector z, z^T J z is a centered Gaussian with variance at most 2j/n, so P(|z^T J z| > t) <= 2 exp(-n t^2/(4j)). A 1/4-net of the unit sphere in R^k has at most 9^k points, and ||B|| <= 2 max_{net} |z^T B z|, so the operator norm of J is at most a fixed M_0 except with probability e^{-cn}. For each row, the truncated square sum sum_{ell in S} J_{i ell}^2 has an exponential tail with rate of order n, and a union over rows and over the C(n,k) sets of size k = floor(alpha n), with log C(n,k) ~ n h(k/n) -> 0, makes the *simultaneous* bound over all small S hold once alpha is small enough. The entry tail max|J_{ij}| <= n^{-1/4} is a standard union bound. Since |tanh z| <= |z| and |tanh z - z| <= |z|^3/3, U and Q inherit these bounds from J after subtracting a matrix of norm n^{-1/4}/3.

### 14.3 The terminal observation gap

**Proposition (terminal gap).** Fix j in (0,1). There are T > 0, c > 0, and events depending only on J, of probability tending to one, such that for every large n and every J in the event, the following holds. Let X have *any* probability law on the cube, and Z an independent standard Gaussian vector. Set Y = T X + sqrt T Z. Then, conditionally on J, except on an event of probability at most e^{-c n}, the inequality Var_{mu_Y}(f) <= 2 D_Y(f) holds simultaneously for every real f.

*Proof mechanism.* Choose the tolerances in order: eps with (1+C)eps <= 1/8 (C the curvature constant), then alpha and M from the small-subset lemma, then T large enough that

$$ \frac{16 M^2}{T^2} \le \frac{\alpha}{2}, \qquad (1+C)M\big(2\sech(T/4) + \sech^2(T/4)\big) \le \tfrac18, \qquad e^{-T/8} \le \frac{\alpha}{4(e-1)}. $$

First, count coordinates whose observation field stays small. Given X, the indicators {|Y_i| < T/2} are independent, each succeeding with probability at most e^{-T/8}; exponential Markov bounds the number B_Y of such coordinates by alpha n/2 except with probability e^{-alpha n/4}.

Second, for every spin configuration x, ||Jx||^2 <= M^2 n, so the set S(x) = {i : |Y_i + (Jx)_i| < T/4} has size at most (16 M^2/T^2) n <= alpha n/2, *simultaneously* for all x. At these coordinates the conditional variance v_i is at most sech^2(T/4) = eta, so both ||a_{S(x)^c}||^2 and ||a||^2 are controlled by the total q_f(x) = sum_i v_i g_i^2, and the dangerous part of a is confined to the small set S(x).

Third, split a^T U a and |a|^T Q |a| between S(x) and its complement. The small-subset bound gives, for this small set, a bound eps q_f(x) (plus a global-norm cross piece M(2 sqrt(eta) + eta) q_f(x)). Feeding this into the curvature inequality yields E[(Hf)^2] >= (1 - Cr_4(U) - (1+C)(eps + M(2 sqrt(eta)+eta))) D_Y(f) >= (1/2) D_Y(f). The use of S(x) is legitimate even though S depends on x, because the small-subset lemma holds on one matrix event for *every* small set simultaneously.

Finally, translate the curvature lower bound into a gap. The kernel of H is exactly the constants (vanishing E[fHf] forces every v_i g_i^2 to vanish, and strict positivity propagates). For an eigenfunction of H with eigenvalue lambda, lambda^2 E f^2 = E[(Hf)^2] >= (1/2) lambda E f^2, so lambda >= 1/2. Expanding in an eigenbasis gives Var_{mu_Y}(f) <= 2 D_Y(f). The failure probability is e^{-alpha n/4}, uniformly in the law of X, so one may take c = alpha/4.

This closes the third and last input. With it, the assembly of Section 7 is complete, and the main theorem follows.

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## 15. What this paper is not doing

It is worth being explicit about the boundaries, because the theorem is easy to overstate.

- *Not at or below beta = 1.* The proof uses j(1+eta)^2 < 1 in the matrix estimates and beta < 1 in the entropy equality curve. Nothing here applies at the spin-glass transition or in the low-temperature phase.
- *Not a sharp constant.* C_beta = 4 e^{C_d T} is absurdly large; it is finite, and that is all that is claimed. The value is not optimized at any step.
- *Not an external-field statement at zero field.* The proof controls posterior measures at nonzero fields and then specializes to mu_0. It does not give a gap statement for mu_h at a fixed nonzero h of comparable generality; the fields here are the random observation fields.
- *Not a mixing-time statement by itself.* The main theorem is a spectral-gap (Poincare) statement. The discrete-chain O(n) attempt bound follows by a one-line conversion, but the paper does not prove a stronger (log-Sobolev or modified-log-Sobolev) inequality.
- *Not sharp about which observables it controls.* The inequality holds for all functions f; the statement is not restricted to linear or low-degree observables, which is the whole point, but the constant is not tailored to any particular f.

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## 16. The high-level "why"

Strip away the constants and the diagnostics and the skeleton of the proof is short.

The gap question is local: does each spin have enough conditional variance left over after seeing its neighbours? At low temperature the answer is not visible from the raw couplings, because the influences are dense and signed. So instead of measuring conditional variance directly, the proof *observes the spins through Gaussian noise*. Observation is a physical operation: it cannot make the dynamics worse (Dirichlet monotonicity), and it makes the posterior law *easier*, because each spin becomes more pinned down. At a large enough observation time the posterior is provably close to a product measure, and the gap there is trivial (Input 3, the curvature).

The only danger is that the variance could be lost *on the way* to the observation time. That is controlled by two complementary facts. Along typical paths, the posterior mean of any function moves at a rate bounded by its own variance plus its Dirichlet form (Input 2a, the directional covariance bound), which is an honest variance-retention statement. On the thin set of bad paths, the version of the observation law induced by weighting the initial law by f^2 is shown to be not too heavy (Input 2b, the weighted mean bound, turned into an entropy bound). The two facts pull in opposite directions: one says variance cannot drop too fast, the other says it cannot pile up on rare events. Together they force the original variance to be comparable to the original Dirichlet form, hence the gap.

The reason the two facts can *both* be true, and the reason the whole scheme needs the specific TAP and entropy machinery, is the positivity of the linearized field map along the observation path (Input 1, stability). That positivity is what makes the field equation have a unique solution, what gives the implicit test vector its inverse margin, and what keeps the root comparison uniform in the recursion depth. In one sentence: *observe the spins, trust that the observation stays invertible, and make sure no variance escapes before the terminal curvature does its job.*

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## 17. Appendix: the main objects in one page

*Model.* Spins x in {-1,1}^n. J symmetric, zero diagonal, J_{ij} ~ N(0, j/n), j = beta^2. mu_h(x) propto exp(½ x^T J x + h^T x). Conditional mean and variance at site i: m_i = tanh(h_i + (Jx)_i), v_i = sech^2(h_i + (Jx)_i) = 1 - m_i^2.

*Dynamics.* P_i^h = conditional average over x_i. Generator L_h = sum_i (P_i^h - I). Unscaled Dirichlet form D_h(f) = sum_i E_{mu_h}(f - P_i^h f)^2 = E sum_i v_i (d_i f)^2, with discrete derivative d_i f = (f(x) - f(x^(i)))/(2x_i). Gap = inf_f D_h(f)/Var_{mu_h}(f). Discrete chain gap = continuous gap / n.

*Test norm and residual.* ||G||_*^2 = E G^2 + D_h(G). Primary recursion m^l = tanh h^l, b_l = <1 - (m^l)^2>, R_l = m^{l-1} - m^l, h^{l+1} = h_0 + J m^l - j b_l m^{l-1}, with S_l = R_l^T m^l / sqrt n and telescoping ||R_l||^2/n = b_l - b_{l-1} - 2 S_l / sqrt n.

*Field equation and Hessian.* F_{J,h_0}(y) = y - h_0 + (B(y)I - J)m(y), B(y) = j(1 - <m(y)^2>), m(y) = tanh y. Root r, t_* = tanh r. H_J(y,A) = I + A^{1/2}(B(y)I - J - (2j/n) m m^T) A^{1/2}.

*Observation channel.* Y(t) = tX + B(t), posterior mu_{Y(t)}. Filtering: d(E_t g) = Cov_t(g,x)^T dI(t). Dirichlet monotonicity: E D_{Y(t)}(f) <= D_0(f). Drift-energy: D(P^f || P) = ½ E^f ∫_0^tau ||a_t||^2 dt with a_t = E_{nu_t} x - E_t x.

*The three inputs.* (1) Stability of fields along observation paths. (2a) ||Cov_{mu_h}(G,x)|| <= C(Var_{mu_h}(G) + D_h(G))^{1/2}. (2b) ||E_nu x - t_*||/sqrt n <= C delta + (C_delta/sqrt n)(1 + D_h(f)/N). (3) Var_{mu_{Y(T)}}(f) <= 2 D_{Y(T)}(f) at a large fixed T.

*The final inequality.* With v = e^{-C_d T}: good paths carry at most 2 d_0, bad paths at most v/4, and the total is at least v - (1-v) d_0; hence d_0 >= v/4, and Var_{mu_0}(f) <= (4/v) D_0(f). The constant is C_beta = 4 e^{C_d T}.

*Choice order.* T (Input 3); then stability constants, a, and the Input 2a diagnostic length (giving C_d and v); then the leading Input 2b constant C; then delta (so 2 C T delta <= v/8); then the Input 2b diagnostic length and C_delta; then n.

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## 18. Closing word

The theorem says that, for every fixed temperature above the spin-glass threshold, a single random SK magnet almost surely has a spectral gap bounded below by a constant independent of its size. The proof does not fight the dense signed couplings head-on; it observes the spins through noise, so the posterior measures become tame, and then it transfers a functional inequality from the tame posterior back to the untamed prior. The three inputs are exactly the three transfers of information needed for that back-and-forth, and the assembly in Section 7 is where they snap shut.

If you have followed the shape of the argument -- observe, stay stable, retain the variance, control the exceptions, finish with curvature -- then you have followed the paper. The technical mass in Sections 9, 11, 12, 13, 14 is the price of making each of those five words honest at every fixed beta < 1, uniformly in n. The beauty of the argument is that it never needs to know the low-temperature spin-glass structure; it only needs the high-temperature phase to be *observably* tame, and that is exactly what stochastic localization delivers.