diff 94d6683..b34550f sk_model_critical_slowing_v2.txt
diff --git a/sk_model_critical_slowing_v2.txt b/sk_model_critical_slowing_v2.txt
index f2d448f..01c4cd3 100644
--- a/sk_model_critical_slowing_v2.txt
+++ b/sk_model_critical_slowing_v2.txt
@@ -1,6 +1,6 @@
# Critical slowing down in the Sherrington--Kirkpatrick model
-*An approachable but complete proof, written so that a motivated high-school reader can follow the main ideas without losing the rigorous steps.*
+*An approachable but complete proof.*
---
@@ -17,7 +17,7 @@ The precise result is that the mixing time is at least of order $n^{2/3}$ in con
Let $n$ be a large positive integer. A spin configuration is a point
$$
-x=(x_1,\dots,x_n)\in\{-1,+1\}^n .
+x=(x_1,...,x_n)\in\{-1,+1\}^n .
$$
The Sherrington--Kirkpatrick (SK) interaction is a random symmetric matrix $W=(W_{ij})_{1\le i,j\le n}$ with
@@ -46,7 +46,7 @@ $$
\sum_{y\in\{-1,+1\}^n}e^{H_W(y)}
}
\propto
-\exp\!\left(\sum_{i<j}W_{ij}x_ix_j\right).
+\exp\!(\sum_{i<j}W_{ij}x_ix_j).
$$
We often write $\mu=\mu_W$ when the disorder $W$ is fixed.
@@ -60,22 +60,22 @@ The number $\beta$ is the inverse temperature. In this paper the inverse tempera
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:
$$
-x_i=\begin{cases}
-+1 & \text{with probability }\dfrac{e^{g_i(x)}}{2\cosh g_i(x)},\\[2mm]
--1 & \text{with probability }\dfrac{e^{-g_i(x)}}{2\cosh g_i(x)},
-\end{cases}
+x_i=
++1 & with probability \dfrac{e^{g_i(x)}}{2cosh g_i(x)},\\[2mm]
+-1 & with probability \dfrac{e^{-g_i(x)}}{2cosh g_i(x)},
+
\qquad
g_i(x)=\sum_{j\ne i}W_{ij}x_j .
$$
-Equivalently, the new spin has conditional mean $\tanh g_i(x)$.
+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
$$
-\mathcal L=\sum_{i=1}^n(P_i-\mathrm{Id}),
+\mathcal L=\sum_{i=1}^n(P_i-Id),
\qquad
P_if(x)=\mathbb E_\mu[f(x)\mid x_{-i}],
$$
@@ -91,7 +91,7 @@ $$
These clocks are related by
$$
-\mathcal L=n(P-\mathrm{Id}).
+\mathcal L=n(P-Id).
$$
So one unit of continuous time corresponds roughly to $n$ attempted discrete updates, but the exact relation is the displayed identity.
@@ -99,31 +99,31 @@ So one unit of continuous time corresponds roughly to $n$ attempted discrete upd
For a starting configuration $x$, let
$$
-d_t^{\mathrm{ct}}(W,x)
-=\|e^{t\mathcal L}(x,\cdot)-\mu_W\|_{\mathrm{TV}},
+d_t^{ct}(W,x)
+=\|e^{t\mathcal L}(x,\cdot)-\mu_W\|_{TV},
\qquad
-d_k^{\mathrm{dt}}(W,x)
-=\|P^k(x,\cdot)-\mu_W\|_{\mathrm{TV}},
+d_k^{dt}(W,x)
+=\|P^k(x,\cdot)-\mu_W\|_{TV},
$$
where total variation is
$$
-\|\nu-\mu\|_{\mathrm{TV}}
+\|\nu-\mu\|_{TV}
=\frac12\sum_x|\nu(x)-\mu(x)|
-=\sup_{A\subseteq\{-1,+1\}^n}|\nu(A)-\mu(A)|.
+=\sup_{A⊆\{-1,+1\}^n}|\nu(A)-\mu(A)|.
$$
The worst-start mixing times are
$$
-t_{\mathrm{mix}}
-=\inf\{t\ge0:\max_x d_t^{\mathrm{ct}}(W,x)\le1/4\},
+t_{mix}
+=\inf\{t\ge0:\max_x d_t^{ct}(W,x)\le1/4\},
$$
$$
-k_{\mathrm{mix}}
-=\min\{k\in\mathbb N\cup\{0\}:\max_x d_k^{\mathrm{dt}}(W,x)\le1/4\}.
+k_{mix}
+=\min\{k\in\mathbb N\cup\{0\}:\max_x d_k^{dt}(W,x)\le1/4\}.
$$
All probabilistic statements below are over the random matrix $W$ (and, when we temporarily add an auxiliary Gaussian diagonal, over that auxiliary randomness as well).
@@ -143,9 +143,9 @@ $$
we have
$$
-\mu_W\{x:d_{t_n}^{\mathrm{ct}}(W,x)>1/4\}
-\longrightarrow 1
-\qquad\text{in probability over }W.
+\mu_W\{x:d_{t_n}^{ct}(W,x)>1/4\}
+→ 1
+\qquadin probability over W.
$$
For every deterministic integer sequence $k_n\ge0$ with
@@ -157,9 +157,9 @@ $$
we have
$$
-\mu_W\{x:d_{k_n}^{\mathrm{dt}}(W,x)>1/4\}
-\longrightarrow 1
-\qquad\text{in probability over }W.
+\mu_W\{x:d_{k_n}^{dt}(W,x)>1/4\}
+→ 1
+\qquadin probability over W.
$$
The theorem tests each realized equilibrium starting configuration separately. If instead one averaged the transition law over the initial configuration drawn from $\mu_W$, stationarity would make the law equal to $\mu_W$ at every time. The theorem says that this averaging hides a split: most fixed starts are still far away.
@@ -169,50 +169,50 @@ The theorem tests each realized equilibrium starting configuration separately. I
Let
$$
-\Sigma=\mathrm{Cov}_{\mu_W}(x)=\mu_W[xx^T],
+Σ=Cov_{\mu_W}(x)=\mu_W[xx^T],
$$
and let
$$
\mathcal D_{\mu_W}(f)
-=\sum_{i=1}^n\mu_W\bigl[\mathrm{Var}(f\mid x_{-i})\bigr]
+=\sum_{i=1}^n\mu_W[Var(f\mid x_{-i})]
$$
be the unscaled Dirichlet form. For linear functions $a^Tx$, define
$$
-\mathcal R_{\mathrm{lin}}(\mu_W)
-=\sup_{a\in\mathbb R^n\setminus\{0\}}
-\frac{\mathrm{Var}_{\mu_W}(a^Tx)}
+\mathcal R_{lin}(\mu_W)
+=\sup_{a\in\mathbb R^n∖\{0\}}
+\frac{Var_{\mu_W}(a^Tx)}
{\mathcal D_{\mu_W}(a^Tx)} .
$$
-For every positive deterministic sequence $M_n\to\infty$, with probability tending to one both $\|\Sigma\|$ and $\mathcal R_{\mathrm{lin}}(\mu_W)$ lie in
+For every positive deterministic sequence $M_n\to∞$, with probability tending to one both $\|Σ\|$ and $\mathcal R_{lin}(\mu_W)$ lie in
$$
-\left[
+[
\frac{n^{2/3}}{M_n},
\ M_n n^{2/3}
-\right].
+].
$$
The factor $M_n$ can diverge as slowly as we like. The theorem only controls linear functions; it does not by itself control every function.
### Theorem C (critical mixing bounds)
-For every fixed $\epsilon>0$,
+For every fixed $ε>0$,
$$
-\mathbb P\left\{
-n^{2/3-\epsilon}\le t_{\mathrm{mix}}\le e^{\epsilon n}
-\right\}\longrightarrow 1,
+\mathbb P{
+n^{2/3-ε}\le t_{mix}\le e^{ε n}
+}→ 1,
$$
$$
-\mathbb P\left\{
-n^{5/3-\epsilon}\le k_{\mathrm{mix}}\le e^{\epsilon n}
-\right\}\longrightarrow 1.
+\mathbb P{
+n^{5/3-ε}\le k_{mix}\le e^{ε n}
+}→ 1.
$$
The lower bounds come from Theorem A. The upper bounds are proved by comparison with one fixed high-temperature model and are exponentially weak. The sharper matching exponents are the subject of the companion paper, but the present paper proves the displayed bounds completely.
@@ -230,8 +230,8 @@ Add an independent Gaussian diagonal to $W$ so that the resulting matrix is GOE.
The key static statement is:
$$
-\mu_W\{x:|v^Tx|\le b_n n^{1/3}\}\longrightarrow 0
-\quad\text{in probability}
+\mu_W\{x:|v^Tx|\le b_n n^{1/3}\}→ 0
+\quadin probability
$$
for every deterministic $b_n\to0$. In words: the top eigenvector projection is rarely small. Its typical size is at least a constant times $n^{1/3}$.
@@ -242,7 +242,7 @@ For $g(x)=v^Tx$,
$$
\mathcal D_{\mu_W}(g)
-=\sum_i v_i^2\,\mu_W[\mathrm{sech}^2 g_i]
+=\sum_i v_i^2\,\mu_W[sech^2 g_i]
\le\sum_i v_i^2=1.
$$
@@ -251,7 +251,7 @@ Reversibility gives a mean-square increment bound along a stationary trajectory:
$$
\mathbb E_\mu[(g(X_t)-g(X_0))^2]\le2t,
\qquad
-\mathbb E_\mu[(g(Y_k)-g(Y_0))^2]\le\frac{2k}{n}.
+\mathbb E_\mu[(g(Y_k)-g(Y_0))^2]\le(2k)/(n).
$$
At a time $t_n=o(n^{2/3})$, the typical change is much smaller than the typical size $n^{1/3}$. Therefore the sign of $g$ persists. For each fixed starting point that is not too close to the hyperplane $g=0$, the half-space $\{y:g(y)g(x)>0\}$ has equilibrium mass at most $1/2$, and the chain is still on the wrong side with high probability. That gives the total-variation lower bound.
@@ -266,7 +266,7 @@ $$
s_0=L_n n^{-2/3},
$$
-with $L_n\to\infty$ arbitrarily slowly. Thus the standard deviation of one coordinate is of order
+with $L_n\to∞$ arbitrarily slowly. Thus the standard deviation of one coordinate is of order
$$
s_0^{-1/2}=n^{1/3}L_n^{-1/2}.
@@ -279,10 +279,10 @@ A Haar-averaging argument then transfers this spherical estimate to the Ising cu
The squared overlap between two independent Gibbs samples controls
$$
-\mathrm{Tr}(\Sigma^2)=n^2\mathbb E_{\mu\otimes\mu}\left[\left(\frac{x^Ty}{n}\right)^2\right].
+Tr(Σ^2)=n^2\mathbb E_{\mu ⊗ \mu}[((x^Ty)/(n))^2].
$$
-The spherical transfer proves that this overlap is of order $n^{-4/3}$, hence $\mathrm{Tr}(\Sigma^2)$ is of order $n^{2/3}$, hence $\|\Sigma\|$ is at most order $n^{2/3}$. The top projection gives the matching lower bound.
+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.
@@ -299,7 +299,7 @@ For every function $f:\{-1,+1\}^n\to\mathbb R$, each $P_i$ is an orthogonal proj
$$
\mathcal D_\mu(f)
-=-\langle f,\mathcal Lf\rangle_\mu
+=-< f,\mathcal Lf>_\mu
=\sum_{i=1}^n\mu[(f-P_if)^2]
=\sum_{i=1}^n\inf_{h_i}\mu[(f-h_i(x_{-i}))^2],
$$
@@ -307,7 +307,7 @@ $$
where the infimum is over all functions $h_i$ of $x_{-i}$. The discrete form is
$$
-\langle f,(\mathrm{Id}-P)f\rangle_\mu=\frac1n\mathcal D_\mu(f),
+< f,(Id-P)f>_\mu=\frac1n\mathcal D_\mu(f),
$$
and the spectrum of $P$ lies in $[0,1]$.
@@ -317,7 +317,7 @@ For a linear function $a^Tx$,
$$
\mathcal D_\mu(a^Tx)=\sum_{i=1}^n a_i^2\alpha_i,
\qquad
-\alpha_i=\mu[\mathrm{sech}^2 g_i],
+\alpha_i=\mu[sech^2 g_i],
\qquad
g_i(x)=\sum_{j\ne i}W_{ij}x_j .
$$
@@ -327,7 +327,7 @@ $$
Conditional expectation onto functions of $x_{-i}$ is the orthogonal projection in $L^2(\mu)$. Therefore
$$
-\langle f,(\mathrm{Id}-P_i)f\rangle_\mu
+< f,(Id-P_i)f>_\mu
=\|f-P_if\|_{L^2(\mu)}^2,
$$
@@ -336,27 +336,27 @@ 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
$$
-\langle f,(\mathrm{Id}-P)f\rangle_\mu
+< f,(Id-P)f>_\mu
=\frac1n\sum_i\|f-P_if\|_{L^2(\mu)}^2
=\frac1n\mathcal D_\mu(f).
$$
-Since each $P_i$ is an orthogonal projection, $0\le\langle f,P_if\rangle_\mu\le\langle f,f\rangle_\mu$. Averaging over $i$ gives the same inequality for $P$, so the spectrum of $P$ lies in $[0,1]$.
+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
+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
$$
-1-\tanh^2 g_i=\mathrm{sech}^2 g_i .
+1-tanh^2 g_i=sech^2 g_i .
$$
Thus the contribution of site $i$ to $\mathcal D_\mu(a^Tx)$ is
$$
-a_i^2\mu[\mathrm{Var}(a_ix_i\mid x_{-i})]
-=a_i^2\mu[\mathrm{sech}^2 g_i].
+a_i^2\mu[Var(a_ix_i\mid x_{-i})]
+=a_i^2\mu[sech^2 g_i].
$$
-Summing proves the formula. $\square$
+Summing proves the formula. $QED$
---
@@ -374,8 +374,8 @@ and
$$
\mathbb E_\mu[(f(Y_k)-f(Y_0))^2]
-\le\frac{2k}{n}\mathcal D_\mu(f),
-\qquad k=0,1,2,\dots .
+\le(2k)/(n)\mathcal D_\mu(f),
+\qquad k=0,1,2,... .
$$
In particular, for $g(x)=v^Tx$ with $\|v\|=1$, the right-hand sides are at most $2t$ and $2k/n$.
@@ -386,7 +386,7 @@ Stationarity gives
$$
\mathbb E_\mu[(f(X_t)-f(X_0))^2]
-=2\langle f,(\mathrm{Id}-e^{t\mathcal L})f\rangle_\mu .
+=2< f,(Id-e^{t\mathcal L})f>_\mu .
$$
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
@@ -398,15 +398,15 @@ $$
Therefore
$$
-2\langle f,(\mathrm{Id}-e^{t\mathcal L})f\rangle_\mu
-\le2t\langle f,-\mathcal Lf\rangle_\mu
+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\langle f,(\mathrm{Id}-P^k)f\rangle_\mu .
+2< f,(Id-P^k)f>_\mu .
$$
For $0\le\theta\le1$,
@@ -418,18 +418,18 @@ $$
Also
$$
-\langle f,(\mathrm{Id}-P)f\rangle_\mu=\frac1n\mathcal D_\mu(f).
+< f,(Id-P)f>_\mu=\frac1n\mathcal D_\mu(f).
$$
This proves the discrete bound. Finally, Lemma 5.1 gives
$$
\mathcal D_\mu(v^Tx)
-=\sum_i v_i^2\mu[\mathrm{sech}^2 g_i]
+=\sum_i v_i^2\mu[sech^2 g_i]
\le\sum_i v_i^2=1.
$$
-$\square$
+$QED$
---
@@ -438,22 +438,22 @@ $\square$
Zero field gives $\mu(x)=\mu(-x)$, so every spin has mean zero. Hence
$$
-\Sigma=\mathrm{Cov}_\mu(x)=\mu[xx^T],
+Σ=Cov_\mu(x)=\mu[xx^T],
\qquad
-\Sigma_{ii}=1.
+Σ_{ii}=1.
$$
Let
$$
-q(x,y)=\frac{x^Ty}{n}
+q(x,y)=(x^Ty)/(n)
$$
be the overlap of two configurations. For independent $x,y\sim\mu$,
$$
-\mathrm{Tr}(\Sigma^2)
-=n^2\mathbb E_{\mu\otimes\mu}[q(x,y)^2].
+Tr(Σ^2)
+=n^2\mathbb E_{\mu ⊗ \mu}[q(x,y)^2].
$$
#### Proof
@@ -461,19 +461,19 @@ $$
Expand the square:
$$
-\mathrm{Tr}(\Sigma^2)
-=\sum_{i,j}\Sigma_{ij}^2
+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\otimes\mu}[q(x,y)^2]
+\mathbb E_{\mu ⊗ \mu}[q(x,y)^2]
=\frac1{n^2}\sum_{i,j}\mu[x_ix_j]^2 .
$$
-Multiplying by $n^2$ proves the identity. $\square$
+Multiplying by $n^2$ proves the identity. $QED$
---
@@ -486,11 +486,11 @@ $$
s_0=L_n n^{-2/3}
$$
-above the edge, where $L_n\to\infty$ very slowly. The slowly growing factor $L_n$ is essential: it gives enough room to prove estimates without destroying the sharp $n^{2/3}$ scale.
+above the edge, where $L_n\to∞$ very slowly. The slowly growing factor $L_n$ is essential: it gives enough room to prove estimates without destroying the sharp $n^{2/3}$ scale.
### 6.1 The scales
-Let $L_n\to\infty$ be deterministic with
+Let $L_n\to∞$ be deterministic with
$$
L_n\le\log n
@@ -507,7 +507,7 @@ $$
$$
h_n=L_n^{1/8},
\qquad
-A_n=h_n+\sqrt{\log(2+L_n)},
+A_n=h_n+√(\log(2+L_n)),
\qquad
\eta_n=A_n L_n^{-3/4}.
$$
@@ -515,18 +515,18 @@ $$
Then $\eta_n\to0$. Define the deterministic reference function
$$
-R(s)=\frac{2+s-\sqrt{s(4+s)}}2,
+R(s)=\frac{2+s-√(s(4+s))}2,
\qquad s>0.
$$
-Note that $R(s)=1-\sqrt s+O(s)$ as $s\downarrow0$.
+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
$$
-\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_n .
+\lambda_1\ge\lambda_2\ge...\ge\lambda_n .
$$
There are events $\mathcal G_n(L_n)$, measurable with respect to the eigenvalues alone, such that
@@ -535,7 +535,7 @@ $$
\mathbb P(\mathcal G_n(L_n)^c)\le C\exp(-cL_n^{1/4})
$$
-for all sufficiently large $n$. On $\mathcal G_n(L_n)$, for every fixed $M<\infty$,
+for all sufficiently large $n$. On $\mathcal G_n(L_n)$, for every fixed $M<∞$,
$$
\lambda_j=2+o(s_0)\quad(1\le j\le3),
@@ -552,7 +552,7 @@ $$
$$
\frac1n\sum_{i=1}^n\frac1{(2+s-\lambda_i)^2}
-\le\frac{C_M}{\sqrt s}.
+\le(C_M)/(\sqrt s).
$$
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
@@ -596,9 +596,9 @@ 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 $\exp(-n\,\mathrm{Tr}(B^2)/4)$. 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.
+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. $\square$
+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. $QED$
---
@@ -607,14 +607,14 @@ Repeat the construction inside the trailing block, then inside each successive b
We need two estimates for centered chi variables. For $m\ge1$, let $\chi_m$ be as above. Then for every real $t$,
$$
-\mathbb E\exp\left(t\frac{\chi_m-\mathbb E\chi_m}{\sqrt n}\right)
+\mathbb E\exp(t(\chi_m-\mathbb E\chi_m)/(\sqrt n))
\le\exp(Ct^2/n),
$$
and
$$
-0\le\sqrt m-\mathbb E\chi_m\le\frac{C}{\sqrt m}.
+0\le\sqrt m-\mathbb E\chi_m\le(C)/(\sqrt m).
$$
#### Proof
@@ -639,12 +639,12 @@ $$
\qquad(r\ge0).
$$
-This tail bound gives $\mathrm{Var}(\chi_m)\le C$. By Jensen's inequality,
+This tail bound gives $Var(\chi_m)\le C$. By Jensen's inequality,
$$
\sqrt m-\mathbb E\chi_m
-=\frac{\mathrm{Var}(\chi_m)}{\sqrt m+\mathbb E\chi_m}
-\le\frac{C}{\sqrt m}.
+=(Var(\chi_m))/(\sqrt m+\mathbb E\chi_m)
+\le(C)/(\sqrt m).
$$
For the exponential bound, let $X,X'$ be independent copies of $\chi_m$ and put $Z=X-X'$. Then $Z$ is symmetric and
@@ -656,17 +656,17 @@ $$
\quad(k\ge1).
$$
-Jensen's inequality conditional on $X$, followed by the power series of $\cosh$, gives
+Jensen's inequality conditional on $X$, followed by the power series of $cosh$, gives
$$
\mathbb E e^{t(X-\mathbb E X)}
\le\mathbb E e^{tZ}
-=\mathbb E\cosh(tZ)
+=\mathbb Ecosh(tZ)
\le1+4\sum_{k\ge1}\frac{8^k k!t^{2k}}{(2k)!}
\le e^{32t^2}.
$$
-Replacing $t$ by $t/\sqrt n$ proves the exponential bound. The centered diagonal entries satisfy the same estimate directly. $\square$
+Replacing $t$ by $t/\sqrt n$ proves the exponential bound. The centered diagonal entries satisfy the same estimate directly. $QED$
---
@@ -678,7 +678,7 @@ We now compare the random tridiagonal matrix $T$ with a deterministic tridiagona
Let $D$ be the deterministic symmetric tridiagonal matrix with zero diagonal and
$$
-D_{i,i+1}=D_{i+1,i}=c_i:=\sqrt{1-i/n}
+D_{i,i+1}=D_{i+1,i}=c_i:=√(1-i/n)
\qquad(1\le i<n).
$$
@@ -709,7 +709,7 @@ There is a universal constant $C_0$ such that, with
$$
\Phi_n(x)
=C_0n^{-1/2}A_n(x+s_0)^{1/4}
-\sqrt{1+\log(1+x/s_0)},
+√(1+\log(1+x/s_0)),
\qquad x\ge0,
$$
@@ -758,12 +758,12 @@ $$
where we set $c_0=c_n=0$.
-The row sums of $D$ are at most $2$, so $0\le E\le4I$. The coefficients $2-c_i-c_{i-1}$ are at least $i/(2n)$: for interior indices this follows from $1-\sqrt{1-t}\ge t/2$, and the boundary indices satisfy the same inequality directly. Also $c_i\ge1/\sqrt2$ when $i\le n/2$.
+The row sums of $D$ are at most $2$, so $0\le E\le4I$. The coefficients $2-c_i-c_{i-1}$ are at least $i/(2n)$: for interior indices this follows from $1-√(1-t)\ge t/2$, and the boundary indices satisfy the same inequality directly. Also $c_i\ge1/\sqrt2$ when $i\le n/2$.
Therefore
$$
-\sum_i\frac{i}{n}u_i^2\le2e,
+\sum_i(i)/(n)u_i^2\le2e,
$$
and
@@ -809,26 +809,26 @@ $$
If $\xi_i$ is either centered entry sequence and a block has length $m\le2d_j$, its partial sums
$$
-S_k=\xi_1+\cdots+\xi_k
+S_k=\xi_1+...+\xi_k
$$
obey
$$
-\mathbb P\left\{\max_{k\le m}|S_k|\ge a\right\}
-\le2\exp\left(-\frac{na^2}{Cd_j}\right).
+\mathbb P{\max_{k\le m}|S_k|\ge a}
+\le2\exp(-(na^2)/(Cd_j)).
$$
Indeed, the exponential bound of Lemma 6.3 makes
$$
-\exp\left(tS_k-\frac{Ckt^2}{n}\right)
+\exp(tS_k-(Ckt^2)/(n))
$$
a nonnegative supermartingale. Stopping it at its first crossing of $a$ bounds the one-sided probability by
$$
-\exp\left(-ta+\frac{Cmt^2}{n}\right).
+\exp(-ta+(Cmt^2)/(n)).
$$
Optimizing $t>0$ and repeating for $-S_k$ proves the displayed estimate.
@@ -836,7 +836,7 @@ 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
$$
-C\left(1+\frac{2^rn\sigma_j}{d_j}\right)
+C(1+(2^rn\sigma_j)/(d_j))
\le C2^rn\sigma_j^{3/2}
=C2^{r+3j/2}.
$$
@@ -846,7 +846,7 @@ For $\sigma_j\le1$, use $d_j\ge\sigma_j^{-1/2}/2$. For $\sigma_j>1$, use $d_j=1$
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
$$
-K\sqrt{d_j/n}\,\bigl(h_n+\sqrt{j+r}\bigr).
+K√(d_j/n)\,(h_n+√(j+r)).
$$
Moreover,
@@ -854,7 +854,7 @@ Moreover,
$$
\mathbb P(\mathcal F_n^c)
\le C\sum_{j,r\ge0}2^{r+3j/2}
-e^{-cK^2(h_n+\sqrt{j+r})^2}
+e^{-cK^2(h_n+√(j+r))^2}
\le Ce^{-c'h_n^2}
=Ce^{-c'L_n^{1/4}}.
$$
@@ -882,7 +882,7 @@ For the weights $w_i=u_i^2$ or $w_i=u_iu_{i+1}$, the sum of their absolute value
For $r\ge1$, every coordinate used is beyond $2^rn\sigma_j$. The energy-tail bound gives total squared mass at most
$$
-\frac{2e}{2^r\sigma_j}\le2^{1-r}.
+(2e)/(2^r\sigma_j)\le2^{1-r}.
$$
The off-diagonal weights obey the same bound because
@@ -909,7 +909,7 @@ for diagonal weights and
$$
u_{i+1}u_{i+2}-u_iu_{i+1}
-=u_{i+1}\bigl[(u_{i+2}-u_{i+1})+(u_{i+1}-u_i)\bigr]
+=u_{i+1}[(u_{i+2}-u_{i+1})+(u_{i+1}-u_i)]
$$
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.
@@ -923,16 +923,16 @@ $$
The prefixes of $\xi_i-\bar\xi$ are at most $2B$ and vanish at the block end. Therefore
$$
-\left|\sum_{i\text{ in block}}\xi_iw_i\right|
-\le\frac{B}{d_j}\sum_{i\text{ in block}}|w_i|
-+2B\sum_{\substack{i,i+1\text{ in block}}}|w_{i+1}-w_i|.
+|\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+\sqrt{j+r})
-\left(d_j^{-1/2}2^{-r}+d_j^{1/2}2^{-r/2}\sqrt e\right)
+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}.
$$
@@ -949,7 +949,7 @@ For $1<\sigma_j<8$, it follows from $d_j=1$ with a changed universal constant. T
The centered entries have now been controlled uniformly in $u$. The only remaining part of $T-D$ is the deterministic error in the off-diagonal means. By Lemma 6.3, their difference from $c_i$ is at most
$$
-\frac{C}{\sqrt n\,\sqrt{n-i}}.
+\frac{C}{\sqrt n\,√(n-i)}.
$$
It is at most $C/n$ for $i\le n/2$ and at most $C/\sqrt n$ thereafter. Thus their quadratic-form contribution is at most
@@ -964,7 +964,7 @@ Finally,
$$
j\le C\log(2+e/s_*)
-\le C\bigl[\log(2+L_n)+\log(1+e/s_0)\bigr].
+\le C[\log(2+L_n)+\log(1+e/s_0)].
$$
Together with $\sigma_j\le2(e+s_*)$ and $s_*\le s_0$, this converts the centered estimate into
@@ -988,10 +988,10 @@ Direct differentiation gives
$$
\Phi_n'(x)
=C_0n^{-1/2}A_n(x+s_0)^{-3/4}
-\left(\frac{\sqrt{q(x)}}4+\frac1{2\sqrt{q(x)}}\right).
+(\frac{√(q(x))}4+\frac1{2√(q(x))}).
$$
-Since $y^{-3/4}\sqrt{1+\log y}$ is bounded for $y\ge1$, its supremum is at most
+Since $y^{-3/4}√(1+\log y)$ is bounded for $y\ge1$, its supremum is at most
$$
Cn^{-1/2}A_ns_0^{-3/4}
@@ -1004,7 +1004,7 @@ $$
\Phi_n(0)/s_0=C_0\eta_n.
$$
-This proves the two scalar bounds. $\square$
+This proves the two scalar bounds. $QED$
---
@@ -1018,7 +1018,7 @@ Lemma 6.4 reduces the random trace problem to the deterministic matrix $D$. Its
Let
$$
-e_1\le e_2\le\cdots\le e_n
+e_1\le e_2\le...\le e_n
$$
be the eigenvalues of
@@ -1030,13 +1030,13 @@ $$
Set
$$
-R_D(s)=\frac1n\mathrm{Tr}(s+E)^{-1},
+R_D(s)=\frac1nTr(s+E)^{-1},
\qquad
-T_D(s)=\frac1n\mathrm{Tr}(s+E)^{-2}
+T_D(s)=\frac1nTr(s+E)^{-2}
=-R_D'(s).
$$
-For every fixed $M<\infty$ and all sufficiently large $n$, uniformly for
+For every fixed $M<∞$ and all sufficiently large $n$, uniformly for
$$
s_0\le s\le M,
@@ -1047,7 +1047,7 @@ we have
$$
T_D(s)\le C_Ms^{-1/2},
\qquad
-0\le R(s)-R_D(s)\le\frac{C_M}{ns}.
+0\le R(s)-R_D(s)\le(C_M)/(ns).
$$
Moreover,
@@ -1062,7 +1062,7 @@ $$
Reverse the coordinates of $D$. Its off-diagonal entries become
$$
-\sqrt{k/n},
+√(k/n),
\qquad
1\le k<n.
$$
@@ -1074,7 +1074,7 @@ P_0(z)=1,
\qquad
P_1(z)=z,
\qquad
-P_{k+1}(z)=zP_k(z)-\frac{k}{n}P_{k-1}(z).
+P_{k+1}(z)=zP_k(z)-(k)/(n)P_{k-1}(z).
$$
Induction gives
@@ -1098,13 +1098,13 @@ $$
Since $P_n$ is the characteristic polynomial of $D$,
$$
-R_D(s)=\frac{P_n'(2+s)}{nP_n(2+s)}.
+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=\frac{T_D(s)}{n}>0.
+R_D(s)^2-(2+s)R_D(s)+1=(T_D(s))/(n)>0.
$$
For every $s>0$, $R_D(s)$ is finite because $E\ge0$.
@@ -1112,16 +1112,16 @@ For every $s>0$, $R_D(s)$ is finite because $E\ge0$.
The two roots of the quadratic on the left are
$$
-R(s)=\frac{2+s-\sqrt{s(4+s)}}2
+R(s)=\frac{2+s-√(s(4+s))}2
$$
and
$$
-R_+(s)=\frac{2+s+\sqrt{s(4+s)}}2.
+R_+(s)=\frac{2+s+√(s(4+s))}2.
$$
-At large $s$, $R_D(s)\to0$ whereas $R_+(s)\to\infty$. Positivity in the Riccati equation thus places $R_D$ below the smaller root there, and continuity prevents it from crossing either root anywhere on $(0,\infty)$.
+At large $s$, $R_D(s)\to0$ whereas $R_+(s)\to∞$. Positivity in the Riccati equation thus places $R_D$ below the smaller root there, and continuity prevents it from crossing either root anywhere on $(0,∞)$.
If
@@ -1133,36 +1133,36 @@ then factoring the quadratic gives
$$
0\le\delta(s)\le
-\frac{T_D(s)}{n\sqrt{s(4+s)}}.
+\frac{T_D(s)}{n√(s(4+s))}.
$$
Termwise comparison of resolvents shows that
$$
-T_D(s)\le\frac2s\bigl[R_D(s/2)-R_D(s)\bigr]
-\le\frac2s\bigl[R(s/2)-R(s)+\delta(s)\bigr].
+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}{\sqrt{s(4+s)}\,
-(2+s+\sqrt{s(4+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-\sqrt{s/2}\le\sqrt s.
+R(s/2)-R(s)\le\sqrt s-√(s/2)\le\sqrt s.
$$
Hence
$$
-T_D(s)\le\frac{C}{\sqrt s}
-+\frac{2T_D(s)}{ns\sqrt{s(4+s)}}.
+T_D(s)\le(C)/(\sqrt s)
++\frac{2T_D(s)}{ns√(s(4+s))}.
$$
For $s\ge s_0$, the coefficient of the last term is at most
@@ -1181,7 +1181,7 @@ $$
N(t)\le4nt^2T_D(t)\le Cnt^{3/2}.
$$
-For $t\ge1$, the same bound follows from $N(t)\le n$. $\square$
+For $t\ge1$, the same bound follows from $N(t)\le n$. $QED$
---
@@ -1204,7 +1204,7 @@ between
$$
f_-(e):=e-\Phi_n(e)
-\qquad\text{and}\qquad
+\qquadand\qquad
f_+(e):=e+\Phi_n(e).
$$
@@ -1239,12 +1239,12 @@ $$
m=\lfloor n^{1/3}\rfloor
$$
-and the span of the first three discrete Dirichlet sine modes on $\{1,\ldots,m\}$, extended by zero.
+and the span of the first three discrete Dirichlet sine modes on $\{1,...,m\}$, extended by zero.
The Dirichlet Laplacian on this segment has eigenvalues
$$
-4\sin^2\left(\frac{k\pi}{2(m+1)}\right),
+4\sin^2((k\pi)/(2(m+1))),
$$
so its quadratic form on this span is at most $C/m^2$. On vectors supported on the segment, the form difference between $E$ and that Laplacian has absolute value at most
@@ -1263,7 +1263,7 @@ $$
The comparison inequalities and $\Phi_n(0)\le C\eta_ns_0$ imply
$$
-\max_{j\le3}\frac{|\lambda_j-2|}{s_0}
+\max_{j\le3}(|\lambda_j-2|)/(s_0)
\le C(L_n^{-1}+\eta_n).
$$
@@ -1280,8 +1280,8 @@ The assumption $L_n\le\log n$ gives $\Phi_n(4)=o(1)$, proving the stated bound b
The eigenvalue comparison also controls every denominator: for every $s\ge s_0$ and every $e\in[0,4]$,
$$
-\frac{\Phi_n(e)}{s+e}
-\le\frac{\Phi_n(0)+C\eta_ne}{s+e}
+(\Phi_n(e))/(s+e)
+\le(\Phi_n(0)+C\eta_ne)/(s+e)
\le C\eta_n.
$$
@@ -1304,9 +1304,9 @@ $$
to obtain
$$
-\left|\frac1n\sum_i\frac1{2+s-\lambda_i}-R_D(s)\right|
+|\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}\sqrt{1+\log(1+e_i/s_0)}.
+(s+e_i)^{-7/4}√(1+\log(1+e_i/s_0)).
$$
We spell out the dyadic sum because its uniformity near $s_0$ is useful.
@@ -1314,8 +1314,8 @@ We spell out the dyadic sum because its uniformity near $s_0$ is useful.
The range $e_i\le s$ contributes at most
$$
-Cs^{-7/4}\frac{N(s)}n\sqrt{1+\log(1+s/s_0)}
-\le Cs^{-1/4}\sqrt{1+\log(1+s/s_0)}.
+Cs^{-7/4}\frac{N(s)}n√(1+\log(1+s/s_0))
+\le Cs^{-1/4}√(1+\log(1+s/s_0)).
$$
For $k\ge0$, the shell
@@ -1328,7 +1328,7 @@ contributes at most
$$
Cs^{-1/4}2^{-k/4}
-\sqrt{1+\log(1+s/s_0)+(k+1)\log2},
+√(1+\log(1+s/s_0)+(k+1)\log2),
$$
by the counting estimate of Lemma 6.5. That counting estimate remains valid above one, so all shells are covered.
@@ -1336,13 +1336,13 @@ by the counting estimate of Lemma 6.5. That counting estimate remains valid abov
The convergent geometric series, with its additional square-root factor, bounds the sum by
$$
-C_Ms^{-1/4}\sqrt{1+\log(1+s/s_0)}.
+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}\sqrt{1+\log(1+r)}
+C_M\eta_nr^{-3/4}√(1+\log(1+r))
\le C_M\eta_n
\qquad(r\ge1).
$$
@@ -1350,12 +1350,12 @@ $$
Finally, Lemma 6.5 gives
$$
-\frac{|R_D(s)-R(s)|}{\sqrt s}
+(|R_D(s)-R(s)|)/(\sqrt s)
\le\frac{C_M}{ns^{3/2}}
\le C_ML_n^{-3/2}.
$$
-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. $\square$
+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. $QED$
---
@@ -1369,7 +1369,7 @@ The Ising cube is hard because its coordinates are discrete. The spherical model
Fix a deterministic ordered spectrum
$$
-\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_n,
+\lambda_1\ge\lambda_2\ge...\ge\lambda_n,
\qquad n\ge3,
\qquad \max_i|\lambda_i|\le3,
$$
@@ -1377,7 +1377,7 @@ $$
and put
$$
-\Lambda=\mathrm{diag}(\lambda_1,\dots,\lambda_n).
+Λ=diag(\lambda_1,...,\lambda_n).
$$
Let $\sigma_n$ be uniform probability measure on
@@ -1389,30 +1389,30 @@ $$
For $0\le a\le2$, define
$$
-S(a)=\int e^{a u^T\Lambda u/2}\,\sigma_n(du),
+S(a)=\int e^{a u^TΛ u/2}\,\sigma_n(du),
$$
and
$$
B(a)=\inf_{z>a\lambda_1}
-\left\{
-\frac{z-1}{2}
+{
+(z-1)/(2)
-\frac1{2n}\sum_{i=1}^n\log(z-a\lambda_i)
-\right\}.
+}.
$$
The associated tilted probability is
$$
-\sigma_{\Lambda,a}(du)
-=S(a)^{-1}e^{a u^T\Lambda u/2}\,\sigma_n(du).
+\sigma_{Λ,a}(du)
+=S(a)^{-1}e^{a u^TΛ u/2}\,\sigma_n(du).
$$
At $a=1$ we write
$$
-\sigma_\Lambda=\sigma_{\Lambda,1}.
+\sigma_Λ=\sigma_{Λ,1}.
$$
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.
@@ -1450,7 +1450,7 @@ $$
where $p_{\chi_n^2}$ is the density of a chi-square variable with $n$ degrees of freedom.
-The conditional law of $Y=(Y_1,\dots,Y_n)$ given $\|Y\|^2=n$, defined by radial disintegration, is $\sigma_{\Lambda,a}$.
+The conditional law of $Y=(Y_1,...,Y_n)$ given $\|Y\|^2=n$, defined by radial disintegration, is $\sigma_{Λ,a}$.
#### Proof
@@ -1466,7 +1466,7 @@ $$
F_a''(z)=\frac1{2n}\sum_i(z-a\lambda_i)^{-2}>0.
$$
-The derivative tends to $-\infty$ at the left endpoint and to $1/2$ at infinity. The function itself tends to $+\infty$ at both ends. This proves existence, uniqueness, and the saddle equation
+The derivative tends to $-∞$ at the left endpoint and to $1/2$ at infinity. The function itself tends to $+∞$ at both ends. This proves existence, uniqueness, and the saddle equation
$$
\sum_i(z_a-a\lambda_i)^{-1}=n.
@@ -1484,14 +1484,14 @@ The lower bound for $k_i(a)$ follows. The upper bound follows from their positiv
Now fix the minimizer $z=z_a$. Put
$$
-Q=zI-a\Lambda,
+Q=zI-aΛ,
$$
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
$$
(\det Q)^{1/2}
-\exp\left(-\frac{z-1}{2}\|y\|^2+\frac a2y^T\Lambda y\right).
+\exp(-(z-1)/(2)\|y\|^2+\frac a2y^TΛ y).
$$
The standard Gaussian has uniform conditional angular law. Integrating this ratio over the sphere of squared radius $n$ gives
@@ -1510,7 +1510,7 @@ S(a)=e^{nB(a)}
\frac{p_a(n)}{p_{\chi_n^2}(n)}.
$$
-The same density ratio, with the radius fixed, proves that the conditional law is $\sigma_{\Lambda,a}$. $\square$
+The same density ratio, with the radius fixed, proves that the conditional law is $\sigma_{Λ,a}$. $QED$
---
@@ -1526,10 +1526,10 @@ satisfies
$$
r_{d,e}(t)
-=\frac1{2\pi\sqrt{de}}
+=\frac1{2\pi√(de)}
\int_0^t
-\frac{e^{-x/(2d)-(t-x)/(2e)}}{\sqrt{x(t-x)}}\,dx
-\le\frac1{2\sqrt{de}},
+\frac{e^{-x/(2d)-(t-x)/(2e)}}{√(x(t-x))}\,dx
+\le\frac1{2√(de)},
\qquad t>0.
$$
@@ -1543,7 +1543,7 @@ The next estimate is needed at the mean of a weighted sum of squares, including
#### Lemma 7.2 (Density at the mean)
-There is a universal constant $c>0$ such that, for every $n\ge3$ and every $k_1,\dots,k_n>0$ with
+There is a universal constant $c>0$ such that, for every $n\ge3$ and every $k_1,...,k_n>0$ with
$$
\sum_{i=1}^n k_i=n,
@@ -1558,7 +1558,7 @@ $$
for independent standard Gaussians $G_i$, satisfies
$$
-p(n)\ge c\left(\sum_{i=1}^n k_i^2\right)^{-1/2}.
+p(n)\ge c(\sum_{i=1}^n k_i^2)^{-1/2}.
$$
#### Proof
@@ -1568,7 +1568,7 @@ After normalization, it is enough to exclude a sequence of densities vanishing a
Set
$$
-\tau=\left(\sum_i k_i^2\right)^{1/2},
+\tau=(\sum_i k_i^2)^{1/2},
\qquad
d_i=k_i/\tau .
$$
@@ -1577,7 +1577,7 @@ The centered random variable
$$
T_d=\sum_i d_i(G_i^2-1)
-=\frac{\sum_i k_iG_i^2-n}{\tau}
+=(\sum_i k_iG_i^2-n)/(\tau)
$$
has density $f_d$ with
@@ -1649,9 +1649,9 @@ give
$$
|\phi_r(t)|
-=\prod_i\left(1+4(d_i^{(r)})^2t^2\right)^{-1/4}
+=\prod_i(1+4(d_i^{(r)})^2t^2)^{-1/4}
\le
-\left(1+4(d_1^{(r)})^2t^2\right)^{-1/(4(d_1^{(r)})^2)}.
+(1+4(d_1^{(r)})^2t^2)^{-1/(4(d_1^{(r)})^2)}.
$$
The function $x\mapsto\log(1+Ax)/x$ is decreasing for $A\ge0$. Once $(d_1^{(r)})^2\le1/8$, the last expression is at most
@@ -1665,7 +1665,7 @@ Fourier inversion and dominated convergence imply
$$
f_{d^{(r)}}(0)
=\frac1{2\pi}\int_{\mathbb R}\phi_r(t)\,dt
-\longrightarrow
+→
\frac1{2\pi}\int_{\mathbb R}e^{-t^2}\,dt
=\frac1{2\sqrt\pi},
$$
@@ -1685,7 +1685,7 @@ We claim that $R_r$ converges in distribution to
$$
R=\sum_{i\ge2}c_i(G_i^2-1)+Z,
\qquad
-Z\sim N\left(0,\,2\left(1-\sum_{i\ge1}c_i^2\right)\right),
+Z\sim N(0,\,2(1-\sum_{i\ge1}c_i^2)),
$$
where $Z$ and the displayed Gaussians are independent.
@@ -1708,11 +1708,11 @@ $$
\log\mathbb E e^{itR_r}
=
\sum_{i=2}^m\log\psi_{d_i^{(r)}}(t)
--t^2\left(1-\sum_{i=1}^m(d_i^{(r)})^2\right)
-+O\left(\frac{|t|^3}{\sqrt{m+1}}\right).
+-t^2(1-\sum_{i=1}^m(d_i^{(r)})^2)
++O(\frac{|t|^3}{√(m+1)}).
$$
-First let $r\to\infty$ and then $m\to\infty$. The resulting characteristic function is exactly that of $R$, which proves the claim.
+First let $r\to∞$ and then $m\to∞$. The resulting characteristic function is exactly that of $R$, which proves the claim.
The residual limit has positive mass arbitrarily close to zero. This is the fact that allows the surviving largest coordinate to control the convolution density.
@@ -1721,7 +1721,7 @@ For any $\delta>0$, choose a finite truncation of its series whose remaining tai
In particular, by convergence in distribution and the open-set inequality,
$$
-\liminf_{r\to\infty}\mathbb P\{|R_r|<c_1/3\}
+liminf_{r\to∞}\mathbb P\{|R_r|<c_1/3\}
\ge\mathbb P\{|R|<c_1/3\}>0.
$$
@@ -1729,7 +1729,7 @@ The density of $d(G^2-1)$, for $d>0$, is
$$
g_d(x)=
-\frac{e^{-(x+d)/(2d)}}{\sqrt{2\pi d(x+d)}}
+\frac{e^{-(x+d)/(2d)}}{√(2\pi d(x+d))}
\mathbf 1_{\{x>-d\}}.
$$
@@ -1738,7 +1738,7 @@ For $d$ in a sufficiently small neighborhood of $c_1$, this density is bounded b
Independence now yields the pointwise convolution identity
$$
-f_{d^{(r)}}(0)=\mathbb E\left[g_{d_1^{(r)}}(-R_r)\right].
+f_{d^{(r)}}(0)=\mathbb E[g_{d_1^{(r)}}(-R_r)].
$$
To justify it at this particular point, rather than only almost everywhere, the residual $R_r$ has a bounded density by the two-square smoothing bound, using two of its positive coordinates. Convolution of that bounded density with the $L^1$ function $g_{d_1^{(r)}}$ is continuous: its change under translation by $h$ is bounded by the residual density supremum times
@@ -1750,13 +1750,13 @@ $$
which tends to zero. It therefore equals the continuous density $f_{d^{(r)}}$ everywhere. Consequently,
$$
-\liminf_{r\to\infty}f_{d^{(r)}}(0)
-\ge c(c_1)\liminf_{r\to\infty}\mathbb P\{|R_r|<c_1/3\}>0,
+liminf_{r\to∞}f_{d^{(r)}}(0)
+\ge c(c_1)liminf_{r\to∞}\mathbb P\{|R_r|<c_1/3\}>0,
$$
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. $\square$
+The two cases prove a universal lower bound and hence the lemma. $QED$
---
@@ -1770,7 +1770,7 @@ The density-at-the-mean lemma lets us compare $S(a)$ with $e^{nB(a)}$.
There are universal constants $c,C>0$ such that, for every $n\ge3$, every ordered spectrum
$$
-\lambda_1\ge\cdots\ge\lambda_n,
+\lambda_1\ge...\ge\lambda_n,
\qquad
\max_i|\lambda_i|\le3,
$$
@@ -1792,7 +1792,7 @@ $$
then
$$
-S_H(a):=\int e^{a u^T\Lambda u/2}\,\sigma_{n,H}(du)
+S_H(a):=\int e^{a u^TΛ u/2}\,\sigma_{n,H}(du)
\le Cn^{1/2}e^{nB(a)}.
$$
@@ -1819,7 +1819,7 @@ for $m=n$ or $m=n-1$ and $n\ge3$; the constants absorb the finitely many small d
By Lemma 7.2 and
$$
-\left(\sum_i k_i(a)^2\right)^{1/2}\le\sum_i k_i(a)=n,
+(\sum_i k_i(a)^2)^{1/2}\le\sum_i k_i(a)=n,
$$
we have
@@ -1836,10 +1836,10 @@ $$
Lemma 7.1 proves the two full-space bounds.
-We now pass to a hyperplane. Let $\Lambda_H$ be the compression of $\Lambda$ to $H$, and write its eigenvalues as
+We now pass to a hyperplane. Let $Λ_H$ be the compression of $Λ$ to $H$, and write its eigenvalues as
$$
-\theta_1\ge\cdots\ge\theta_{n-1}.
+\theta_1\ge...\ge\theta_{n-1}.
$$
The min--max principle gives
@@ -1851,7 +1851,7 @@ $$
At the full-space minimizer $z=z_a$, the matrix
$$
-Q_H=z_aI_H-a\Lambda_H
+Q_H=z_aI_H-aΛ_H
$$
is positive definite. Since $a\ge0$, interlacing gives
@@ -1859,11 +1859,11 @@ is positive definite. Since $a\ge0$, interlacing gives
$$
(\det Q_H)^{-1/2}
\le\prod_{i=1}^{n-1}(z_a-a\lambda_i)^{-1/2}
-=\left(\prod_{i=1}^n(z_a-a\lambda_i)^{-1/2}\right)
-\sqrt{z_a-a\lambda_n}.
+=(\prod_{i=1}^n(z_a-a\lambda_i)^{-1/2})
+√(z_a-a\lambda_n).
$$
-The last square root is at most $\sqrt{13}$.
+The last square root is at most $√(13)$.
Let $p_{H,a}$ be the squared-norm density of a Gaussian on $H$ with covariance $Q_H^{-1}$. The same polar-coordinate calculation as in Lemma 7.1, now in dimension $n-1$ but still at squared radius $n$, yields
@@ -1887,13 +1887,13 @@ $$
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. $\square$
+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 $L_n\to\infty$ with $L_n\le\log n$, put
+Now impose Proposition 6.1. Fix a deterministic $L_n\to∞$ with $L_n\le\log n$, put
$$
s_0=L_n n^{-2/3},
@@ -1908,7 +1908,7 @@ At $a=1$, abbreviate
$$
k_i=k_i(1),
\qquad
-Y_i\sim N(0,k_i)\ \text{independently},
+Y_i\sim N(0,k_i)\ independently,
\qquad
p=p_1.
$$
@@ -1928,7 +1928,7 @@ $$
$$
$$
-\left(\sum_i k_i^2\right)^{1/2}\le Cn\sqrt{s_0},
+(\sum_i k_i^2)^{1/2}\le Cn√(s_0),
$$
and
@@ -1941,7 +1941,7 @@ $$
Moreover,
$$
-p(n)\ge\frac{c}{n\sqrt{s_0}},
+p(n)\ge\frac{c}{n√(s_0)},
\qquad
\sup_{r\in\mathbb R}p_{-i}(r)\le Cs_0
\qquad(1\le i\le n).
@@ -1963,14 +1963,14 @@ whose denominators are positive by the edge estimates. Since
$$
R(s)=1-\sqrt s+O(s)
-\qquad(s\downarrow0),
+\qquad(s↓0),
$$
Proposition 6.1 gives, uniformly over the good spectra,
$$
\sum_i h_i
-=n\bigl(1-\sqrt{s_0}+o(\sqrt{s_0})\bigr)<n.
+=n(1-√(s_0)+o(√(s_0)))<n.
$$
The function
@@ -1979,7 +1979,7 @@ $$
z\mapsto\sum_i(z-\lambda_i)^{-1}
$$
-is strictly decreasing from infinity on $(\lambda_1,\infty)$. Its value at $z_1$ is $n$; hence
+is strictly decreasing from infinity on $(\lambda_1,∞)$. Its value at $z_1$ is $n$; hence
$$
\lambda_1<z_1<2+s_0
@@ -1996,17 +1996,17 @@ Because $\sum_i k_i=n$, the total increase is exactly
$$
\sum_i(k_i-h_i)
=n-\sum_i h_i
-=n\sqrt{s_0}\bigl(1+o(1)\bigr).
+=n√(s_0)(1+o(1)).
$$
A nonnegative vector has Euclidean norm at most the sum of its coordinates. The triangle inequality and the second resolvent bound therefore give
$$
-\left(\sum_i k_i^2\right)^{1/2}
-\le\left(\sum_i h_i^2\right)^{1/2}
+(\sum_i k_i^2)^{1/2}
+\le(\sum_i h_i^2)^{1/2}
+\sum_i(k_i-h_i)
-\le C\sqrt n\,s_0^{-1/4}+Cn\sqrt{s_0}
-\le Cn\sqrt{s_0}.
+\le C\sqrt n\,s_0^{-1/4}+Cn√(s_0)
+\le Cn√(s_0).
$$
For the last step,
@@ -2031,7 +2031,7 @@ $$
\sup_r p_{-i}(r)\le Cs_0.
$$
-$\square$
+$QED$
---
@@ -2042,11 +2042,11 @@ $\square$
For every $b>0$, the preceding good spectra satisfy
$$
-\sigma_\Lambda\{|u_1|\le bn^{1/3}\}
+\sigma_Λ\{|u_1|\le bn^{1/3}\}
\le CL_n^2b.
$$
-In particular, given any positive deterministic $b_n\to0$, one can choose $L_n\to\infty$ 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.
+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
@@ -2055,10 +2055,10 @@ Conditioning a single coordinate makes the role of the radial estimates transpar
Let $\varphi_k$ be the density of $N(0,k)$. Lemma 7.1 gives, for $T>0$,
$$
-\sigma_\Lambda\{|u_1|\le T\}
+\sigma_Λ\{|u_1|\le T\}
=\frac1{p(n)}
\int_{|y|\le T}\varphi_{k_1}(y)p_{-1}(n-y^2)\,dy
-\le\frac{Cs_0}{p(n)}\mathbb P\{|Y_1|\le T\}.
+\le(Cs_0)/(p(n))\mathbb P\{|Y_1|\le T\}.
$$
The densities of the nonnegative residual sums are taken to be zero at negative arguments.
@@ -2066,7 +2066,7 @@ The densities of the nonnegative residual sums are taken to be zero at negative
By Lemma 7.4 and the Gaussian density bound,
$$
-\frac{Cs_0}{p(n)}
+(Cs_0)/(p(n))
\le Cns_0^{3/2},
$$
@@ -2074,9 +2074,9 @@ and
$$
\mathbb P\{|Y_1|\le bn^{1/3}\}
-\le\frac{Cbn^{1/3}}{\sqrt{k_1}}
-\le Cbn^{1/3}\sqrt{s_0}
-=Cb\sqrt{L_n}.
+\le\frac{Cbn^{1/3}}{√(k_1)}
+\le Cbn^{1/3}√(s_0)
+=Cb√(L_n).
$$
Their product is $CbL_n^2$, proving the proposition.
@@ -2087,7 +2087,7 @@ $$
L_n=\min\{\log n,b_n^{-1/4}\}
$$
-for all sufficiently large $n$ has the required properties. $\square$
+for all sufficiently large $n$ has the required properties. $QED$
---
@@ -2095,18 +2095,18 @@ for all sufficiently large $n$ has the required properties. $\square$
#### Lemma 7.6 (Squared overlap on the sphere)
-The probability $\sigma_\Lambda$ has mean zero and diagonal covariance in the eigenbasis of $\Lambda$. For every coordinate,
+The probability $\sigma_Λ$ has mean zero and diagonal covariance in the eigenbasis of $Λ$. For every coordinate,
$$
-\int u_i^2\,\sigma_\Lambda(du)
-\le\frac{Cs_0k_i}{p(n)}.
+\int u_i^2\,\sigma_Λ(du)
+\le(Cs_0k_i)/(p(n)).
$$
-Consequently, for independent $U,V$ with law $\sigma_\Lambda$,
+Consequently, for independent $U,V$ with law $\sigma_Λ$,
$$
-\mathbb E\left(\frac{U^TV}{n}\right)^2
-\le\frac{Cs_0^2}{n^2p(n)^2}\sum_i k_i^2
+\mathbb E((U^TV)/(n))^2
+\le(Cs_0^2)/(n^2p(n)^2)\sum_i k_i^2
\le Cn^{-2/3}L_n^4.
$$
@@ -2114,30 +2114,30 @@ $$
We reuse the one-coordinate conditioning, now with the weight $y^2$.
-First, the defining density of $\sigma_\Lambda$ is unchanged by changing the sign of any individual coordinate. This proves the assertions about its mean and off-diagonal covariances.
+First, the defining density of $\sigma_Λ$ is unchanged by changing the sign of any individual coordinate. This proves the assertions about its mean and off-diagonal covariances.
Conditioning as in the preceding proof, now with the nonnegative integrand $y^2$, gives
$$
-\int u_i^2\,\sigma_\Lambda(du)
+\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\frac{Cs_0}{p(n)}\mathbb E Y_i^2
-=\frac{Cs_0k_i}{p(n)}.
+\le(Cs_0)/(p(n))\mathbb E Y_i^2
+=(Cs_0k_i)/(p(n)).
$$
Write
$$
-m_i=\int u_i^2\,\sigma_\Lambda(du).
+m_i=\int u_i^2\,\sigma_Λ(du).
$$
Independence of $U,V$ and diagonality give
$$
-\mathbb E\left(\frac{U^TV}{n}\right)^2
+\mathbb E((U^TV)/(n))^2
=\frac1{n^2}\sum_i m_i^2
-\le\frac{Cs_0^2}{n^2p(n)^2}\sum_i k_i^2.
+\le(Cs_0^2)/(n^2p(n)^2)\sum_i k_i^2.
$$
Lemma 7.2 bounds $p(n)^{-2}$ by
@@ -2152,7 +2152,7 @@ $$
Cn^2s_0^4=Cn^{-2/3}L_n^4,
$$
-as claimed. $\square$
+as claimed. $QED$
---
@@ -2161,7 +2161,7 @@ as claimed. $\square$
We now transfer spherical estimates to the Ising cube. The bridge is Haar averaging over eigenvector orientations.
-Throughout this section, let $L_n\to\infty$ with $L_n\le\log n$, put
+Throughout this section, let $L_n\to∞$ with $L_n\le\log n$, put
$$
s_0=L_n n^{-2/3},
@@ -2172,9 +2172,9 @@ and let $\mathcal G_n=\mathcal G_n(L_n)$ be the eigenvalue-only good event of Pr
We fix a deterministic sequence of spectra in $\mathcal G_n$ and use the notation
$$
-\Lambda=\mathrm{diag}(\lambda_1,\dots,\lambda_n),
+Λ=diag(\lambda_1,...,\lambda_n),
\qquad
-S(a),\quad B(a),\quad z_a,\quad \sigma_\Lambda
+S(a),\quad B(a),\quad z_a,\quad \sigma_Λ
$$
from Section 7.
@@ -2186,18 +2186,18 @@ The error bounds supplied by Proposition 6.1 are deterministic on $\mathcal G_n$
Let $O$ be Haar distributed on the orthogonal group, independent of the ordered spectrum, and put
$$
-W=O\Lambda O^T.
+W=OΛ O^T.
$$
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 $\Lambda$ fixed.
+We write $\mathbb E_O$ and $\mathbb P_O$ for expectation and probability over $O$ with $Λ$ fixed.
Define the normalized Ising partition sum
$$
I=2^{-n}\sum_{x\in\{-1,+1\}^n}
-\exp\left(\frac{x^TO\Lambda O^Tx}{2}\right).
+\exp((x^TOΛ O^Tx)/(2)).
$$
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
@@ -2226,17 +2226,17 @@ 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 $-1\le q\le1$,
$$
-M(q)=\mathbb E\exp\left(
-\frac{(1+q)U^T\Lambda U+(1-q)V^T\Lambda V}{2}
-\right).
+M(q)=\mathbb E\exp(
+((1+q)U^TΛ U+(1-q)V^TΛ V)/(2)
+).
$$
If cube points $x,y$ have overlap $q\in(-1,1)$, their normalized sum and difference
$$
-\frac{x+y}{\sqrt{2(1+q)}},
+\frac{x+y}{√(2(1+q))},
\qquad
-\frac{x-y}{\sqrt{2(1-q)}}
+\frac{x-y}{√(2(1-q))}
$$
are orthogonal and both have norm $\sqrt n$. Rotating them by $O^T$ gives the pair in the definition of $M$.
@@ -2257,17 +2257,17 @@ $$
The exchangeability of $U,V$ shows that $M$ is even and nondecreasing in $|q|$. More explicitly, if
$$
-A=U^T\Lambda U,
+A=U^TΛ U,
\qquad
-D=V^T\Lambda V,
+D=V^TΛ V,
$$
then
$$
-M(q)=\mathbb E\left[
-e^{(A+D)/2}\cosh\left(\frac{q(A-D)}2\right)
-\right],
+M(q)=\mathbb E[
+e^{(A+D)/2}cosh(\frac{q(A-D)}2)
+],
$$
whose integrand is nondecreasing in $|q|$.
@@ -2277,7 +2277,7 @@ For two independent uniform sphere points the overlap has density
$$
f_S(q)=c_n(1-q^2)^{(n-3)/2},
\qquad
-c_n=\frac{\Gamma(n/2)}{\sqrt\pi\,\Gamma((n-1)/2)},
+c_n=(\Gamma(n/2))/(\sqrt\pi\,\Gamma((n-1)/2)),
\qquad -1<q<1.
$$
@@ -2305,8 +2305,8 @@ First consider the size of the pair integral. By evenness it suffices to take $q
Dividing by the lower bound for $S(1)^2$ from the same lemma gives
$$
-\frac{M(q)}{S(1)^2}
-\le Cn^2\exp\left(n\bigl[B(1+|q|)+B(1-|q|)-2B(1)\bigr]\right).
+(M(q))/(S(1)^2)
+\le Cn^2\exp(n[B(1+|q|)+B(1-|q|)-2B(1)]).
$$
This remains valid at $|q|=1$, since the integral with parameter zero equals one.
@@ -2317,7 +2317,7 @@ $$
p_I(q)\le e^{-nq^2/2}.
$$
-Indeed, for a mean of $n$ independent uniform signs, the exponential moment bound $\cosh t\le e^{t^2/2}$ follows by integrating $(\log\cosh t)''\le1$ twice from zero. Optimizing the resulting Chernoff bound and using symmetry gives the stated bound on each lattice mass.
+Indeed, for a mean of $n$ independent uniform signs, the exponential moment bound $cosh t\le e^{t^2/2}$ follows by integrating $(\logcosh t)''\le1$ twice from zero. Optimizing the resulting Chernoff bound and using symmetry gives the stated bound on each lattice mass.
Thus a margin below $q^2/2$ in the exponent of the pair bound will suppress large overlaps.
@@ -2330,11 +2330,11 @@ $$
The proof uses three scale relations:
$$
-nq_0^3=n^{1/8}\longrightarrow\infty,
+nq_0^3=n^{1/8}→∞,
\qquad
-nq_0^4=n^{-1/6}\longrightarrow0,
+nq_0^4=n^{-1/6}→0,
\qquad
-\frac{s_0}{q_0^2}=L_n n^{-1/12}\longrightarrow0.
+(s_0)/(q_0^2)=L_n n^{-1/12}→0.
$$
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 $L_n\le\log n$.
@@ -2351,7 +2351,7 @@ There is a universal constant $c>0$ such that, for all sufficiently large $n$, u
$$
B(1+q)+B(1-q)-2B(1)
-\le\frac{q^2}{2}-cq^3.
+\le(q^2)/(2)-cq^3.
$$
One may take $c=1/24$.
@@ -2364,9 +2364,9 @@ The saddle equation in Lemma 7.1 and implicit differentiation show that $z_a$ is
$$
B'(a)=\frac1{2n}\sum_{i=1}^n
-\frac{\lambda_i}{z_a-a\lambda_i},
+(\lambda_i)/(z_a-a\lambda_i),
\qquad
-B'(a)=\frac{z_a-1}{2a}
+B'(a)=(z_a-1)/(2a)
\quad(a>0).
$$
@@ -2392,14 +2392,14 @@ $$
At $s=s_0$, the spectral estimates give
$$
-m_n(s_0)=1-\sqrt{s_0}+o(\sqrt{s_0})<1.
+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\frac{1+u}{2}-\frac{u^2}{2(1+u)}+\frac{s_0}{2}.
+\le(1+u)/(2)-(u^2)/(2(1+u))+(s_0)/(2).
$$
The upper derivative estimate must be paired with a lower estimate below $a=1$. The small-overlap range requires the following quantitative comparison.
@@ -2407,13 +2407,13 @@ The upper derivative estimate must be paired with a lower estimate below $a=1$.
If
$$
-4\sqrt{s_0}\le u\le1/4,
+4√(s_0)\le u\le1/4,
$$
set
$$
-t=\frac{u^2}{1-u}.
+t=(u^2)/(1-u).
$$
Then $t\le1/12$ and $0.9t\ge14.4s_0$. The function $R$ from Proposition 6.1 satisfies
@@ -2425,8 +2425,8 @@ $$
and
$$
--R'(s)=\frac{2+s}{2\sqrt{s(4+s)}}-\frac12
-\ge\frac{c_1}{\sqrt s}
+-R'(s)=\frac{2+s}{2√(s(4+s))}-\frac12
+\ge(c_1)/(\sqrt s)
\qquad(0<s\le1/12)
$$
@@ -2453,11 +2453,11 @@ $$
Substitution in the derivative formula yields
$$
-B'(1-u)\ge\frac{1-u}{2}-\frac{u^2}{20(1-u)}
-\qquad(4\sqrt{s_0}\le u\le1/4).
+B'(1-u)\ge(1-u)/(2)-(u^2)/(20(1-u))
+\qquad(4√(s_0)\le u\le1/4).
$$
-For $0\le u\le4\sqrt{s_0}$, use instead
+For $0\le u\le4√(s_0)$, use instead
$$
r_{1-u}>\lambda_1=2+o(s_0)
@@ -2467,8 +2467,8 @@ to obtain
$$
B'(1-u)
-\ge\frac{1-u}{2}-\frac{u^2}{2(1-u)}-o(s_0)
-\ge\frac{1-u}{2}-Cs_0.
+\ge(1-u)/(2)-(u^2)/(2(1-u))-o(s_0)
+\ge(1-u)/(2)-Cs_0.
$$
Write
@@ -2483,25 +2483,25 @@ $$
\frac1{2(1+u)}-\frac1{20(1-u)}\ge\frac13.
$$
-Combining the preceding bounds, and absorbing $u^2\le16s_0$ on the short interval $u\le4\sqrt{s_0}$ into the error, gives
+Combining the preceding bounds, and absorbing $u^2\le16s_0$ on the short interval $u\le4√(s_0)$ into the error, gives
$$
-F'(u)\le u-\frac{u^2}{3}+Cs_0
+F'(u)\le u-(u^2)/(3)+Cs_0
\qquad(0\le u\le1/4).
$$
Since $F(0)=0$,
$$
-F(q)\le\frac{q^2}{2}-\frac{q^3}{9}+Cs_0q
+F(q)\le(q^2)/(2)-(q^3)/(9)+Cs_0q
\qquad(0\le q\le1/4).
$$
For $q\ge q_0$, the last error divided by $q^3$ is at most
$$
-C\frac{s_0}{q_0^2}
-=CL_n n^{-1/12}\longrightarrow0.
+C(s_0)/(q_0^2)
+=CL_n n^{-1/12}→0.
$$
Thus the cubic deficit holds on $[q_0,1/4]$, even with $c=1/18$.
@@ -2517,9 +2517,9 @@ $$
the strict monotonicity of $R$, and the spectral resolvent estimate imply
$$
-r_a\longrightarrow a+a^{-1},
+r_a→ a+a^{-1},
\qquad
-B'(a)\longrightarrow a/2.
+B'(a)→ a/2.
$$
Indeed, evaluating the empirical resolvent at
@@ -2528,28 +2528,28 @@ $$
a+a^{-1}\pm\delta,
$$
-with $\delta>0$ small enough that the lower point is above $2$, traps $r_a$ between these two points for all large $n$. The upper resolvent parameter needed here is finite for each fixed $a$. No estimate uniform as $a\downarrow0$ is used.
+with $\delta>0$ small enough that the lower point is above $2$, traps $r_a$ between these two points for all large $n$. The upper resolvent parameter needed here is finite for each fixed $a$. No estimate uniform as $a↓0$ is used.
The bound $|B'|\le3/2$ now gives, by dominated convergence,
$$
-\rho_n:=\int_0^1|B'(a)-a/2|\,da\longrightarrow0.
+\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\frac{q^2}{2}
--\int_0^q\frac{u^2}{2(1+u)}\,du+\rho_n+\frac{s_0}{2}
-\le\frac{q^2}{2}-\frac{q^3}{12}+\rho_n+\frac{s_0}{2}.
+\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).
$$
When $q\ge1/4$, the cubic term is bounded away from zero, so the last two errors are absorbed with $c=1/24$. This also covers $q=1$.
The conclusion must hold uniformly over the entire deterministic spectral event. The estimates in the shrinking range already have that uniformity. The same is true of $\rho_n\to0$: otherwise one could select spectra from $\mathcal G_n$ along a violating subsequence. Their deterministic spectral bounds would still imply the pointwise convergence and bounded domination just proved, a contradiction.
-Thus the two ranges give the claimed uniform statement. $\square$
+Thus the two ranges give the claimed uniform statement. $QED$
---
@@ -2567,15 +2567,15 @@ $$
and
$$
-\sup_{\Lambda\in\mathcal G_n}
-\mathbb E_O\left[\left(\frac{I}{S(1)}-1\right)^2\right]
-\longrightarrow0.
+\sup_{Λ\in\mathcal G_n}
+\mathbb E_O[((I)/(S(1))-1)^2]
+→0.
$$
In particular, under the GOE law,
$$
-\frac{I}{S(1)}\to1
+(I)/(S(1))\to1
$$
in probability.
@@ -2589,7 +2589,7 @@ The first-moment identity was established above. We now bound the exact second m
Combining the pair bound, the Chernoff bound $p_I(q)\le e^{-nq^2/2}$, and Lemma 8.1 gives
$$
-p_I(q)\frac{M(q)}{S(1)^2}
+p_I(q)(M(q))/(S(1)^2)
\le Cn^2e^{-cn|q|^3}
\qquad(q\in\mathcal Q_n,\ |q|\ge q_0).
$$
@@ -2598,7 +2598,7 @@ There are at most $n+1$ lattice points. Since $nq_0^3=n^{1/8}$,
$$
\sum_{\substack{q\in\mathcal Q_n\\ |q|\ge q_0}}
-p_I(q)\frac{M(q)}{S(1)^2}
+p_I(q)(M(q))/(S(1)^2)
\le Cn^3e^{-cn^{1/8}}
=:\tau_n.
$$
@@ -2619,8 +2619,8 @@ Stirling's formula, uniformly for $|q|<q_0$, gives
$$
p_I(q)=
-\sqrt{\frac{2}{\pi n(1-q^2)}}\,
-e^{-nJ(q)}\bigl(1+O(n^{-1})\bigr),
+√((2)/(\pi n(1-q^2)))\,
+e^{-nJ(q)}(1+O(n^{-1})),
$$
where
@@ -2632,7 +2632,7 @@ $$
Here $J(q)=q^2/2+O(q^4)$. Likewise
$$
-c_n=\sqrt{n/(2\pi)}(1+O(n^{-1}))
+c_n=√(n/(2\pi))(1+O(n^{-1}))
$$
and
@@ -2644,8 +2644,8 @@ $$
Dividing the two expressions shows that
$$
-\frac{p_I(q)}{\Delta f_S(q)}
-=\exp\bigl(O(n^{-1}+q^2+nq^4)\bigr)
+(p_I(q))/(\Delta f_S(q))
+=\exp(O(n^{-1}+q^2+nq^4))
=1+o(1),
\qquad |q|<q_0,
$$
@@ -2656,17 +2656,17 @@ Assign to each retained lattice point the outward bin
$$
\mathcal B_q=
-\begin{cases}
+
[q,q+\Delta],&q>0,\\
[q-\Delta,q],&q<0,\\
[-\Delta/2,\Delta/2],&q=0.
-\end{cases}
+
$$
For even $n$, the lattice near zero is
$$
-\ldots,-2\Delta,-\Delta,0,\Delta,2\Delta,\ldots .
+...,-2\Delta,-\Delta,0,\Delta,2\Delta,... .
$$
The centered zero bin leaves gaps between $\Delta/2$ and $\Delta$, and between $-\Delta$ and $-\Delta/2$.
@@ -2684,7 +2684,7 @@ for large $n$.
On each such bin,
$$
-\left|\log\frac{f_S(r)}{f_S(q)}\right|
+|\log(f_S(r))/(f_S(q))|
\le Cn(q_0+\Delta)\Delta
\le C(q_0+\Delta),
$$
@@ -2728,19 +2728,19 @@ This estimate does not depend on how rapidly $h$ varies. The outermost bin may e
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
$$
-\frac{\mathbb E_O I^2}{S(1)^2}
+(\mathbb E_O I^2)/(S(1)^2)
\le1+\delta_n+\tau_n.
$$
The first-moment identity therefore gives
$$
-\mathbb E_O\left[\left(\frac{I}{S(1)}-1\right)^2\right]
+\mathbb E_O[((I)/(S(1))-1)^2]
\le\delta_n+\tau_n
-\longrightarrow0.
+→0.
$$
-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 $\mathbb P(\mathcal G_n^c)\to0$. $\square$
+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 $\mathbb P(\mathcal G_n^c)\to0$. $QED$
---
@@ -2752,8 +2752,8 @@ Let $W$ be the augmented GOE matrix and let $v$ be a unit top eigenvector. For e
$$
\mu\{x:|v^Tx|\le b_nn^{1/3}\}
-\longrightarrow0
-\qquad\text{in probability}.
+→0
+\qquadin probability.
$$
The probability here includes the auxiliary diagonal used to define $v$; the Gibbs measure itself does not depend on that diagonal.
@@ -2762,13 +2762,13 @@ The probability here includes the auxiliary diagonal used to define $v$; the Gib
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 $L_n\to\infty$ with $L_n\le\log n$ and $L_n^2b_n\to0$. For example, for all sufficiently large $n$ one may take
+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 $W=O\Lambda O^T$, take $v=Oe_1$, and define the restricted normalized sum
+In the representation $W=OΛ O^T$, take $v=Oe_1$, and define the restricted normalized sum
$$
I_b=2^{-n}\sum_x
@@ -2778,25 +2778,25 @@ $$
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 $\mathcal G_n$,
$$
-\mathbb E_O\frac{I_b}{S(1)}
-=\sigma_\Lambda\{|U_1|\le b_nn^{1/3}\}
+\mathbb E_O(I_b)/(S(1))
+=\sigma_Λ\{|U_1|\le b_nn^{1/3}\}
\le CL_n^2b_n.
$$
-For every fixed $\varepsilon>0$, Markov's inequality and Proposition 8.2 imply
+For every fixed $ε>0$, Markov's inequality and Proposition 8.2 imply
$$
-\mathbb P_O\left(\frac{I_b}{I}>\varepsilon\right)
-\le\mathbb P_O\left(\frac{I}{S(1)}<\frac12\right)
-+\frac{2}{\varepsilon}\mathbb E_O\frac{I_b}{S(1)}
-\longrightarrow0
+\mathbb P_O((I_b)/(I)>ε)
+\le\mathbb P_O((I)/(S(1))<\frac12)
++(2)/(ε)\mathbb E_O(I_b)/(S(1))
+→0
$$
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. $\square$
+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. $QED$
---
@@ -2807,13 +2807,13 @@ GOE has a simple spectrum almost surely: the set of matrices with repeated eigen
Let
$$
-Q_n=\mathbb E_{\mu^{\otimes2}}q(x,y)^2.
+Q_n=\mathbb E_{\mu^{ ⊗ 2}}q(x,y)^2.
$$
-Then $n^{2/3}Q_n$ is bounded in probability. Equivalently, for every positive deterministic sequence $T_n\to\infty$,
+Then $n^{2/3}Q_n$ is bounded in probability. Equivalently, for every positive deterministic sequence $T_n\to∞$,
$$
-\mathbb P\{n^{2/3}Q_n>T_n\}\longrightarrow0.
+\mathbb P\{n^{2/3}Q_n>T_n\}→0.
$$
This assertion concerns the original off-diagonal SK disorder.
@@ -2825,7 +2825,7 @@ To prove tightness rather than a bound with a fixed logarithmic loss, keep $L_n$
The same second-moment comparison works with the nonnegative overlap weight $q^2$. Expanding two Gibbs weights gives the exact identity
$$
-\mathbb E_O\left[\left(\frac{I}{S(1)}\right)^2Q_n\right]
+\mathbb E_O[((I)/(S(1)))^2Q_n]
=\frac1{S(1)^2}\sum_{q\in\mathcal Q_n}p_I(q)q^2M(q).
$$
@@ -2835,8 +2835,8 @@ The central terms therefore satisfy the same bin comparison. The tails are at mo
$$
\frac1{S(1)^2}\int_{-1}^1q^2M(q)f_S(q)\,dq
-=\mathbb E_{\sigma_\Lambda^{\otimes2}}
-\left(\frac{U^TV}{n}\right)^2.
+=\mathbb E_{\sigma_Λ^{ ⊗ 2}}
+((U^TV)/(n))^2.
$$
Here $U,V$ are independent under the tilted spherical law. Lemma 7.6 bounds this expectation by
@@ -2848,20 +2848,20 @@ $$
Thus
$$
-\mathbb E_O\left[\left(\frac{I}{S(1)}\right)^2n^{2/3}Q_n\right]
+\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
$$
for all large $n$, uniformly on $\mathcal G_n$.
-The weighted estimate is now available for every admissible $L_n$. Given an arbitrary deterministic $T_n\to\infty$, choose
+The weighted estimate is now available for every admissible $L_n$. Given an arbitrary deterministic $T_n\to∞$, choose
$$
L_n=\min\{\log n,T_n^{1/8}\}
$$
-for all sufficiently large $n$. Then $L_n\to\infty$ and
+for all sufficiently large $n$. Then $L_n\to∞$ and
$$
L_n^4/T_n\le T_n^{-1/2}\to0.
@@ -2879,24 +2879,24 @@ $$
\mathbb P_O(n^{2/3}Q_n>T_n)
\le\mathbb P_O(\widehat I<1/2)
+\frac4{T_n}\mathbb E_O[\widehat I^{\,2}n^{2/3}Q_n]
-\le o(1)+\frac{CL_n^4}{T_n}.
+\le o(1)+(CL_n^4)/(T_n).
$$
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 $\varepsilon>0$ and increasing indices $n_j$ with
+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
$$
-\mathbb P(n_j^{2/3}Q_{n_j}>j)>\varepsilon.
+\mathbb P(n_j^{2/3}Q_{n_j}>j)>ε.
$$
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. $\square$
+Finally, because the Gibbs law is invariant under adding a diagonal to $W$, integrating out the auxiliary diagonal preserves the asserted probabilities. $QED$
---
@@ -2904,12 +2904,12 @@ Finally, because the Gibbs law is invariant under adding a diagonal to $W$, inte
#### Corollary 8.5 (Covariance scale)
-For every positive deterministic sequence $M_n\to\infty$,
+For every positive deterministic sequence $M_n\to∞$,
$$
-\mathbb P\left\{
-\frac{n^{2/3}}{M_n}\le\|\Sigma\|\le M_nn^{2/3}
-\right\}\longrightarrow1.
+\mathbb P{
+\frac{n^{2/3}}{M_n}\le\|Σ\|\le M_nn^{2/3}
+}→1.
$$
#### Proof
@@ -2919,22 +2919,22 @@ The upper and lower bounds use different outputs of the transfer: squared overla
Spin-flip symmetry gives $\mu[x]=0$, so
$$
-\Sigma=\mu[xx^T]
+Σ=\mu[xx^T]
$$
and
$$
-\mathrm{Tr}(\Sigma^2)
+Tr(Σ^2)
=\sum_{i,j}\mu[x_ix_j]^2
-=n^2\mathbb E_{\mu^{\otimes2}}q(x,y)^2
+=n^2\mathbb E_{\mu^{ ⊗ 2}}q(x,y)^2
=n^2Q_n.
$$
-Since $\Sigma$ is positive semidefinite,
+Since $Σ$ is positive semidefinite,
$$
-\|\Sigma\|^2\le\mathrm{Tr}(\Sigma^2).
+\|Σ\|^2\leTr(Σ^2).
$$
The upper bound follows from Proposition 8.4 with threshold $T_n=M_n^2$.
@@ -2954,7 +2954,7 @@ $$
Then $B_n\to0$ in probability and
$$
-\|\Sigma\|\ge v^T\Sigma v
+\|Σ\|\ge v^TΣ v
=\mu[(v^Tx)^2]
\ge b_n^2n^{2/3}(1-B_n).
$$
@@ -2962,10 +2962,10 @@ $$
On $B_n\le1/2$, for all sufficiently large $n$ this is at least
$$
-n^{2/3}/(2\sqrt{M_n})\ge n^{2/3}/M_n.
+n^{2/3}/(2√(M_n))\ge n^{2/3}/M_n.
$$
-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. $\square$
+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. $QED$
---
@@ -2985,7 +2985,7 @@ $$
Choose a positive deterministic $b_n\to0$ such that
$$
-\frac{t_n}{b_n^2n^{2/3}}\longrightarrow0.
+\frac{t_n}{b_n^2n^{2/3}}→0.
$$
For example,
@@ -3035,20 +3035,20 @@ $$
Hence Lemma 5.2 and Markov's inequality give
$$
-\mathbb P_{\mu_{W_0}}\bigl(
-g(X_0),g(X_{t_n})\text{ do not have the same nonzero sign}
+\mathbb P_{\mu_{W_0}}(
+g(X_0),g(X_{t_n}) do not have the same nonzero sign
\mid W_0,\Delta
-\bigr)
-\le\mu_{W_0}\{|g|\le a_n\}+\frac{2t_n}{a_n^2}.
+)
+\le\mu_{W_0}\{|g|\le a_n\}+(2t_n)/(a_n^2).
$$
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
$$
-\varepsilon_n
+ε_n
:=\mathbb E_{W_0,\Delta}\mu_{W_0}\{|g|\le a_n\}
-+\frac{2t_n}{a_n^2}
-\longrightarrow0
++(2t_n)/(a_n^2)
+→0
$$
bounds the average sign-failure probability.
@@ -3080,7 +3080,7 @@ This definition includes every zero projection in the failure event, and the sig
$$
\mathbb E_{W_0,\Delta}\mu_{W_0}[r_n(W_0,\Delta,\cdot)]
-\le\varepsilon_n.
+\leε_n.
$$
Spin-flip symmetry gives
@@ -3092,14 +3092,14 @@ $$
If $g(x)\ne0$, testing total variation against this half-space yields
$$
-d_{t_n}^{\mathrm{ct}}(W_0,x)
+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}^{\mathrm{ct}}(W_0,x)\le1/4\}}
+\mathbf 1_{\{d_{t_n}^{ct}(W_0,x)\le1/4\}}
\le4r_n(W_0,\Delta,x).
$$
@@ -3109,21 +3109,21 @@ The test half-space depends on $\Delta$, but the bad-state indicator on the left
$$
\mathbb E_{W_0}\mu_{W_0}
-\{x:d_{t_n}^{\mathrm{ct}}(W_0,x)\le1/4\}
+\{x:d_{t_n}^{ct}(W_0,x)\le1/4\}
=
\mathbb E_{W_0,\Delta}\mu_{W_0}
-\{x:d_{t_n}^{\mathrm{ct}}(W_0,x)\le1/4\}
-\le4\varepsilon_n.
+\{x:d_{t_n}^{ct}(W_0,x)\le1/4\}
+\le4ε_n.
$$
In particular, for every fixed $\delta>0$,
$$
-\mathbb P_{W_0}\left(
-\mu_{W_0}\{x:d_{t_n}^{\mathrm{ct}}(W_0,x)>1/4\}<1-\delta
-\right)
-\le\frac{4\varepsilon_n}{\delta}
-\longrightarrow0.
+\mathbb P_{W_0}(
+\mu_{W_0}\{x:d_{t_n}^{ct}(W_0,x)>1/4\}<1-\delta
+)
+\le(4ε_n)/(\delta)
+→0.
$$
This is the asserted convergence in probability over the original off-diagonal disorder.
@@ -3139,12 +3139,12 @@ $$
and the discrete increment bound. The same proof replaces $2t_n/a_n^2$ by
$$
-\frac{2k_n}{na_n^2}
+(2k_n)/(na_n^2)
$$
-and proves the displayed conclusion with $d_{k_n}^{\mathrm{dt}}$.
+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. $\square$
+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. $QED$
---
@@ -3160,11 +3160,11 @@ We use the standard spin-deletion cavity reweighting. The calculation below uses
With probability tending to one,
$$
-\|\Sigma\|\le\mathcal R_{\mathrm{lin}}(\mu)
-\le16\|\Sigma\|+n^{13/25}.
+\|Σ\|\le\mathcal R_{lin}(\mu)
+\le16\|Σ\|+n^{13/25}.
$$
-In particular, $n^{-2/3}\mathcal R_{\mathrm{lin}}(\mu)$ is bounded in probability.
+In particular, $n^{-2/3}\mathcal R_{lin}(\mu)$ is bounded in probability.
#### Proof
@@ -3177,7 +3177,7 @@ For each site define its local field and average conditional variance by
$$
g_i(x)=\sum_{j\ne i}W_{ij}x_j,
\qquad
-\alpha_i=\mu[\mathrm{sech}^2g_i].
+\alpha_i=\mu[sech^2g_i].
$$
Thus $0<\alpha_i\le1$, and Lemma 5.1 gives
@@ -3185,29 +3185,29 @@ Thus $0<\alpha_i\le1$, and Lemma 5.1 gives
$$
\mathcal D_\mu(a^Tx)=\sum_i\alpha_i a_i^2,
\qquad
-\mathrm{Var}_\mu(a^Tx)=a^T\Sigma a.
+Var_\mu(a^Tx)=a^TΣ a.
$$
It follows by the change of variables
$$
-b_i=\sqrt{\alpha_i}a_i
+b_i=√(\alpha_i)a_i
$$
that
$$
-\mathcal R_{\mathrm{lin}}(\mu)=\|\Gamma\|,
+\mathcal R_{lin}(\mu)=\|\Gamma\|,
\qquad
-\Gamma=\mathrm{diag}(\alpha_i^{-1/2})\Sigma
-\mathrm{diag}(\alpha_i^{-1/2}).
+\Gamma=diag(\alpha_i^{-1/2})Σ
+diag(\alpha_i^{-1/2}).
$$
For every nonzero $a$, the inequalities $\alpha_i\le1$ and the nonnegativity of the variance give
$$
-\frac{a^T\Sigma a}{\sum_i\alpha_i a_i^2}
-\ge\frac{a^T\Sigma a}{\|a\|^2}.
+(a^TΣ a)/(\sum_i\alpha_i a_i^2)
+\ge(a^TΣ a)/(\|a\|^2).
$$
Taking the supremum proves the lower bound in the proposition.
@@ -3217,28 +3217,28 @@ Taking the supremum proves the lower bound in the proposition.
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
$$
-Z_i=\nu_i[\cosh g_i],
+Z_i=\nu_i[cosh g_i],
\qquad
-R_i=\mathbb E_{\nu_i\otimes\nu_i}
-\left[\left(\frac{x_{-i}^Ty_{-i}}{n}\right)^2\right].
+R_i=\mathbb E_{\nu_i ⊗ \nu_i}
+[(\frac{x_{-i}^Ty_{-i}}{n})^2].
$$
Summing over $x_i$ gives the exact identities
$$
-\frac{d\bar\mu_i}{d\nu_i}=\frac{\cosh g_i}{Z_i},
+(d\bar\mu_i)/(d\nu_i)=(cosh g_i)/(Z_i),
$$
and
$$
-\alpha_i=\frac{\nu_i[(\cosh g_i)^{-1}]}{Z_i}
+\alpha_i=\frac{\nu_i[(cosh g_i)^{-1}]}{Z_i}
\ge Z_i^{-2},
$$
where the last step is Jensen's inequality for the reciprocal function.
-Since $\cosh g_i\ge1$, the product marginal comparison and
+Since $cosh g_i\ge1$, the product marginal comparison and
$$
(q-x_iy_i/n)^2\le2q^2+2n^{-2}
@@ -3248,37 +3248,37 @@ imply
$$
R_i
-\le Z_i^2\mathbb E_{\bar\mu_i\otimes\bar\mu_i}
-\left[\left(\frac{x_{-i}^Ty_{-i}}n\right)^2\right]
+\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_{\mathrm{loc}}^2=(n-1)/n.
+\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_{\mathrm{loc}}^2)}[\cosh^mG]
+\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\left(\max_iZ_i>n^{1/100}\right)\longrightarrow0.
+\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)\longrightarrow1,
+\mathbb P(E_n)→1,
\qquad
-E_n=\left\{\max_iR_i\le r_n\right\},
+E_n={\max_iR_i\le r_n},
\qquad
r_n=n^{-7/12}.
$$
@@ -3313,30 +3313,30 @@ $$
c=x_{-i}^Ty_{-i}/n.
$$
-Their two local fields have variances $\sigma_{\mathrm{loc}}^2$ and covariance $c$, so
+Their two local fields have variances $\sigma_{loc}^2$ and covariance $c$, so
$$
-\mathbb E[\cosh g_i(x)\cosh g_i(y)\mid\mathcal F_i]
-=e^{\sigma_{\mathrm{loc}}^2}\cosh c.
+\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_{\mathrm{loc}}^2/2},
+=e^{\sigma_{loc}^2/2},
$$
and
$$
-\mathrm{Var}(Z_i\mid\mathcal F_i)
-=e^{\sigma_{\mathrm{loc}}^2}
-\mathbb E_{\nu_i\otimes\nu_i}[\cosh c-1]
+Var(Z_i\mid\mathcal F_i)
+=e^{\sigma_{loc}^2}
+\mathbb E_{\nu_i ⊗ \nu_i}[cosh c-1]
\le CR_i,
$$
-because $|c|\le1$ and $\cosh c-1\le Cc^2$ on this interval.
+because $|c|\le1$ and $cosh c-1\le Cc^2$ on this interval.
The event $E_n$ depends on many incident rows. To use the variance estimate with its stated conditioning, introduce the clipped count
@@ -3348,16 +3348,16 @@ $$
The variable $R_i$ is $\mathcal F_i$-measurable, and
$$
-4-e^{\sigma_{\mathrm{loc}}^2/2}\ge4-e^{1/2}>0.
+4-e^{\sigma_{loc}^2/2}\ge4-e^{1/2}>0.
$$
Thus conditional Chebyshev gives
$$
\mathbb E N_n
-=\sum_i\mathbb E\left[
+=\sum_i\mathbb E[
\mathbf 1_{\{R_i\le r_n\}}
-\mathbb P(Z_i>4\mid\mathcal F_i)\right]
+\mathbb P(Z_i>4\mid\mathcal F_i)]
\le Cnr_n=Cn^{5/12}.
$$
@@ -3373,7 +3373,7 @@ $$
\mathbb P(|B|>n^{1/2})
\le\mathbb P(E_n^c)+\mathbb P(N_n>n^{1/2})
\le\mathbb P(E_n^c)+Cn^{-1/12}
-\longrightarrow0.
+→0.
$$
No independence between sites, or conditioning on $E_n$ in the Gaussian calculation, is used.
@@ -3397,19 +3397,19 @@ $$
Thus the principal blocks of the positive semidefinite matrix $\Gamma$ satisfy
$$
-\|\Gamma_{GG}\|\le16\|\Sigma\|,
+\|\Gamma_{GG}\|\le16\|Σ\|,
$$
and
$$
-\|\Gamma_{BB}\|\le\mathrm{Tr}\Gamma_{BB}
+\|\Gamma_{BB}\|\leTr\Gamma_{BB}
=\sum_{i\in B}\alpha_i^{-1}
\le|B|\max_iZ_i^2
\le n^{13/25}.
$$
-Here $\Sigma_{ii}=1$ by spin-flip symmetry.
+Here $Σ_{ii}=1$ by spin-flip symmetry.
Positivity also controls the cross block: for $u\in\mathbb R^G$ and $w\in\mathbb R^B$,
@@ -3439,30 +3439,30 @@ $$
The same bound holds when either block is empty. Therefore
$$
-\|\Gamma\|\le16\|\Sigma\|+n^{13/25},
+\|\Gamma\|\le16\|Σ\|+n^{13/25},
$$
-proving the proposition. Corollary 8.5 and $13/25<2/3$ give the final assertion. $\square$
+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 $M_n\to\infty$ 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,
+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,
$$
-\frac{32n^{2/3}}{M_n}\le\|\Sigma\|
+\frac{32n^{2/3}}{M_n}\le\|Σ\|
\le\frac{M_nn^{2/3}}{32},
$$
and
$$
-\|\Sigma\|\le\mathcal R_{\mathrm{lin}}(\mu)
+\|Σ\|\le\mathcal R_{lin}(\mu)
\le\frac{M_nn^{2/3}}2+n^{13/25}.
$$
-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. $\square$
+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. $QED$
---
@@ -3483,17 +3483,17 @@ $$
and let
$$
-\nu_J(x)=Z_J^{-1}\exp\left(\frac12x^TJx\right),
+\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\left\{
-\mathrm{Var}_{\nu_J}(f)\le C_\beta\mathcal D_{\nu_J}(f)
-\text{ for every }f:\{-1,+1\}^n\to\mathbb R
-\right\}\longrightarrow1.
+\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.
$$
Here $\mathcal D_{\nu_J}$ is the unscaled sum of conditional variances. Consequently, the discrete-time heat-bath chain choosing one uniform site has spectral gap at least $1/(nC_\beta)$ with probability tending to one.
@@ -3505,23 +3505,23 @@ This theorem is supplied by the separate high-temperature companion paper. The c
Let $\rho,\nu$ be probability measures with full support on $\{-1,+1\}^n$. Suppose
$$
-0<m\le\frac{\rho(x)}{\nu(x)}\le M<\infty
-\qquad\text{for every }x.
+0<m\le(\rho(x))/(\nu(x))\le M<∞
+\qquadfor every x.
$$
If
$$
-\mathrm{Var}_\nu(f)\le C_\nu\mathcal D_\nu(f)
-\qquad\text{for every }f,
+Var_\nu(f)\le C_\nu\mathcal D_\nu(f)
+\qquadfor every f,
$$
then
$$
-\mathrm{Var}_\rho(f)
-\le\frac{M}{m}C_\nu\mathcal D_\rho(f)
-\qquad\text{for every }f.
+Var_\rho(f)
+\le(M)/(m)C_\nu\mathcal D_\rho(f)
+\qquadfor every f.
$$
In particular, for
@@ -3537,7 +3537,7 @@ $$
one may take
$$
-M/m=e^{(1-\beta)\mathrm{osc}\,H}.
+M/m=e^{(1-\beta)osc\,H}.
$$
#### Proof
@@ -3547,7 +3547,7 @@ The comparison works directly with least-squares characterizations; one need not
For any full-support probability measure $\eta$,
$$
-\mathrm{Var}_\eta(f)=\inf_{c\in\mathbb R}\eta[(f-c)^2],
+Var_\eta(f)=\inf_{c\in\mathbb R}\eta[(f-c)^2],
$$
and
@@ -3572,7 +3572,7 @@ The first identity follows by the same decomposition with constants.
The density bounds and the variational identities imply
$$
-\mathrm{Var}_\rho(f)\le M\mathrm{Var}_\nu(f),
+Var_\rho(f)\le MVar_\nu(f),
$$
and
@@ -3583,28 +3583,28 @@ $$
which prove the comparison.
-For the exponential densities, $\rho(x)/\nu(x)$ is a constant times $e^{(1-\beta)H(x)}$; the common normalization cancels in the ratio of its maximum and minimum. $\square$
+For the exponential densities, $\rho(x)/\nu(x)$ is a constant times $e^{(1-\beta)H(x)}$; the common normalization cancels in the ratio of its maximum and minimum. $QED$
#### Proposition 9.4 (Subexponential upper bound)
-For every fixed $\epsilon>0$,
+For every fixed $ε>0$,
$$
-\mathbb P\left\{
-t_{\mathrm{mix}}\le e^{\epsilon n},
+\mathbb P{
+t_{mix}\le e^{ε n},
\quad
-k_{\mathrm{mix}}\le e^{\epsilon n}
-\right\}\longrightarrow1.
+k_{mix}\le e^{ε n}
+}→1.
$$
#### Proof
We keep the order of parameters explicit: first the requested exponential tolerance, then a fixed subcritical temperature, then the dimension.
-Fix $\epsilon>0$ and choose $\beta\in(0,1)$ such that
+Fix $ε>0$ and choose $\beta\in(0,1)$ such that
$$
-3(1-\beta)<\epsilon.
+3(1-\beta)<ε.
$$
Let
@@ -3616,9 +3616,9 @@ $$
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,
$$
-\mathrm{Var}_{\mu_\beta}(f)
+Var_{\mu_\beta}(f)
\le C_\beta\mathcal D_{\mu_\beta}(f)
-\qquad\text{for every }f.
+\qquadfor every f.
$$
Intersect this event with $\|W\|\le3$, whose probability tends to one by Proposition 6.1.
@@ -3626,21 +3626,21 @@ Intersect this event with $\|W\|\le3$, whose probability tends to one by Proposi
On their intersection,
$$
-\mathrm{osc}\,H_W\le3n,
+osc\,H_W\le3n,
$$
and
$$
\mu_{\min}:=\min_x\mu(x)
-\ge2^{-n}e^{-\mathrm{osc}\,H_W}
+\ge2^{-n}e^{-osc\,H_W}
\ge e^{-(3+\log2)n}.
$$
Lemma 9.3 now gives
$$
-\mathrm{Var}_\mu(f)\le C_n\mathcal D_\mu(f),
+Var_\mu(f)\le C_n\mathcal D_\mu(f),
\qquad
C_n=C_\beta e^{3(1-\beta)n}.
$$
@@ -3680,14 +3680,14 @@ $$
Reversibility identifies the transition density relative to $\mu$, minus one, with $e^{t\mathcal L}h_x$ or $P^kh_x$. The $L^1$ expression for total variation and Cauchy--Schwarz give
$$
-\max_x d_t^{\mathrm{ct}}(W,x)
+\max_x d_t^{ct}(W,x)
\le\tfrac12\mu_{\min}^{-1/2}e^{-t/C_n},
$$
and
$$
-\max_x d_k^{\mathrm{dt}}(W,x)
+\max_x d_k^{dt}(W,x)
\le\tfrac12\mu_{\min}^{-1/2}e^{-k/(nC_n)}.
$$
@@ -3700,25 +3700,25 @@ $$
these bounds imply
$$
-t_{\mathrm{mix}}\le C_nB_n,
+t_{mix}\le C_nB_n,
\qquad
-k_{\mathrm{mix}}\le\lceil nC_nB_n\rceil.
+k_{mix}\le\lceil nC_nB_n\rceil.
$$
-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 $3(1-\beta)<\epsilon$, both are at most $e^{\epsilon n}$ for all sufficiently large $n$.
+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. $\square$
+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 $\epsilon>0$ and apply Theorem A to
+Fix $ε>0$ and apply Theorem A to
$$
-t_n=n^{2/3-\epsilon},
+t_n=n^{2/3-ε},
\qquad
-k_n=\lfloor n^{5/3-\epsilon}\rfloor.
+k_n=\lfloor n^{5/3-ε}\rfloor.
$$
For each clock, with probability tending to one there is a starting configuration whose distance at the displayed time exceeds $1/4$.
@@ -3726,24 +3726,24 @@ For each clock, with probability tending to one there is a starting configuratio
Worst-start distance is nonincreasing in time by contraction of total variation under a Markov kernel. Hence
$$
-t_{\mathrm{mix}}\ge t_n
+t_{mix}\ge t_n
$$
and
$$
-k_{\mathrm{mix}}>k_n
+k_{mix}>k_n
$$
-on those events. Since $k_{\mathrm{mix}}$ is an integer, the latter implies
+on those events. Since $k_{mix}$ is an integer, the latter implies
$$
-k_{\mathrm{mix}}\ge n^{5/3-\epsilon},
+k_{mix}\ge n^{5/3-ε},
$$
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. $\square$
+Intersect these two lower-bound events with the event in Proposition 9.4 to obtain all four asserted bounds simultaneously. $QED$
---
@@ -3756,7 +3756,7 @@ A third calculation distinguishes high-probability asymptotic bounds from essent
### Proposition 10.1 (Arbitrary refresh times)
-Fix any realization of the SK measure on $\Omega_n=\{-1,+1\}^n$, $n\ge2$. 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 $r$.
+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
@@ -3765,7 +3765,7 @@ At each step, refresh the whole configuration from $\mu$ with probability $p$ an
$$
\Pi(x,y)=\mu(y),
\qquad
-K_p=(1-p)\mathrm{Id}+p\Pi,
+K_p=(1-p)Id+p\Pi,
\qquad
0<p\le1.
$$
@@ -3779,13 +3779,13 @@ $$
The identity $\Pi^2=\Pi$ gives, for $0<p<1$,
$$
-K_p^k=\Pi+(1-p)^k(\mathrm{Id}-\Pi),
+K_p^k=\Pi+(1-p)^k(Id-\Pi),
$$
and
$$
-\max_x\|K_p^k(x,\cdot)-\mu\|_{\mathrm{TV}}
+\max_x\|K_p^k(x,\cdot)-\mu\|_{TV}
=(1-p)^k(1-\mu_{\min}).
$$
@@ -3798,15 +3798,15 @@ $$
Choosing
$$
-1-p=\left[\frac{1}{4(1-\mu_{\min})}\right]^{1/r}
+1-p=[\frac{1}{4(1-\mu_{\min})}]^{1/r}
$$
-gives distance exactly $1/4$ at $r$ and strictly larger at every earlier integer. The case $p=1$ also has mixing time one. $\square$
+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
$$
-(1-p)\mathrm{Id}+pP
+(1-p)Id+pP
$$
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
@@ -3815,7 +3815,7 @@ $$
(1-p)^k-\mu_{\min},
$$
-by the event that no update was attempted. Its mixing time consequently diverges as $p\downarrow0$. Allowing $p$ to depend on $n$ changes possible growth classes as well.
+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.
@@ -3838,19 +3838,19 @@ Put $w=W_{12}$. Spin-flip symmetry gives $\mu[g]=0$, and conditional expectation
$$
Pg=\theta g,
\qquad
-\theta=\frac{1+\tanh w}{2}.
+\theta=(1+tanh w)/(2).
$$
-Starting at $(1,1)$, the expectation of $g$ after $k$ steps is $2\theta^k$. Since $\|g\|_\infty=2$, the dual total-variation bound yields
+Starting at $(1,1)$, the expectation of $g$ after $k$ steps is $2\theta^k$. Since $\|g\|_∞=2$, the dual total-variation bound yields
$$
-\|P^k((1,1),\cdot)-\mu\|_{\mathrm{TV}}
+\|P^k((1,1),\cdot)-\mu\|_{TV}
\ge\tfrac12\theta^k.
$$
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. $\square$
+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.