\chapter{Fixed high temperature}\label{sec:hot-fixed}

In this section, we prove the second case of \cref{thm:intro-main}. Fix $c\in(0,1)$ and let $g=(1-c)^2$. We show in \cref{hp-fixed-profile} that $D_N$ has the cutoff profile $\Psi$ of \eqref{hp-universal-profile}, the Gaussian profile that also describes the random walk on the hypercube \cite[Theorem~1]{DiaconisGrahamMorrison1990}, with window $1/g$ and center
\[
T_N=\frac{\log N-\frac32\log\log N+\log A_c}{4g},\qquad A_c=\frac{\sqrt c\,(1-c)^3}{4\sqrt\pi\,(1+c)^2}.
\]
We combine three results from \cref{part:tools}. First, \cref{ht-bingham-bridge} implies that the time-$t$ law $\nu_NP_t$ is close to the Bingham comparison law $\mu_t$ of \cref{sec:heat} once $t$ is large. Second, both $\mu_t$ and $\pi_{N,c}$ are Gaussian laws conditioned on $S_N$, with variances $v_i(t)$ and $q_i$ of equal sum, and \cref{gp-pinned-lan} implies that the distance between them is close to $2\Phi(\sqrt{\cI_N(t)}/2)-1$, where $\cI_N(t)$ is the information of the covariance evolution (\cref{def:model-information}). The new work in this section is the computation of $\cI_N(t)$ at the times $T_N+u/g$. We compare the covariance evolution of the GOE spectrum with the same evolution for the semicircle law, which we solve exactly. Its relative errors are of order $e^{-2gt}$, and their amplitude is of order one only for eigenvalues within distance $O(1/t)$ of the spectral edge. The square-root vanishing of the semicircle density at the edge then produces a factor $t^{-3/2}$ in the information, and this factor is the source of the term $-\frac32\log\log N$ in $T_N$.

We use the rigidity estimate of \cref{sp-goe-rigidity}, \cref{gp-pinned-lan}, the covariance evolution of \cref{sec:heat}, \cref{ht-bingham-bridge}, and the results of \cref{sec:model-profiles} on profiles. The computations for the semicircle law (\cref{prop:hotf-exact,prop:hotf-edge}), the comparison of the finite and continuum evolutions (\cref{lem:hotf-transfer}), and the uniform form of the likelihood limit (\cref{lem:hotf-lan-uniform}) are stated for all $c$ in an interval $(0,c_1]$ with $c_1<1$, with constants depending only on $c_1$. They are used again in \cref{sec:infinite}, where $c=c_N\to0$. The comparison bounds the difference between the finite and continuum relative errors in $L^2(0,1)$ by a multiple of $e^{-2gt}$, and it gives a separate bound on the largest relative error. Because of the factor $e^{-2gt}$, we need the rigidity estimate only in a weak form: it suffices that the eigenvalues lie within $o((\log N)^{-7/4})$ of the semicircle quantiles, uniformly.

The continuum evolution of \cref{sec:hotf-continuum} is a deterministic approximation of the covariance evolution of \cref{sec:heat}. It is not a description of the diffusion, which enters the proof only through \cref{ht-bingham-bridge}. In the thermodynamic limit, the normalization of the covariance evolution is the solution of the scalar Volterra equation of Cugliandolo and Dean \cite[(II.4)--(II.7)]{CugliandoloDean1995}, which they solved by Laplace transform \cite[Appendix~A]{CugliandoloDean1995}. The power of $t$ produced by the edge of the semicircle law in such solutions also appears in the work of Zippold, K\"uhn, and Horner \cite[(14), (17)--(18)]{ZippoldKuhnHorner2000}. We follow the same route in \cref{prop:hotf-exact}.

\paragraph{Notation.}
Throughout this section, we fix $c\in(0,1)$ and write
\begin{equation}\label{eq:hotf-constants}
g=(1-c)^2,\qquad K_c=\frac{\sqrt c}{16\sqrt\pi\,(1+c)^2},\qquad A_c=4K_cg^{3/2}=\frac{\sqrt c\,(1-c)^3}{4\sqrt\pi\,(1+c)^2}.
\end{equation}
Let $\gamma:[0,1]\to[-2,2]$ be the decreasing quantile function of the semicircle law, defined by $\int_{\gamma(\omega)}^2\rho_{\mathrm{sc}}(\lambda)\,d\lambda=\omega$, so that $\gamma(0)=2$ and $\gamma(1)=-2$. Then $\gamma(i/N)=\gamma_i$ is the $i$th semicircle quantile of \cref{sec:spectral}, and $\int_0^1F(\gamma(\omega))\,d\omega=\int F(\lambda)\rho_{\mathrm{sc}}(\lambda)\,d\lambda$ for every bounded measurable $F$. We identify a vector $f\in\R^N$ with the step function on $(0,1)$ equal to $f_i$ on $((i-1)/N,i/N]$, so that $N^{-1}\sum_if_ig_i=\int_0^1fg\,d\omega$. For real numbers $\lambda_1\ge\cdots\ge\lambda_N$, we write $\ell_N$ for the step function equal to $\lambda_i$ on $((i-1)/N,i/N]$.

For $I\in[0,\infty)$, let
\begin{equation}\label{eq:hotf-gaussian-tv}
\mathcal G(I)=2\Phi\big(\sqrt I/2\big)-1=\big\|\cN(\sqrt I,1)-\cN(0,1)\big\|_{\TV},
\end{equation}
where the second equality is the last part of \cref{lem:model-tv}. The function $\mathcal G$ is continuous and increasing, $\mathcal G(I)\to1$ as $I\to\infty$, and $\Psi(u)=\mathcal G(e^{-4u})$. The derivative of $x\mapsto2\Phi(x/2)$ is at most $(2\pi)^{-1/2}$, and we have
\begin{equation}\label{eq:hotf-G-lipschitz}
|\mathcal G(I)-\mathcal G(I')|\le\big|\sqrt I-\sqrt{I'}\big|\qquad(I,I'\ge0).
\end{equation}

For a real symmetric matrix $A$, we use the Lagrange multiplier $z$, the precisions $p_i$, the variances $q_i$, and the smallest precision $d$ of \cref{def:model-multiplier}, together with the covariance evolution of \cref{sec:heat}: the variances $v_i(t)$, the relative errors $e_i(t)=v_i(t)/q_i-1$, the information $\cI_N(t)=\frac12\sum_ie_i(t)^2$, and the Bingham comparison law $\mu_t$. Recall that $\sum_iv_i(t)=\sum_iq_i=N$, and that $\mu_t=\mu_{v(t)}$ and $\pi_A=\mu_q$ are the Gaussian laws with variances $v_i(t)$ and $q_i$ conditioned on $S_N$ (\cref{def:cond-finite,prop:cond-bingham}). For the model, $A=cW_N$, and we write $z_N$ and $d_N$ as in \eqref{sp-canonical-definition}.

\section{The main result}\label{sec:hotf-main}

\begin{theorem}[Fixed high temperature]\label{hp-fixed-profile}
Fix $0<c<1$, let $g=(1-c)^2$, and define
\begin{equation}\label{hp-fixed-center}
A_c=\frac{\sqrt c\,(1-c)^3}{4\sqrt\pi\,(1+c)^2},\qquad
T_N=\frac{\log N-\frac32\log\log N+\log A_c}{4g}.
\end{equation}
Then for every compact set $K\subset\R$, in probability,
\[
\sup_{u\in K}\big|D_N(T_N+u/g)-\Psi(u)\big|\longrightarrow0 .
\]
For each $\eps\in(0,1)$, we have $g\,(t_N(\eps)-T_N)\to u_\eps$ in probability, where $u_\eps=\Psi^{-1}(\eps)$ is given in \eqref{hp-universal-profile}. In particular, $t_N(\eps)/\log N\to(4g)^{-1}$ in probability, and there is cutoff with window $1/g$. For finitely many fixed values $c\in(0,1)$, these conclusions hold simultaneously for the same matrices $W_N$.
\end{theorem}

As in \cref{def:model-cutoff}, a supremum over $u\in K$ of the distance at a time such as $T_N+u/g$ is taken over those $u$ for which the time is nonnegative, and an empty supremum is zero. This convention is used throughout this section and \cref{sec:infinite}; for each compact $K$, the restriction is void for $N$ large or with probability tending to one.

The constant in the center comes from the following computation, which is made precise in \cref{prop:hotf-edge}. At large times, the continuum approximation to the information of the covariance evolution is asymptotic to $\frac N2K_ct^{-3/2}e^{-4gt}$, with $K_c$ as in \eqref{eq:hotf-constants}. At $t=T_N+u/g$, we have $Ne^{-4gt}=(\log N)^{3/2}e^{-4u}/A_c$ and $t^{-3/2}\sim(4g)^{3/2}(\log N)^{-3/2}$, so that the information tends to $\frac12K_c(4g)^{3/2}A_c^{-1}e^{-4u}=e^{-4u}$. Then $2\Phi(\sqrt{e^{-4u}}/2)-1=\Psi(u)$.

The proof of \cref{hp-fixed-profile} is given in \cref{sec:hotf-proof}. Its steps are as follows.
\begin{enumerate}
\item By \cref{ht-bingham-bridge}, $\|\nu_NP_t-\mu_t\|_{\TV}\le Ce^{-gt/4}$ on an event of probability tending to one, on which $d_N\ge g/2$.
\item By \cref{lem:hotf-lan-uniform}, a uniform form of \cref{gp-pinned-lan}, the distance $\|\mu_t-\pi_{N,c}\|_{\TV}$ is close to $\mathcal G(\cI_N(t))$, because $\max_i|e_i(t)|$ and $\max_iq_i/(2\sum_iq_i^2)^{1/2}$ tend to zero.
\item By the first bound of \cref{lem:hotf-transfer}, which compares the relative errors with those of the semicircle law in $L^2(0,1)$, $\cI_N(t)$ is close to $\frac N2\int\eta_t^2\rho_{\mathrm{sc}}$, where $\eta_t$ is the relative error of the covariance evolution of the semicircle law (\cref{sec:hotf-continuum}). \Cref{prop:hotf-edge} evaluates this integral.
\item The pointwise profile implies the locally uniform profile and the convergence of the mixing times, by \cref{hp-monotone-profile-transfer}.
\end{enumerate}

\section{Total variation between Gaussian laws with equal traces}\label{sec:hotf-lan}

The following lemma is a uniform version of \cref{gp-pinned-lan}. It covers all values of the information at once, including large values.

\begin{lemma}[A uniform form of the likelihood limit]\label{lem:hotf-lan-uniform}
For every $\kappa>0$, there exists $\delta>0$ with the following property. Let $n\ge1$, and let $v,\tilde v\in(0,\infty)^n$ satisfy $\sum_iv_i=\sum_i\tilde v_i$. Write $\tilde v_i=v_i(1+e_i)$ and $I=\frac12\sum_ie_i^2$, and let $\eta_v$ be as in \eqref{gp-standardization}. If $\max_i|e_i|\le\delta$ and $\eta_v\le\delta$, then
\[
\Big|\big\|\mu_{\tilde v}-\mu_v\big\|_{\TV}-\mathcal G(I)\Big|\le\kappa .
\]
\end{lemma}

\begin{proof}
Suppose that the claim fails for some $\kappa>0$. Then for every $k\ge1$, there exist $n_k\ge1$ and sequences $v^{(k)},\tilde v^{(k)}\in(0,\infty)^{n_k}$ with equal sums, whose relative differences $e^{(k)}$ and information $I_k$ satisfy $\max_i|e_i^{(k)}|\le1/k$ and $\eta_{v^{(k)}}\le1/k$, and
\begin{equation}\label{eq:hotf-lan-contra}
\Big|\big\|\mu_{\tilde v^{(k)}}-\mu_{v^{(k)}}\big\|_{\TV}-\mathcal G(I_k)\Big|>\kappa .
\end{equation}
Passing to a subsequence, we may assume that $I_k\to I\in[0,\infty]$. The sequences $v^{(k)}$ and $\tilde v^{(k)}$ satisfy the hypotheses \eqref{gp-lan-assumptions} of \cref{gp-pinned-lan}, with $k$ in place of $N$. If $I<\infty$, then \eqref{gp-lan-tv} implies that $\|\mu_{\tilde v^{(k)}}-\mu_{v^{(k)}}\|_{\TV}\to\mathcal G(I)$, and $\mathcal G(I_k)\to\mathcal G(I)$ by continuity. If $I=\infty$, then the last part of \cref{gp-pinned-lan} implies that $\|\mu_{\tilde v^{(k)}}-\mu_{v^{(k)}}\|_{\TV}\to1$, and $\mathcal G(I_k)\to1$. In both cases, the left side of \eqref{eq:hotf-lan-contra} tends to zero, a contradiction.
\end{proof}

\section{The covariance evolution for the semicircle law}\label{sec:hotf-continuum}

We first write the covariance evolution of \cref{sec:heat} as an evolution of functions on $(0,1)$, so that a finite spectrum and the semicircle law can be compared directly. Let $L^\infty(0,1)$ be the space of essentially bounded functions on $(0,1)$, with the norm $\|\cdot\|_\infty$. For $a\in L^\infty(0,1)$, let $\mathcal L_a$ be the bounded operator on $L^\infty(0,1)$ defined by
\[
\mathcal L_af=2c\,af+2\Big(\int_0^1f\,d\omega\Big)\ind,
\]
and set
\begin{equation}\label{hp-linear-system}
M^a_t=e^{t\mathcal L_a}\ind,\qquad \bar M^a_t=\int_0^1M^a_t\,d\omega,\qquad v^a_t=\frac{M^a_t}{\bar M^a_t}\qquad(t\ge0).
\end{equation}
Equivalently, $M^a$ is the unique solution of $\partial_tM^a_t=2c\,aM^a_t+2\bar M^a_t$ with $M^a_0=\ind$.

Two choices of $a$ are used. First, let $\lambda_1\ge\cdots\ge\lambda_N$ be real and $a=\ell_N$. Let $A=\diag(c\lambda_i)$. The operator $\mathcal L_{\ell_N}$ maps step functions to step functions, and on them it acts as the matrix $2\diag(c\lambda_i)+(2/N)\ind\ind^\T$ of \eqref{ht-covariance-ode}, with $\alpha_i=c\lambda_i$. We conclude that $M^{\ell_N}_t$ is the step function of the solution $m(t)$ of \eqref{ht-covariance-ode}, and that $v^{\ell_N}_t$ is the step function of the variances $v_i(t)=Nm_i(t)/\sum_jm_j(t)$ of the covariance evolution of $A$ (\cref{def:heat-covariance}).

Second, let $a=\gamma$. Functions of the form $F\circ\gamma$ form an invariant subspace of $\mathcal L_\gamma$, and we have $M^\gamma_t=M_t\circ\gamma$ and $v^\gamma_t=v_t\circ\gamma$, where $M_t$ and $v_t$ are functions on $[-2,2]$ with
\[
\partial_tM_t(\lambda)=2c\lambda M_t(\lambda)+2\bar M_t,\qquad \bar M_t=\int M_t(\lambda)\rho_{\mathrm{sc}}(\lambda)\,d\lambda,\qquad M_0=1,\qquad v_t=\frac{M_t}{\bar M_t}.
\]
We call this the continuum covariance evolution. The Lagrange multiplier of the semicircle law is $1+c^2$. Indeed, with the Stieltjes transform $\mathfrak m_{\mathrm{sc}}$ of \eqref{eq:spec-stieltjes},
\begin{equation}\label{eq:hotf-sc-multiplier}
\int\frac{\rho_{\mathrm{sc}}(\lambda)\,d\lambda}{1+c^2-c\lambda}=\frac1c\,\mathfrak m_{\mathrm{sc}}\Big(c+\frac1c\Big)=\frac1c\cdot\frac{c+c^{-1}-(c^{-1}-c)}2=1,
\end{equation}
since $(c+c^{-1})^2-4=(c^{-1}-c)^2$. Accordingly, we define the continuum precision and relative error by
\begin{equation}\label{eq:hotf-continuum-precision}
p(\lambda)=1+c^2-c\lambda,\qquad \eta_t(\lambda)=p(\lambda)v_t(\lambda)-1 .
\end{equation}
Note that $p(\lambda)\ge p(2)=g>0$ on $[-2,2]$.

The multiplication operator $f\mapsto2c\,af$ does not preserve positivity when $a$ takes negative values. The positivity of $e^{t\mathcal L_a}$ comes instead from the positivity of $e^{2cat}$ and of the averaging term, as follows.

\begin{lemma}[Positivity and comparison]\label{lem:hotf-positive}
Let $a,a'\in L^\infty(0,1)$.
\begin{enumerate}
\item For every $t\ge0$, the operator $e^{t\mathcal L_a}$ maps nonnegative functions to nonnegative functions, and $M^a_t\ge e^{2cat}>0$.
\item Suppose that $\|a-a'\|_\infty\le E$. Then for all $t\ge0$,
\begin{equation}\label{hp-positive-comparison}
e^{-2cEt}M^{a'}_t\le M^a_t\le e^{2cEt}M^{a'}_t,\qquad e^{-4cEt}v^{a'}_t\le v^a_t\le e^{4cEt}v^{a'}_t .
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
For the first part, write $\mathcal L_a=\mathcal D+\mathcal P$ with $\mathcal Df=2c\,af$ and $\mathcal Pf=2(\int_0^1f\,d\omega)\ind$. The operator $e^{s\mathcal D}$ is multiplication by $e^{2cas}>0$, and $\mathcal P$ maps nonnegative functions to nonnegative constants. Define $S_0(t)=e^{t\mathcal D}$ and $S_{n+1}(t)=\int_0^te^{(t-s)\mathcal D}\mathcal PS_n(s)\,ds$. By induction, each $S_n(t)$ maps nonnegative functions to nonnegative functions, and $\|S_n(t)\|\le e^{t\|\mathcal D\|}(2t)^n/n!$. The Dyson expansion $e^{t\mathcal L_a}=\sum_{n\ge0}S_n(t)$ converges in operator norm, which proves the first claim, and $M^a_t\ge S_0(t)\ind=e^{2cat}$.

For the second part, let $U_t=e^{2cEt}M^{a'}_t$. Then
\[
\partial_tU_t=\mathcal L_{a'}U_t+2cEU_t=\mathcal L_aU_t+F_t,\qquad F_t=2c\,(E+a'-a)\,U_t\ge0,
\]
where $F_t\ge0$ because $U_t\ge0$ and $E+a'-a\ge0$. Since $U_0=M^a_0=\ind$, we find by variation of constants that
\[
U_t-M^a_t=\int_0^te^{(t-s)\mathcal L_a}F_s\,ds\ge0,
\]
by the first part. This is the upper bound for $M^a_t$, and exchanging the roles of $a$ and $a'$ proves the lower bound. Integrating over $(0,1)$ yields the same bounds for $\bar M^a_t$, and the bounds for $v^a_t$ follow by division.
\end{proof}

We now solve the continuum covariance evolution. Let
\begin{equation}\label{eq:hotf-j-phi}
\begin{gathered}
\varphi(t)=\int e^{2c\lambda t}\rho_{\mathrm{sc}}(\lambda)\,d\lambda,\qquad
j(t)=c^2\int\frac{e^{-2p(\lambda)t}}{p(\lambda)}\,\rho_{\mathrm{sc}}(\lambda)\,d\lambda,\\
K(w)=\int e^{-2(2-\lambda)w}\rho_{\mathrm{sc}}(\lambda)\,d\lambda,
\end{gathered}
\end{equation}
for $t,w\ge0$. Note that $j(0)=c^2$ by \eqref{eq:hotf-sc-multiplier}, and that $0\le j(t)\le c^2e^{-2gt}$ because $p\ge g$.

\begin{proposition}[The continuum covariance evolution]\label{prop:hotf-exact}
Let $c\in(0,1)$. For all $t\ge0$ and $\lambda\in[-2,2]$,
\begin{gather}
\bar M=\varphi+2\varphi*\bar M,\qquad
v_t(\lambda)=\frac{e^{2c\lambda t}+2\int_0^te^{2c\lambda(t-s)}\bar M_s\,ds}{\bar M_t},\label{hp-renewal}\\
\bar M_t=e^{2(1+c^2)t}\big(1-c^2+j(t)\big),\label{hp-exact-renewal}\\
\eta_t(\lambda)=\frac{e^{-2p(\lambda)t}}{1-c^2+j(t)}\Big[p(\lambda)-1+2c^2\int_0^te^{2p(\lambda)s}\int e^{-2p(a)s}\rho_{\mathrm{sc}}(a)\,da\,ds\Big],\label{hp-exact-deficit}
\end{gather}
where $(\varphi*\bar M)(t)=\int_0^t\varphi(t-s)\bar M_s\,ds$. Moreover, with $\tilde t=ct$, for all $y\in[0,4]$,
\begin{equation}\label{wf-rescaled-edge-deficit}
\eta_t(2-y)=\frac{c\,e^{-2gt}e^{-2y\tilde t}}{1-c^2+j(t)}\Big[c-2+y+2\int_0^{\tilde t}e^{2yw}K(w)\,dw\Big].
\end{equation}
\end{proposition}

The first identity in \eqref{hp-renewal} is the scalar Volterra equation of \cite[(II.7)]{CugliandoloDean1995}, and \eqref{hp-exact-renewal} is its explicit solution for the semicircle law.

\begin{proof}
The proof has three steps: the renewal equation, its solution by Laplace transform, and an integration by parts.

\paragraph{Step 1: the renewal equation.}
Solving the equation for $M_t(\lambda)$ by variation of constants, with the forcing $2\bar M_t$, we find
\[
M_t(\lambda)=e^{2c\lambda t}+2\int_0^te^{2c\lambda(t-s)}\bar M_s\,ds .
\]
The first identity in \eqref{hp-renewal} follows by integrating against $\rho_{\mathrm{sc}}$, and the second by dividing by $\bar M_t$.

\paragraph{Step 2: the Laplace transform.}
The operator $\mathcal L_\gamma$ has norm at most $4c+2$, so $0<\bar M_t\le e^{(4c+2)t}$, and $\varphi(t)\le e^{4ct}$. For $\zeta>4c+2$, let $\hat f(\zeta)=\int_0^\infty e^{-\zeta t}f(t)\,dt$. By Tonelli's theorem,
\[
\hat\varphi(\zeta)=\int\frac{\rho_{\mathrm{sc}}(\lambda)\,d\lambda}{\zeta-2c\lambda}=\frac1{2c}\,\mathfrak m_{\mathrm{sc}}\Big(\frac\zeta{2c}\Big)\le\frac1{\zeta-4c}<\frac12 .
\]
Recall that $\mathfrak m=\mathfrak m_{\mathrm{sc}}(E)$ solves $\mathfrak m^2-E\mathfrak m+1=0$. With $E=\zeta/(2c)$ and $\mathfrak m=2c\hat\varphi(\zeta)$, this equation becomes
\begin{equation}\label{eq:hotf-quadratic}
4c^2\hat\varphi(\zeta)^2-\zeta\hat\varphi(\zeta)+1=0 .
\end{equation}
The Laplace transform of the renewal equation is $\hat{\bar M}=\hat\varphi+2\hat\varphi\hat{\bar M}$, that is, $\hat{\bar M}=\hat\varphi/(1-2\hat\varphi)$. Next, since $(\zeta-2c\lambda)-2p(\lambda)=\zeta-2(1+c^2)$, we have the partial fraction decomposition
\[
\frac1{p(\lambda)(\zeta-2c\lambda)}=\frac1{\zeta-2(1+c^2)}\Big[\frac1{p(\lambda)}-\frac2{\zeta-2c\lambda}\Big].
\]
Integrating against $\rho_{\mathrm{sc}}$ and using \eqref{eq:hotf-sc-multiplier}, we obtain
\[
\int\frac{\rho_{\mathrm{sc}}(\lambda)\,d\lambda}{p(\lambda)(\zeta-2c\lambda)}=\frac{1-2\hat\varphi(\zeta)}{\zeta-2(1+c^2)} .
\]
We conclude that
\begin{equation}\label{hp-renewal-transform}
\frac{1-c^2}{\zeta-2(1+c^2)}+c^2\int\frac{\rho_{\mathrm{sc}}(\lambda)\,d\lambda}{p(\lambda)(\zeta-2c\lambda)}
=\frac{1-2c^2\hat\varphi}{\zeta-2(1+c^2)}
=\frac{\hat\varphi}{1-2\hat\varphi}=\hat{\bar M}(\zeta).
\end{equation}
For the second equality, cross-multiplication shows that it is equivalent to $\zeta\hat\varphi=1+4c^2\hat\varphi^2$, which is \eqref{eq:hotf-quadratic}. Note that $e^{2(1+c^2)t}j(t)=c^2\int p(\lambda)^{-1}e^{2c\lambda t}\rho_{\mathrm{sc}}(\lambda)\,d\lambda$, because $2(1+c^2)-2p(\lambda)=2c\lambda$. This implies that the left side of \eqref{hp-renewal-transform} is the Laplace transform of $e^{2(1+c^2)t}(1-c^2+j(t))$. Both this function and $\bar M$ are continuous and bounded by $e^{(4c+2)t}$, and their Laplace transforms agree for $\zeta>4c+2$. We now check that they are equal, using the compactification-and-moments proof of uniqueness of Laplace transforms in \cite[Chapter~XIII, Section~1, Theorem~1 and its second proof, pp.~429--430]{Feller1971}. Let $f$ be their difference, and fix $b>4c+2$. The push-forward of the finite signed measure $e^{-bt}f(t)\,dt$ under the map $t\mapsto x=e^{-t}$ is a finite signed measure on $[0,1]$, whose $n$th moment is $\hat f(b+n)=0$ for every integer $n\ge0$. Since polynomials are dense in $C([0,1])$, this measure is zero. Then $f=0$ almost everywhere, and $f=0$ everywhere by continuity. This proves \eqref{hp-exact-renewal}.

\paragraph{Step 3: the relative error.}
Fix $\lambda$ and write $p=p(\lambda)$. Inserting \eqref{hp-exact-renewal} into \eqref{hp-renewal}, dividing the numerator and the denominator by $e^{2(1+c^2)t}$, and using $2c\lambda-2(1+c^2)=-2p$, we find
\[
\big(1-c^2+j(t)\big)v_t(\lambda)=e^{-2pt}+2\int_0^te^{-2p(t-s)}\big(1-c^2+j(s)\big)\,ds .
\]
We have $2\int_0^te^{-2p(t-s)}ds=(1-e^{-2pt})/p$, and integrating by parts, we find
\[
2\int_0^te^{-2p(t-s)}j(s)\,ds=\frac{j(t)-e^{-2pt}j(0)}p-\frac1p\int_0^te^{-2p(t-s)}j'(s)\,ds .
\]
Multiplying by $p$, subtracting $1-c^2+j(t)$, and using $j(0)=c^2$, we obtain
\[
\big(1-c^2+j(t)\big)\eta_t(\lambda)=e^{-2pt}\Big[p-1-\int_0^te^{2ps}j'(s)\,ds\Big].
\]
This is \eqref{hp-exact-deficit}, because $j'(s)=-2c^2\int e^{-2p(a)s}\rho_{\mathrm{sc}}(a)\,da$.

Finally, let $\lambda=2-y$. Then $p(2-y)=g+cy$ and $p(2-y)-1=c(c-2+y)$. For every $a$, we have $p(a)=g+c(2-a)$, which gives $\int e^{-2p(a)s}\rho_{\mathrm{sc}}(a)\,da=e^{-2gs}K(cs)$. The integral in \eqref{hp-exact-deficit} becomes
\[
2c^2\int_0^te^{2(g+cy)s}e^{-2gs}K(cs)\,ds=2c\int_0^{\tilde t}e^{2yw}K(w)\,dw,
\]
where we substituted $w=cs$. Together with $e^{-2p(2-y)t}=e^{-2gt}e^{-2y\tilde t}$, this yields \eqref{wf-rescaled-edge-deficit}.
\end{proof}

The kernel $K$ in \eqref{wf-rescaled-edge-deficit} carries the edge behavior of the semicircle law. We record its two properties that are used below.

\begin{lemma}[The edge kernel]\label{lem:hotf-kernel}
For all $w\ge0$, we have $0\le K(w)\le3(1+w)^{-3/2}$, and $\int_0^\infty K(w)\,dw=1/2$.
\end{lemma}

\begin{proof}
We have $K(w)\le1$, because $e^{-2(2-\lambda)w}\le1$ on $[-2,2]$. For $0\le y\le4$, we have $\rho_{\mathrm{sc}}(2-y)=(2\pi)^{-1}\sqrt{y(4-y)}\le\pi^{-1}\sqrt y$. For $w>0$, this gives
\[
K(w)=\int_0^4e^{-2yw}\rho_{\mathrm{sc}}(2-y)\,dy\le\frac1\pi\int_0^\infty\sqrt y\,e^{-2yw}\,dy=\frac{\sqrt\pi}{2\pi(2w)^{3/2}}=\frac1{2\sqrt\pi\,(2w)^{3/2}} .
\]
We use $K(w)\le1\le2^{3/2}(1+w)^{-3/2}$ for $w\le1$, and $2w\ge1+w$ for $w\ge1$. Finally, Tonelli's theorem and monotone convergence give $\int_0^\infty K(w)\,dw=\frac12\int(2-\lambda)^{-1}\rho_{\mathrm{sc}}(\lambda)\,d\lambda=\frac12\lim_{E\downarrow2}\mathfrak m_{\mathrm{sc}}(E)=\frac12$.
\end{proof}

\begin{lemma}[An exponential convolution bound]\label{hp-edge-convolution-bound}
There exists a universal constant $C$ such that for all $b,T\ge0$,
\[
\int_0^Te^{bs}(1+s)^{-3/2}\,ds\le C\Big[1+\frac{e^{bT}(1+T)^{-1/2}}{1+bT}\Big].
\]
\end{lemma}

\begin{proof}
We consider three cases. If $bT\le2$, then $e^{bs}\le e^2$ on $[0,T]$, and the integral is at most $e^2\int_0^\infty(1+s)^{-3/2}ds=2e^2$.

If $T\le1$ and $bT>2$, then the integral is at most $\int_0^Te^{bs}ds\le e^{bT}/b=Te^{bT}/(bT)\le e^{bT}/(bT)$. Using $1+bT\le2bT$ and $(1+T)^{-1/2}\ge2^{-1/2}$, we see that this is at most $2^{3/2}e^{bT}(1+T)^{-1/2}/(1+bT)$.

Let $T>1$ and $bT>2$. Then $1/b<T/2$. We split the integral at $s=1/b$. On $[0,1/b]$, we have $e^{bs}\le e$, and this part is at most $2e$. On $[1/b,T]$, we substitute $s=x/b$ and use $(1+x/b)^{-3/2}\le b^{3/2}x^{-3/2}$, to find
\[
\int_{1/b}^Te^{bs}(1+s)^{-3/2}\,ds\le b^{1/2}\int_1^{bT}e^xx^{-3/2}\,dx .
\]
For $S\ge2$, we have $\int_{S/2}^Se^xx^{-3/2}dx\le2^{3/2}e^SS^{-3/2}$ and $\int_1^{S/2}e^xx^{-3/2}dx\le2e^{S/2}$. Both are at most $C_0e^SS^{-3/2}$ with $C_0=2^{3/2}+2(3/e)^{3/2}$, because $\sup_{x>0}x^{3/2}e^{-x/2}=(3/e)^{3/2}$. With $S=bT$, the second part of the integral is at most $C_0e^{bT}/(bT^{3/2})$. Since $T>1$ and $bT>2$, we have $(1+T)^{1/2}\le(2T)^{1/2}$ and $1+bT\le2bT$, which give $e^{bT}/(bT^{3/2})\le2^{3/2}e^{bT}(1+T)^{-1/2}/(1+bT)$. This completes the proof.
\end{proof}

We now evaluate the relative errors of the continuum evolution. The following proposition holds uniformly in $c\in(0,c_1]$ for every fixed $c_1<1$, so that it applies both at a fixed temperature and in \cref{sec:infinite}, where $c\to0$. All limits are taken as $\tilde t=ct\to\infty$. For a fixed $c$, this means $t\to\infty$.

\begin{proposition}[Edge asymptotics of the continuum relative errors]\label{prop:hotf-edge}
Let $0<c_1<1$. There exists $C_1$ depending only on $c_1$ such that, for all $c\in(0,c_1]$, $t\ge0$ and $y\in[0,4]$, with $\tilde t=ct$,
\begin{equation}\label{hp-deficit-envelope}
|\eta_t(2-y)|\le C_1c\,e^{-2gt}\Big[e^{-2y\tilde t}+\frac{(1+\tilde t)^{-1/2}}{1+y\tilde t}\Big].
\end{equation}
In particular, $\sup_\lambda|\eta_t(\lambda)|\le2C_1ce^{-2gt}$. Moreover, for all $L,\kappa>0$, there exists $T_0$ depending only on $c_1$, $L$ and $\kappa$ such that, for all $c\in(0,c_1]$ and all $t$ with $\tilde t=ct\ge T_0$,
\begin{equation}\label{hp-edge-amplitude}
\sup_{0\le y\le L/\tilde t}\Big|\frac{1+c}c\,e^{2gt}\eta_t(2-y)+e^{-2y\tilde t}\Big|\le\kappa,
\end{equation}
and, for all $\kappa>0$, there exists $T_0'$ depending only on $c_1$ and $\kappa$ such that, for all $c\in(0,c_1]$ and $ct\ge T_0'$,
\begin{equation}\label{hp-continuum-information}
\bigg|\frac{\int\eta_t(\lambda)^2\rho_{\mathrm{sc}}(\lambda)\,d\lambda}{K_c\,t^{-3/2}e^{-4gt}}-1\bigg|\le\kappa .
\end{equation}
\end{proposition}

For a fixed $c$, \eqref{hp-edge-amplitude} states that $e^{2gt}\eta_t(2-y)=-c(1+c)^{-1}e^{-2cyt}+o(1)$ as $t\to\infty$, uniformly for $yt$ in compact sets. The relative errors of eigenvalues at distance of order $1/t$ from the edge are of order $e^{-2gt}$, while \eqref{hp-deficit-envelope} shows that the other relative errors are smaller. The computation in \eqref{hp-continuum-information} is of Laplace type, as in Watson's lemma \cite[\S2.3(ii)]{OlverLozierBoisvertClark2010}.

\begin{proof}
Throughout the proof, $C$ denotes a constant depending only on $c_1$, whose value may change from line to line. We use \eqref{wf-rescaled-edge-deficit}, whose prefactor is at most $Cce^{-2gt}e^{-2y\tilde t}$, because $1-c^2+j(t)\ge1-c_1^2$. Note also that $t\ge\tilde t$ and $g\ge(1-c_1)^2$, which give
\begin{equation}\label{eq:hotf-j-small}
0\le j(t)\le c^2e^{-2gt}\le e^{-2(1-c_1)^2\tilde t}.
\end{equation}

\paragraph{Step 1: the global bound.}
In the bracket of \eqref{wf-rescaled-edge-deficit}, we have $|c-2+y|\le3$. By \cref{lem:hotf-kernel,hp-edge-convolution-bound} with $b=2y$,
\[
2\int_0^{\tilde t}e^{2yw}K(w)\,dw\le6\int_0^{\tilde t}e^{2yw}(1+w)^{-3/2}\,dw\le C\Big[1+\frac{e^{2y\tilde t}(1+\tilde t)^{-1/2}}{1+2y\tilde t}\Big].
\]
Multiplying by the prefactor and using $1+2y\tilde t\ge1+y\tilde t$, we obtain \eqref{hp-deficit-envelope}. The bound on $\sup_\lambda|\eta_t(\lambda)|$ follows because both terms in the bracket of \eqref{hp-deficit-envelope} are at most one.

\paragraph{Step 2: the amplitude at the edge.}
Let $\mathcal E(\tilde t,y)=2\int_0^{\tilde t}e^{2yw}K(w)\,dw-1$. Since $2\int_0^\infty K=1$ by \cref{lem:hotf-kernel}, we have, for $0<R\le\tilde t$,
\[
\mathcal E(\tilde t,y)=2\int_0^R\big(e^{2yw}-1\big)K(w)\,dw+2\int_R^{\tilde t}e^{2yw}K(w)\,dw-2\int_R^\infty K(w)\,dw .
\]
For $0\le y\le L/\tilde t$, we have $e^{2yw}-1\le e^{2LR/\tilde t}-1$ for $w\le R$, and $e^{2yw}\le e^{2L}$ for $w\le\tilde t$. We obtain
\begin{equation}\label{eq:hotf-D-bound}
\sup_{0\le y\le L/\tilde t}|\mathcal E(\tilde t,y)|\le\big(e^{2LR/\tilde t}-1\big)+2\big(e^{2L}+1\big)\int_R^\infty K(w)\,dw .
\end{equation}
Taking first $R$ large and then $\tilde t$ large, we see that the right side of \eqref{eq:hotf-D-bound} tends to zero as $\tilde t\to\infty$, at a rate depending only on $L$. The bracket in \eqref{wf-rescaled-edge-deficit} equals $-(1-c)+y+\mathcal E(\tilde t,y)$, and we find
\[
\frac{1+c}c\,e^{2gt}\eta_t(2-y)+e^{-2y\tilde t}
=e^{-2y\tilde t}\Big[\frac{(1+c)\big(y+\mathcal E(\tilde t,y)\big)}{1-c^2+j(t)}+\frac{j(t)}{1-c^2+j(t)}\Big],
\]
where we used $(1+c)(1-c)=1-c^2$. For $0\le y\le L/\tilde t$, the absolute value of the right side is at most $(1-c_1^2)^{-1}[2(L/\tilde t+|\mathcal E(\tilde t,y)|)+j(t)]$. Using \eqref{eq:hotf-D-bound} and \eqref{eq:hotf-j-small}, we see that this tends to zero as $\tilde t\to\infty$, uniformly in $c\in(0,c_1]$. This proves \eqref{hp-edge-amplitude}.

\paragraph{Step 3: the information.}
Let $\Theta=c^2(1+c)^{-2}e^{-4gt}\tilde t^{-3/2}$. Note that $K_ct^{-3/2}e^{-4gt}=\Theta/(16\sqrt\pi)$, since $c^2\tilde t^{-3/2}=\sqrt c\,t^{-3/2}$. Also
\[
\frac1\pi\int_0^\infty\sqrt x\,e^{-4x}\,dx=\frac{\sqrt\pi}{2\cdot4^{3/2}\pi}=\frac1{16\sqrt\pi} .
\]
It therefore suffices to show that $\Theta^{-1}\int\eta_t^2\rho_{\mathrm{sc}}\to(16\sqrt\pi)^{-1}$ as $\tilde t\to\infty$, uniformly in $c\in(0,c_1]$. Fix $L>0$ and $\kappa\in(0,1]$, let $\tilde t\ge\max\{T_0,L\}$ with $T_0$ as in \eqref{hp-edge-amplitude}, and split $\int\eta_t^2\rho_{\mathrm{sc}}=\int_0^4\eta_t(2-y)^2\rho_{\mathrm{sc}}(2-y)\,dy$ at $y=L/\tilde t$.

On $[0,L/\tilde t]$, we write $\eta_t(2-y)=-c(1+c)^{-1}e^{-2gt}(e^{-2y\tilde t}+\varrho)$, where $|\varrho|\le\kappa$ by \eqref{hp-edge-amplitude}, and $\rho_{\mathrm{sc}}(2-y)=\pi^{-1}\sqrt y\,\sqrt{1-y/4}$. The substitution $y=x/\tilde t$ yields
\[
\Theta^{-1}\int_0^{L/\tilde t}\eta_t(2-y)^2\rho_{\mathrm{sc}}(2-y)\,dy=\frac1\pi\int_0^L\big(e^{-2x}+\varrho\big)^2\sqrt x\,\sqrt{1-x/(4\tilde t)}\,dx .
\]
For $x\le L$, we have $|(e^{-2x}+\varrho)^2-e^{-4x}|\le3\kappa$ and $1-\sqrt{1-x/(4\tilde t)}\le L/(4\tilde t)$. It follows that the right side differs from $\pi^{-1}\int_0^L\sqrt xe^{-4x}dx$ by at most $C_L(\kappa+\tilde t^{-1})$, with $C_L$ depending only on $L$.

On $[L/\tilde t,4]$, we use \eqref{hp-deficit-envelope}, $(\alpha+\beta)^2\le2\alpha^2+2\beta^2$, and $\rho_{\mathrm{sc}}(2-y)\le\pi^{-1}\sqrt y$. Substituting $y=x/\tilde t$ again, we have
\[
\begin{gathered}
\int_{L/\tilde t}^4e^{-4y\tilde t}\sqrt y\,dy\le\tilde t^{-3/2}\int_L^\infty\sqrt x\,e^{-4x}\,dx,\\
\int_0^4\frac{\sqrt y\,dy}{(1+y\tilde t)^2}\le\tilde t^{-3/2}\int_0^\infty\frac{\sqrt x\,dx}{(1+x)^2}=\frac\pi2\,\tilde t^{-3/2} .
\end{gathered}
\]
Since $(1+c)^2\le4$, this gives
\[
\Theta^{-1}\int_{L/\tilde t}^4\eta_t(2-y)^2\rho_{\mathrm{sc}}(2-y)\,dy\le\frac{8C_1^2}\pi\Big[\int_L^\infty\sqrt x\,e^{-4x}\,dx+\frac\pi2(1+\tilde t)^{-1}\Big].
\]
Combining the two parts, we have shown that
\[
\limsup_{\tilde t\to\infty}\,\sup_{c\in(0,c_1]}\Big|\Theta^{-1}\int\eta_t^2\rho_{\mathrm{sc}}-\frac1{16\sqrt\pi}\Big|\le C_L\kappa+C\int_L^\infty\sqrt x\,e^{-4x}\,dx .
\]
Letting $\kappa\to0$ and then $L\to\infty$, we obtain \eqref{hp-continuum-information}.
\end{proof}

\section{Comparison of the finite and continuum evolutions}\label{sec:hotf-transfer}

We now compare the covariance evolution of a finite spectrum with the continuum evolution, at times up to a multiple of $\log N$. The comparison is made on the common parameter space $(0,1)$. Let $E=\|\ell_N-\gamma\|_\infty$. In \cref{lem:hotf-transfer}, we show that the difference between the finite and continuum relative errors has norm at most $C_2cE(1+t)e^{-2gt}$ in $L^2(0,1)$, and we bound the largest relative error by a separate argument. We first bound $E$ for the GOE. We then prove two lemmas on operators on $L^2(0,1)$: the gap of the operator that generates the covariance evolution, on the orthogonal complement of its stationary vector (\cref{lem:hotf-energy}), and a comparison of the decaying parts of two such evolutions (\cref{lem:hotf-transient}).

\begin{lemma}[Semicircle quantiles and rigidity]\label{lem:hotf-grid}
For all $\omega,\omega'\in[0,1]$, we have $|\gamma(\omega)-\gamma(\omega')|\le(4\sqrt2\,\pi|\omega-\omega'|)^{2/3}$. Let $\lambda_1\ge\cdots\ge\lambda_N$ be the eigenvalues of $W_N$, and fix $C\ge1$. There exists $N_C$ depending only on $C$ such that, on the event $F_C$ of \eqref{sgl-rigidity-input} and for $N\ge N_C$,
\[
\|\ell_N-\gamma\|_\infty\le(C+7)N^{-5/12},\qquad \max_i|\lambda_i|\le3,\qquad \Big|\frac1N\sum_i\lambda_i\Big|\le(C+7)N^{-5/12}.
\]
For each fixed confidence $1-\eta$, \Cref{sp-goe-rigidity} supplies such a $C$ for all sufficiently large $N$.
\end{lemma}

\begin{proof}
Let $J=[\alpha,\alpha+h]\subset[-2,2]$. The set of $\lambda\in J$ with $2-|\lambda|<h/4$ has length at most $h/2$. For the other points of $J$, we have $\rho_{\mathrm{sc}}(\lambda)=(2\pi)^{-1}\sqrt{(2-\lambda)(2+\lambda)}\ge(2\pi)^{-1}\sqrt{2(2-|\lambda|)}\ge(2\pi)^{-1}\sqrt{h/2}$. Hence the semicircle mass of $J$ is at least $(h/2)^{3/2}/(2\pi)=h^{3/2}/(4\sqrt2\,\pi)$. If $|\gamma(\omega)-\gamma(\omega')|=h$, then $|\omega-\omega'|$ is the mass of an interval of length $h$, which proves the first claim.

On $F_C$, we have $|\lambda_i-\gamma_i|\le C N^{-5/12}$ for all $i$. For $\omega\in((i-1)/N,i/N]$, we have $\gamma_i=\gamma(i/N)$, and by the first claim,
\[
\begin{split}
|\ell_N(\omega)-\gamma(\omega)|&\le|\lambda_i-\gamma_i|+|\gamma(i/N)-\gamma(\omega)|\\
&\le C N^{-5/12}+(4\sqrt2\,\pi)^{2/3}N^{-2/3}\le(C+7)N^{-5/12}.
\end{split}
\]
This is at most $1/2$ for $N$ large depending on $C$. The other two bounds follow from $|\gamma|\le2$, $\int_0^1\gamma\,d\omega=\int\lambda\rho_{\mathrm{sc}}(\lambda)\,d\lambda=0$, and $N^{-1}\sum_i\lambda_i=\int_0^1\ell_N\,d\omega$.
\end{proof}

For the two lemmas on operators, we write $\langle f,f'\rangle=\int_0^1ff'\,d\omega$ and $\|f\|_2=\langle f,f\rangle^{1/2}$ for the inner product and the norm of $L^2(0,1)$, and $\|\cdot\|$ for the norm of an operator on it. For background on orthogonal projections in Hilbert spaces, see \cite[Section~5.1]{Brezis2011}, and for exponentials of bounded operators, see \cite[Chapter~I, Section~3]{EngelNagel2000}. For $p\in L^\infty(0,1)$, let $H_p$ be the operator on $L^2(0,1)$ defined by
\[
H_pf=-2pf+2\langle\ind,f\rangle\ind .
\]
If $p$ is the step function of precisions $p_1,\dots,p_N$ satisfying \eqref{eq:heat-trace}, then $H_p$ maps step functions to step functions, and on them it acts as the heat matrix $H$ of \eqref{eq:heat-objects}. In the proof of \cref{lem:hotf-transfer}, we write the covariance evolutions of $\ell_N$ and of $\gamma$ through two such operators.

\begin{lemma}[The gap on the orthogonal complement of the stationary vector]\label{lem:hotf-energy}
Let $0<d\le s$, and let $p$ be a measurable function on $(0,1)$ such that $d\le p\le s$ almost everywhere and $\int_0^1p^{-1}\,d\omega=1$. Let $q=p^{-1}$, and let $P_q$ be the orthogonal projection onto the span of $q$ in $L^2(0,1)$. Then $H_p$ is a bounded self-adjoint operator with $\|H_p\|\le2s+2$, and its kernel is spanned by $q$. For every $f\in L^2(0,1)$ with $\langle f,q\rangle=0$,
\[
\langle f,H_pf\rangle\le-2d\|f\|_2^2 .
\]
For every $t\ge0$,
\[
\|e^{tH_p}\|\le1,\qquad\|e^{tH_p}(I-P_q)\|\le e^{-2dt},\qquad\|H_pe^{tH_p}\|\le(2s+2)e^{-2dt}.
\]
Finally, $1\le\|q\|_2\le d^{-1}$, and for every $t\ge0$,
\[
\langle\ind,e^{tH_p}\ind\rangle=\big\|e^{tH_p/2}\ind\big\|_2^2\ge\|q\|_2^{-2}\ge d^2 .
\]
\end{lemma}

\begin{proof}
Multiplication by $p$ is self-adjoint with norm at most $s$, and the map $f\mapsto\langle\ind,f\rangle\ind$ is self-adjoint with norm $\|\ind\|_2^2=1$. This proves the first claim. Since $pq=1$ and $\langle\ind,q\rangle=\int_0^1p^{-1}\,d\omega=1$, we have $H_pq=-2\ind+2\ind=0$. Conversely, if $H_pf=0$, then $pf=\langle\ind,f\rangle\ind$, that is, $f=\langle\ind,f\rangle q$.

Next, let $\langle f,q\rangle=0$, let $\bar f=\langle\ind,f\rangle$, and let $\tilde f=f-\bar fq$. All integrals below are finite, because $p$ and $q$ are bounded. Using $pq=1$, $\langle\ind,q\rangle=1$, and $\langle f,q\rangle=0$, we find
\[
\int_0^1p\tilde f^2\,d\omega=\int_0^1pf^2\,d\omega-2\bar f^2+\bar f^2=\int_0^1pf^2\,d\omega-\bar f^2,\qquad
\|\tilde f\|_2^2=\|f\|_2^2+\bar f^2\|q\|_2^2 .
\]
By the definition of $H_p$, we have $\langle f,H_pf\rangle=-2\int_0^1pf^2\,d\omega+2\bar f^2$, and the first identity gives
\[
\langle f,H_pf\rangle=-2\int_0^1p\tilde f^2\,d\omega\le-2d\|\tilde f\|_2^2\le-2d\|f\|_2^2 .
\]
In the last inequality, we used the second identity.

For the bounds on the exponential, we use only that $H_p$ is bounded and self-adjoint, that $H_pq=0$, and the inequality just proved. Since $H_p$ is self-adjoint and $H_pq=0$, the orthogonal complement $q^\perp$ of $q$ is invariant under $H_p$ and under $e^{tH_p}$, which is defined by a power series that converges in norm. Let $f\in q^\perp$ and $f_t=e^{tH_p}f$. The map $t\mapsto e^{tH_p}$ is differentiable in the operator norm, with derivative $H_pe^{tH_p}$ \cite[Chapter~I, Proposition~3.5]{EngelNagel2000}, and we have
\[
\frac d{dt}\Big(e^{4dt}\|f_t\|_2^2\Big)=e^{4dt}\Big(4d\|f_t\|_2^2+2\langle f_t,H_pf_t\rangle\Big)\le0 .
\]
We conclude that $\|f_t\|_2\le e^{-2dt}\|f\|_2$, which is the second bound. Since $e^{tH_p}q=q$, the orthogonal decomposition $f=P_qf+(I-P_q)f$ of an arbitrary $f\in L^2(0,1)$ gives $\|e^{tH_p}f\|_2^2=\|P_qf\|_2^2+\|e^{tH_p}(I-P_q)f\|_2^2\le\|f\|_2^2$, which is the first bound. Further, $H_pe^{tH_p}P_q=e^{tH_p}H_pP_q=0$, so that $H_pe^{tH_p}=H_pe^{tH_p}(I-P_q)$, and the third bound follows from the second and $\|H_p\|\le2s+2$.

Finally, $1=\langle\ind,q\rangle\le\|q\|_2$ by the Cauchy--Schwarz inequality, and $\|q\|_2\le d^{-1}$ because $0<q\le d^{-1}$. Since $e^{tH_p/2}$ is self-adjoint and $e^{tH_p/2}q=q$, we have
\[
\langle\ind,e^{tH_p}\ind\rangle=\big\|e^{tH_p/2}\ind\big\|_2^2,\qquad
\big\langle e^{tH_p/2}\ind,q\big\rangle=\langle\ind,q\rangle=1 .
\]
By the Cauchy--Schwarz inequality, we have $1\le\|e^{tH_p/2}\ind\|_2\|q\|_2$, which proves the last claim.
\end{proof}

The next lemma compares the decaying parts of two such evolutions whose stationary vectors are close. It is stated on a real Hilbert space, with inner product $\langle\cdot,\cdot\rangle$ and norm $\|\cdot\|$, and its constants do not depend on the space.

\begin{lemma}[Comparison of the decaying parts of two evolutions]\label{lem:hotf-transient}
Let $H_1$ and $H_2$ be bounded self-adjoint operators on a real Hilbert space, and let $u_1,u_2$ be unit vectors with $\langle u_1,u_2\rangle>0$ and $H_1u_1=H_2u_2=0$. Suppose that there exist $d>0$ and $B\ge0$ such that $\|H_1\|,\|H_2\|\le B$ and, for $j=1,2$,
\[
\langle f,H_jf\rangle\le-2d\|f\|^2\qquad\text{for all }f\text{ with }\langle f,u_j\rangle=0 .
\]
Let $a=\|H_1-H_2\|$ and $b=\|u_1-u_2\|$. Then for all $t\ge0$,
\[
\big\|H_1e^{tH_1}-H_2e^{tH_2}\big\|\le20(1+B)^2(a+b)(1+t)e^{-2dt}.
\]
Moreover, let $x$ be a unit vector, let $t\ge0$, and suppose that $m_j(t)=\langle x,e^{tH_j}x\rangle$ satisfies $m_j(t)\ge\delta$ for $j=1,2$, for some $\delta>0$. Then
\[
\Big\|\frac{H_1e^{tH_1}x}{m_1(t)}-\frac{H_2e^{tH_2}x}{m_2(t)}\Big\|\le\Big(\frac{20(1+B)^2(a+b)}\delta+\frac{Ba}{\delta^2}\Big)(1+t)e^{-2dt}.
\]
\end{lemma}

\begin{proof}
The proof of the bounds on the exponential in \cref{lem:hotf-energy} uses only that the operator is bounded and self-adjoint, that it annihilates a given vector, and the inequality on the orthogonal complement of that vector. The same argument applies to $H_1$ and $H_2$. For $j=1,2$ and $t\ge0$, we obtain that $e^{tH_j}$ is a contraction, that $\|e^{tH_j}f\|\le e^{-2dt}\|f\|$ whenever $\langle f,u_j\rangle=0$, and that $\|H_je^{tH_j}\|\le Be^{-2dt}$.

We begin by rotating $u_1$ onto $u_2$. Let $\alpha=\langle u_1,u_2\rangle\in(0,1]$. If $u_1=u_2$, let $U=I$. Otherwise, $\alpha<1$, and we let $\beta=(1-\alpha^2)^{1/2}>0$ and $w=\beta^{-1}(u_2-\alpha u_1)$, which is a unit vector orthogonal to $u_1$. We define $U$ by
\[
Uu_1=\alpha u_1+\beta w=u_2,\qquad Uw=-\beta u_1+\alpha w,
\]
and $Uf=f$ for $f$ orthogonal to $u_1$ and $w$. Then $U$ is orthogonal, that is, $U^*=U^{-1}$. On the span of $u_1$ and $w$, we have $(U-I)^*(U-I)=2(1-\alpha)I$, and $U-I$ vanishes on the orthogonal complement of this span. This implies that
\[
\|U-I\|=\|U^{-1}-I\|=\sqrt{2(1-\alpha)}=\|u_1-u_2\|\le2b .
\]
This is the case of one-dimensional subspaces of the classical construction of a unitary operator that maps one projection onto a nearby one \cite[Chapter~I, Section~6.8, Theorem~6.32]{Kato1995}; we use only the explicit bound above.

Let $\tilde H_1=UH_1U^{-1}$. Then $\tilde H_1$ is self-adjoint, $\|\tilde H_1\|\le B$, and $\tilde H_1u_2=UH_1u_1=0$. If $\langle f,u_2\rangle=0$, then $\langle U^{-1}f,u_1\rangle=\langle f,Uu_1\rangle=0$, and the inequality for $H_1$ implies that $\langle f,\tilde H_1f\rangle=\langle U^{-1}f,H_1U^{-1}f\rangle\le-2d\|f\|^2$. We have shown that $\tilde H_1$ satisfies the hypotheses imposed on $H_2$, with the same vector $u_2$. From the identity $\tilde H_1-H_1=(U-I)H_1U^{-1}+H_1(U^{-1}-I)$, we find
\[
\|\tilde H_1-H_1\|\le4Bb,\qquad\|\tilde H_1-H_2\|\le a+4Bb .
\]

Next, we compare $\tilde H_1$ with $H_2$, which annihilate the same vector $u_2$. The operator $\tilde H_1-H_2$ maps every vector into the orthogonal complement of $u_2$, since $\langle(\tilde H_1-H_2)f,u_2\rangle=\langle f,(\tilde H_1-H_2)u_2\rangle=0$. Let $\langle f,u_2\rangle=0$. Differentiating $s\mapsto e^{(t-s)H_2}e^{s\tilde H_1}f$ and integrating over $[0,t]$, we obtain the variation of constants formula \cite[Chapter~III, Corollary~1.7]{EngelNagel2000}
\[
\big(e^{t\tilde H_1}-e^{tH_2}\big)f=\int_0^te^{(t-s)H_2}\big(\tilde H_1-H_2\big)e^{s\tilde H_1}f\,ds .
\]
The integrand is continuous in $s$. The vector $e^{s\tilde H_1}f$ is orthogonal to $u_2$ and has norm at most $e^{-2ds}\|f\|$, and $(\tilde H_1-H_2)e^{s\tilde H_1}f$ is orthogonal to $u_2$. The integrand has norm at most $e^{-2d(t-s)}\|\tilde H_1-H_2\|e^{-2ds}\|f\|$, and we find $\|(e^{t\tilde H_1}-e^{tH_2})f\|\le t\|\tilde H_1-H_2\|e^{-2dt}\|f\|$. Now
\[
\tilde H_1e^{t\tilde H_1}-H_2e^{tH_2}=\big(\tilde H_1-H_2\big)e^{t\tilde H_1}+H_2\big(e^{t\tilde H_1}-e^{tH_2}\big),
\]
and both sides vanish at $u_2$. Applying the bounds above to $f-\langle f,u_2\rangle u_2$, whose norm is at most $\|f\|$, we obtain
\[
\big\|\tilde H_1e^{t\tilde H_1}-H_2e^{tH_2}\big\|\le\|\tilde H_1-H_2\|(1+Bt)e^{-2dt}\le(a+4Bb)(1+Bt)e^{-2dt}.
\]

We now return to $H_1$. By the power series of the exponential, we have $\tilde H_1e^{t\tilde H_1}=UH_1e^{tH_1}U^{-1}$. As above, $\|UXU^{-1}-X\|\le4b\|X\|$ for $X=H_1e^{tH_1}$, whose norm is at most $Be^{-2dt}$. Combining this with the last display, we find
\[
\big\|H_1e^{tH_1}-H_2e^{tH_2}\big\|\le\big(4Bb+(a+4Bb)(1+Bt)\big)e^{-2dt}.
\]
Using $4Bb\le4(1+B)(a+b)$, $a+4Bb\le4(1+B)(a+b)$, and $1+Bt\le(1+B)(1+t)$ bounds the prefactor by $8(1+B)^2(a+b)(1+t)$, and hence by the stated $20(1+B)^2(a+b)(1+t)$.

For the second claim, the variation of constants formula for $H_1$ and $H_2$ on the whole space, together with the contraction bound, gives $\|e^{tH_1}-e^{tH_2}\|\le ta$, which implies that $|m_1(t)-m_2(t)|\le ta$. Let $X_j=H_je^{tH_j}x$, so that $\|X_2\|\le Be^{-2dt}$. We have
\[
\frac{X_1}{m_1(t)}-\frac{X_2}{m_2(t)}=\frac{X_1-X_2}{m_1(t)}+\frac{m_2(t)-m_1(t)}{m_1(t)m_2(t)}\,X_2 ,
\]
and the right side has norm at most $\delta^{-1}\|X_1-X_2\|+\delta^{-2}taBe^{-2dt}$. The first claim and $t\le1+t$ complete the proof.
\end{proof}

\begin{lemma}[Comparison with the continuum evolution]\label{lem:hotf-transfer}
Let $0<c_1<1$. There exists $C_2$ depending only on $c_1$ such that the following holds. Let $c\in(0,c_1]$, $N\ge1$, and $\lambda_1\ge\cdots\ge\lambda_N$, and suppose that $E=\|\ell_N-\gamma\|_\infty\le(1-c_1)^2/4$. Let $A=\diag(c\lambda_i)$, with Lagrange multiplier $z$ and smallest precision $d$, and let $e_i(t)$ and $\cI_N(t)$ be the relative errors and the information of its covariance evolution. Then $|z-(1+c^2)|\le cE$ and $d\ge g/2$. Moreover, let $e_N(t,\cdot)$ be the step function of $(e_i(t))_i$. For all $t\ge0$ with $cEt\le1/4$,
\begin{equation}\label{hp-finite-transfer}
\big\|e_N(t,\cdot)-\eta_t\circ\gamma\big\|_2\le C_2\,cE\,(1+t)\,e^{-2gt}.
\end{equation}
Further, for such $t$,
\begin{equation}\label{eq:hotf-information-transfer}
\begin{gathered}
\Big|\sqrt{2\cI_N(t)}-\Big(N\int\eta_t^2\rho_{\mathrm{sc}}\Big)^{1/2}\Big|\le C_2\,cE\,(1+t)\sqrt N\,e^{-2gt},\\
\max_i|e_i(t)|\le C_2\,cE\,(1+t)+2C_1ce^{-2gt},
\end{gathered}
\end{equation}
with $C_1$ as in \cref{prop:hotf-edge}.
\end{lemma}

\begin{proof}
We first compare the Lagrange multipliers. Let $\Upsilon_N(y)=\int_0^1(y-c\ell_N)^{-1}d\omega$, which is decreasing on $(c\lambda_1,\infty)$, and $\Upsilon(y)=\int_0^1(y-c\gamma)^{-1}d\omega$, which is decreasing on $(2c,\infty)$. Then $\Upsilon_N(z)=1$, and $\Upsilon(1+c^2)=1$ by \eqref{eq:hotf-sc-multiplier}. The inequalities $c\ell_N\le c\gamma+cE$ and $1+c^2>2c$ imply that $\Upsilon_N(1+c^2+cE)\le\Upsilon(1+c^2)=1$, and then $z\le1+c^2+cE$. If $1+c^2-cE\le c\lambda_1$, then $z>c\lambda_1\ge1+c^2-cE$. Otherwise, $c\ell_N\ge c\gamma-cE$ gives $\Upsilon_N(1+c^2-cE)\ge\Upsilon(1+c^2)=1$, and again $z\ge1+c^2-cE$. This proves $|z-(1+c^2)|\le cE$. For small $\omega$, we have $\lambda_1=\ell_N(\omega)\le\gamma(\omega)+E\le2+E$. Together with $z\ge1+c^2-cE$, this gives $d=z-c\lambda_1\ge g-2cE\ge g/2$, where we used $2cE\le(1-c_1)^2/2\le g/2$. Let $p_N(\omega)=z-c\ell_N(\omega)$, the step function of the precisions $p_i$. Then
\begin{equation}\label{eq:hotf-precision-transfer}
|p_N(\omega)-p(\gamma(\omega))|\le|z-(1+c^2)|+c|\ell_N(\omega)-\gamma(\omega)|\le2cE .
\end{equation}

Next, we write the relative errors through the operators of \cref{lem:hotf-energy}. Let $d_0=(1-c_1)^2/2$ and $d_*=g-2cE$. Since $2cE\le(1-c_1)^2/2\le g/2$, we have $d_*\ge g/2\ge d_0$. On $(0,1)$, we have $g\le p\circ\gamma\le(1+c)^2\le4$, and by \eqref{eq:hotf-precision-transfer}, we have $d_*\le p_N\le4+2cE\le5$. Further, $\int_0^1p_N^{-1}\,d\omega=N^{-1}\sum_ip_i^{-1}=1$ by the definition of $z$, and $\int_0^1(p\circ\gamma)^{-1}\,d\omega=\int p^{-1}\rho_{\mathrm{sc}}=1$ by \eqref{eq:hotf-sc-multiplier}. By these bounds, \cref{lem:hotf-energy} applies to $p_N$ and to $p\circ\gamma$, with $d_*$ in place of $d$ and with $s=5$. Let $H_1=H_{p_N}$ and $H_2=H_{p\circ\gamma}$. Since $2c\ell_N=2z-2p_N$ and $2c\gamma=2(1+c^2)-2p\circ\gamma$, the operators of \eqref{hp-linear-system} satisfy $\mathcal L_{\ell_N}f=H_1f+2zf$ and $\mathcal L_\gamma f=H_2f+2(1+c^2)f$ for $f\in L^\infty(0,1)$. The partial sums of the exponential series of $t\mathcal L_{\ell_N}$, applied to $\ind$, converge in $L^\infty(0,1)$ to $M^{\ell_N}_t$. Since the norm of $L^\infty(0,1)$ dominates that of $L^2(0,1)$, they also converge to $M^{\ell_N}_t$ in $L^2(0,1)$, where their limit is $e^{2zt}e^{tH_1}\ind$. Let $h_j(t)=e^{tH_j}\ind$ and $\bar h_j(t)=\langle\ind,h_j(t)\rangle$ for $j=1,2$. By \eqref{hp-linear-system}, we have
\[
M^{\ell_N}_t=e^{2zt}h_1(t),\qquad\bar M^{\ell_N}_t=e^{2zt}\bar h_1(t),\qquad v^{\ell_N}_t=\frac{h_1(t)}{\bar h_1(t)},
\]
and in the same way $v^\gamma_t=h_2(t)/\bar h_2(t)$. By the definition of $H_1$, we have $p_Nh_1(t)=-H_1h_1(t)/2+\bar h_1(t)\ind$, and similarly for $H_2$. Recall from the discussion before \cref{lem:hotf-positive} that $v^{\ell_N}_t$ is the step function of $(v_i(t))_i$, and that $v^\gamma_t=v_t\circ\gamma$. Using also the definition \eqref{eq:hotf-continuum-precision} of $\eta_t$, we find
\[
e_N(t,\cdot)=p_Nv^{\ell_N}_t-\ind=-\frac{H_1e^{tH_1}\ind}{2\bar h_1(t)},\qquad
\eta_t\circ\gamma=(p\circ\gamma)\,v^\gamma_t-\ind=-\frac{H_2e^{tH_2}\ind}{2\bar h_2(t)} .
\]

We now check the hypotheses of \cref{lem:hotf-transient}. Let $q_1=p_N^{-1}$ and $q_2=(p\circ\gamma)^{-1}$, and let $u_j=q_j/\|q_j\|_2$. By \cref{lem:hotf-energy}, the operators $H_j$ are self-adjoint, $\|H_j\|\le12$, $H_ju_j=0$, and $\langle f,H_jf\rangle\le-2d_*\|f\|_2^2$ whenever $\langle f,u_j\rangle=0$. Also $1\le\|q_j\|_2\le d_0^{-1}$, and $\bar h_j(t)\ge\|q_j\|_2^{-2}\ge d_0^2$ for all $t\ge0$. Since $q_1$ and $q_2$ are positive, we have $\langle u_1,u_2\rangle>0$. By \eqref{eq:hotf-precision-transfer},
\[
\|H_1-H_2\|\le2\|p_N-p\circ\gamma\|_\infty\le4cE,\qquad
\|q_1-q_2\|_2\le\Big\|\frac{p\circ\gamma-p_N}{p_N\,(p\circ\gamma)}\Big\|_\infty\le\frac{2cE}{d_0^2} .
\]
Further,
\[
u_1-u_2=\frac{q_1-q_2}{\|q_1\|_2}+\frac{\|q_2\|_2-\|q_1\|_2}{\|q_1\|_2}\,u_2 ,
\]
and the triangle inequality and $\|q_1\|_2\ge1$ give $\|u_1-u_2\|_2\le2\|q_1-q_2\|_2\le4cE/d_0^2$.

We apply \cref{lem:hotf-transient} with $B=12$, with $d_*$ in place of $d$, with $x=\ind$, for which $m_j(t)=\bar h_j(t)$, and with $\delta=d_0^2$. Using the formulas for $e_N(t,\cdot)$ and $\eta_t\circ\gamma$ above, we find
\[
\begin{split}
\big\|e_N(t,\cdot)-\eta_t\circ\gamma\big\|_2&\le\frac12\Big(\frac{20\cdot13^2\cdot4cE\,(1+d_0^{-2})}{d_0^2}+\frac{12\cdot4cE}{d_0^4}\Big)(1+t)e^{-2d_*t}\\
&\le\frac{2^{14}cE}{d_0^4}\,(1+t)\,e^{-2d_*t}.
\end{split}
\]
In the last inequality, we used $d_0\le1/2$. Since $4cEt\le1$, we have $e^{-2d_*t}=e^{-2gt}e^{4cEt}\le e\cdot e^{-2gt}$. This proves \eqref{hp-finite-transfer} for every $C_2\ge2^{14}e\,d_0^{-4}$. The lower bound $d_*=g-2cE$ for the precisions, rather than $g/2$, is what gives the factor $e^{-2gt}$.

Next, note that $2\cI_N(t)=\sum_ie_i(t)^2=N\|e_N(t,\cdot)\|_2^2$ and $\int\eta_t^2\rho_{\mathrm{sc}}=\|\eta_t\circ\gamma\|_2^2$. The first bound in \eqref{eq:hotf-information-transfer} follows from \eqref{hp-finite-transfer} and the triangle inequality in $L^2(0,1)$.

It remains to prove the second bound in \eqref{eq:hotf-information-transfer}. We do not deduce it from \eqref{hp-finite-transfer}, which controls only the norm in $L^2(0,1)$, but from the positivity of the evolution. Equation \eqref{hp-positive-comparison}, with $a=\ell_N$ and $a'=\gamma$, implies that $|v^{\ell_N}_t/v^\gamma_t-1|\le e^{4cEt}-1\le4e\,cEt$, where we used $4cEt\le1$. Further, \eqref{eq:hotf-precision-transfer} and $p\ge g$ give $|p_N/(p\circ\gamma)-1|\le2cE/g\le1/2$. Now
\[
1+e_N(t,\cdot)=p_Nv^{\ell_N}_t=\big(1+\eta_t\circ\gamma\big)\,\frac{p_N}{p\circ\gamma}\,\frac{v^{\ell_N}_t}{v^\gamma_t},
\]
and $|(1+\alpha)(1+\beta)-1|\le|\alpha|+2|\beta|$ for $|\alpha|\le1$. Using $|\eta_t|\le2C_1$ from \cref{prop:hotf-edge}, we find, for every $\omega\in(0,1)$,
\[
|e_N(t,\omega)-\eta_t(\gamma(\omega))|\le(1+2C_1)\Big(\frac{2cE}g+8e\,cEt\Big)\le C_\infty\,cE\,(1+t),
\]
with $C_\infty=(1+2C_1)\max\{2(1-c_1)^{-2},8e\}$. Together with $\sup_\lambda|\eta_t(\lambda)|\le2C_1ce^{-2gt}$ from \cref{prop:hotf-edge}, this gives the second bound in \eqref{eq:hotf-information-transfer} for every $C_2\ge C_\infty$. We take $C_2=\max\{2^{14}e\,d_0^{-4},C_\infty\}$, which depends only on $c_1$.
\end{proof}

The comparison in \cref{lem:hotf-transfer} is made at the actual times $t$, which grow with $N$; no limit $N\to\infty$ at fixed time is interchanged with the limit $t\to\infty$. At the times $t=T_N+u/g$ of \cref{hp-fixed-profile}, the information is of order one, and $\sqrt N\,e^{-2gt}$ is of order $(\log N)^{3/4}$. The first error in \eqref{eq:hotf-information-transfer} is then $O(E(\log N)^{7/4})$, and the second is $O(E\log N)+o(1)$, with constants depending on $c$ and $u$. Without the factor $e^{-2gt}$, the first error would be of order $\sqrt N\,E\log N$. For the GOE, at every fixed confidence we have $E\le(C+7)N^{-5/12}$ on the event $F_C$ for $N$ large, by \cref{lem:hotf-grid}. The argument in \cref{sec:hotf-proof} uses the eigenvalues only through this bound, and it applies on any event on which $\|\ell_N-\gamma\|_\infty\le E_N$, for a deterministic sequence $E_N$ with $E_N(\log N)^{7/4}\to0$.

\section{Proof of the main result}\label{sec:hotf-proof}

Fix $c\in(0,1)$, a compact set $K\subset\R$, and $\eta>0$. Choose $C\ge1$ using \cref{sp-goe-rigidity} so that $\PP(F_C)\ge1-\eta$ for all sufficiently large $N$. All constants below depend only on $c$, $K$ and $C$, and they are chosen before $N\to\infty$. All estimates on $F_C$ below are deterministic consequences of the bounds of \cref{lem:hotf-grid}. Let $E_N=(C+7)N^{-5/12}$ and $t_N(u)=T_N+u/g$ for $u\in K$. Since $T_N\sim\log N/(4g)$, for $N$ large we have $1\le t_N(u)\le\log N/(2g)$ for all $u\in K$.

\paragraph{Step 1: the information.}
On $F_C$, we have $\|\ell_N-\gamma\|_\infty\le E_N$ by \cref{lem:hotf-grid}, and for $N$ large, $E_N\le(1-c)^2/4$ and $cE_Nt_N(u)\le1/4$. By the definition \eqref{hp-fixed-center} of $T_N$, we have $\sqrt N\,e^{-2gt_N(u)}=A_c^{-1/2}(\log N)^{3/4}e^{-2u}$. Let $U=\sup\bigl(\{0\}\cup\{|u|:u\in K\}\bigr)$. We apply the first bound in \eqref{eq:hotf-information-transfer} with $c_1=c$, and use $1+t_N(u)\le(1+(2g)^{-1})\log N$ for $N\ge3$. We find
\[
\sup_{u\in K}\Big|\sqrt{2\cI_N(t_N(u))}-\Big(N\int\eta_{t_N(u)}^2\rho_{\mathrm{sc}}\Big)^{1/2}\Big|\le C_2c\big(1+(2g)^{-1}\big)A_c^{-1/2}e^{2U}\,E_N(\log N)^{7/4},
\]
which tends to zero, because $E_N=(C+7)N^{-5/12}$. Next, we evaluate the integral with \cref{prop:hotf-edge}, again with $c_1=c$. We have
\[
4gt_N(u)=\log N-\frac32\log\log N+\log A_c+4u,\qquad Ne^{-4gt_N(u)}=\frac{(\log N)^{3/2}e^{-4u}}{A_c}.
\]
In particular, $4gt_N(u)/\log N\to1$ uniformly in $u\in K$, and $t_N(u)^{-3/2}=(4g)^{3/2}(\log N)^{-3/2}(1+o(1))$. Using \eqref{hp-continuum-information}, we find
\[
\begin{split}
\frac N2\int\eta_{t_N(u)}^2\rho_{\mathrm{sc}}&=\frac N2K_ct_N(u)^{-3/2}e^{-4gt_N(u)}\big(1+o(1)\big)\\
&=\frac{K_c(4g)^{3/2}}{2A_c}\,e^{-4u}\big(1+o(1)\big)=e^{-4u}\big(1+o(1)\big),
\end{split}
\]
uniformly in $u\in K$, where in the last step we used $A_c=4K_cg^{3/2}$. Combining the two displays, we conclude that, on $F_C$,
\begin{equation}\label{eq:hotf-info-limit}
\sup_{u\in K}\big|\cI_N(t_N(u))-e^{-4u}\big|\to0,\qquad \sup_{u\in K}\max_i|e_i(t_N(u))|\to0,
\end{equation}
where the second limit follows from the second, maximum-coordinate bound in \eqref{eq:hotf-information-transfer}, since $E_N\log N\to0$ and $e^{-2gt_N(u)}\to0$ uniformly in $u\in K$.

\paragraph{Step 2: the Bingham comparison law and the Gibbs measure.}
We have $\pi_{N,c}=\mu_q$ and $\mu_t=\mu_{v(t)}$, with $\sum_iq_i=\sum_iv_i(t)=N$ and $v_i(t)=q_i(1+e_i(t))$. Recall from \cref{lem:hotf-transfer} that $d_N\ge g/2$, that is, $\max_iq_i\le2/g$. Together with $\sum_iq_i^2\ge N^{-1}(\sum_iq_i)^2=N$, this gives
\[
\eta_q=\frac{\max_iq_i}{(2\sum_iq_i^2)^{1/2}}\le\frac2{g\sqrt{2N}} .
\]
Combining \eqref{eq:hotf-info-limit} with \cref{lem:hotf-lan-uniform}, we have $\sup_{u\in K}|\|\mu_{t_N(u)}-\pi_{N,c}\|_{\TV}-\mathcal G(\cI_N(t_N(u)))|\to0$ on $F_C$. Further, \eqref{eq:hotf-G-lipschitz} and \eqref{eq:hotf-info-limit} imply that $\mathcal G(\cI_N(t_N(u)))\to\mathcal G(e^{-4u})=\Psi(u)$ uniformly in $u\in K$. We conclude that
\[
\sup_{u\in K}\big|\|\mu_{t_N(u)}-\pi_{N,c}\|_{\TV}-\Psi(u)\big|\to0\qquad\text{on }F_C .
\]

\paragraph{Step 3: the time-$t$ law.}
On $F_C$, we have $\|cW_N\|\le3c\le3$ and $d_N\ge g/2$, so that $Nd_N^2\ge Ng^2/4\ge C_{\mathrm w}(3)$ for $N$ large, where $C_{\mathrm w}$ is the constant of \eqref{ht-weighted-bridge-dimension-condition}. Then \cref{ht-bingham-bridge}, with $\Lambda=3$, implies that for all $t\ge0$,
\[
\|\nu_NP_t-\mu_t\|_{\TV}\le C_3d_N^{-2}e^{-d_Nt/2}\le4C_3g^{-2}e^{-gt/4},
\]
where $C_3$ is the constant $C_\Lambda$ of \eqref{ht-bingham-bridge-bound} with $\Lambda=3$. This tends to zero uniformly in $u\in K$, because $t_N(u)\ge T_N-U/g\to\infty$. By the triangle inequality and Step~2, we conclude that $\sup_{u\in K}|D_N(t_N(u))-\Psi(u)|\to0$ on $F_C$.

\paragraph{Step 4: conclusion.}
Every bound in Steps 1--3 is deterministic on $F_C$ and holds for $N$ larger than a constant depending on $c$, $K$ and $C$. Thus, for every $\varepsilon>0$, the limsup of the probability that $\sup_{u\in K}|D_N(T_N+u/g)-\Psi(u)|>\varepsilon$ is at most $\eta$. Sending $\eta\downarrow0$, we obtain $\sup_{u\in K}|D_N(T_N+u/g)-\Psi(u)|\to0$ in probability. Because $gT_N\to\infty$, the distance $D_N$ has cutoff profile $\Psi$ with center $T_N$ and window $1/g$ in the sense of \cref{def:model-cutoff}. By \cref{cot-profile} and \cref{cor:model-profile-cutoff}, we have $g(t_N(\eps)-T_N)\to\Psi^{-1}(\eps)=u_\eps$ in probability for each $\eps\in(0,1)$, and there is cutoff. Finally, $t_N(\eps)/\log N\to(4g)^{-1}$ in probability, because $T_N/\log N\to(4g)^{-1}$.

For finitely many values $c\in(0,1)$, the event $F_C$ does not depend on $c$, and on $F_C$ the bounds above hold for all of them once $N$ exceeds the largest of the corresponding thresholds. This proves the simultaneous statement and completes the proof.