# Critical slowing down in the Sherrington--Kirkpatrick model

*An approachable but complete proof.*

---

## 0. The paper in one sentence

At the critical temperature of a mean-field spin glass, there is a direction in configuration space along which the Gibbs measure is spread out on a scale much larger than a single spin. Flipping one spin at a time changes that direction very slowly. Therefore most equilibrium starting configurations remain far from equilibrium for a long time.

The precise result is that the mixing time is at least of order $n^{2/3}$ in continuous time, or $n^{5/3}$ attempted spin flips in discrete time, up to arbitrarily small power losses. The same static estimates also pin down the covariance matrix and the slowest linear observable.

---

## 1. The model

Let $n$ be a large positive integer. A spin configuration is a point

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

The Sherrington--Kirkpatrick (SK) interaction is a random symmetric matrix $W=(W_{ij})_{1\le i,j\le n}$ with

$$
W_{ij}\sim N(0,1/n)\quad(i<j),
\qquad
W_{ii}\sim N(0,2/n),
\qquad
W_{ij}=W_{ji}.
$$

Thus $W$ is a Gaussian orthogonal ensemble (GOE) matrix. The diagonal entries do not affect the spin model, because $x_i^2=1$ for every spin. They are included only because they make the matrix spectrum behave exactly like GOE; we will remove them from the Gibbs law whenever convenient.

The Hamiltonian is

$$
H_W(x)=\frac12 x^T W x
=\sum_{i<j}W_{ij}x_ix_j+\frac12\sum_i W_{ii}.
$$

The last term is the same for every configuration $x$, so it cancels from every probability ratio. The Gibbs measure is

$$
\mu_W(x)=\frac{e^{H_W(x)}}{
\sum_{y\in\{-1,+1\}^n}e^{H_W(y)}
}
\propto
\exp\!(\sum_{i<j}W_{ij}x_ix_j).
$$

We often write $\mu=\mu_W$ when the disorder $W$ is fixed.

The number $\beta$ is the inverse temperature. In this paper the inverse temperature is exactly $\beta=1$, the critical value. For $\beta<1$ the model is in the high-temperature phase and mixes quickly; at $\beta=1$ the spectral edge is critical and slow modes appear on scale $n^{-2/3}$.

---

## 2. Heat-bath dynamics

A heat-bath update at site $i$ replaces $x_i$ by a fresh spin chosen from its conditional law given all other spins. Since a spin has only two possible values, this means:

$$
x_i=
+1 & with probability \dfrac{e^{g_i(x)}}{2cosh g_i(x)},\\[2mm]
-1 & with probability \dfrac{e^{-g_i(x)}}{2cosh g_i(x)},

\qquad
g_i(x)=\sum_{j\ne i}W_{ij}x_j .
$$

Equivalently, the new spin has conditional mean $tanh g_i(x)$.

We use two clocks.

* **Continuous time.** Each site has an independent rate-one exponential clock. When its clock rings, that site is refreshed. The generator is

$$
\mathcal L=\sum_{i=1}^n(P_i-Id),
\qquad
P_if(x)=\mathbb E_\mu[f(x)\mid x_{-i}],
$$

where $x_{-i}$ means all spins except $x_i$.

* **Discrete time.** At each step one site is chosen uniformly at random and refreshed. The transition operator is

$$
P=\frac1n\sum_{i=1}^n P_i .
$$

These clocks are related by

$$
\mathcal L=n(P-Id).
$$

So one unit of continuous time corresponds roughly to $n$ attempted discrete updates, but the exact relation is the displayed identity.

For a starting configuration $x$, let

$$
d_t^{ct}(W,x)
=\|e^{t\mathcal L}(x,\cdot)-\mu_W\|_{TV},
\qquad
d_k^{dt}(W,x)
=\|P^k(x,\cdot)-\mu_W\|_{TV},
$$

where total variation is

$$
\|\nu-\mu\|_{TV}
=\frac12\sum_x|\nu(x)-\mu(x)|
=\sup_{A⊆\{-1,+1\}^n}|\nu(A)-\mu(A)|.
$$

The worst-start mixing times are

$$
t_{mix}
=\inf\{t\ge0:\max_x d_t^{ct}(W,x)\le1/4\},
$$

$$
k_{mix}
=\min\{k\in\mathbb N\cup\{0\}:\max_x d_k^{dt}(W,x)\le1/4\}.
$$

All probabilistic statements below are over the random matrix $W$ (and, when we temporarily add an auxiliary Gaussian diagonal, over that auxiliary randomness as well).

---

## 3. The three main results

### Theorem A (typical equilibrium starts are slow)

For every deterministic sequence $t_n\ge0$ with

$$
t_n=o(n^{2/3}),
$$

we have

$$
\mu_W\{x:d_{t_n}^{ct}(W,x)>1/4\}
→ 1
\qquadin probability over W.
$$

For every deterministic integer sequence $k_n\ge0$ with

$$
k_n=o(n^{5/3}),
$$

we have

$$
\mu_W\{x:d_{k_n}^{dt}(W,x)>1/4\}
→ 1
\qquadin probability over W.
$$

The theorem tests each realized equilibrium starting configuration separately. If instead one averaged the transition law over the initial configuration drawn from $\mu_W$, stationarity would make the law equal to $\mu_W$ at every time. The theorem says that this averaging hides a split: most fixed starts are still far away.

### Theorem B (covariance and linear tests)

Let

$$
Σ=Cov_{\mu_W}(x)=\mu_W[xx^T],
$$

and let

$$
\mathcal D_{\mu_W}(f)
=\sum_{i=1}^n\mu_W[Var(f\mid x_{-i})]
$$

be the unscaled Dirichlet form. For linear functions $a^Tx$, define

$$
\mathcal R_{lin}(\mu_W)
=\sup_{a\in\mathbb R^n∖\{0\}}
\frac{Var_{\mu_W}(a^Tx)}
{\mathcal D_{\mu_W}(a^Tx)} .
$$

For every positive deterministic sequence $M_n\to∞$, with probability tending to one both $\|Σ\|$ and $\mathcal R_{lin}(\mu_W)$ lie in

$$
[
\frac{n^{2/3}}{M_n},
\ M_n n^{2/3}
].
$$

The factor $M_n$ can diverge as slowly as we like. The theorem only controls linear functions; it does not by itself control every function.

### Theorem C (critical mixing bounds)

For every fixed $ε>0$,

$$
\mathbb P{
n^{2/3-ε}\le t_{mix}\le e^{ε n}
}→ 1,
$$

$$
\mathbb P{
n^{5/3-ε}\le k_{mix}\le e^{ε n}
}→ 1.
$$

The lower bounds come from Theorem A. The upper bounds are proved by comparison with one fixed high-temperature model and are exponentially weak. The sharper matching exponents are the subject of the companion paper, but the present paper proves the displayed bounds completely.

---

## 4. Roadmap of the proof

There are four layers.

### 4.1 Static spectral input

Add an independent Gaussian diagonal to $W$ so that the resulting matrix is GOE. Let $v$ be a unit eigenvector for its largest eigenvalue. Since the diagonal is constant on the cube, the Gibbs measure is unchanged.

The key static statement is:

$$
\mu_W\{x:|v^Tx|\le b_n n^{1/3}\}→ 0
\quadin probability
$$

for every deterministic $b_n\to0$. In words: the top eigenvector projection is rarely small. Its typical size is at least a constant times $n^{1/3}$.

### 4.2 From static to dynamic

For $g(x)=v^Tx$,

$$
\mathcal D_{\mu_W}(g)
=\sum_i v_i^2\,\mu_W[sech^2 g_i]
\le\sum_i v_i^2=1.
$$

Reversibility gives a mean-square increment bound along a stationary trajectory:

$$
\mathbb E_\mu[(g(X_t)-g(X_0))^2]\le2t,
\qquad
\mathbb E_\mu[(g(Y_k)-g(Y_0))^2]\le(2k)/(n).
$$

At a time $t_n=o(n^{2/3})$, the typical change is much smaller than the typical size $n^{1/3}$. Therefore the sign of $g$ persists. For each fixed starting point that is not too close to the hyperplane $g=0$, the half-space $\{y:g(y)g(x)>0\}$ has equilibrium mass at most $1/2$, and the chain is still on the wrong side with high probability. That gives the total-variation lower bound.

### 4.3 Where the $n^{1/3}$ scale comes from

The top eigenvector projection is controlled by the spherical model. After conditioning independent Gaussians on their radius, the tilted spherical law has coordinate variances of order

$$
\frac1{s_0},
\qquad
s_0=L_n n^{-2/3},
$$

with $L_n\to∞$ arbitrarily slowly. Thus the standard deviation of one coordinate is of order

$$
s_0^{-1/2}=n^{1/3}L_n^{-1/2}.
$$

A Haar-averaging argument then transfers this spherical estimate to the Ising cube.

### 4.4 Covariance and linear functions

The squared overlap between two independent Gibbs samples controls

$$
Tr(Σ^2)=n^2\mathbb E_{\mu ⊗ \mu}[((x^Ty)/(n))^2].
$$

The spherical transfer proves that this overlap is of order $n^{-4/3}$, hence $Tr(Σ^2)$ is of order $n^{2/3}$, hence $\|Σ\|$ is at most order $n^{2/3}$. The top projection gives the matching lower bound.

For linear inverse Rayleigh quotients, the Dirichlet denominator weights each coefficient by a conditional variance. A cavity deletion argument shows that only a small set of sites can have unusually large weight, and those sites contribute a negligible covariance block.


---

## 5. Elementary identities

Before proving anything hard, we record the bookkeeping that makes every later step work.

### Lemma 5.1 (Heat-bath forms)

For every function $f:\{-1,+1\}^n\to\mathbb R$, each $P_i$ is an orthogonal projection in $L^2(\mu)$. The unscaled Dirichlet form is

$$
\mathcal D_\mu(f)
=-< f,\mathcal Lf>_\mu
=\sum_{i=1}^n\mu[(f-P_if)^2]
=\sum_{i=1}^n\inf_{h_i}\mu[(f-h_i(x_{-i}))^2],
$$

where the infimum is over all functions $h_i$ of $x_{-i}$. The discrete form is

$$
< f,(Id-P)f>_\mu=\frac1n\mathcal D_\mu(f),
$$

and the spectrum of $P$ lies in $[0,1]$.

For a linear function $a^Tx$,

$$
\mathcal D_\mu(a^Tx)=\sum_{i=1}^n a_i^2\alpha_i,
\qquad
\alpha_i=\mu[sech^2 g_i],
\qquad
g_i(x)=\sum_{j\ne i}W_{ij}x_j .
$$

#### Proof

Conditional expectation onto functions of $x_{-i}$ is the orthogonal projection in $L^2(\mu)$. Therefore

$$
< f,(Id-P_i)f>_\mu
=\|f-P_if\|_{L^2(\mu)}^2,
$$

and this squared distance is the infimum over all functions of $x_{-i}$.

Summing over $i$ gives the formula for $\mathcal D_\mu$. Averaging instead gives

$$
< f,(Id-P)f>_\mu
=\frac1n\sum_i\|f-P_if\|_{L^2(\mu)}^2
=\frac1n\mathcal D_\mu(f).
$$

Since each $P_i$ is an orthogonal projection, $0\le< f,P_if>_\mu\le< f,f>_\mu$. Averaging over $i$ gives the same inequality for $P$, so the spectrum of $P$ lies in $[0,1]$.

For a linear function, everything except $a_ix_i$ is fixed once $x_{-i}$ is fixed. Under $\mu(\cdot\mid x_{-i})$, the spin $x_i$ has mean $tanh g_i$ and variance

$$
1-tanh^2 g_i=sech^2 g_i .
$$

Thus the contribution of site $i$ to $\mathcal D_\mu(a^Tx)$ is

$$
a_i^2\mu[Var(a_ix_i\mid x_{-i})]
=a_i^2\mu[sech^2 g_i].
$$

Summing proves the formula. $QED$

---

### Lemma 5.2 (Stationary increments)

Fix $W$ and a function $f:\{-1,+1\}^n\to\mathbb R$. Let $(X_t)_{t\ge0}$ be the continuous-time chain and $(Y_k)_{k\ge0}$ the discrete-time chain, each started from $\mu$. Then

$$
\mathbb E_\mu[(f(X_t)-f(X_0))^2]
\le2t\,\mathcal D_\mu(f),
\qquad t\ge0,
$$

and

$$
\mathbb E_\mu[(f(Y_k)-f(Y_0))^2]
\le(2k)/(n)\mathcal D_\mu(f),
\qquad k=0,1,2,... .
$$

In particular, for $g(x)=v^Tx$ with $\|v\|=1$, the right-hand sides are at most $2t$ and $2k/n$.

#### Proof

Stationarity gives

$$
\mathbb E_\mu[(f(X_t)-f(X_0))^2]
=2< f,(Id-e^{t\mathcal L})f>_\mu .
$$

The operator $-\mathcal L$ is self-adjoint and nonnegative. Diagonalize it in an orthonormal eigenbasis with eigenvalues $\lambda\ge0$. The multiplier of the $f$-component in eigenvalue $\lambda$ is

$$
1-e^{-t\lambda}\le t\lambda .
$$

Therefore

$$
2< f,(Id-e^{t\mathcal L})f>_\mu
\le2t< f,-\mathcal Lf>_\mu
=2t\mathcal D_\mu(f).
$$

For the discrete chain, $P$ is self-adjoint with spectrum in $[0,1]$. The stationary increment is

$$
2< f,(Id-P^k)f>_\mu .
$$

For $0\le\theta\le1$,

$$
1-\theta^k\le k(1-\theta).
$$

Also

$$
< f,(Id-P)f>_\mu=\frac1n\mathcal D_\mu(f).
$$

This proves the discrete bound. Finally, Lemma 5.1 gives

$$
\mathcal D_\mu(v^Tx)
=\sum_i v_i^2\mu[sech^2 g_i]
\le\sum_i v_i^2=1.
$$

$QED$

---

### Lemma 5.3 (Overlap controls covariance)

Zero field gives $\mu(x)=\mu(-x)$, so every spin has mean zero. Hence

$$
Σ=Cov_\mu(x)=\mu[xx^T],
\qquad
Σ_{ii}=1.
$$

Let

$$
q(x,y)=(x^Ty)/(n)
$$

be the overlap of two configurations. For independent $x,y\sim\mu$,

$$
Tr(Σ^2)
=n^2\mathbb E_{\mu ⊗ \mu}[q(x,y)^2].
$$

#### Proof

Expand the square:

$$
Tr(Σ^2)
=\sum_{i,j}Σ_{ij}^2
=\sum_{i,j}\mu[x_ix_j]^2 .
$$

For independent $x,y\sim\mu$,

$$
\mathbb E_{\mu ⊗ \mu}[q(x,y)^2]
=\frac1{n^2}\sum_{i,j}\mu[x_ix_j]^2 .
$$

Multiplying by $n^2$ proves the identity. $QED$


---

## 6. Spectral input near the critical edge

This section proves the matrix estimates that feed the spherical model. The reader should keep one picture in mind: GOE eigenvalues near the edge $2$ are spaced like $n^{-2/3}$. We will look a tiny distance

$$
s_0=L_n n^{-2/3}
$$

above the edge, where $L_n\to∞$ very slowly. The slowly growing factor $L_n$ is essential: it gives enough room to prove estimates without destroying the sharp $n^{2/3}$ scale.

### 6.1 The scales

Let $L_n\to∞$ be deterministic with

$$
L_n\le\log n
$$

for all sufficiently large $n$. Define

$$
s_*=n^{-2/3},
\qquad
s_0=L_n s_*=L_n n^{-2/3},
$$

$$
h_n=L_n^{1/8},
\qquad
A_n=h_n+√(\log(2+L_n)),
\qquad
\eta_n=A_n L_n^{-3/4}.
$$

Then $\eta_n\to0$. Define the deterministic reference function

$$
R(s)=\frac{2+s-√(s(4+s))}2,
\qquad s>0.
$$

Note that $R(s)=1-\sqrt s+O(s)$ as $s↓0$.

### Proposition 6.1 (An eigenvalue-only good event)

Let $W$ be the GOE matrix of Section 1, with eigenvalues

$$
\lambda_1\ge\lambda_2\ge...\ge\lambda_n .
$$

There are events $\mathcal G_n(L_n)$, measurable with respect to the eigenvalues alone, such that

$$
\mathbb P(\mathcal G_n(L_n)^c)\le C\exp(-cL_n^{1/4})
$$

for all sufficiently large $n$. On $\mathcal G_n(L_n)$, for every fixed $M<∞$,

$$
\lambda_j=2+o(s_0)\quad(1\le j\le3),
\qquad
\max_i|\lambda_i|\le3,
$$

and, for $s_0\le s\le M$,

$$
\frac1n\sum_{i=1}^n\frac1{2+s-\lambda_i}
=R(s)+o(\sqrt s),
$$

$$
\frac1n\sum_{i=1}^n\frac1{(2+s-\lambda_i)^2}
\le(C_M)/(\sqrt s).
$$

All denominators are positive. The little-oh errors are deterministic on the event: the first three eigenvalue errors divided by $s_0$ are at most

$$
C(L_n^{-1}+\eta_n),
$$

and the first resolvent error divided by $\sqrt s$ is at most

$$
C_M(L_n^{-3/2}+\eta_n).
$$

The event does not depend on $M$.

The rest of Section 6 proves this proposition.

---

### 6.2 Tridiagonal reduction

The first step is classical: a GOE matrix has the same eigenvalues as a random tridiagonal matrix. Tridiagonal matrices are much easier because their entries are independent and because their resolvents satisfy a three-term recurrence.

#### Lemma 6.2 (Tridiagonal representation)

The eigenvalues of $W$ have the same law as the eigenvalues of a symmetric tridiagonal matrix $T$ with independent entries

$$
T_{ii}\sim N(0,2/n),
\qquad
T_{i,i+1}=T_{i+1,i}\sim\frac{\chi_{n-i}}{\sqrt n}
\quad(1\le i<n),
$$

where $\chi_m$ is the Euclidean norm of $m$ independent standard normal variables.

#### Proof

Expose one column of $W$ at a time.

The first diagonal entry, the off-diagonal entries in the first column, and the trailing $(n-1)\times(n-1)$ block are independent. The first column has $n-1$ independent $N(0,1/n)$ off-diagonal entries. Choose an orthogonal transformation of the trailing $n-1$ coordinates that sends this column vector to its nonnegative length times the first coordinate vector. Conjugating by this transformation preserves eigenvalues and produces the desired first off-diagonal entry, whose law is $\chi_{n-1}/\sqrt n$.

The trailing block has density proportional to $\exp(-n\,Tr(B^2)/4)$. This density and Euclidean volume are invariant under orthogonal conjugation. Conditional on the column vector, the chosen transformation is fixed, so the conjugated trailing block is still GOE and is independent of the entries already exposed.

Repeat the construction inside the trailing block, then inside each successive block. At step $i$, the fresh column has $n-i$ independent $N(0,1/n)$ entries. This gives the stated distributions and their joint independence. $QED$

---

### 6.3 Concentration of chi entries

We need two estimates for centered chi variables. For $m\ge1$, let $\chi_m$ be as above. Then for every real $t$,

$$
\mathbb E\exp(t(\chi_m-\mathbb E\chi_m)/(\sqrt n))
\le\exp(Ct^2/n),
$$

and

$$
0\le\sqrt m-\mathbb E\chi_m\le(C)/(\sqrt m).
$$

#### Proof

The Gaussian-square moment-generating function and Chernoff's inequality give

$$
\mathbb P\{\chi_m^2\ge ma\}\le(ae^{1-a})^{m/2}
\qquad(a>1),
$$

and the same right-hand side for the lower-tail event when $0<a<1$. Since

$$
a-1-\log a\ge(\sqrt a-1)^2,
$$

we obtain

$$
\mathbb P\{|\chi_m-\sqrt m|\ge r\}\le2e^{-r^2/2}
\qquad(r\ge0).
$$

This tail bound gives $Var(\chi_m)\le C$. By Jensen's inequality,

$$
\sqrt m-\mathbb E\chi_m
=(Var(\chi_m))/(\sqrt m+\mathbb E\chi_m)
\le(C)/(\sqrt m).
$$

For the exponential bound, let $X,X'$ be independent copies of $\chi_m$ and put $Z=X-X'$. Then $Z$ is symmetric and

$$
\mathbb P\{|Z|\ge r\}\le4e^{-r^2/8},
\qquad
\mathbb E Z^{2k}\le4\cdot8^k k!
\quad(k\ge1).
$$

Jensen's inequality conditional on $X$, followed by the power series of $cosh$, gives

$$
\mathbb E e^{t(X-\mathbb E X)}
\le\mathbb E e^{tZ}
=\mathbb Ecosh(tZ)
\le1+4\sum_{k\ge1}\frac{8^k k!t^{2k}}{(2k)!}
\le e^{32t^2}.
$$

Replacing $t$ by $t/\sqrt n$ proves the exponential bound. The centered diagonal entries satisfy the same estimate directly. $QED$


---

### 6.4 The noise is controlled by a deterministic energy

We now compare the random tridiagonal matrix $T$ with a deterministic tridiagonal matrix $D$ whose off-diagonal entries are the means of the random ones.

Let $D$ be the deterministic symmetric tridiagonal matrix with zero diagonal and

$$
D_{i,i+1}=D_{i+1,i}=c_i:=√(1-i/n)
\qquad(1\le i<n).
$$

Set

$$
E=2I-D.
$$

Then $E$ is tridiagonal, has diagonal entries $2$, and has off-diagonal entries $-c_i$. It is positive definite, and $0\le E\le4I$.

The random matrix is $T$ from Lemma 6.2. Put

$$
H=T-D.
$$

The entries of $H$ are centered: the diagonal entries are centered Gaussian, and the off-diagonal entries are

$$
\frac{\chi_{n-i}}{\sqrt n}-c_i .
$$

#### Lemma 6.4 (Noise controlled by deterministic energy)

There is a universal constant $C_0$ such that, with

$$
\Phi_n(x)
=C_0n^{-1/2}A_n(x+s_0)^{1/4}
√(1+\log(1+x/s_0)),
\qquad x\ge0,
$$

the following holds. On an event $\mathcal F_n$ with

$$
\mathbb P(\mathcal F_n^c)\le C\exp(-cL_n^{1/4}),
$$

we have, simultaneously for every unit vector $u\in\mathbb R^n$,

$$
|u^THu|\le\Phi_n(u^TEu).
$$

Moreover,

$$
\sup_{x\ge0}|\Phi_n'(x)|\le C\eta_n,
\qquad
\Phi_n(0)\le C\eta_ns_0.
$$

#### Proof

The proof has three steps.

1. The deterministic energy $u^TEu$ controls where a unit vector can concentrate and how fast it can oscillate.
2. One high-probability event controls block prefixes of the random entries at every relevant scale.
3. Summation by parts turns those prefix bounds into a quadratic-form bound.

##### Step 1: energy controls mass and gradients

For a unit vector $u$, write

$$
e=u^TEu.
$$

Expanding the quadratic form gives

$$
e=\sum_{i<n}c_i(u_i-u_{i+1})^2
+\sum_{i=1}^n(2-c_i-c_{i-1})u_i^2,
$$

where we set $c_0=c_n=0$.

The row sums of $D$ are at most $2$, so $0\le E\le4I$. The coefficients $2-c_i-c_{i-1}$ are at least $i/(2n)$: for interior indices this follows from $1-√(1-t)\ge t/2$, and the boundary indices satisfy the same inequality directly. Also $c_i\ge1/\sqrt2$ when $i\le n/2$.

Therefore

$$
\sum_i(i)/(n)u_i^2\le2e,
$$

and

$$
\sum_{i>n/2}u_i^2+\sum_{i<n}(u_i-u_{i+1})^2\le Ce.
$$

For the second inequality, control gradients in the first half by the weighted gradient term, and use

$$
(u_i-u_{i+1})^2\le2u_i^2+2u_{i+1}^2
$$

for the remaining edges.

##### Step 2: one event for every energy scale

The vector $u$ will be chosen after the matrix is realized, so the probability estimate must be simultaneous over all energy scales and all spatial shells.

Set

$$
\sigma_j=2^js_*,
\qquad
0\le j\le J_n:=\left\lceil\log_2(4/s_*)\right\rceil,
$$

and

$$
d_j=\max\{1,\lfloor\sigma_j^{-1/2}\rfloor\}.
$$

For each of the two independent entry sequences (diagonal entries and centered off-diagonal entries), partition its index set into consecutive blocks of lengths between $d_j$ and $2d_j$. Such a partition is obtained by merging a last short block with its predecessor.

Assign label $r=0$ when the first index of a block is at most $2n\sigma_j$. Otherwise assign label $r\ge1$ when the first index lies in

$$
(2^rn\sigma_j,2^{r+1}n\sigma_j].
$$

If $\xi_i$ is either centered entry sequence and a block has length $m\le2d_j$, its partial sums

$$
S_k=\xi_1+...+\xi_k
$$

obey

$$
\mathbb P{\max_{k\le m}|S_k|\ge a}
\le2\exp(-(na^2)/(Cd_j)).
$$

Indeed, the exponential bound of Lemma 6.3 makes

$$
\exp(tS_k-(Ckt^2)/(n))
$$

a nonnegative supermartingale. Stopping it at its first crossing of $a$ bounds the one-sided probability by

$$
\exp(-ta+(Cmt^2)/(n)).
$$

Optimizing $t>0$ and repeating for $-S_k$ proves the displayed estimate.

At a fixed pair $(j,r)$, the number of blocks is at most

$$
C(1+(2^rn\sigma_j)/(d_j))
\le C2^rn\sigma_j^{3/2}
=C2^{r+3j/2}.
$$

For $\sigma_j\le1$, use $d_j\ge\sigma_j^{-1/2}/2$. For $\sigma_j>1$, use $d_j=1$. The additive constant is absorbed because $n\sigma_j^{3/2}\ge1$.

Consequently, for a sufficiently large fixed constant $K$, there is an event $\mathcal F_n$ on which every block of both sequences has all its centered partial sums bounded by

$$
K√(d_j/n)\,(h_n+√(j+r)).
$$

Moreover,

$$
\mathbb P(\mathcal F_n^c)
\le C\sum_{j,r\ge0}2^{r+3j/2}
e^{-cK^2(h_n+√(j+r))^2}
\le Ce^{-c'h_n^2}
=Ce^{-c'L_n^{1/4}}.
$$

Here $K$ is chosen so that the factors in $j$ and $r$ form summable geometric series. The bound does not acquire a factor from the number of scales.

##### Step 3: from block prefixes to quadratic forms

We now fix the noise event $\mathcal F_n$ and choose an arbitrary unit vector $u$.

Select the first $j$ for which

$$
e\le\sigma_j.
$$

Since $e\le4$, this index exists, and

$$
\sigma_j\le2(e+s_*).
$$

For the weights $w_i=u_i^2$ or $w_i=u_iu_{i+1}$, the sum of their absolute values over all blocks with label $r$ is at most $C2^{-r}$.

For $r\ge1$, every coordinate used is beyond $2^rn\sigma_j$. The energy-tail bound gives total squared mass at most

$$
(2e)/(2^r\sigma_j)\le2^{1-r}.
$$

The off-diagonal weights obey the same bound because

$$
|u_iu_{i+1}|\le\frac{u_i^2+u_{i+1}^2}{2}.
$$

For $r=0$, use the unit norm.

The summed variation of these weights within those blocks is at most

$$
C2^{-r/2}\sqrt e.
$$

To verify this, use

$$
u_{i+1}^2-u_i^2=(u_{i+1}-u_i)(u_{i+1}+u_i)
$$

for diagonal weights and

$$
u_{i+1}u_{i+2}-u_iu_{i+1}
=u_{i+1}[(u_{i+2}-u_{i+1})+(u_{i+1}-u_i)]
$$

for off-diagonal weights. Cauchy--Schwarz combines the tail mass bound with the total gradient bound. Each coordinate or edge is counted at most a fixed number of times.

For clarity, the summation-by-parts step can be made blockwise. Suppose all centered entry prefixes in a block are bounded by $B$, and let $\bar\xi$ be the entry average on that block. Then

$$
|\bar\xi|\le B/d_j.
$$

The prefixes of $\xi_i-\bar\xi$ are at most $2B$ and vanish at the block end. Therefore

$$
|\sum_{i in block}\xi_iw_i|
\le(B)/(d_j)\sum_{i in block}|w_i|
+2B\sum_{\substack{i,i+1 in block}}|w_{i+1}-w_i|.
$$

Using the prefix bound and summing by labels proves that either centered weighted sum is bounded by

$$
Cn^{-1/2}\sum_{r\ge0}(h_n+√(j+r))
(d_j^{-1/2}2^{-r}+d_j^{1/2}2^{-r/2}\sqrt e)
\le C(h_n+\sqrt j)n^{-1/2}\sigma_j^{1/4}.
$$

For $\sigma_j\le1$, the last step uses

$$
d_j\asymp\sigma_j^{-1/2},
\qquad
e\le\sigma_j.
$$

For $1<\sigma_j<8$, it follows from $d_j=1$ with a changed universal constant. The series in $r$ converges geometrically, including its $\sqrt r$ factor.

The centered entries have now been controlled uniformly in $u$. The only remaining part of $T-D$ is the deterministic error in the off-diagonal means. By Lemma 6.3, their difference from $c_i$ is at most

$$
\frac{C}{\sqrt n\,√(n-i)}.
$$

It is at most $C/n$ for $i\le n/2$ and at most $C/\sqrt n$ thereafter. Thus their quadratic-form contribution is at most

$$
C/n+Ce/\sqrt n,
$$

using again the tail mass bound.

Finally,

$$
j\le C\log(2+e/s_*)
\le C[\log(2+L_n)+\log(1+e/s_0)].
$$

Together with $\sigma_j\le2(e+s_*)$ and $s_*\le s_0$, this converts the centered estimate into

$$
|u^THu|\le\Phi_n(e).
$$

The mean errors are absorbed into the same bound because $0\le e\le4$ and $L_n\ge1$ for large $n$. Both entry sequences, and the factor two from off-diagonal entries, are absorbed by choosing $C_0$ once and for all.

##### Step 4: smallness of the comparison function

Finally, put

$$
q(x)=1+\log(1+x/s_0).
$$

Direct differentiation gives

$$
\Phi_n'(x)
=C_0n^{-1/2}A_n(x+s_0)^{-3/4}
(\frac{√(q(x))}4+\frac1{2√(q(x))}).
$$

Since $y^{-3/4}√(1+\log y)$ is bounded for $y\ge1$, its supremum is at most

$$
Cn^{-1/2}A_ns_0^{-3/4}
=C\eta_n.
$$

Also

$$
\Phi_n(0)/s_0=C_0\eta_n.
$$

This proves the two scalar bounds. $QED$


---

### 6.5 The deterministic resolvent

Lemma 6.4 reduces the random trace problem to the deterministic matrix $D$. Its entries were chosen so that its characteristic polynomials satisfy a Hermite recurrence. That recurrence lets us compute its resolvent very accurately near the edge.

#### Lemma 6.5 (Hermite resolvent estimates)

Let

$$
e_1\le e_2\le...\le e_n
$$

be the eigenvalues of

$$
E=2I-D.
$$

Set

$$
R_D(s)=\frac1nTr(s+E)^{-1},
\qquad
T_D(s)=\frac1nTr(s+E)^{-2}
=-R_D'(s).
$$

For every fixed $M<∞$ and all sufficiently large $n$, uniformly for

$$
s_0\le s\le M,
$$

we have

$$
T_D(s)\le C_Ms^{-1/2},
\qquad
0\le R(s)-R_D(s)\le(C_M)/(ns).
$$

Moreover,

$$
N(t):=\#\{i:e_i\le t\}\le Cnt^{3/2}
\qquad(t\ge s_0).
$$

#### Proof

Reverse the coordinates of $D$. Its off-diagonal entries become

$$
√(k/n),
\qquad
1\le k<n.
$$

Let $P_k(z)$ be the characteristic polynomial of the $k\times k$ principal minor obtained from the first $k$ reversed coordinates. Then

$$
P_0(z)=1,
\qquad
P_1(z)=z,
\qquad
P_{k+1}(z)=zP_k(z)-(k)/(n)P_{k-1}(z).
$$

Induction gives

$$
P_k'=kP_{k-1}.
$$

Differentiating the recurrence and using the recurrence one step earlier shows that

$$
P_{k+1}'=P_k+kP_k.
$$

Consequently,

$$
P_n''-nzP_n'+n^2P_n=0.
$$

Since $P_n$ is the characteristic polynomial of $D$,

$$
R_D(s)=(P_n'(2+s))/(nP_n(2+s)).
$$

The differential equation therefore yields

$$
R_D(s)^2-(2+s)R_D(s)+1=(T_D(s))/(n)>0.
$$

For every $s>0$, $R_D(s)$ is finite because $E\ge0$.

The two roots of the quadratic on the left are

$$
R(s)=\frac{2+s-√(s(4+s))}2
$$

and

$$
R_+(s)=\frac{2+s+√(s(4+s))}2.
$$

At large $s$, $R_D(s)\to0$ whereas $R_+(s)\to∞$. Positivity in the Riccati equation thus places $R_D$ below the smaller root there, and continuity prevents it from crossing either root anywhere on $(0,∞)$.

If

$$
\delta(s)=R(s)-R_D(s),
$$

then factoring the quadratic gives

$$
0\le\delta(s)\le
\frac{T_D(s)}{n√(s(4+s))}.
$$

Termwise comparison of resolvents shows that

$$
T_D(s)\le\frac2s[R_D(s/2)-R_D(s)]
\le\frac2s[R(s/2)-R(s)+\delta(s)].
$$

Differentiating the explicit expression for $R$ gives

$$
0<-R'(s)
=\frac{2}{√(s(4+s))\,
(2+s+√(s(4+s)))}
\le\frac1{2\sqrt s}.
$$

Integrating from $s/2$ to $s$ therefore gives

$$
R(s/2)-R(s)\le\sqrt s-√(s/2)\le\sqrt s.
$$

Hence

$$
T_D(s)\le(C)/(\sqrt s)
+\frac{2T_D(s)}{ns√(s(4+s))}.
$$

For $s\ge s_0$, the coefficient of the last term is at most

$$
(ns_0^{3/2})^{-1}=L_n^{-3/2}\to0.
$$

Absorbing it proves the bound on $T_D$. The displayed bound on $\delta$ then follows from the root-difference inequality. Notice that $R_D(s/2)\le R(s/2)$ was established for every positive $s/2$; it does not require $s/2\ge s_0$.

The second-trace estimate also gives the counting estimate needed for the later dyadic transfer.

For $s_0\le t\le1$, each $e_i\le t$ contributes at least $(2t)^{-2}$ to $nT_D(t)$. Thus

$$
N(t)\le4nt^2T_D(t)\le Cnt^{3/2}.
$$

For $t\ge1$, the same bound follows from $N(t)\le n$. $QED$


---

### 6.6 Eigenvalue comparison and proof of Proposition 6.1

We now transfer the quadratic-form bound of Lemma 6.4 and the deterministic resolvent estimates of Lemma 6.5 to the actual eigenvalues of $W$.

#### Step 1: ordered eigenvalues

Work first in the tridiagonal realization $T$.

On the event $\mathcal F_n$ from Lemma 6.4, the form bound controls the Rayleigh quotient of

$$
E-H=2I-T
$$

between

$$
f_-(e):=e-\Phi_n(e)
\qquadand\qquad
f_+(e):=e+\Phi_n(e).
$$

By $\sup|\Phi_n'|\le C\eta_n$ and $\eta_n\to0$, both functions are increasing for large $n$.

The span of the first $i$ eigenvectors of $E$ has every unit-vector energy at most $e_i$. The min--max principle therefore gives

$$
2-\lambda_i\le f_+(e_i).
$$

On the span of eigenvectors $i$ through $n$, every energy is at least $e_i$. The complementary max--min principle gives

$$
2-\lambda_i\ge f_-(e_i).
$$

Therefore

$$
|2-\lambda_i-e_i|\le\Phi_n(e_i),
\qquad 1\le i\le n.
$$

Define $\mathcal G_n(L_n)$ to be precisely the event on which this inequality holds, with the deterministic $e_i$ and $\Phi_n$ constructed above. It depends only on the ordered eigenvalues. The tridiagonal event $\mathcal F_n$ implies it, so Lemma 6.2 transfers the probability bound to $W$. All remaining conclusions are deterministic on $\mathcal G_n(L_n)$.

#### Step 2: the first three eigenvalues

The conditioning argument will need three large coordinate variances. To locate the corresponding eigenvalues, take

$$
m=\lfloor n^{1/3}\rfloor
$$

and the span of the first three discrete Dirichlet sine modes on $\{1,...,m\}$, extended by zero.

The Dirichlet Laplacian on this segment has eigenvalues

$$
4\sin^2((k\pi)/(2(m+1))),
$$

so its quadratic form on this span is at most $C/m^2$. On vectors supported on the segment, the form difference between $E$ and that Laplacian has absolute value at most

$$
2\max_{1\le i<m}|1-c_i|\sum_{i<m}|u_iu_{i+1}|
\le Cm/n.
$$

The min--max principle therefore gives

$$
0\le e_1,e_2,e_3\le C(m^{-2}+m/n)\le Cn^{-2/3}.
$$

The comparison inequalities and $\Phi_n(0)\le C\eta_ns_0$ imply

$$
\max_{j\le3}(|\lambda_j-2|)/(s_0)
\le C(L_n^{-1}+\eta_n).
$$

Also every $e_i\in[0,4]$, so

$$
\max_i|\lambda_i|\le2+\Phi_n(4).
$$

The assumption $L_n\le\log n$ gives $\Phi_n(4)=o(1)$, proving the stated bound by $3$.

#### Step 3: the resolvent traces

The eigenvalue comparison also controls every denominator: for every $s\ge s_0$ and every $e\in[0,4]$,

$$
(\Phi_n(e))/(s+e)
\le(\Phi_n(0)+C\eta_ne)/(s+e)
\le C\eta_n.
$$

Thus

$$
2+s-\lambda_i=(s+e_i)(1+O(\eta_n)),
$$

uniformly in both $i$ and $s\ge s_0$.

The normalized second trace is consequently at most a constant times $T_D(s)$, proving its bound in Proposition 6.1.

For the first trace, subtract term by term and use

$$
(e_i+s_0)^{1/4}\le(s+e_i)^{1/4}
$$

to obtain

$$
|\frac1n\sum_i\frac1{2+s-\lambda_i}-R_D(s)|
\le Cn^{-1/2}A_n\frac1n\sum_i
(s+e_i)^{-7/4}√(1+\log(1+e_i/s_0)).
$$

We spell out the dyadic sum because its uniformity near $s_0$ is useful.

The range $e_i\le s$ contributes at most

$$
Cs^{-7/4}\frac{N(s)}n√(1+\log(1+s/s_0))
\le Cs^{-1/4}√(1+\log(1+s/s_0)).
$$

For $k\ge0$, the shell

$$
2^ks<e_i\le2^{k+1}s
$$

contributes at most

$$
Cs^{-1/4}2^{-k/4}
√(1+\log(1+s/s_0)+(k+1)\log2),
$$

by the counting estimate of Lemma 6.5. That counting estimate remains valid above one, so all shells are covered.

The convergent geometric series, with its additional square-root factor, bounds the sum by

$$
C_Ms^{-1/4}√(1+\log(1+s/s_0)).
$$

Dividing the resulting trace error by $\sqrt s$ and writing $r=s/s_0$ gives

$$
C_M\eta_nr^{-3/4}√(1+\log(1+r))
\le C_M\eta_n
\qquad(r\ge1).
$$

Finally, Lemma 6.5 gives

$$
(|R_D(s)-R(s)|)/(\sqrt s)
\le\frac{C_M}{ns^{3/2}}
\le C_ML_n^{-3/2}.
$$

This proves the quantitative first-trace error and completes all of Proposition 6.1. The event was defined before $M$ was chosen, and each displayed error is deterministic on that event. Consequently every deterministic sequence of spectra in $\mathcal G_n(L_n)$ satisfies the estimates for every fixed finite $M$, without any condition on eigenvector orientation. $QED$


---

## 7. The spherical model

The Ising cube is hard because its coordinates are discrete. The spherical model replaces the cube by a sphere of radius $\sqrt n$, where Gaussian conditioning gives exact formulas. We then transfer spherical estimates back to the cube in Section 8.

### 7.1 Tilting the sphere

Fix a deterministic ordered spectrum

$$
\lambda_1\ge\lambda_2\ge...\ge\lambda_n,
\qquad n\ge3,
\qquad \max_i|\lambda_i|\le3,
$$

and put

$$
Λ=diag(\lambda_1,...,\lambda_n).
$$

Let $\sigma_n$ be uniform probability measure on

$$
\{u\in\mathbb R^n:\|u\|^2=n\}.
$$

For $0\le a\le2$, define

$$
S(a)=\int e^{a u^TΛ u/2}\,\sigma_n(du),
$$

and

$$
B(a)=\inf_{z>a\lambda_1}
{
(z-1)/(2)
-\frac1{2n}\sum_{i=1}^n\log(z-a\lambda_i)
}.
$$

The associated tilted probability is

$$
\sigma_{Λ,a}(du)
=S(a)^{-1}e^{a u^TΛ u/2}\,\sigma_n(du).
$$

At $a=1$ we write

$$
\sigma_Λ=\sigma_{Λ,1}.
$$

Quadratic exponential laws on a sphere are Bingham distributions. The key fact is that they can be represented by independent Gaussian coordinates conditioned on their squared radius.

#### Lemma 7.1 (Gaussian representation)

For every $a\in[0,2]$, the infimum defining $B(a)$ has a unique minimizer $z_a$. With

$$
k_i(a)=(z_a-a\lambda_i)^{-1},
\qquad
\sum_{i=1}^n k_i(a)=n,
$$

one has

$$
1+a\lambda_n\le z_a\le1+a\lambda_1,
\qquad
\frac1{13}\le k_i(a)\le n.
$$

Let $Y_i$ be independent $N(0,k_i(a))$ variables, and let $p_a$ be the density of

$$
\sum_{i=1}^n Y_i^2 .
$$

Then

$$
S(a)=e^{nB(a)}
\frac{p_a(n)}{p_{\chi_n^2}(n)},
$$

where $p_{\chi_n^2}$ is the density of a chi-square variable with $n$ degrees of freedom.

The conditional law of $Y=(Y_1,...,Y_n)$ given $\|Y\|^2=n$, defined by radial disintegration, is $\sigma_{Λ,a}$.

#### Proof

Denote the objective in the definition of $B(a)$ by $F_a(z)$. On $z>a\lambda_1$,

$$
F_a'(z)=\frac12-\frac1{2n}\sum_i(z-a\lambda_i)^{-1},
$$

and

$$
F_a''(z)=\frac1{2n}\sum_i(z-a\lambda_i)^{-2}>0.
$$

The derivative tends to $-∞$ at the left endpoint and to $1/2$ at infinity. The function itself tends to $+∞$ at both ends. This proves existence, uniqueness, and the saddle equation

$$
\sum_i(z_a-a\lambda_i)^{-1}=n.
$$

The average of the reciprocals of $z_a-a\lambda_i$ is one. Their minimum is therefore at most one, and their maximum is at least one. This gives the stated bounds on $z_a$. Consequently,

$$
0<z_a-a\lambda_i
\le1+a(\lambda_1-\lambda_n)\le13.
$$

The lower bound for $k_i(a)$ follows. The upper bound follows from their positivity and sum $n$. At $a=0$, these formulas give $z_0=1$, $k_i(0)=1$, and $B(0)=0$.

Now fix the minimizer $z=z_a$. Put

$$
Q=zI-aΛ,
$$

and let $Y^{z,a}\sim N(0,Q^{-1})$. Its density relative to a standard Gaussian vector at $y\in\mathbb R^n$ is

$$
(\det Q)^{1/2}
\exp(-(z-1)/(2)\|y\|^2+\frac a2y^TΛ y).
$$

The standard Gaussian has uniform conditional angular law. Integrating this ratio over the sphere of squared radius $n$ gives

$$
\frac{p_{z,a}(n)}{p_{\chi_n^2}(n)}
=(\det Q)^{1/2}e^{-n(z-1)/2}S(a),
$$

where $p_{z,a}$ is the density of $\|Y^{z,a}\|^2$.

This is a polar-coordinate identity between continuous radial densities at a positive radius. At $z=z_a$, its rearrangement is

$$
S(a)=e^{nB(a)}
\frac{p_a(n)}{p_{\chi_n^2}(n)}.
$$

The same density ratio, with the radius fixed, proves that the conditional law is $\sigma_{Λ,a}$. $QED$

---

### 7.2 Two-square smoothing

We will repeatedly use a simple density bound. If $G,H$ are independent standard Gaussians and $d,e>0$, then the density $r_{d,e}$ of

$$
dG^2+eH^2
$$

satisfies

$$
r_{d,e}(t)
=\frac1{2\pi√(de)}
\int_0^t
\frac{e^{-x/(2d)-(t-x)/(2e)}}{√(x(t-x))}\,dx
\le\frac1{2√(de)},
\qquad t>0.
$$

The integral of $[x(t-x)]^{-1/2}$ over $(0,t)$ is $\pi$. Convolving with further independent random variables preserves this upper bound.

---

### 7.3 Density at the mean

The next estimate is needed at the mean of a weighted sum of squares, including arrays with a dominant weight.

#### Lemma 7.2 (Density at the mean)

There is a universal constant $c>0$ such that, for every $n\ge3$ and every $k_1,...,k_n>0$ with

$$
\sum_{i=1}^n k_i=n,
$$

the density $p$ of

$$
\sum_{i=1}^n k_iG_i^2,
$$

for independent standard Gaussians $G_i$, satisfies

$$
p(n)\ge c(\sum_{i=1}^n k_i^2)^{-1/2}.
$$

#### Proof

After normalization, it is enough to exclude a sequence of densities vanishing at zero. There are two ways such a sequence of weights can behave: all weights vanish individually, giving a Gaussian limit, or a positive largest weight remains and supplies a density bounded below near zero.

Set

$$
\tau=(\sum_i k_i^2)^{1/2},
\qquad
d_i=k_i/\tau .
$$

The centered random variable

$$
T_d=\sum_i d_i(G_i^2-1)
=(\sum_i k_iG_i^2-n)/(\tau)
$$

has density $f_d$ with

$$
f_d(0)=\tau p(n),
\qquad
\sum_i d_i^2=1.
$$

Thus it suffices to bound $f_d(0)$ below uniformly over all such finite arrays. Each array has at least three positive weights. Its characteristic function is integrable, since its absolute value decays as $O(|t|^{-n/2})$ at infinity; we always take the resulting continuous version of its density.

Suppose, to the contrary, that arrays $d^{(r)}$ satisfy

$$
f_{d^{(r)}}(0)\to0.
$$

Sort each array in nonincreasing order and extend it by zeros. A diagonal subsequence has

$$
d_i^{(r)}\to c_i
$$

for every $i$, with

$$
\sum_i c_i^2\le1.
$$

We consider the two possible values of $c_1$.

##### Case 1: $c_1=0$

For $d\ge0$, define

$$
\psi_d(t)=\mathbb E e^{itd(G^2-1)}
=e^{-itd}(1-2itd)^{-1/2}.
$$

For logarithms of characteristic functions, use the continuous branch equal to zero at $t=0$. For fixed $t$ and $|td|$ small, expansion gives

$$
\log\psi_d(t)=-t^2d^2+O(|t|^3d^3).
$$

It follows that the characteristic function

$$
\phi_r(t)=\prod_i\psi_{d_i^{(r)}}(t)
$$

converges pointwise to $e^{-t^2}$, because

$$
\sum_i(d_i^{(r)})^3
\le d_1^{(r)}\sum_i(d_i^{(r)})^2
=d_1^{(r)}.
$$

Weak convergence alone does not control a density at one point. To pass through Fourier inversion we need an integrable bound uniform in $r$. Concavity of $\log(1+x)$ and

$$
\sum_i(d_i^{(r)})^2=1
$$

give

$$
|\phi_r(t)|
=\prod_i(1+4(d_i^{(r)})^2t^2)^{-1/4}
\le
(1+4(d_1^{(r)})^2t^2)^{-1/(4(d_1^{(r)})^2)}.
$$

The function $x\mapsto\log(1+Ax)/x$ is decreasing for $A\ge0$. Once $(d_1^{(r)})^2\le1/8$, the last expression is at most

$$
(1+t^2/2)^{-2}.
$$

Fourier inversion and dominated convergence imply

$$
f_{d^{(r)}}(0)
=\frac1{2\pi}\int_{\mathbb R}\phi_r(t)\,dt
→
\frac1{2\pi}\int_{\mathbb R}e^{-t^2}\,dt
=\frac1{2\sqrt\pi},
$$

a contradiction.

##### Case 2: $c_1>0$

Remove the largest coordinate and write

$$
R_r=\sum_{i\ge2}d_i^{(r)}(G_i^2-1).
$$

We claim that $R_r$ converges in distribution to

$$
R=\sum_{i\ge2}c_i(G_i^2-1)+Z,
\qquad
Z\sim N(0,\,2(1-\sum_{i\ge1}c_i^2)),
$$

where $Z$ and the displayed Gaussians are independent.

The series converges in $L^2$, since the tail beyond $m$ has variance

$$
2\sum_{i>m}c_i^2\to0.
$$

To identify the limit, note that the sorted weights satisfy

$$
d_{m+1}^{(r)}\le(m+1)^{-1/2}.
$$

For fixed $t$ and sufficiently large $m$, the same logarithmic expansion as above gives, uniformly in $r$,

$$
\log\mathbb E e^{itR_r}
=
\sum_{i=2}^m\log\psi_{d_i^{(r)}}(t)
-t^2(1-\sum_{i=1}^m(d_i^{(r)})^2)
+O(\frac{|t|^3}{√(m+1)}).
$$

First let $r\to∞$ and then $m\to∞$. The resulting characteristic function is exactly that of $R$, which proves the claim.

The residual limit has positive mass arbitrarily close to zero. This is the fact that allows the surviving largest coordinate to control the convolution density.

For any $\delta>0$, choose a finite truncation of its series whose remaining tail variance is less than $\delta^2/16$. Each nonzero summand in the finite part has a density positive near zero, so the finite part together with $Z$ has positive probability in $(-\delta/2,\delta/2)$. Zero summands or a degenerate $Z$ cause no difficulty. The independent tail has positive probability in the same interval by Chebyshev's inequality. Intersecting these events proves the assertion. This argument uses square summability of $(c_i)$, not summability.

In particular, by convergence in distribution and the open-set inequality,

$$
liminf_{r\to∞}\mathbb P\{|R_r|<c_1/3\}
\ge\mathbb P\{|R|<c_1/3\}>0.
$$

The density of $d(G^2-1)$, for $d>0$, is

$$
g_d(x)=
\frac{e^{-(x+d)/(2d)}}{√(2\pi d(x+d))}
\mathbf 1_{\{x>-d\}}.
$$

For $d$ in a sufficiently small neighborhood of $c_1$, this density is bounded below by a positive constant depending on $c_1$ whenever $|x|\le c_1/3$. This interval stays strictly inside its support.

Independence now yields the pointwise convolution identity

$$
f_{d^{(r)}}(0)=\mathbb E[g_{d_1^{(r)}}(-R_r)].
$$

To justify it at this particular point, rather than only almost everywhere, the residual $R_r$ has a bounded density by the two-square smoothing bound, using two of its positive coordinates. Convolution of that bounded density with the $L^1$ function $g_{d_1^{(r)}}$ is continuous: its change under translation by $h$ is bounded by the residual density supremum times

$$
\|g_{d_1^{(r)}}(\cdot+h)-g_{d_1^{(r)}}\|_{L^1},
$$

which tends to zero. It therefore equals the continuous density $f_{d^{(r)}}$ everywhere. Consequently,

$$
liminf_{r\to∞}f_{d^{(r)}}(0)
\ge c(c_1)liminf_{r\to∞}\mathbb P\{|R_r|<c_1/3\}>0,
$$

again a contradiction. The constant may depend on the subsequential limit: any positive liminf contradicts the chosen sequence tending to zero.

The two cases prove a universal lower bound and hence the lemma. $QED$


---

### 7.4 Partition bounds on the sphere and on hyperplanes

The density-at-the-mean lemma lets us compare $S(a)$ with $e^{nB(a)}$.

#### Lemma 7.3 (Partition bounds)

There are universal constants $c,C>0$ such that, for every $n\ge3$, every ordered spectrum

$$
\lambda_1\ge...\ge\lambda_n,
\qquad
\max_i|\lambda_i|\le3,
$$

and every $a\in[0,2]$,

$$
cn^{-1/2}e^{nB(a)}
\le S(a)\le
Cn^{1/2}e^{nB(a)}.
$$

If $H\subset\mathbb R^n$ is any subspace of dimension $n-1$, and $\sigma_{n,H}$ is uniform probability on

$$
\{u\in H:\|u\|^2=n\},
$$

then

$$
S_H(a):=\int e^{a u^TΛ u/2}\,\sigma_{n,H}(du)
\le Cn^{1/2}e^{nB(a)}.
$$

The same constant $C$ works for all $H$ and all $a\in[0,2]$.

#### Proof

The chi-square density is

$$
p_{\chi_m^2}(r)
=\frac{r^{m/2-1}e^{-r/2}}{2^{m/2}\Gamma(m/2)},
\qquad r>0.
$$

Stirling's formula gives

$$
cn^{-1/2}\le p_{\chi_m^2}(n)\le Cn^{-1/2}
$$

for $m=n$ or $m=n-1$ and $n\ge3$; the constants absorb the finitely many small dimensions.

By Lemma 7.2 and

$$
(\sum_i k_i(a)^2)^{1/2}\le\sum_i k_i(a)=n,
$$

we have

$$
p_a(n)\ge c/n.
$$

By the two-square smoothing bound and $k_i(a)\ge1/13$, we also have

$$
\sup_r p_a(r)\le C.
$$

Lemma 7.1 proves the two full-space bounds.

We now pass to a hyperplane. Let $Λ_H$ be the compression of $Λ$ to $H$, and write its eigenvalues as

$$
\theta_1\ge...\ge\theta_{n-1}.
$$

The min--max principle gives

$$
\lambda_i\ge\theta_i\ge\lambda_{i+1}.
$$

At the full-space minimizer $z=z_a$, the matrix

$$
Q_H=z_aI_H-aΛ_H
$$

is positive definite. Since $a\ge0$, interlacing gives

$$
(\det Q_H)^{-1/2}
\le\prod_{i=1}^{n-1}(z_a-a\lambda_i)^{-1/2}
=(\prod_{i=1}^n(z_a-a\lambda_i)^{-1/2})
√(z_a-a\lambda_n).
$$

The last square root is at most $√(13)$.

Let $p_{H,a}$ be the squared-norm density of a Gaussian on $H$ with covariance $Q_H^{-1}$. The same polar-coordinate calculation as in Lemma 7.1, now in dimension $n-1$ but still at squared radius $n$, yields

$$
S_H(a)
=e^{n(z_a-1)/2}(\det Q_H)^{-1/2}
\frac{p_{H,a}(n)}{p_{\chi_{n-1}^2}(n)}.
$$

Every eigenvalue of $Q_H$ is at most

$$
z_a-a\lambda_n\le13,
$$

so all its Gaussian variances are at least $1/13$. There are at least two of them, and the two-square smoothing bound gives

$$
p_{H,a}(n)\le C.
$$

Combining this bound with the determinant estimate proves the assertion.

Only an upper density bound is used: the compressed Gaussian need not have mean squared radius $n$, and $z_a$ need not minimize its own variational problem. $QED$

---

### 7.5 Critical parameters

Now impose Proposition 6.1. Fix a deterministic $L_n\to∞$ with $L_n\le\log n$, put

$$
s_0=L_n n^{-2/3},
$$

and take spectra in the good event $\mathcal G_n(L_n)$.

All statements below hold for all sufficiently large $n$, uniformly over these spectra.

At $a=1$, abbreviate

$$
k_i=k_i(1),
\qquad
Y_i\sim N(0,k_i)\ independently,
\qquad
p=p_1.
$$

For each $i$, let $p_{-i}$ be the density of

$$
\sum_{j\ne i}Y_j^2.
$$

#### Lemma 7.4 (Gaussian parameters at criticality)

With the preceding notation,

$$
\lambda_1<z_1<2+s_0,
$$

$$
(\sum_i k_i^2)^{1/2}\le Cn√(s_0),
$$

and

$$
k_j\ge c/s_0
\qquad(1\le j\le3).
$$

Moreover,

$$
p(n)\ge\frac{c}{n√(s_0)},
\qquad
\sup_{r\in\mathbb R}p_{-i}(r)\le Cs_0
\qquad(1\le i\le n).
$$

#### Proof

We compare the exact saddle with the explicit point $2+s_0$.

The first trace locates the saddle and controls the total increase in variances when moving to it; the second trace bounds the variances at the explicit point.

Set

$$
h_i=(2+s_0-\lambda_i)^{-1},
$$

whose denominators are positive by the edge estimates. Since

$$
R(s)=1-\sqrt s+O(s)
\qquad(s↓0),
$$

Proposition 6.1 gives, uniformly over the good spectra,

$$
\sum_i h_i
=n(1-√(s_0)+o(√(s_0)))<n.
$$

The function

$$
z\mapsto\sum_i(z-\lambda_i)^{-1}
$$

is strictly decreasing from infinity on $(\lambda_1,∞)$. Its value at $z_1$ is $n$; hence

$$
\lambda_1<z_1<2+s_0
$$

and

$$
k_i\ge h_i.
$$

Because $\sum_i k_i=n$, the total increase is exactly

$$
\sum_i(k_i-h_i)
=n-\sum_i h_i
=n√(s_0)(1+o(1)).
$$

A nonnegative vector has Euclidean norm at most the sum of its coordinates. The triangle inequality and the second resolvent bound therefore give

$$
(\sum_i k_i^2)^{1/2}
\le(\sum_i h_i^2)^{1/2}
+\sum_i(k_i-h_i)
\le C\sqrt n\,s_0^{-1/4}+Cn√(s_0)
\le Cn√(s_0).
$$

For the last step,

$$
n^{-1/2}s_0^{-3/4}=L_n^{-3/4}\to0.
$$

For $j\le3$, the edge estimates also give

$$
0<z_1-\lambda_j\le2+s_0-\lambda_j\le Cs_0,
$$

which proves the lower bounds on the first three variances.

Lemma 7.2 now gives the asserted lower bound on $p(n)$.

The three leading variances are needed for a uniform deletion estimate: after deleting any one index $i$, at least two of the indices $1,2,3$ remain. Applying the two-square smoothing bound to their squares, and then convolving with the other squares, yields

$$
\sup_r p_{-i}(r)\le Cs_0.
$$

$QED$

---

### 7.6 Small coordinate projections on the tilted sphere

#### Proposition 7.5 (Small coordinate)

For every $b>0$, the preceding good spectra satisfy

$$
\sigma_Λ\{|u_1|\le bn^{1/3}\}
\le CL_n^2b.
$$

In particular, given any positive deterministic $b_n\to0$, one can choose $L_n\to∞$ with $L_n\le\log n$ and $L_n^2b_n\to0$, so that the probability above tends to zero uniformly over the resulting good spectra.

#### Proof

Conditioning a single coordinate makes the role of the radial estimates transparent. Its unconditioned Gaussian mass is small, and the ratio of the residual radial density to the full radial density quantifies the cost of conditioning.

Let $\varphi_k$ be the density of $N(0,k)$. Lemma 7.1 gives, for $T>0$,

$$
\sigma_Λ\{|u_1|\le T\}
=\frac1{p(n)}
\int_{|y|\le T}\varphi_{k_1}(y)p_{-1}(n-y^2)\,dy
\le(Cs_0)/(p(n))\mathbb P\{|Y_1|\le T\}.
$$

The densities of the nonnegative residual sums are taken to be zero at negative arguments.

By Lemma 7.4 and the Gaussian density bound,

$$
(Cs_0)/(p(n))
\le Cns_0^{3/2},
$$

and

$$
\mathbb P\{|Y_1|\le bn^{1/3}\}
\le\frac{Cbn^{1/3}}{√(k_1)}
\le Cbn^{1/3}√(s_0)
=Cb√(L_n).
$$

Their product is $CbL_n^2$, proving the proposition.

For the final assertion, for example,

$$
L_n=\min\{\log n,b_n^{-1/4}\}
$$

for all sufficiently large $n$ has the required properties. $QED$

---

### 7.7 Coordinate moments and squared overlap

#### Lemma 7.6 (Squared overlap on the sphere)

The probability $\sigma_Λ$ has mean zero and diagonal covariance in the eigenbasis of $Λ$. For every coordinate,

$$
\int u_i^2\,\sigma_Λ(du)
\le(Cs_0k_i)/(p(n)).
$$

Consequently, for independent $U,V$ with law $\sigma_Λ$,

$$
\mathbb E((U^TV)/(n))^2
\le(Cs_0^2)/(n^2p(n)^2)\sum_i k_i^2
\le Cn^{-2/3}L_n^4.
$$

#### Proof

We reuse the one-coordinate conditioning, now with the weight $y^2$.

First, the defining density of $\sigma_Λ$ is unchanged by changing the sign of any individual coordinate. This proves the assertions about its mean and off-diagonal covariances.

Conditioning as in the preceding proof, now with the nonnegative integrand $y^2$, gives

$$
\int u_i^2\,\sigma_Λ(du)
=\frac1{p(n)}
\int_{\mathbb R}y^2\varphi_{k_i}(y)p_{-i}(n-y^2)\,dy
\le(Cs_0)/(p(n))\mathbb E Y_i^2
=(Cs_0k_i)/(p(n)).
$$

Write

$$
m_i=\int u_i^2\,\sigma_Λ(du).
$$

Independence of $U,V$ and diagonality give

$$
\mathbb E((U^TV)/(n))^2
=\frac1{n^2}\sum_i m_i^2
\le(Cs_0^2)/(n^2p(n)^2)\sum_i k_i^2.
$$

Lemma 7.2 bounds $p(n)^{-2}$ by

$$
C\sum_i k_i^2,
$$

and Lemma 7.4 bounds the latter sum by $Cn^2s_0$. The resulting upper bound is

$$
Cn^2s_0^4=Cn^{-2/3}L_n^4,
$$

as claimed. $QED$


---

## 8. Transfer from the sphere to the cube

We now transfer spherical estimates to the Ising cube. The bridge is Haar averaging over eigenvector orientations.

Throughout this section, let $L_n\to∞$ with $L_n\le\log n$, put

$$
s_0=L_n n^{-2/3},
$$

and let $\mathcal G_n=\mathcal G_n(L_n)$ be the eigenvalue-only good event of Proposition 6.1.

We fix a deterministic sequence of spectra in $\mathcal G_n$ and use the notation

$$
Λ=diag(\lambda_1,...,\lambda_n),
\qquad
S(a),\quad B(a),\quad z_a,\quad \sigma_Λ
$$

from Section 7.

The error bounds supplied by Proposition 6.1 are deterministic on $\mathcal G_n$ and hold there for every fixed finite upper resolvent parameter.

### 8.1 Haar averaging and the two overlap laws

Let $O$ be Haar distributed on the orthogonal group, independent of the ordered spectrum, and put

$$
W=OΛ O^T.
$$

Orthogonal invariance of GOE justifies this representation: Haar averaging a test function of a symmetric matrix depends only on its eigenvalues, and gives the invariant probability measure on their conjugacy orbit.

We write $\mathbb E_O$ and $\mathbb P_O$ for expectation and probability over $O$ with $Λ$ fixed.

Define the normalized Ising partition sum

$$
I=2^{-n}\sum_{x\in\{-1,+1\}^n}
\exp((x^TOΛ O^Tx)/(2)).
$$

For every fixed cube point $x$, the vector $O^Tx$ is uniform on the sphere of radius $\sqrt n$. Averaging each summand therefore gives the exact first-moment identity

$$
\mathbb E_O I=S(1).
$$

To control the normalization of the Gibbs law, we will compare the second moment of $I$ with $S(1)^2$. Both quantities can be expressed using the same pair integral; only their overlap laws differ.

Let

$$
\mathcal Q_n=\{-1+2j/n:0\le j\le n\},
$$

and

$$
p_I(q)=2^{-n}\binom{n}{n(1+q)/2}
\qquad(q\in\mathcal Q_n).
$$

This is the overlap law of two independent uniform cube points.

Let $(U,V)$ be a uniformly oriented orthogonal pair, each vector having norm $\sqrt n$, and define, for $-1\le q\le1$,

$$
M(q)=\mathbb E\exp(
((1+q)U^TΛ U+(1-q)V^TΛ V)/(2)
).
$$

If cube points $x,y$ have overlap $q\in(-1,1)$, their normalized sum and difference

$$
\frac{x+y}{√(2(1+q))},
\qquad
\frac{x-y}{√(2(1-q))}
$$

are orthogonal and both have norm $\sqrt n$. Rotating them by $O^T$ gives the pair in the definition of $M$.

At $q=1$ or $q=-1$, the points satisfy $y=x$ or $y=-x$, respectively, and the combined energy is $x^TWx$; accordingly

$$
M(1)=M(-1)=S(2).
$$

Thus the second-moment identity holds at every lattice point:

$$
\mathbb E_O I^2
=\sum_{q\in\mathcal Q_n}p_I(q)M(q).
$$

The exchangeability of $U,V$ shows that $M$ is even and nondecreasing in $|q|$. More explicitly, if

$$
A=U^TΛ U,
\qquad
D=V^TΛ V,
$$

then

$$
M(q)=\mathbb E[
e^{(A+D)/2}cosh(\frac{q(A-D)}2)
],
$$

whose integrand is nondecreasing in $|q|$.

For two independent uniform sphere points the overlap has density

$$
f_S(q)=c_n(1-q^2)^{(n-3)/2},
\qquad
c_n=(\Gamma(n/2))/(\sqrt\pi\,\Gamma((n-1)/2)),
\qquad -1<q<1.
$$

To see this, fix the first point by rotation and integrate the remaining $n-1$ coordinates of the second point; the resulting one-coordinate density is proportional to $(1-q^2)^{(n-3)/2}$, with the displayed normalization given by the beta integral.

Conditional on this overlap, the normalized sum and difference are again the orthogonal pair above. Independence of the two spherical points consequently gives the exact identity

$$
\int_{-1}^1 M(q)f_S(q)\,dq=S(1)^2.
$$

The moment identities reduce concentration to a comparison of the two overlap laws. Because $\mathbb E_O I=S(1)$, it is enough to prove

$$
\sum_{q\in\mathcal Q_n}p_I(q)M(q)
\le(1+o(1))\int_{-1}^1 M(q)f_S(q)\,dq
$$

with a deterministic $o(1)$ uniform over $\mathcal G_n$.

We will use a variational estimate for large overlaps and a direct comparison of the unweighted overlap laws near zero.

First consider the size of the pair integral. By evenness it suffices to take $q\ge0$. Conditional on $U$, the vector $V$ is uniform on the radius-$\sqrt n$ sphere in $U^\perp$. The codimension-one bound of Lemma 7.3 controls its integral at parameter $1-q$, uniformly in $U$. The full-space bound then controls the remaining integral at parameter $1+q$.

Dividing by the lower bound for $S(1)^2$ from the same lemma gives

$$
(M(q))/(S(1)^2)
\le Cn^2\exp(n[B(1+|q|)+B(1-|q|)-2B(1)]).
$$

This remains valid at $|q|=1$, since the integral with parameter zero equals one.

The cube overlap mass satisfies

$$
p_I(q)\le e^{-nq^2/2}.
$$

Indeed, for a mean of $n$ independent uniform signs, the exponential moment bound $cosh t\le e^{t^2/2}$ follows by integrating $(\logcosh t)''\le1$ twice from zero. Optimizing the resulting Chernoff bound and using symmetry gives the stated bound on each lattice mass.

Thus a margin below $q^2/2$ in the exponent of the pair bound will suppress large overlaps.

We split the overlap range at

$$
q_0=n^{-7/24}.
$$

The proof uses three scale relations:

$$
nq_0^3=n^{1/8}→∞,
\qquad
nq_0^4=n^{-1/6}→0,
\qquad
(s_0)/(q_0^2)=L_n n^{-1/12}→0.
$$

The first turns a cubic variational margin into a negligible tail. The second permits a relative local comparison of the unweighted cube and sphere overlap laws. The third absorbs the spectral error in the cubic estimate; it uses $L_n\le\log n$.

---

### 8.2 The cubic deficit

The next lemma supplies the margin needed in the pair bound for $|q|\ge q_0$.

#### Lemma 8.1 (Cubic deficit)

There is a universal constant $c>0$ such that, for all sufficiently large $n$, uniformly for spectra in $\mathcal G_n$ and $q_0\le q\le1$,

$$
B(1+q)+B(1-q)-2B(1)
\le(q^2)/(2)-cq^3.
$$

One may take $c=1/24$.

#### Proof

We estimate the derivative of the spherical variational value on both sides of the critical parameter $a=1$. Close to that parameter the resolvent estimates give a quantitative margin; away from it a bounded-convergence argument suffices.

The saddle equation in Lemma 7.1 and implicit differentiation show that $z_a$ is continuously differentiable in $a$. Differentiating the objective at its minimizer gives

$$
B'(a)=\frac1{2n}\sum_{i=1}^n
(\lambda_i)/(z_a-a\lambda_i),
\qquad
B'(a)=(z_a-1)/(2a)
\quad(a>0).
$$

The first identity also holds at $a=0$, with a one-sided derivative at the endpoint. Since the positive weights $(z_a-a\lambda_i)^{-1}$ sum to $n$,

$$
|B'(a)|\le\frac12\max_i|\lambda_i|\le\frac32
\qquad(0\le a\le2).
$$

For $a>0$, put $r_a=z_a/a$ and

$$
m_n(s)=\frac1n\sum_i(2+s-\lambda_i)^{-1}.
$$

The saddle equation is equivalently

$$
\frac1n\sum_i(r_a-\lambda_i)^{-1}=a.
$$

At $s=s_0$, the spectral estimates give

$$
m_n(s_0)=1-√(s_0)+o(√(s_0))<1.
$$

Thus, for $a=1+u$ with $0\le u\le1$, monotonicity of this resolvent gives $r_a<2+s_0$ and hence

$$
B'(1+u)
\le(1+u)/(2)-(u^2)/(2(1+u))+(s_0)/(2).
$$

The upper derivative estimate must be paired with a lower estimate below $a=1$. The small-overlap range requires the following quantitative comparison.

If

$$
4√(s_0)\le u\le1/4,
$$

set

$$
t=(u^2)/(1-u).
$$

Then $t\le1/12$ and $0.9t\ge14.4s_0$. The function $R$ from Proposition 6.1 satisfies

$$
R(t)=1-u,
$$

and

$$
-R'(s)=\frac{2+s}{2√(s(4+s))}-\frac12
\ge(c_1)/(\sqrt s)
\qquad(0<s\le1/12)
$$

for a universal $c_1>0$.

Integrating the last bound from $0.9t$ to $t$ therefore gives

$$
R(0.9t)-(1-u)\ge c_2\sqrt t\ge c_2u.
$$

The error in

$$
m_n(0.9t)=R(0.9t)+o(\sqrt t)
$$

is uniform over the indicated range. It is smaller than this fixed positive margin for all sufficiently large $n$. Consequently $m_n(0.9t)>1-u$, so monotonicity implies

$$
r_{1-u}\ge2+0.9t.
$$

Substitution in the derivative formula yields

$$
B'(1-u)\ge(1-u)/(2)-(u^2)/(20(1-u))
\qquad(4√(s_0)\le u\le1/4).
$$

For $0\le u\le4√(s_0)$, use instead

$$
r_{1-u}>\lambda_1=2+o(s_0)
$$

to obtain

$$
B'(1-u)
\ge(1-u)/(2)-(u^2)/(2(1-u))-o(s_0)
\ge(1-u)/(2)-Cs_0.
$$

Write

$$
F(q)=B(1+q)+B(1-q)-2B(1).
$$

On $0\le u\le1/4$,

$$
\frac1{2(1+u)}-\frac1{20(1-u)}\ge\frac13.
$$

Combining the preceding bounds, and absorbing $u^2\le16s_0$ on the short interval $u\le4√(s_0)$ into the error, gives

$$
F'(u)\le u-(u^2)/(3)+Cs_0
\qquad(0\le u\le1/4).
$$

Since $F(0)=0$,

$$
F(q)\le(q^2)/(2)-(q^3)/(9)+Cs_0q
\qquad(0\le q\le1/4).
$$

For $q\ge q_0$, the last error divided by $q^3$ is at most

$$
C(s_0)/(q_0^2)
=CL_n n^{-1/12}→0.
$$

Thus the cubic deficit holds on $[q_0,1/4]$, even with $c=1/18$.

It remains to cover overlaps bounded away from zero. Here a uniform additive error tending to zero is enough.

Fix $a\in(0,1)$. The identity

$$
R(a+a^{-1}-2)=a,
$$

the strict monotonicity of $R$, and the spectral resolvent estimate imply

$$
r_a→ a+a^{-1},
\qquad
B'(a)→ a/2.
$$

Indeed, evaluating the empirical resolvent at

$$
a+a^{-1}\pm\delta,
$$

with $\delta>0$ small enough that the lower point is above $2$, traps $r_a$ between these two points for all large $n$. The upper resolvent parameter needed here is finite for each fixed $a$. No estimate uniform as $a↓0$ is used.

The bound $|B'|\le3/2$ now gives, by dominated convergence,

$$
\rho_n:=\int_0^1|B'(a)-a/2|\,da→0.
$$

Integrating the supercritical derivative bound and bounding the subcritical error by $\rho_n$ gives, uniformly for $0\le q\le1$,

$$
F(q)
\le(q^2)/(2)
-\int_0^q(u^2)/(2(1+u))\,du+\rho_n+(s_0)/(2)
\le(q^2)/(2)-(q^3)/(12)+\rho_n+(s_0)/(2).
$$

When $q\ge1/4$, the cubic term is bounded away from zero, so the last two errors are absorbed with $c=1/24$. This also covers $q=1$.

The conclusion must hold uniformly over the entire deterministic spectral event. The estimates in the shrinking range already have that uniformity. The same is true of $\rho_n\to0$: otherwise one could select spectra from $\mathcal G_n$ along a violating subsequence. Their deterministic spectral bounds would still imply the pointwise convergence and bounded domination just proved, a contradiction.

Thus the two ranges give the claimed uniform statement. $QED$


---

### 8.3 Concentration of the partition sum

#### Proposition 8.2 (Conditional partition concentration)

For spectra in $\mathcal G_n$,

$$
\mathbb E_O I=S(1),
$$

and

$$
\sup_{Λ\in\mathcal G_n}
\mathbb E_O[((I)/(S(1))-1)^2]
→0.
$$

In particular, under the GOE law,

$$
(I)/(S(1))\to1
$$

in probability.

#### Proof

The first-moment identity was established above. We now bound the exact second moment by treating the two overlap regions separately.

##### Large overlaps

Combining the pair bound, the Chernoff bound $p_I(q)\le e^{-nq^2/2}$, and Lemma 8.1 gives

$$
p_I(q)(M(q))/(S(1)^2)
\le Cn^2e^{-cn|q|^3}
\qquad(q\in\mathcal Q_n,\ |q|\ge q_0).
$$

There are at most $n+1$ lattice points. Since $nq_0^3=n^{1/8}$,

$$
\sum_{\substack{q\in\mathcal Q_n\\ |q|\ge q_0}}
p_I(q)(M(q))/(S(1)^2)
\le Cn^3e^{-cn^{1/8}}
=:\tau_n.
$$

In particular, $\tau_n$ is smaller than every fixed negative power of $n$.

##### Central overlaps

The remaining obstacle is that $M(q)$ may vary too rapidly for a derivative-based Riemann-sum estimate. We compare the unweighted overlap laws locally, then move each lattice mass outward, where $M$ is larger.

Let

$$
\Delta=2/n.
$$

Stirling's formula, uniformly for $|q|<q_0$, gives

$$
p_I(q)=
√((2)/(\pi n(1-q^2)))\,
e^{-nJ(q)}(1+O(n^{-1})),
$$

where

$$
J(q)=\frac{(1+q)\log(1+q)+(1-q)\log(1-q)}2.
$$

Here $J(q)=q^2/2+O(q^4)$. Likewise

$$
c_n=√(n/(2\pi))(1+O(n^{-1}))
$$

and

$$
\log(1-q^2)=-q^2+O(q^4).
$$

Dividing the two expressions shows that

$$
(p_I(q))/(\Delta f_S(q))
=\exp(O(n^{-1}+q^2+nq^4))
=1+o(1),
\qquad |q|<q_0,
$$

with a uniform relative error, since $nq_0^4\to0$.

Assign to each retained lattice point the outward bin

$$
\mathcal B_q=

[q,q+\Delta],&q>0,\\
[q-\Delta,q],&q<0,\\
[-\Delta/2,\Delta/2],&q=0.

$$

For even $n$, the lattice near zero is

$$
...,-2\Delta,-\Delta,0,\Delta,2\Delta,... .
$$

The centered zero bin leaves gaps between $\Delta/2$ and $\Delta$, and between $-\Delta$ and $-\Delta/2$.

For odd $n$, the nearest points are $-\Delta/2,\Delta/2$, whose outward bins leave the gap $(-\Delta/2,\Delta/2)$.

In both cases all bin interiors are disjoint. Every $r\in\mathcal B_q$ satisfies $|r|\ge|q|$, and all these bins lie in

$$
[-q_0-\Delta,q_0+\Delta]\subset(-1,1)
$$

for large $n$.

On each such bin,

$$
|\log(f_S(r))/(f_S(q))|
\le Cn(q_0+\Delta)\Delta
\le C(q_0+\Delta),
$$

because

$$
(\log f_S)'(r)=-(n-3)r/(1-r^2).
$$

It follows from the local-density comparison that there is a deterministic $\delta_n\to0$ such that, for every nonnegative even function $h$ nondecreasing in $|q|$,

$$
p_I(q)h(q)
\le(1+\delta_n)\int_{\mathcal B_q}h(r)f_S(r)\,dr
\qquad(q\in\mathcal Q_n,\ |q|<q_0).
$$

Indeed, the integral is at least

$$
\Delta h(q)f_S(q)e^{-C(q_0+\Delta)},
$$

whereas the lattice mass is at most

$$
(1+e_n)\Delta f_S(q)
$$

for a deterministic $e_n\to0$. One can take

$$
1+\delta_n=(1+e_n)e^{C(q_0+\Delta)}.
$$

This estimate does not depend on how rapidly $h$ varies. The outermost bin may extend beyond $q_0$, which causes no difficulty because the comparison integral is over the full interval $[-1,1]$.

##### Combining the two regions

Apply the bin comparison with $h=M$. Because the bin interiors are disjoint, their sum is bounded by the full spherical integral. Together with the tail estimate and the spherical second-moment identity, it yields

$$
(\mathbb E_O I^2)/(S(1)^2)
\le1+\delta_n+\tau_n.
$$

The first-moment identity therefore gives

$$
\mathbb E_O[((I)/(S(1))-1)^2]
\le\delta_n+\tau_n
→0.
$$

The bin error is independent of the spectrum, and the tail estimate is uniform on $\mathcal G_n$ by Lemma 8.1. Thus the relative variance bound is uniform there. The unconditional assertion follows by Chebyshev's inequality and $\mathbb P(\mathcal G_n^c)\to0$. $QED$

---

### 8.4 Small projections on the Ising cube

#### Theorem 8.3 (Small projections in the Ising law)

Let $W$ be the augmented GOE matrix and let $v$ be a unit top eigenvector. For every positive deterministic sequence $b_n\to0$,

$$
\mu\{x:|v^Tx|\le b_nn^{1/3}\}
→0
\qquadin probability.
$$

The probability here includes the auxiliary diagonal used to define $v$; the Gibbs measure itself does not depend on that diagonal.

#### Proof

The restricted partition sum has a small first moment, while the unrestricted sum is close to its spherical mean.

We choose the auxiliary spectral event after the prescribed threshold $b_n$: take $L_n\to∞$ with $L_n\le\log n$ and $L_n^2b_n\to0$. For example, for all sufficiently large $n$ one may take

$$
L_n=\min\{\log n,b_n^{-1/4}\}.
$$

In the representation $W=OΛ O^T$, take $v=Oe_1$, and define the restricted normalized sum

$$
I_b=2^{-n}\sum_x
\mathbf 1_{\{|v^Tx|\le b_nn^{1/3}\}}e^{x^TWx/2}.
$$

For every fixed $x$, $O^Tx$ is uniform on the sphere and its first coordinate is $v^Tx$. Consequently Proposition 7.5 gives, conditional on any spectrum in $\mathcal G_n$,

$$
\mathbb E_O(I_b)/(S(1))
=\sigma_Λ\{|U_1|\le b_nn^{1/3}\}
\le CL_n^2b_n.
$$

For every fixed $ε>0$, Markov's inequality and Proposition 8.2 imply

$$
\mathbb P_O((I_b)/(I)>ε)
\le\mathbb P_O((I)/(S(1))<\frac12)
+(2)/(ε)\mathbb E_O(I_b)/(S(1))
→0
$$

uniformly over the good spectra.

The ratio $I_b/I$ is the desired Gibbs mass, so adding $\mathbb P(\mathcal G_n^c)$ proves the claim.

GOE has a simple spectrum almost surely: the set of matrices with repeated eigenvalues is the zero set of the nonzero discriminant polynomial and has Lebesgue measure zero, whereas GOE has a density. Thus any choice of the unit top eigenvector differs from $Oe_1$ only by sign, which leaves the asserted event unchanged. $QED$

---

### 8.5 Squared overlap on the cube

#### Proposition 8.4 (Squared overlap)

Let

$$
Q_n=\mathbb E_{\mu^{ ⊗ 2}}q(x,y)^2.
$$

Then $n^{2/3}Q_n$ is bounded in probability. Equivalently, for every positive deterministic sequence $T_n\to∞$,

$$
\mathbb P\{n^{2/3}Q_n>T_n\}→0.
$$

This assertion concerns the original off-diagonal SK disorder.

#### Proof

To prove tightness rather than a bound with a fixed logarithmic loss, keep $L_n$ free until the tail threshold has been specified.

The same second-moment comparison works with the nonnegative overlap weight $q^2$. Expanding two Gibbs weights gives the exact identity

$$
\mathbb E_O[((I)/(S(1)))^2Q_n]
=\frac1{S(1)^2}\sum_{q\in\mathcal Q_n}p_I(q)q^2M(q).
$$

The function $q^2M(q)$ is nonnegative, even, and nondecreasing in $|q|$.

The central terms therefore satisfy the same bin comparison. The tails are at most $\tau_n$ because $q^2\le1$. Moreover,

$$
\frac1{S(1)^2}\int_{-1}^1q^2M(q)f_S(q)\,dq
=\mathbb E_{\sigma_Λ^{ ⊗ 2}}
((U^TV)/(n))^2.
$$

Here $U,V$ are independent under the tilted spherical law. Lemma 7.6 bounds this expectation by

$$
Cn^{-2/3}L_n^4.
$$

Thus

$$
\mathbb E_O[((I)/(S(1)))^2n^{2/3}Q_n]
\le CL_n^4+n^{2/3}\tau_n
\le C'L_n^4
$$

for all large $n$, uniformly on $\mathcal G_n$.

The weighted estimate is now available for every admissible $L_n$. Given an arbitrary deterministic $T_n\to∞$, choose

$$
L_n=\min\{\log n,T_n^{1/8}\}
$$

for all sufficiently large $n$. Then $L_n\to∞$ and

$$
L_n^4/T_n\le T_n^{-1/2}\to0.
$$

Writing

$$
\widehat I=I/S(1),
$$

conditional Markov's inequality gives

$$
\mathbb P_O(n^{2/3}Q_n>T_n)
\le\mathbb P_O(\widehat I<1/2)
+\frac4{T_n}\mathbb E_O[\widehat I^{\,2}n^{2/3}Q_n]
\le o(1)+(CL_n^4)/(T_n).
$$

The first error is uniform on $\mathcal G_n$ by Proposition 8.2. Adding $\mathbb P(\mathcal G_n^c)\to0$ proves the tail estimate.

The random variable $n^{2/3}Q_n$ has one fixed law for each $n$ throughout this argument: it is a function of the off-diagonal disorder alone. Changing $L_n$ changes only the auxiliary good event and estimates used to prove the assertion, not this law.

To verify explicitly that the tail estimate implies boundedness in probability, suppose tightness failed. Then there would be an $ε>0$ and increasing indices $n_j$ with

$$
\mathbb P(n_j^{2/3}Q_{n_j}>j)>ε.
$$

Define a deterministic staircase by $T_n=j$ for $n_j\le n<n_{j+1}$, with an arbitrary positive initial segment. This sequence tends to infinity and contradicts the tail estimate at $n=n_j$.

Conversely, tightness directly implies the tail estimate by comparing a divergent threshold with any fixed threshold.

Finally, because the Gibbs law is invariant under adding a diagonal to $W$, integrating out the auxiliary diagonal preserves the asserted probabilities. $QED$

---

### 8.6 Covariance scale

#### Corollary 8.5 (Covariance scale)

For every positive deterministic sequence $M_n\to∞$,

$$
\mathbb P{
\frac{n^{2/3}}{M_n}\le\|Σ\|\le M_nn^{2/3}
}→1.
$$

#### Proof

The upper and lower bounds use different outputs of the transfer: squared overlap bounds the covariance norm from above, while the top projection bounds it from below.

Spin-flip symmetry gives $\mu[x]=0$, so

$$
Σ=\mu[xx^T]
$$

and

$$
Tr(Σ^2)
=\sum_{i,j}\mu[x_ix_j]^2
=n^2\mathbb E_{\mu^{ ⊗ 2}}q(x,y)^2
=n^2Q_n.
$$

Since $Σ$ is positive semidefinite,

$$
\|Σ\|^2\leTr(Σ^2).
$$

The upper bound follows from Proposition 8.4 with threshold $T_n=M_n^2$.

For the lower bound use Theorem 8.3 with

$$
b_n=M_n^{-1/4}
$$

and put

$$
B_n=\mu\{|v^Tx|\le b_nn^{1/3}\}.
$$

Then $B_n\to0$ in probability and

$$
\|Σ\|\ge v^TΣ v
=\mu[(v^Tx)^2]
\ge b_n^2n^{2/3}(1-B_n).
$$

On $B_n\le1/2$, for all sufficiently large $n$ this is at least

$$
n^{2/3}/(2√(M_n))\ge n^{2/3}/M_n.
$$

Both bounds hold simultaneously with probability tending to one by a union bound. The covariance is unchanged by the auxiliary diagonal, so these probabilities are also those of the original SK disorder. $QED$


---

## 9. Dynamics: from static estimates to slow mixing

We now use the static estimates to prove slow mixing. The core idea is the sign test from the roadmap.

### 9.1 Proof of Theorem A: typical equilibrium starts are slow

We prove the continuous-time statement first. Fix a deterministic sequence $t_n\ge0$ with

$$
t_n=o(n^{2/3}).
$$

Choose a positive deterministic $b_n\to0$ such that

$$
\frac{t_n}{b_n^2n^{2/3}}→0.
$$

For example,

$$
b_n=(t_n/n^{2/3})^{1/4}+n^{-1}
$$

works. Put

$$
a_n=b_nn^{1/3}.
$$

#### A sign that persists

Separate the auxiliary randomness by writing

$$
W=W_0+\Delta,
$$

where $W_0$ has zero diagonal and $\Delta$ is an independent Gaussian diagonal.

The Gibbs law $\mu_{W_0}=\mu_W$ and the heat-bath transition laws depend only on $W_0$. The unit top eigenvector

$$
v=v(W_0+\Delta)
$$

is allowed to depend on both matrices.

Set

$$
g(x)=v^Tx.
$$

Conditional on $W_0,\Delta$, start the trajectory from $\mu_{W_0}$.

If $|g(X_0)|>a_n$, failure of the two endpoint projections to have the same nonzero sign requires

$$
|g(X_{t_n})-g(X_0)|\ge a_n.
$$

Hence Lemma 5.2 and Markov's inequality give

$$
\mathbb P_{\mu_{W_0}}(
g(X_0),g(X_{t_n}) do not have the same nonzero sign
\mid W_0,\Delta
)
\le\mu_{W_0}\{|g|\le a_n\}+(2t_n)/(a_n^2).
$$

By Theorem 8.3, the first term tends to zero in probability over $W_0,\Delta$. It also tends to zero in expectation, since it lies in $[0,1]$. Therefore the deterministic quantity

$$
ε_n
:=\mathbb E_{W_0,\Delta}\mu_{W_0}\{|g|\le a_n\}
+(2t_n)/(a_n^2)
→0
$$

bounds the average sign-failure probability.

#### Testing each realized state

We now fix the initial configuration before defining its test event.

For $x$ with $g(x)\ne0$, let

$$
H_{v,x}=\{y:g(y)g(x)>0\},
$$

and

$$
r_n(W_0,\Delta,x)
=1-(e^{t_n\mathcal L}\mathbf 1_{H_{v,x}})(x).
$$

For $g(x)=0$, set

$$
r_n(W_0,\Delta,x)=1.
$$

This definition includes every zero projection in the failure event, and the sign-failure estimate implies

$$
\mathbb E_{W_0,\Delta}\mu_{W_0}[r_n(W_0,\Delta,\cdot)]
\leε_n.
$$

Spin-flip symmetry gives

$$
\mu_{W_0}(H_{v,x})\le1/2.
$$

If $g(x)\ne0$, testing total variation against this half-space yields

$$
d_{t_n}^{ct}(W_0,x)
\ge\frac12-r_n(W_0,\Delta,x).
$$

When $g(x)=0$, the following consequence remains valid because $r_n=1$:

$$
\mathbf 1_{\{d_{t_n}^{ct}(W_0,x)\le1/4\}}
\le4r_n(W_0,\Delta,x).
$$

#### Removing the auxiliary diagonal

The test half-space depends on $\Delta$, but the bad-state indicator on the left does not. We can therefore integrate the last inequality to obtain

$$
\mathbb E_{W_0}\mu_{W_0}
\{x:d_{t_n}^{ct}(W_0,x)\le1/4\}
=
\mathbb E_{W_0,\Delta}\mu_{W_0}
\{x:d_{t_n}^{ct}(W_0,x)\le1/4\}
\le4ε_n.
$$

In particular, for every fixed $\delta>0$,

$$
\mathbb P_{W_0}(
\mu_{W_0}\{x:d_{t_n}^{ct}(W_0,x)>1/4\}<1-\delta
)
\le(4ε_n)/(\delta)
→0.
$$

This is the asserted convergence in probability over the original off-diagonal disorder.

The argument integrates an intrinsic distance; it does not require choosing one auxiliary diagonal that works for a prescribed collection of initial states.

For deterministic integers $k_n=o(n^{5/3})$, use

$$
b_n=(k_n/n^{5/3})^{1/4}+n^{-1}
$$

and the discrete increment bound. The same proof replaces $2t_n/a_n^2$ by

$$
(2k_n)/(na_n^2)
$$

and proves the displayed conclusion with $d_{k_n}^{dt}$.

These are distances conditional on the realized initial configuration. Averaging the transition law itself over a Gibbs initial configuration would instead leave it equal to $\mu_{W_0}$ by stationarity. $QED$


---

### 9.2 Proof of Theorem B: covariance and linear tests

We now prove the linear inverse Rayleigh bounds. The covariance estimate controls variances of linear functions. To control their inverse Rayleigh quotients, we must also account for sites with small average conditional variance.

We use the standard spin-deletion cavity reweighting. The calculation below uses only exact deletion identities and conditional Gaussian second moments, which hold at the critical temperature.

#### Proposition 9.1 (Linear tests)

With probability tending to one,

$$
\|Σ\|\le\mathcal R_{lin}(\mu)
\le16\|Σ\|+n^{13/25}.
$$

In particular, $n^{-2/3}\mathcal R_{lin}(\mu)$ is bounded in probability.

#### Proof

The denominator in the inverse Rayleigh quotient of a linear function is diagonal in its coefficients. We first rewrite the quotient as a weighted covariance norm, then show that the sites with large weights form a sufficiently small exceptional set.

##### Weighted covariance

For each site define its local field and average conditional variance by

$$
g_i(x)=\sum_{j\ne i}W_{ij}x_j,
\qquad
\alpha_i=\mu[sech^2g_i].
$$

Thus $0<\alpha_i\le1$, and Lemma 5.1 gives

$$
\mathcal D_\mu(a^Tx)=\sum_i\alpha_i a_i^2,
\qquad
Var_\mu(a^Tx)=a^TΣ a.
$$

It follows by the change of variables

$$
b_i=√(\alpha_i)a_i
$$

that

$$
\mathcal R_{lin}(\mu)=\|\Gamma\|,
\qquad
\Gamma=diag(\alpha_i^{-1/2})Σ
diag(\alpha_i^{-1/2}).
$$

For every nonzero $a$, the inequalities $\alpha_i\le1$ and the nonnegativity of the variance give

$$
(a^TΣ a)/(\sum_i\alpha_i a_i^2)
\ge(a^TΣ a)/(\|a\|^2).
$$

Taking the supremum proves the lower bound in the proposition.

##### Deleting one spin

To control a site's weight, let $\nu_i$ be the Gibbs law on the other $n-1$ spins after deleting site $i$ and all its incident couplings, without rescaling the remaining couplings. Write $\bar\mu_i$ for the corresponding marginal of $\mu$ and set

$$
Z_i=\nu_i[cosh g_i],
\qquad
R_i=\mathbb E_{\nu_i ⊗ \nu_i}
[(\frac{x_{-i}^Ty_{-i}}{n})^2].
$$

Summing over $x_i$ gives the exact identities

$$
(d\bar\mu_i)/(d\nu_i)=(cosh g_i)/(Z_i),
$$

and

$$
\alpha_i=\frac{\nu_i[(cosh g_i)^{-1}]}{Z_i}
\ge Z_i^{-2},
$$

where the last step is Jensen's inequality for the reciprocal function.

Since $cosh g_i\ge1$, the product marginal comparison and

$$
(q-x_iy_i/n)^2\le2q^2+2n^{-2}
$$

imply

$$
R_i
\le Z_i^2\mathbb E_{\bar\mu_i ⊗ \bar\mu_i}
[(\frac{x_{-i}^Ty_{-i}}n)^2]
\le2Z_i^2(Q_n+n^{-2}).
$$

Let $\mathcal F_i$ be the sigma-field generated by the matrix with row and column $i$ deleted. Conditional on $\mathcal F_i$, the law $\nu_i$ is fixed and the incident couplings are independent $N(0,1/n)$ variables. For a fixed spin configuration, $g_i$ is consequently a centered Gaussian with variance

$$
\sigma_{loc}^2=(n-1)/n.
$$

For every fixed integer $m\ge1$, Jensen's inequality yields

$$
\mathbb E[Z_i^m\mid\mathcal F_i]
\le\mathbb E_{G\sim N(0,\sigma_{loc}^2)}[cosh^mG]
\le2e^{m^2/2}.
$$

Taking, for example, $m=200$ and using a union bound gives

$$
\mathbb P(\max_iZ_i>n^{1/100})→0.
$$

Proposition 8.4 says that $n^{2/3}Q_n$ is bounded in probability. Combining it with the cavity overlap bound and the max-normalization estimate gives

$$
\mathbb P(E_n)→1,
\qquad
E_n={\max_iR_i\le r_n},
\qquad
r_n=n^{-7/12}.
$$

For clarity, the exponent slack here is

$$
1/12-1/50=19/300>0.
$$

Indeed, on

$$
\max_iZ_i\le n^{1/100}
$$

and

$$
Q_n\le\tfrac14n^{-2/3+19/300},
$$

the first term on the right of the cavity overlap bound is at most $r_n/2$, and its second term is at most $r_n/2$ for all sufficiently large $n$.

##### Counting exceptional sites

The uniform polynomial bound on $Z_i$ is not by itself enough. We also need most $Z_i$ to stay below a fixed constant.

Conditional on $\mathcal F_i$, for two configurations of the remaining spins put

$$
c=x_{-i}^Ty_{-i}/n.
$$

Their two local fields have variances $\sigma_{loc}^2$ and covariance $c$, so

$$
\mathbb E[cosh g_i(x)cosh g_i(y)\mid\mathcal F_i]
=e^{\sigma_{loc}^2}cosh c.
$$

It follows that

$$
\mathbb E[Z_i\mid\mathcal F_i]
=e^{\sigma_{loc}^2/2},
$$

and

$$
Var(Z_i\mid\mathcal F_i)
=e^{\sigma_{loc}^2}
\mathbb E_{\nu_i ⊗ \nu_i}[cosh c-1]
\le CR_i,
$$

because $|c|\le1$ and $cosh c-1\le Cc^2$ on this interval.

The event $E_n$ depends on many incident rows. To use the variance estimate with its stated conditioning, introduce the clipped count

$$
N_n=\sum_{i=1}^n
\mathbf 1_{\{R_i\le r_n\}}\mathbf 1_{\{Z_i>4\}}.
$$

The variable $R_i$ is $\mathcal F_i$-measurable, and

$$
4-e^{\sigma_{loc}^2/2}\ge4-e^{1/2}>0.
$$

Thus conditional Chebyshev gives

$$
\mathbb E N_n
=\sum_i\mathbb E[
\mathbf 1_{\{R_i\le r_n\}}
\mathbb P(Z_i>4\mid\mathcal F_i)]
\le Cnr_n=Cn^{5/12}.
$$

On $E_n$, the clipped count equals $|B|$, where

$$
B=\{i:Z_i>4\}.
$$

Consequently,

$$
\mathbb P(|B|>n^{1/2})
\le\mathbb P(E_n^c)+\mathbb P(N_n>n^{1/2})
\le\mathbb P(E_n^c)+Cn^{-1/12}
→0.
$$

No independence between sites, or conditioning on $E_n$ in the Gaussian calculation, is used.

##### Combining the covariance blocks

The probabilistic work is complete. Work on the event

$$
\{\max_iZ_i\le n^{1/100},\ |B|\le n^{1/2}\},
$$

whose probability tends to one by the max-normalization and exceptional-count estimates.

Let $G=B^c$. On $G$,

$$
\alpha_i^{-1/2}\le Z_i\le4.
$$

Thus the principal blocks of the positive semidefinite matrix $\Gamma$ satisfy

$$
\|\Gamma_{GG}\|\le16\|Σ\|,
$$

and

$$
\|\Gamma_{BB}\|\leTr\Gamma_{BB}
=\sum_{i\in B}\alpha_i^{-1}
\le|B|\max_iZ_i^2
\le n^{13/25}.
$$

Here $Σ_{ii}=1$ by spin-flip symmetry.

Positivity also controls the cross block: for $u\in\mathbb R^G$ and $w\in\mathbb R^B$,

$$
|u^T\Gamma_{GB}w|^2
\le(u^T\Gamma_{GG}u)(w^T\Gamma_{BB}w).
$$

This follows by applying nonnegativity to the quadratic form at $(u,sw)$ for all real $s$.

If

$$
a=\|\Gamma_{GG}\|,
\qquad
b=\|\Gamma_{BB}\|,
$$

it follows that

$$
(u,w)^T\Gamma(u,w)
\le(\sqrt a\,\|u\|+\sqrt b\,\|w\|)^2
\le(a+b)(\|u\|^2+\|w\|^2).
$$

The same bound holds when either block is empty. Therefore

$$
\|\Gamma\|\le16\|Σ\|+n^{13/25},
$$

proving the proposition. Corollary 8.5 and $13/25<2/3$ give the final assertion. $QED$

#### Proof of Theorem B

The factor in the linear comparison is fixed, so it can be absorbed while preserving every divergent threshold.

Let $M_n\to∞$ be positive and deterministic. Apply Corollary 8.5 with the divergent threshold $M_n/32$, and intersect its event with that of Proposition 9.1. With probability tending to one,

$$
\frac{32n^{2/3}}{M_n}\le\|Σ\|
\le\frac{M_nn^{2/3}}{32},
$$

and

$$
\|Σ\|\le\mathcal R_{lin}(\mu)
\le\frac{M_nn^{2/3}}2+n^{13/25}.
$$

Since $13/25<2/3$, the final term is at most $M_nn^{2/3}/2$ for all sufficiently large $n$. These inequalities give the claimed common interval for both quantities. $QED$


---

### 9.3 The subexponential upper bound and Theorem C

We now prove the upper bounds in Theorem C. The strategy is to compare the critical Gibbs measure with one fixed subcritical high-temperature model.

#### Theorem 9.2 (High-temperature spectral gap)

Fix $0<\beta<1$. Let $J$ be a symmetric matrix with zero diagonal and independent entries

$$
J_{ij}\sim N(0,\beta^2/n)
\qquad(i<j),
$$

and let

$$
\nu_J(x)=Z_J^{-1}\exp(\frac12x^TJx),
\qquad x\in\{-1,+1\}^n.
$$

There is a finite constant $C_\beta$, depending only on $\beta$, such that

$$
\mathbb P_J{
Var_{\nu_J}(f)\le C_\beta\mathcal D_{\nu_J}(f)
 for every f:\{-1,+1\}^n\to\mathbb R
}→1.
$$

Here $\mathcal D_{\nu_J}$ is the unscaled sum of conditional variances. Consequently, the discrete-time heat-bath chain choosing one uniform site has spectral gap at least $1/(nC_\beta)$ with probability tending to one.

This theorem is supplied by the separate high-temperature companion paper. The constant $C_\beta$ is required only for each fixed $\beta<1$: we select $\beta$ after the desired exponential tolerance and before taking the limit in $n$.

#### Lemma 9.3 (Comparison of conditional-variance forms)

Let $\rho,\nu$ be probability measures with full support on $\{-1,+1\}^n$. Suppose

$$
0<m\le(\rho(x))/(\nu(x))\le M<∞
\qquadfor every x.
$$

If

$$
Var_\nu(f)\le C_\nu\mathcal D_\nu(f)
\qquadfor every f,
$$

then

$$
Var_\rho(f)
\le(M)/(m)C_\nu\mathcal D_\rho(f)
\qquadfor every f.
$$

In particular, for

$$
\rho(x)\propto e^{H(x)},
\qquad
\nu(x)\propto e^{\beta H(x)},
\qquad
\beta<1,
$$

one may take

$$
M/m=e^{(1-\beta)osc\,H}.
$$

#### Proof

The comparison works directly with least-squares characterizations; one need not compare the conditional kernels pointwise.

For any full-support probability measure $\eta$,

$$
Var_\eta(f)=\inf_{c\in\mathbb R}\eta[(f-c)^2],
$$

and

$$
\mathcal D_\eta(f)
=\sum_i\inf_{h_i}\eta[(f-h_i(x_{-i}))^2],
$$

where every $h_i$ ranges independently over all functions of the other coordinates.

For the second identity, conditional expectation gives the orthogonal decomposition

$$
\eta[(f-h_i)^2]
=\eta[(f-\eta[f\mid x_{-i}])^2]
+\eta[(\eta[f\mid x_{-i}]-h_i)^2].
$$

The first identity follows by the same decomposition with constants.

The density bounds and the variational identities imply

$$
Var_\rho(f)\le MVar_\nu(f),
$$

and

$$
\mathcal D_\rho(f)\ge m\mathcal D_\nu(f),
$$

which prove the comparison.

For the exponential densities, $\rho(x)/\nu(x)$ is a constant times $e^{(1-\beta)H(x)}$; the common normalization cancels in the ratio of its maximum and minimum. $QED$

#### Proposition 9.4 (Subexponential upper bound)

For every fixed $ε>0$,

$$
\mathbb P{
t_{mix}\le e^{ε n},
\quad
k_{mix}\le e^{ε n}
}→1.
$$

#### Proof

We keep the order of parameters explicit: first the requested exponential tolerance, then a fixed subcritical temperature, then the dimension.

Fix $ε>0$ and choose $\beta\in(0,1)$ such that

$$
3(1-\beta)<ε.
$$

Let

$$
\mu_\beta(x)\propto e^{\beta H_W(x)}.
$$

The zero-diagonal part of $\beta W$ satisfies exactly the hypotheses of Theorem 9.2; omitting its diagonal changes the Hamiltonian by a constant. With probability tending to one,

$$
Var_{\mu_\beta}(f)
\le C_\beta\mathcal D_{\mu_\beta}(f)
\qquadfor every f.
$$

Intersect this event with $\|W\|\le3$, whose probability tends to one by Proposition 6.1.

On their intersection,

$$
osc\,H_W\le3n,
$$

and

$$
\mu_{\min}:=\min_x\mu(x)
\ge2^{-n}e^{-osc\,H_W}
\ge e^{-(3+\log2)n}.
$$

Lemma 9.3 now gives

$$
Var_\mu(f)\le C_n\mathcal D_\mu(f),
\qquad
C_n=C_\beta e^{3(1-\beta)n}.
$$

We may increase $C_\beta$ to be at least one.

The comparison has produced a full Poincare inequality. To obtain worst-start total variation for both clocks, combine its spectral decay with the lower bound on the smallest Gibbs mass. We give the conversion explicitly so that the factor $n$ in attempted updates remains visible.

By Lemma 5.1, the Poincare inequality gives a continuous-time gap at least $1/C_n$ and a discrete-time gap at least $1/(nC_n)$.

Since $P$ is an average of orthogonal projections, its spectrum is nonnegative. Spectral decomposition on the mean-zero subspace therefore yields

$$
\|e^{t\mathcal L}h\|_{L^2(\mu)}
\le e^{-t/C_n}\|h\|_{L^2(\mu)},
$$

and

$$
\|P^kh\|_{L^2(\mu)}
\le e^{-k/(nC_n)}\|h\|_{L^2(\mu)}.
$$

For a starting configuration $x$, take

$$
h_x(y)=\frac{\mathbf 1_{\{y=x\}}}{\mu(x)}-1,
$$

whose squared norm is

$$
\mu(x)^{-1}-1.
$$

Reversibility identifies the transition density relative to $\mu$, minus one, with $e^{t\mathcal L}h_x$ or $P^kh_x$. The $L^1$ expression for total variation and Cauchy--Schwarz give

$$
\max_x d_t^{ct}(W,x)
\le\tfrac12\mu_{\min}^{-1/2}e^{-t/C_n},
$$

and

$$
\max_x d_k^{dt}(W,x)
\le\tfrac12\mu_{\min}^{-1/2}e^{-k/(nC_n)}.
$$

With

$$
B_n=\log2+\tfrac12(3+\log2)n,
$$

these bounds imply

$$
t_{mix}\le C_nB_n,
\qquad
k_{mix}\le\lceil nC_nB_n\rceil.
$$

The two right-hand sides grow at most as a fixed constant depending on $\beta$ times $n^2e^{3(1-\beta)n}$, plus the rounding term. Since $\beta$ is fixed and $3(1-\beta)<ε$, both are at most $e^{ε n}$ for all sufficiently large $n$.

The two events used above have an intersection of probability tending to one without any independence assumption. Finally, the mixing times depend only on the off-diagonal disorder, so averaging out the auxiliary diagonal preserves the conclusion. $QED$

#### Proof of Theorem C

The independent upper comparison is now proved. It remains to extract worst-start lower bounds from the stronger realized-start statement.

Fix $ε>0$ and apply Theorem A to

$$
t_n=n^{2/3-ε},
\qquad
k_n=\lfloor n^{5/3-ε}\rfloor.
$$

For each clock, with probability tending to one there is a starting configuration whose distance at the displayed time exceeds $1/4$.

Worst-start distance is nonincreasing in time by contraction of total variation under a Markov kernel. Hence

$$
t_{mix}\ge t_n
$$

and

$$
k_{mix}>k_n
$$

on those events. Since $k_{mix}$ is an integer, the latter implies

$$
k_{mix}\ge n^{5/3-ε},
$$

including when $k_n=0$.

Intersect these two lower-bound events with the event in Proposition 9.4 to obtain all four asserted bounds simultaneously. $QED$


---

## 10. Appendix: why the equilibrium model alone has no mixing time

The main theorems specify both the update rule and the clock. Two elementary constructions show why these are mathematical assumptions, even if the equilibrium law and reversibility are already fixed.

A third calculation distinguishes high-probability asymptotic bounds from essential bounds over every disorder realization.

### Proposition 10.1 (Arbitrary refresh times)

Fix any realization of the SK measure on $Ω_n=\{-1,+1\}^n$, $n\ge2$. For every positive integer $r$, there is an irreducible, aperiodic, reversible Markov kernel with stationary law $\mu$ whose worst-state threshold-$1/4$ mixing time equals $r$.

#### Proof

At each step, refresh the whole configuration from $\mu$ with probability $p$ and otherwise leave it fixed. In operator notation,

$$
\Pi(x,y)=\mu(y),
\qquad
K_p=(1-p)Id+p\Pi,
\qquad
0<p\le1.
$$

All entries of $K_p$ are positive, and

$$
\mu(x)K_p(x,y)=\mu(y)K_p(y,x).
$$

The identity $\Pi^2=\Pi$ gives, for $0<p<1$,

$$
K_p^k=\Pi+(1-p)^k(Id-\Pi),
$$

and

$$
\max_x\|K_p^k(x,\cdot)-\mu\|_{TV}
=(1-p)^k(1-\mu_{\min}).
$$

Here

$$
\mu_{\min}\le2^{-n}\le1/4.
$$

Choosing

$$
1-p=[\frac{1}{4(1-\mu_{\min})}]^{1/r}
$$

gives distance exactly $1/4$ at $r$ and strictly larger at every earlier integer. The case $p=1$ also has mixing time one. $QED$

The same freedom remains if transitions are required to change at most one spin. Given the uniform-site heat-bath kernel $P$, the kernel

$$
(1-p)Id+pP
$$

has the same stationary law, detailed balance, irreducibility, and transition restriction. Starting from a minimum-mass state, its distance after $k$ steps is at least

$$
(1-p)^k-\mu_{\min},
$$

by the event that no update was attempted. Its mixing time consequently diverges as $p↓0$. Allowing $p$ to depend on $n$ changes possible growth classes as well.

We next fix the dynamics completely. The remaining distinction is between typical disorder as $n$ grows and all disorder at a fixed dimension. Unbounded Gaussian couplings already separate those notions for two spins.

### Proposition 10.2 (A two-spin example)

For $n=2$ and the discrete uniform-site heat-bath kernel, the mixing time has no finite deterministic essential upper bound over the Gaussian coupling, although it is finite for every realization.

#### Proof

A large positive coupling keeps the two aligned states persistent.

This can be quantified using the linear observable

$$
g(x)=x_1+x_2.
$$

Put $w=W_{12}$. Spin-flip symmetry gives $\mu[g]=0$, and conditional expectation computes the eigenvalue exactly:

$$
Pg=\theta g,
\qquad
\theta=(1+tanh w)/(2).
$$

Starting at $(1,1)$, the expectation of $g$ after $k$ steps is $2\theta^k$. Since $\|g\|_∞=2$, the dual total-variation bound yields

$$
\|P^k((1,1),\cdot)-\mu\|_{TV}
\ge\tfrac12\theta^k.
$$

For each fixed $k$, this exceeds $1/4$ when $w$ is sufficiently large and positive. Every such Gaussian tail has positive probability.

The chain for any finite $w$ remains finite, irreducible, and aperiodic, so its own mixing time is finite. $QED$

Proposition 10.2 concerns essential bounds over all disorder realizations at fixed $n$. It does not contradict the typical-disorder asymptotic statements in Theorem C.

---

## 11. End of proof

The static spectral estimates of Section 6, the spherical estimates of Section 7, and the Haar transfer of Section 8 prove the small-projection theorem and the squared-overlap bound. Section 9 uses those inputs to prove Theorem A, Theorem B, and Theorem C. This completes the proof.