\chapter{Reduction to three propositions}\label{s:reduction}

In this section, we prove \zcref{t:main} assuming three propositions. The first, \zcref{p:pairsl}, counts pairs of eigenvalues at a scale slightly above the critical scale $N^{-3/2}$; we prove it in \zcref{s:pairs}. The second, \zcref{p:llt}, is a local limit theorem for the linearized increments produced by resampling a small number of entries of $H$, which we prove in \zcref{s:llt}. The third, \zcref{p:linearization}, states that the small gaps of the resampled matrix are close to the linearized gaps; its proof is given in \zcref{s:linearization}. We state the propositions in \zcref{s:reduction-pairs} and \zcref{s:reduction-resampling}, and we deduce \zcref{t:main} from them in \zcref{s:reduction-proof}.

Throughout the rest of the paper, we use the scales
\begin{equation}\label{e:reduction-scales}
d = N^{-3/2}, \qquad \tau = N^{1/100}, \qquad \ell = d \tau^{1/2} = N^{-299/200}, \qquad s = N^{-7/5}.
\end{equation}
The smallest gaps in \zcref{t:main} are of order $d$. Recall the scales $\ell_\alpha = N^{-1 - \alpha}$ and $M_\alpha = N^{1 - 2\alpha}$ from \eqref{e:prelim-scales}. We set $\alpha_* = 99/200$, so that $\ell = \ell_{\alpha_*}$ and $M_{\alpha_*} = N^3 \ell^2 = \tau$. Heuristically, a bulk interval contains of order $\tau$ pairs of eigenvalues at distance at most $\ell$ from each other. The scale $s$ is the threshold below which two consecutive eigenvalues are treated together in \zcref{p:linearization}. These scales satisfy
\begin{equation}\label{e:reduction-ratios}
\frac{d}{\ell} = N^{-1/200}, \qquad \frac{\ell}{s} = N^{-19/200}, \qquad N s = N^{-2/5}.
\end{equation}
Recall that $g_i = \lambda_{i + 1} - \lambda_i$ and $E_i = (\lambda_i + \lambda_{i + 1})/2$ for $i \in [N - 1]$.

\section{Pair counts at the scale \texorpdfstring{$\ell$}{l}}\label{s:reduction-pairs}

Recall the functions $f_p(y) = (1 + y^2)^{-p}$ from \eqref{e:prelim-fp}. We set $p_* = 61$. This value, like the lower bound $\alpha \ge 1/10$ in \zcref{p:induction}, is chosen for the bound \eqref{e:pairs-cross-bound} in the proof of \zcref{p:induction}, which requires $2p(\alpha' - \kappa) > 1 + 2\alpha'$ for all integers $p \ge p_*$, where $\kappa = \alpha - \alpha'/6$; this holds because $\alpha' - \kappa > \alpha/6 \ge 1/60$ and $\alpha' < 1/2$.

\begin{proposition}\label{p:pairsl}
\begin{enumerate}[label=(\roman*)]
\item For every continuous function $f \colon [0, \infty) \times (-2, 2) \to \R$ with compact support, we have
\begin{equation}\label{e:reduction-pairs-lln}
\frac{1}{\tau} \sum_{1 \le i < j \le N} f \Big( \frac{\lambda_j - \lambda_i}{\ell}, \frac{\lambda_i + \lambda_j}{2} \Big) \longrightarrow \int_{-2}^{2} \int_0^\infty c(E) r f(r, E) \dd r \dd E
\end{equation}
in probability.
\item For every bulk interval $I$, there exists a constant $C_I > 0$ such that
\begin{equation}\label{e:reduction-pairs-upper}
\lim_{N \to \infty} \Prob \bigg( \sum_{\substack{1 \le i < j \le N \\ (\lambda_i + \lambda_j)/2 \in I}} f_{p_*} \Big( \frac{\lambda_j - \lambda_i}{\ell} \Big) \le C_I \tau \bigg) = 1.
\end{equation}
\end{enumerate}
\end{proposition}

Compared to \zcref{t:pairs} with $\alpha = \alpha_*$, the sum in the first part runs over all pairs $i < j$ instead of consecutive eigenvalues, and the test function may depend on the energy. The second part controls pairs at all distances, with a weight that decays only polynomially in $(\lambda_j - \lambda_i)/\ell$. We will use it to bound the number of pairs with $g_i \le R \ell$ by $O_R(\tau)$, and to control the pairs with $g_i \gg \ell$, which the first part does not see.

\section{The resampling and the linearized increments}\label{s:reduction-resampling}

\paragraph{The resampling.}
Set
\begin{equation}\label{e:reduction-m}
m = m_N = 2 \lfloor \tau/8 \rfloor, \qquad \text{so that} \qquad \lim_{N \to \infty} \frac{4m}{\tau} = 1.
\end{equation}
Let $(a_1, b_1), \dots, (a_m, b_m)$ be independent random pairs, each uniformly distributed on the set $\{ (a, b) \in [N]^2 : a < b \}$. Further, let $Y_1, \dots, Y_m$ be independent copies of $X$. We assume that $H$, the pairs $(a_l, b_l)$, and the variables $Y_l$ are mutually independent. We call $(a_l, b_l)$ the $l$-th \emph{selected position}, and we call $(a_l, b_l, Y_l)_{l \in [m]}$ the \emph{resampling variables}. Define real symmetric matrices $H^{(0)}, \dots, H^{(m)}$ by $H^{(0)} = H$ and
\begin{equation}\label{e:reduction-step}
H^{(l)} = H^{(l - 1)} + \big( N^{-1/2} Y_l - H^{(l - 1)}_{a_l b_l} \big) V^{a_l b_l}, \qquad l \in [m],
\end{equation}
where $V^{ab}$ is as in \eqref{e:prelim-V}. Thus $H^{(l)}$ is obtained from $H^{(l - 1)}$ by setting its $(a_l, b_l)$ and $(b_l, a_l)$ entries equal to $N^{-1/2} Y_l$. We set
\begin{equation}\label{e:reduction-Htilde}
\widetilde H = H^{(m)}, \qquad \Delta = \widetilde H - H,
\end{equation}
and we call $\widetilde H$ the \emph{resampled matrix}. We also write $\tilde\lambda_i = \lambda_i(\widetilde H)$ for $i \in [N]$, and $\tilde g_i = \tilde\lambda_{i + 1} - \tilde\lambda_i$ and $\widetilde E_i = (\tilde\lambda_i + \tilde\lambda_{i + 1})/2$ for $i \in [N - 1]$. Since $a_l < b_l$, the resampling leaves the diagonal entries of $H$ unchanged. A position may be selected more than once; in that case, the last replacement determines the final entry in that position. (The fact that $m$ is even is used in \zcref{s:llt}.)

\begin{lemma}\label{l:stationary}
The resampled matrix $\widetilde H$ has the same law as $H$.
\end{lemma}

\begin{proof}
Let $\sigma = ((a'_l, b'_l))_{l \in [m]}$ be a deterministic sequence of pairs of integers with $1 \le a'_l < b'_l \le N$, and define $H^{(l)}_\sigma$ by \eqref{e:reduction-step} with $(a'_l, b'_l)$ in place of $(a_l, b_l)$ and $H^{(0)}_\sigma = H$. We show by induction on $l$ that $H^{(l)}_\sigma$ has the law of $H$, the case $l = 0$ being trivial.

Suppose that this holds for $l - 1$. Then the entries of $H^{(l - 1)}_\sigma$ on and above the diagonal are independent with the laws of the corresponding entries of $H$, and they are independent of $Y_l$, since $H^{(l - 1)}_\sigma$ is a function of $H$ and $Y_1, \dots, Y_{l - 1}$. The matrix $H^{(l)}_\sigma$ is obtained by replacing the entry at $(a'_l, b'_l)$ by $N^{-1/2} Y_l$, which has the law of $H_{a'_l b'_l}$, so the claim holds for $l$.

The selected positions are independent of $H$ and of $(Y_l)_{l \in [m]}$, and $\widetilde H = H^{(m)}_\sigma$ on the event that $(a_l, b_l) = (a'_l, b'_l)$ for all $l \in [m]$. Therefore, for every Borel set $\mathfrak B$ of real symmetric $N \times N$ matrices, we have
\begin{equation*}
\Prob \big( \widetilde H \in \mathfrak B \big) = \sum_{\sigma} \Prob \big( (a_l, b_l)_{l \in [m]} = \sigma \big) \, \Prob \big( H^{(m)}_\sigma \in \mathfrak B \big) = \Prob \big( H \in \mathfrak B \big),
\end{equation*}
where the sum runs over all sequences $\sigma$ as above, and the second equality follows from the claim.
\end{proof}

\paragraph{Conditioning on $H$.}
Recall that the eigenvectors $u_1, \dots, u_N$ are measurable functions of $H$. Since the resampling variables are independent of $H$, Fubini's theorem shows that integrating over the resampling variables with $H$ held fixed gives a version of the conditional expectation given $H$, and we always use this version. An event $\mathcal A$ that is measurable with respect to $H$ has the form $\{ H \in \mathfrak A \}$ for a Borel set $\mathfrak A$ of real symmetric matrices. A statement such as ``on $\mathcal A$, we have $\Prob(\mathcal B \mid H) \ge 1 - o(1)$'' means that, for every $A \in \mathfrak A$, the probability of $\mathcal B$ computed with $H$ replaced by $A$ is bounded below by $1 - o(1)$. In \zcref{p:llt} and \zcref{p:linearization} below, the error terms are deterministic quantities in the sense of our conventions. They may depend on fixed parameters, such as $k$ and the test functions, but not on $A$, on the resampling variables, or on the eigenvalue labels involved.

\paragraph{Linearized increments.}
For $i \in [N - 1]$ and a real symmetric $N \times N$ matrix $B$, define
\begin{equation}\label{e:reduction-Li}
L_i(B) = \Big( \frac{u_i^T B u_i - u_{i + 1}^T B u_{i + 1}}{2}, \, u_i^T B u_{i + 1} \Big) \in \R^2.
\end{equation}
The map $L_i$ is linear, and $L_i(H) = ((\lambda_i - \lambda_{i + 1})/2, 0)$. We set
\begin{equation}\label{e:reduction-zi}
z_i = d^{-1} L_i(H) = \Big( - \frac{g_i}{2d}, 0 \Big), \qquad \text{so that} \qquad |z_i| = \frac{g_i}{2d}.
\end{equation}
The map $L_i$ is relevant because of an elementary identity for $2 \times 2$ matrices. For $y_1, y_2, y_3 \in \R$, the formula for the roots of the characteristic polynomial gives
\begin{equation}\label{e:reduction-gap2x2}
\lambda_2 \begin{pmatrix} y_1 & y_3 \\ y_3 & y_2 \end{pmatrix} - \lambda_1 \begin{pmatrix} y_1 & y_3 \\ y_3 & y_2 \end{pmatrix} = \big( (y_1 - y_2)^2 + 4 y_3^2 \big)^{1/2} = 2 \, \Big| \Big( \frac{y_1 - y_2}{2}, y_3 \Big) \Big|.
\end{equation}
Applying \eqref{e:reduction-gap2x2} to the compression of $H + B$ to the span of $u_i$ and $u_{i + 1}$, we find that the $2 \times 2$ matrix $( u_j^T (H + B) u_{j'} )_{j, j' \in \{i, i + 1\}}$ has eigenvalue gap $2 |L_i(H + B)|$.

For $1 \le a < b \le N$ and $i \in [N - 1]$, define
\begin{equation}\label{e:reduction-D}
D^{(i)}_{ab} = N L_i(V^{ab}) = N \big( u_i(a) u_i(b) - u_{i + 1}(a) u_{i + 1}(b), \, u_i(a) u_{i + 1}(b) + u_i(b) u_{i + 1}(a) \big) \in \R^2,
\end{equation}
where the second equality follows from \eqref{e:prelim-V}. For $i \in [N - 1]$ and $l \in [m]$, define
\begin{equation}\label{e:reduction-S}
J_{i, l} = (Y_l - X_{a_l b_l}) D^{(i)}_{a_l b_l}, \qquad S_i = \sum_{l = 1}^m J_{i, l}.
\end{equation}
We call $S_i$ the \emph{linearized increment of the pair $(i, i + 1)$}. Note that $X_{a_l b_l} = N^{1/2} H_{a_l b_l}$ is the entry of the original matrix $H$, also when the position $(a_l, b_l)$ was selected before the $l$-th step. Conditionally on $H$, the triples $(a_l, b_l, Y_l)$, $l \in [m]$, are independent and identically distributed, so $S_i$ is a sum of $m$ conditionally independent and identically distributed random vectors.

For several pairs, we use the following notation. For an integer $k \ge 1$ and $\mathbf i = (i_1, \dots, i_k) \in [N - 1]^k$, let
\begin{equation}\label{e:reduction-tuples}
z_{\mathbf i} = (z_{i_1}, \dots, z_{i_k}), \qquad D^{(\mathbf i)}_{ab} = \big( D^{(i_1)}_{ab}, \dots, D^{(i_k)}_{ab} \big), \qquad S_{\mathbf i} = (S_{i_1}, \dots, S_{i_k}),
\end{equation}
which are vectors in $\R^{2k}$. Then $S_{\mathbf i} = \sum_{l = 1}^m (Y_l - X_{a_l b_l}) D^{(\mathbf i)}_{a_l b_l}$. The vectors $S_{i_1}, \dots, S_{i_k}$ are built from the same resampling variables, so they are not conditionally independent given $H$. The next lemma explains the name of $S_i$.

\begin{lemma}\label{l:reduction-identity}
On the event that the selected positions $(a_1, b_1), \dots, (a_m, b_m)$ are distinct, we have
\begin{equation}\label{e:reduction-identity}
\Delta = N^{-1/2} \sum_{l = 1}^m (Y_l - X_{a_l b_l}) V^{a_l b_l} \qquad \text{and} \qquad d^{-1} L_i(\Delta) = S_i \quad \text{for all } i \in [N - 1].
\end{equation}
In particular, on this event, the $2 \times 2$ matrix $( u_j^T \widetilde H u_{j'} )_{j, j' \in \{i, i + 1\}}$ has eigenvalue gap $2 d |z_i + S_i|$ for every $i \in [N - 1]$.
\end{lemma}

\begin{proof}
Suppose that the selected positions are distinct. Then for every $l \in [m]$, the entry of $H^{(l - 1)}$ in position $(a_l, b_l)$ is not changed in the first $l - 1$ steps. Using \eqref{e:reduction-step}, we deduce that
\begin{equation*}
H^{(l - 1)}_{a_l b_l} = H_{a_l b_l} = N^{-1/2} X_{a_l b_l}, \qquad H^{(l)} - H^{(l - 1)} = N^{-1/2} (Y_l - X_{a_l b_l}) V^{a_l b_l}.
\end{equation*}
Summing over $l \in [m]$ gives the first identity in \eqref{e:reduction-identity}. Using the linearity of $L_i$, \eqref{e:reduction-D}, and \eqref{e:reduction-zi}, we then obtain
\begin{align*}
L_i(\Delta) & = N^{-1/2} \sum_{l = 1}^m (Y_l - X_{a_l b_l}) L_i(V^{a_l b_l}) = N^{-3/2} \sum_{l = 1}^m J_{i, l} = d S_i, \\
L_i(\widetilde H) & = L_i(H) + L_i(\Delta) = d (z_i + S_i).
\end{align*}
The last claim now follows from the consequence of \eqref{e:reduction-gap2x2} stated above, applied with $B = \Delta$.
\end{proof}

We call $2d|z_i + S_i|$ the \emph{linearized gap} of the pair $(i, i + 1)$. It is the gap that first-order perturbation theory predicts for the pair $(i, i + 1)$ when the interaction with the other eigenvalues is neglected.

\paragraph{A local limit theorem.}
We now state the local limit theorem for the linearized increments. The heuristic behind its normalization is explained in \zcref{s:ov-resampling}, and the conditional covariance of the summands of $S_{\mathbf i}$, which is close to $4 \Id_{2k}$, is computed in \zcref{l:moments}. The proposition resolves the conditional law of $S_{\mathbf i}$ on the unit scale, which is much smaller than the scale $\tau^{1/2}$ of its fluctuations.

\begin{proposition}\label{p:llt}
Fix an integer $k \ge 1$. There exist events $\mathcal A_{N, k}$, measurable with respect to $H$, and a constant $c > 0$ such that $\lim_{N \to \infty} \Prob(\mathcal A_{N, k}) = 1$ and the following statements hold.
\begin{enumerate}[label=(\roman*)]
\item Let $\Psi \subset C_c^\infty(\R^{2k})$ be a family of functions whose supports are contained in a common compact set, and such that every partial derivative of every order is bounded uniformly over $\Psi$. Then there exist numbers $\mathfrak e_N = \mathfrak e_N(k, \Psi) \ge 0$ such that $\lim_{N \to \infty} \mathfrak e_N = 0$ and, on $\mathcal A_{N, k}$,
\begin{equation}\label{e:reduction-llt}
\sup_{\mathbf i, \, x, \, \psi} \bigg| \tau^k \, \E \big[ \psi(S_{\mathbf i} + x) \bigm| H \big] - \frac{e^{-|x|^2/(2\tau)}}{(2\pi)^k} \int_{\R^{2k}} \psi(y) \dd y \bigg| \le \mathfrak e_N,
\end{equation}
where the supremum is over all $\psi \in \Psi$, all $x \in \R^{2k}$, and all $\mathbf i = (i_1, \dots, i_k) \in [N - 1]^k$ such that $i_{j + 1} \ge i_j + 2$ for all $j \in [k - 1]$.
\item Suppose that $k = 1$. For every $x_0 > 0$, there exists $C_{x_0} > 0$ such that on $\mathcal A_{N, 1}$, we have
\begin{equation}\label{e:reduction-largegap}
\Prob \big( 2 |z_i + S_i| \le x_0 \bigm| H \big) \le C_{x_0} \tau^{-1} e^{-c g_i/\ell}
\end{equation}
for all $i \in [N - 1]$ such that $g_i \ge \ell$.
\end{enumerate}
\end{proposition}

The uniformity in $\mathbf i$ and $x$ in the first part is essential, since in \zcref{s:reduction-proof} we apply \zcref{p:llt} with $x = z_{\mathbf i}$ to tuples $\mathbf i$ of pairs with small gaps, which are selected according to $H$. The second part shows that a pair with $g_i \gg \ell$, for which $|z_i| \gg \tau^{1/2}$, has a linearized gap of order $d$ with probability much smaller than $\tau^{-1}$.

\paragraph{Linearization of the small gaps.}
Finally, we compare the gaps of $\widetilde H$ with the linearized gaps. By \zcref{p:three} and a union bound over $i$, the event that $\lambda_{i + 2} - \lambda_i > 4s$ for all $i \in [N - 2]$ has probability at least $1 - CN(Ns)^3 = 1 - CN^{-1/5}$. On this event, every pair $(i, i + 1)$ with $g_i \le s$ is separated from the other eigenvalues by more than $3s$. To state the error terms in the comparison, we define
\begin{equation}\label{e:reduction-lin-parameters}
w_N = \tau N^{-3/2 + 3/100}, \qquad \mathscr L_N = 2 s^{-1} \log N, \qquad \delta_N = 8 w_N^2 \mathscr L_N.
\end{equation}
These quantities are also used in \zcref{s:linearization}, where $w_N$ bounds the entries of $\Delta$ in the eigenbasis of $H$, $\mathscr L_N$ bounds the sums in \eqref{e:prelim-cluster-harmonic}, and $\delta_N/2$ is the resulting error in \eqref{e:prelim-cluster-ordered}. By \eqref{e:reduction-scales}, we have
\begin{equation}\label{e:reduction-lin-rates}
w_N = N^{-73/50}, \qquad \frac{\delta_N}{d} = 16 N^{-1/50} \log N,
\end{equation}
so $w_N$ and $\delta_N/d$ tend to zero.

\begin{proposition}\label{p:linearization}
There exist events $\mathcal A_N$, measurable with respect to $H$, and events $\mathcal B_N$, measurable with respect to $H$ and the resampling variables, such that $\lim_{N \to \infty} \Prob(\mathcal A_N) = 1$ and the following statements hold for all sufficiently large $N$.
\begin{enumerate}[label=(\roman*)]
\item On $\mathcal A_N$, we have $\lambda_{i + 2} - \lambda_i > 4s$ for all $i \in [N - 2]$, and $\Prob(\mathcal B_N \mid H) \ge 1 - o(1)$.
\item On $\mathcal A_N \cap \mathcal B_N$, we have for all $i \in [N - 1]$ that
\begin{align}
\big| d^{-1} \tilde g_i - 2 |z_i + S_i| \big| \le \frac{\delta_N}{d} \quad \text{and} \quad \big| \widetilde E_i - E_i \big| \le w_N + \frac{\delta_N}{2} & \qquad \text{if } g_i \le s, \label{e:reduction-lin-small}\\
\tilde g_i > s/2 & \qquad \text{if } g_i > s. \label{e:reduction-lin-large}
\end{align}
\end{enumerate}
\end{proposition}

Thus every gap of $\widetilde H$ of order $d$ is, up to an error $\delta_N = o(d)$, the linearized gap of a pair with $g_i \le s$. We prove \zcref{p:linearization} in \zcref{s:linearization} by applying \zcref{l:cluster} in the eigenbasis of $H$. Note that $\Prob(\mathcal B_N^c \mid H)$ is only $o(1)$ on $\mathcal A_N$, so in the proof of \zcref{t:main} we use it only to bound differences of bounded functionals, and we never multiply it by a number of pairs.

\section{Proof of the main theorem}\label{s:reduction-proof}

The heuristic behind the proof is explained in \zcref{s:ov-resampling}, and we proceed as follows. We introduce a point process $\widehat \Xi_N$ of rescaled linearized gaps and its truncation $\widehat \Xi_{N, R}$ to pairs with $g_i \le R\ell$. In \zcref{l:reduction-laplace}, which is the main step, we compute the limit of the Laplace functional of $\widehat \Xi_{N, R}$ for smooth test functions, using the Bonferroni inequalities and \zcref{p:llt}. The truncation is needed because the errors in \zcref{p:llt} are summed over $j$-tuples of pairs, and \zcref{p:pairsl} bounds the number of pairs with $g_i \le R \ell$ by $O_R(\tau)$. In \zcref{l:reduction-continuous} and \zcref{l:reduction-truncation}, we extend this limit to continuous test functions and remove the truncation using the second part of \zcref{p:llt}. In the proof of \zcref{t:main}, we pass to the gaps of $\widetilde H$ using \zcref{p:linearization} and to $H$ using \zcref{l:stationary}, and we deduce the law of the smallest gap from the void probabilities of the Poisson limit, which are the subject of \zcref{l:reduction-void}.

\paragraph{Setup.}
Let $C_c^+$ denote the set of continuous functions $f \colon [0, \infty) \times (-2, 2) \to [0, \infty)$ with compact support. For $f \in C_c^+$, set
\begin{equation}\label{e:reduction-Lfun}
\mathsf L(f) = \int \big( 1 - e^{-f} \big) \dd \nu = \int_{-2}^{2} \int_0^\infty c(E) r \big( 1 - e^{-f(r, E)} \big) \dd r \dd E.
\end{equation}
For $R > 0$, define the measure $\nu_R = (1 - e^{-R^2/8}) \nu$ and set
\begin{equation}\label{e:reduction-LR}
\mathsf L_R(f) = \int \big( 1 - e^{-f} \big) \dd \nu_R = \big( 1 - e^{-R^2/8} \big) \mathsf L(f).
\end{equation}
Let $\Xi$ be a Poisson point process on $[0, \infty) \times (-2, 2)$ with intensity $\nu$. Then $\E e^{-\Xi(f)} = e^{-\mathsf L(f)}$ for all $f \in C_c^+$; see \cite[Chapter 3]{kallenberg2017random}.

Until the proof of \zcref{t:main}, $I$ denotes an arbitrary bulk interval; in that proof, it will be chosen according to the test function. Let $I'$ be the closed interval with the same midpoint as $I$ and with length
\begin{equation}\label{e:reduction-intervals}
|I'| = |I| + \dist(I, \{-2, 2\}),
\end{equation}
where $|I|$ denotes the length of $I$. Then $I'$ is a bulk interval and $I \subset \operatorname{int}(I')$, where $\operatorname{int}$ denotes the interior. Define the sets of eigenvalue labels
\begin{equation}\label{e:reduction-C}
\mathcal C = \{ i \in [N - 1] : g_i \le s, \ E_i \in I' \}, \qquad \mathcal C_R = \{ i \in \mathcal C : g_i \le R \ell \} \quad (R > 0),
\end{equation}
and the point processes
\begin{equation}\label{e:reduction-Xihat}
\widehat \Xi_N = \sum_{i \in \mathcal C} \delta_{(2|z_i + S_i|, \, E_i)}, \qquad \widehat \Xi_{N, R} = \sum_{i \in \mathcal C_R} \delta_{(2|z_i + S_i|, \, E_i)}
\end{equation}
on $[0, \infty) \times (-2, 2)$.

Let $\mathcal A_N$ and $\mathcal B_N$ be the events of \zcref{p:linearization}, and for every integer $k \ge 1$, let $\mathcal A_{N, k}$ be the events of \zcref{p:llt}. Let $C_{I'}$ be the constant of the second part of \zcref{p:pairsl} for the interval $I'$. Define
\begin{equation}\label{e:reduction-Astar}
\mathcal A^\star_N = \mathcal A_N \cap \mathcal A_{N, 1} \cap \bigg\{ \sum_{\substack{1 \le i < j \le N \\ (\lambda_i + \lambda_j)/2 \in I'}} f_{p_*} \Big( \frac{\lambda_j - \lambda_i}{\ell} \Big) \le C_{I'} \tau \bigg\}.
\end{equation}
All objects introduced in this paragraph, except $\widehat \Xi_N$, $\widehat \Xi_{N, R}$, and $\mathcal B_N$, are measurable with respect to $H$.

\begin{lemma}\label{l:reduction-structure}
We have $\lim_{N \to \infty} \Prob(\mathcal A^\star_N) = 1$. Further, for every $R' > 0$, there exists $N_0 > 0$, depending on $R'$, such that for $N \ge N_0$ the following statements hold on $\mathcal A^\star_N$.
\begin{enumerate}[label=(\roman*)]
\item Distinct elements of $\mathcal C$ differ by at least $2$.
\item We have $\sum_{i \in \mathcal C} f_{p_*}(g_i/\ell) \le C_{I'} \tau$, and $|\mathcal C_R| \le C_{I'} (1 + R^2)^{p_*} \tau$ for all $R > 0$.
\item For every function $Q \colon [0, \infty) \times (-2, 2) \to \R$ such that $Q(y, E) = 0$ for $y \ge R'$, we have
\begin{equation}\label{e:reduction-adjacent}
\sum_{1 \le i < j \le N} Q \Big( \frac{\lambda_j - \lambda_i}{\ell}, \frac{\lambda_i + \lambda_j}{2} \Big) = \sum_{i = 1}^{N - 1} Q \Big( \frac{g_i}{\ell}, E_i \Big).
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
The first claim follows from \zcref{p:linearization}, \zcref{p:llt}, the second part of \zcref{p:pairsl}, and a union bound. Now suppose that $\mathcal A^\star_N$ holds. The first part of \zcref{p:linearization} gives
\begin{equation}\label{e:reduction-nothree}
\lambda_{i + 2} - \lambda_i > 4s \qquad \text{for all } i \in [N - 2].
\end{equation}

To prove (i), suppose that $i, i + 1 \in \mathcal C$ for some $i$. Then
\begin{equation*}
\lambda_{i + 2} - \lambda_i = g_i + g_{i + 1} \le 2s,
\end{equation*}
which contradicts \eqref{e:reduction-nothree}.

To prove (ii), note that for every $i \in \mathcal C$, the pair $(i, i + 1)$ appears in the sum in \eqref{e:reduction-Astar}, since its midpoint is $E_i \in I'$. All terms in this sum are nonnegative, so
\begin{equation*}
\sum_{i \in \mathcal C} f_{p_*} \Big( \frac{g_i}{\ell} \Big) \le C_{I'} \tau.
\end{equation*}
The bound on $|\mathcal C_R|$ follows, because $f_{p_*}(g_i/\ell) \ge (1 + R^2)^{-p_*}$ for $i \in \mathcal C_R$.

Finally, to prove (iii), observe that for $j \ge i + 2$,
\begin{equation*}
\lambda_j - \lambda_i \ge \lambda_{i + 2} - \lambda_i > 4s \ge R' \ell,
\end{equation*}
where we used \eqref{e:reduction-nothree} in the second inequality, and the last inequality holds for large $N$ by \eqref{e:reduction-ratios}. Hence the terms with $j \ge i + 2$ on the left side of \eqref{e:reduction-adjacent} vanish, and the remaining terms, with $j = i + 1$, are the terms on the right side.
\end{proof}

\paragraph{The Bonferroni inequalities.}
For a finite set $\mathcal T$, real numbers $y = (y_i)_{i \in \mathcal T}$, and an integer $j \ge 0$, let
\begin{equation}\label{e:reduction-Pj}
P_j(y) = \sum_{\substack{\mathcal T' \subset \mathcal T \\ |\mathcal T'| = j}} \prod_{i \in \mathcal T'} y_i
\end{equation}
denote the $j$-th elementary symmetric polynomial of $y$. In particular, $P_0(y) = 1$, and $P_j(y) = 0$ for $j > |\mathcal T|$.

\begin{lemma}\label{l:reduction-bonferroni}
Let $\mathcal T$ be a finite set, and let $\mathsf w = (\mathsf w_i)_{i \in \mathcal T} \in [0, 1]^{\mathcal T}$. Then for every integer $n \ge 0$,
\begin{equation}\label{e:reduction-bonferroni}
\sum_{j = 0}^{2n + 1} (-1)^j P_j(\mathsf w) \le \prod_{i \in \mathcal T} (1 - \mathsf w_i) \le \sum_{j = 0}^{2n} (-1)^j P_j(\mathsf w).
\end{equation}
\end{lemma}

\begin{proof}
Let $(\mathsf b_i)_{i \in \mathcal T}$ be independent random variables such that $\Prob(\mathsf b_i = 1) = \mathsf w_i$ and $\Prob(\mathsf b_i = 0) = 1 - \mathsf w_i$. We set
\begin{equation*}
\mathsf M = \sum_{i \in \mathcal T} \mathsf b_i.
\end{equation*}
Expanding the products and using independence, we find
\begin{equation*}
P_j(\mathsf w) = \E \sum_{\substack{\mathcal T' \subset \mathcal T \\ |\mathcal T'| = j}} \prod_{i \in \mathcal T'} \mathsf b_i = \E \binom{\mathsf M}{j}, \qquad \prod_{i \in \mathcal T} (1 - \mathsf w_i) = \E \prod_{i \in \mathcal T} (1 - \mathsf b_i) = \Prob(\mathsf M = 0).
\end{equation*}
For integers $\mathsf M_0 \ge 1$ and $n_0 \ge 0$, Pascal's rule gives the telescoping identity
\begin{equation*}
\sum_{j = 0}^{n_0} (-1)^j \binom{\mathsf M_0}{j} = (-1)^{n_0} \binom{\mathsf M_0 - 1}{n_0},
\end{equation*}
whose right side is nonnegative for even $n_0$ and nonpositive for odd $n_0$, while for $\mathsf M_0 = 0$ the left side equals $1$. Applying this with $\mathsf M_0 = \mathsf M$ and $n_0 \in \{2n, 2n + 1\}$ and taking expectations, we obtain \eqref{e:reduction-bonferroni}. (This is the classical Bonferroni inequality for the events $\{ \mathsf b_i = 1 \}$.)
\end{proof}

\paragraph{Counts of small linearized gaps.}
\begin{lemma}\label{l:reduction-counts}
Let $x_0 > 0$. There exist constants $C, c > 0$ and $N_0 > 0$, depending on $I$ and $x_0$, such that for $N \ge N_0$, on $\mathcal A^\star_N$, we have
\begin{equation}\label{e:reduction-counts}
\sum_{\substack{i \in \mathcal C \\ g_i \ge R \ell}} \Prob \big( 2|z_i + S_i| \le x_0 \bigm| H \big) \le C \sup_{y \ge R} \big( (1 + y^2)^{p_*} e^{-c y} \big) \qquad \text{for all } R \ge 0.
\end{equation}
In particular, for $N \ge N_0$, on $\mathcal A^\star_N$, we have
\begin{align}
\E \big[ \widehat \Xi_N([0, x_0] \times I') \bigm| H \big] & \le C', \label{e:reduction-counts-total}\\
\E \big[ (\widehat \Xi_N - \widehat \Xi_{N, R})([0, x_0] \times I') \bigm| H \big] & \le C \sup_{y \ge R} \big( (1 + y^2)^{p_*} e^{-cy} \big) \qquad \text{for all } R > 0, \label{e:reduction-counts-tail}
\end{align}
where $C' = C \sup_{y \ge 0} ( (1 + y^2)^{p_*} e^{-cy} )$.
\end{lemma}

\begin{proof}
Let $c > 0$ be the constant of \zcref{p:llt} for $k = 1$. Fix $\psi_0 \in C_c^\infty(\R^2)$ such that $0 \le \psi_0 \le 1$ and $\psi_0(y) = 1$ for $|y| \le x_0/2$. Suppose that $N$ is large and that $\mathcal A^\star_N$ holds. Let $i \in \mathcal C$. Applying the first part of \zcref{p:llt} with $k = 1$, the family $\Psi = \{\psi_0\}$, and $x = z_i$ on the event $\mathcal A_{N, 1} \supset \mathcal A^\star_N$, we obtain
\begin{equation*}
\Prob \big( 2|z_i + S_i| \le x_0 \bigm| H \big) \le \E \big[ \psi_0(S_i + z_i) \bigm| H \big] \le \frac{1}{\tau} \Big( \frac{1}{2\pi} \int_{\R^2} \psi_0(y) \dd y + \mathfrak e_N \Big) \le \frac{C}{\tau},
\end{equation*}
where $\mathfrak e_N = \mathfrak e_N(1, \{\psi_0\})$ is the sequence of \zcref{p:llt}, which is bounded.

For $g_i < \ell$, we have $e^{-c g_i/\ell} \ge e^{-c}$, while for $g_i \ge \ell$, the second part of \zcref{p:llt} applies. After enlarging $C$, we therefore find that
\begin{equation*}
\Prob \big( 2|z_i + S_i| \le x_0 \bigm| H \big) \le C \tau^{-1} e^{-c g_i/\ell} \qquad \text{for all } i \in \mathcal C.
\end{equation*}
The sum over the pairs with $g_i \ge R\ell$ can then be bounded using the second part of \zcref{l:reduction-structure}. For $R \ge 0$, we obtain
\begin{align*}
\sum_{\substack{i \in \mathcal C \\ g_i \ge R \ell}} \Prob \big( 2|z_i + S_i| \le x_0 \bigm| H \big) & \le \frac{C}{\tau} \sup_{y \ge R} \big( (1 + y^2)^{p_*} e^{-cy} \big) \sum_{i \in \mathcal C} f_{p_*} \Big( \frac{g_i}{\ell} \Big) \\
& \le C C_{I'} \sup_{y \ge R} \big( (1 + y^2)^{p_*} e^{-cy} \big),
\end{align*}
which is \eqref{e:reduction-counts}.

Observe that every point of $\widehat \Xi_N$ has its second coordinate in $I'$. Using the fact that $\mathcal C$ is measurable with respect to $H$, we therefore find
\begin{align*}
\E \big[ \widehat \Xi_N([0, x_0] \times I') \bigm| H \big] & = \sum_{i \in \mathcal C} \Prob \big( 2|z_i + S_i| \le x_0 \bigm| H \big), \\
\E \big[ (\widehat \Xi_N - \widehat \Xi_{N, R})([0, x_0] \times I') \bigm| H \big] & = \sum_{\substack{i \in \mathcal C \\ g_i > R \ell}} \Prob \big( 2|z_i + S_i| \le x_0 \bigm| H \big).
\end{align*}
Hence \eqref{e:reduction-counts} implies \eqref{e:reduction-counts-total} and \eqref{e:reduction-counts-tail}.
\end{proof}

\paragraph{Smooth test functions.}
The local limit theorem applies only to smooth test functions, but the function $y \mapsto f(2|y|, E)$ on $\R^2$ is in general not smooth at $y = 0$, even for smooth $f$. For this reason, we first consider test functions of the form $f(r, E) = h(r^2, E)$ with $h$ smooth.

\begin{lemma}\label{l:reduction-laplace}
Let $R > 0$, and let $h \in C_c^\infty(\R^2)$ be a nonnegative function with $\supp h \subset \R \times \operatorname{int}(I')$. Define $f(r, E) = h(r^2, E)$ for $(r, E) \in [0, \infty) \times (-2, 2)$. Then $f \in C_c^+$, and
\begin{equation}\label{e:reduction-laplace}
\E \big[ e^{-\widehat \Xi_{N, R}(f)} \bigm| H \big] \longrightarrow e^{-\mathsf L_R(f)}
\end{equation}
in probability.
\end{lemma}

\begin{proof}
Fix $u_h > 0$ and a compact interval $I_h \subset \operatorname{int}(I')$ such that $\supp h \subset [-u_h, u_h] \times I_h$. Then $f$ is continuous and nonnegative, and $\supp f \subset [0, u_h^{1/2}] \times I_h$, so $f \in C_c^+$.

We begin by introducing planar test functions. For $E \in (-2, 2)$ and $y \in \R^2$, define
\begin{equation}\label{e:reduction-psi}
\psi_E(y) = 1 - \exp \big( - h(4|y|^2, E) \big), \qquad A_h(E) = \int_{\R^2} \psi_E(y) \dd y.
\end{equation}
Since $f(2|y|, E) = h(4|y|^2, E)$, we have
\begin{equation}\label{e:reduction-psi-f}
1 - e^{-f(2|y|, E)} = \psi_E(y) \qquad \text{for all } y \in \R^2 \text{ and } E \in (-2, 2).
\end{equation}
We record three properties of $\psi_E$ and $A_h$. First, $0 \le \psi_E \le 1$, and $\psi_E(y) = 0$ if $|y| > u_h^{1/2}/2$ or $E \notin I_h$. Second, $\psi_E \in C_c^\infty(\R^2)$, and the chain rule shows that every partial derivative of $\psi_E$ is bounded uniformly in $E \in (-2, 2)$. Third, $A_h$ is continuous by dominated convergence, it vanishes outside $I_h$, and
\begin{equation}\label{e:reduction-polar}
A_h(E) = \frac{\pi}{2} \int_0^\infty r \big( 1 - e^{-f(r, E)} \big) \dd r,
\end{equation}
which follows from \eqref{e:reduction-psi-f} by using polar coordinates in $\R^2$ with radius $|y| = r/2$.

We now apply the Bonferroni inequalities. For $i \in \mathcal C_R$, define
\begin{equation}\label{e:reduction-wi}
\mathsf w_i = \psi_{E_i}(z_i + S_i) \in [0, 1],
\end{equation}
and let $\mathsf w = (\mathsf w_i)_{i \in \mathcal C_R}$. Using \eqref{e:reduction-psi-f}, we find
\begin{equation*}
e^{-\widehat \Xi_{N, R}(f)} = \prod_{i \in \mathcal C_R} e^{-f(2|z_i + S_i|, E_i)} = \prod_{i \in \mathcal C_R} (1 - \mathsf w_i).
\end{equation*}
Since $\mathcal C_R$ is measurable with respect to $H$, \zcref{l:reduction-bonferroni} with $\mathcal T = \mathcal C_R$ implies that for every integer $n \ge 0$,
\begin{equation}\label{e:reduction-bonf-cond}
\sum_{j = 0}^{2n + 1} (-1)^j \, \E \big[ P_j(\mathsf w) \bigm| H \big] \le \E \big[ e^{-\widehat \Xi_{N, R}(f)} \bigm| H \big] \le \sum_{j = 0}^{2n} (-1)^j \, \E \big[ P_j(\mathsf w) \bigm| H \big].
\end{equation}
We claim that for every integer $j \ge 1$,
\begin{equation}\label{e:reduction-Pj-limit}
\E \big[ P_j(\mathsf w) \bigm| H \big] \longrightarrow \frac{\mathsf L_R(f)^j}{j!} \qquad \text{in probability}.
\end{equation}

We first show that \eqref{e:reduction-Pj-limit} implies the lemma. Let $\varpi > 0$. The alternating series $\sum_{j \ge 0} (-1)^j \mathsf L_R(f)^j/j!$ converges to $e^{-\mathsf L_R(f)}$, so there exists $n \ge 0$ such that
\begin{equation*}
\bigg| \sum_{j = 0}^{2n} \frac{(-1)^j \mathsf L_R(f)^j}{j!} - e^{-\mathsf L_R(f)} \bigg| \le \varpi, \qquad \bigg| \sum_{j = 0}^{2n + 1} \frac{(-1)^j \mathsf L_R(f)^j}{j!} - e^{-\mathsf L_R(f)} \bigg| \le \varpi.
\end{equation*}
Then \eqref{e:reduction-Pj-limit} implies that the event
\begin{equation*}
\Big| \E \big[ P_j(\mathsf w) \bigm| H \big] - \frac{\mathsf L_R(f)^j}{j!} \Big| \le \frac{\varpi}{2n + 2} \qquad \text{for all } j \in [2n + 1]
\end{equation*}
has probability tending to one. On this event, \eqref{e:reduction-bonf-cond} and the choice of $n$ give
\begin{equation*}
\Big| \E \big[ e^{-\widehat \Xi_{N, R}(f)} \bigm| H \big] - e^{-\mathsf L_R(f)} \Big| \le 2 \varpi.
\end{equation*}
The number $\varpi > 0$ was arbitrary, so \eqref{e:reduction-laplace} follows.

It remains to prove \eqref{e:reduction-Pj-limit}. Fix $j \ge 1$. For $i \in \mathcal C_R$, define
\begin{equation}\label{e:reduction-wbar}
\bar{\mathsf w}_i = \frac{1}{2\pi\tau} \exp \Big( - \frac{g_i^2}{8 \ell^2} \Big) A_h(E_i),
\end{equation}
and let $\bar{\mathsf w} = (\bar{\mathsf w}_i)_{i \in \mathcal C_R}$. We compare $\E [ P_j(\mathsf w) \mid H ]$ with $P_j(\bar{\mathsf w})$, then $P_j(\bar{\mathsf w})$ with $(\sum_{i \in \mathcal C_R} \bar{\mathsf w}_i)^j/j!$, and finally we show that $\sum_{i \in \mathcal C_R} \bar{\mathsf w}_i$ converges to $\mathsf L_R(f)$ in probability.

For the first comparison, let $\Psi_j$ be the family of all functions of the form $\psi_{E'_1} \otimes \dots \otimes \psi_{E'_j}$ with $E'_1, \dots, E'_j \in (-2, 2)$, where
\begin{equation*}
\big( \psi_{E'_1} \otimes \dots \otimes \psi_{E'_j} \big) (y_1, \dots, y_j) = \psi_{E'_1}(y_1) \cdots \psi_{E'_j}(y_j), \qquad y_1, \dots, y_j \in \R^2.
\end{equation*}
By the first two properties of $\psi_E$, the members of $\Psi_j$ are supported in a product of $j$ disks of radius $u_h^{1/2}/2$, and each of their partial derivatives is bounded uniformly over $\Psi_j$. Thus $\Psi_j$ satisfies the hypotheses of the first part of \zcref{p:llt} with $k = j$. Let $N$ be large enough that \zcref{l:reduction-structure} applies with $R' = R + 1$, and suppose that $\mathcal A^\star_N \cap \mathcal A_{N, j}$ holds.

Let $\mathcal T' \subset \mathcal C_R$ with $|\mathcal T'| = j$. We list the elements of $\mathcal T'$ in increasing order as $\mathbf i = (i_1, \dots, i_j)$. The first part of \zcref{l:reduction-structure} shows that consecutive entries of $\mathbf i$ differ by at least $2$. We set
\begin{equation*}
\psi = \psi_{E_{i_1}} \otimes \dots \otimes \psi_{E_{i_j}} \in \Psi_j.
\end{equation*}
Then we have
\begin{equation*}
\prod_{i \in \mathcal T'} \mathsf w_i = \psi(S_{\mathbf i} + z_{\mathbf i}), \qquad \int_{\R^{2j}} \psi(y) \dd y = \prod_{i \in \mathcal T'} A_h(E_i), \qquad \frac{|z_{\mathbf i}|^2}{2\tau} = \sum_{i \in \mathcal T'} \frac{g_i^2}{8 d^2 \tau} = \sum_{i \in \mathcal T'} \frac{g_i^2}{8 \ell^2},
\end{equation*}
where we used \eqref{e:reduction-zi} and $\ell^2 = d^2 \tau$ in the last two equalities. Therefore the first part of \zcref{p:llt}, applied with $k = j$ and $x = z_{\mathbf i}$, gives
\begin{equation*}
\Big| \E \Big[ \prod_{i \in \mathcal T'} \mathsf w_i \Bigm| H \Big] - \prod_{i \in \mathcal T'} \bar{\mathsf w}_i \Big| \le \mathfrak e_N \tau^{-j},
\end{equation*}
where $\mathfrak e_N = \mathfrak e_N(j, \Psi_j)$ is the sequence of \zcref{p:llt}, which does not depend on $\mathcal T'$.

We now sum over the subsets $\mathcal T' \subset \mathcal C_R$ with $|\mathcal T'| = j$. By the second part of \zcref{l:reduction-structure}, their number is bounded by
\begin{equation*}
\frac{|\mathcal C_R|^j}{j!} \le \frac{(K_R \tau)^j}{j!}, \qquad \text{where} \qquad K_R = C_{I'} (1 + R^2)^{p_*}.
\end{equation*}
Hence, on $\mathcal A^\star_N \cap \mathcal A_{N, j}$,
\begin{equation}\label{e:reduction-first-comp}
\big| \E \big[ P_j(\mathsf w) \bigm| H \big] - P_j(\bar{\mathsf w}) \big| \le \frac{(K_R \tau)^j}{j!} \, \mathfrak e_N \tau^{-j} = \frac{K_R^j \mathfrak e_N}{j!}.
\end{equation}

Next, we compare $P_j(\bar{\mathsf w})$ with $(\sum_{i \in \mathcal C_R} \bar{\mathsf w}_i)^j/j!$. We may assume that $j \ge 2$, since the two quantities coincide for $j = 1$. We expand $(\sum_{i \in \mathcal C_R} \bar{\mathsf w}_i)^j$ as a sum over ordered $j$-tuples of elements of $\mathcal C_R$. The tuples with distinct entries contribute $j! P_j(\bar{\mathsf w})$. Every other tuple has equal entries in at least two positions, and for each of the $\binom{j}{2}$ choices of two positions, the tuples with equal entries in these positions contribute
\begin{equation*}
\Big( \sum_{i \in \mathcal C_R} \bar{\mathsf w}_i^2 \Big) \Big( \sum_{i \in \mathcal C_R} \bar{\mathsf w}_i \Big)^{j - 2} \le \Big( \max_{i \in \mathcal C_R} \bar{\mathsf w}_i \Big) \Big( \sum_{i \in \mathcal C_R} \bar{\mathsf w}_i \Big)^{j - 1}.
\end{equation*}

Let $\|A_h\|_\infty = \sup_E A_h(E)$, which is finite by the first property of $\psi_E$. Then
\begin{equation*}
\max_{i \in \mathcal C_R} \bar{\mathsf w}_i \le \frac{\|A_h\|_\infty}{2\pi\tau}, \qquad \sum_{i \in \mathcal C_R} \bar{\mathsf w}_i \le \frac{K_R \|A_h\|_\infty}{2\pi},
\end{equation*}
where the second bound follows from the second part of \zcref{l:reduction-structure}. We conclude that
\begin{equation}\label{e:reduction-second-comp}
0 \le \Big( \sum_{i \in \mathcal C_R} \bar{\mathsf w}_i \Big)^j - j! \, P_j(\bar{\mathsf w}) \le \binom{j}{2} \frac{\|A_h\|_\infty}{2\pi\tau} \Big( \frac{K_R \|A_h\|_\infty}{2\pi} \Big)^{j - 1} = O(\tau^{-1}).
\end{equation}

Finally, we show that
\begin{equation}\label{e:reduction-third-comp}
\sum_{i \in \mathcal C_R} \bar{\mathsf w}_i \longrightarrow \mathsf L_R(f) \qquad \text{in probability}.
\end{equation}
For $\varpi \in (0, 1)$, define functions on $[0, \infty) \times (-2, 2)$ by
\begin{align*}
Q^+_\varpi(y, E) & = \frac{1}{2\pi} \min \Big\{ 1, \frac{(R + \varpi - y)_+}{\varpi} \Big\} e^{-y^2/8} A_h(E), \\
Q^-_\varpi(y, E) & = \frac{1}{2\pi} \min \Big\{ 1, \frac{(R - y)_+}{\varpi} \Big\} e^{-y^2/8} A_h(E).
\end{align*}
They are continuous with support in $[0, R + 1] \times I_h$, and they satisfy
\begin{equation*}
Q^-_\varpi(y, E) \le \frac{1}{2\pi} \1\{ y \le R \} e^{-y^2/8} A_h(E) \le Q^+_\varpi(y, E).
\end{equation*}

Note that $R \ell \le s$ for large $N$ by \eqref{e:reduction-ratios}. Since $A_h(E_i) = 0$ unless $E_i \in I_h \subset I'$, we then have
\begin{equation*}
\sum_{i \in \mathcal C_R} \bar{\mathsf w}_i = \frac{1}{\tau} \sum_{i = 1}^{N - 1} \frac{1}{2\pi} \1\{ g_i \le R \ell \} \exp \Big( - \frac{g_i^2}{8 \ell^2} \Big) A_h(E_i).
\end{equation*}
Using \eqref{e:reduction-adjacent} with $R' = R + 1$, we deduce that on $\mathcal A^\star_N$, for large $N$,
\begin{equation*}
\frac{1}{\tau} \sum_{1 \le i < j' \le N} Q^-_\varpi \Big( \frac{\lambda_{j'} - \lambda_i}{\ell}, \frac{\lambda_i + \lambda_{j'}}{2} \Big) \le \sum_{i \in \mathcal C_R} \bar{\mathsf w}_i \le \frac{1}{\tau} \sum_{1 \le i < j' \le N} Q^+_\varpi \Big( \frac{\lambda_{j'} - \lambda_i}{\ell}, \frac{\lambda_i + \lambda_{j'}}{2} \Big).
\end{equation*}
The first part of \zcref{p:pairsl} implies that the left and right sides converge in probability to
\begin{equation*}
\int_{-2}^{2} \int_0^\infty c(E) y Q^-_\varpi(y, E) \dd y \dd E \qquad \text{and} \qquad \int_{-2}^{2} \int_0^\infty c(E) y Q^+_\varpi(y, E) \dd y \dd E,
\end{equation*}
respectively. As $\varpi \to 0$, dominated convergence shows that these integrals converge to
\begin{equation*}
\frac{1}{2\pi} \int_0^R y e^{-y^2/8} \dd y \int_{-2}^{2} c(E) A_h(E) \dd E = \frac{1}{2\pi} \cdot 4 \big( 1 - e^{-R^2/8} \big) \cdot \frac{\pi}{2} \mathsf L(f) = \mathsf L_R(f),
\end{equation*}
where in the first equality we used \eqref{e:reduction-polar} and \eqref{e:reduction-Lfun}. Now let $\varpi' > 0$, and choose $\varpi \in (0, 1)$ such that each of these integrals differs from $\mathsf L_R(f)$ by at most $\varpi'$. Since $\Prob(\mathcal A^\star_N) \to 1$, the two-sided bound on $\sum_{i \in \mathcal C_R} \bar{\mathsf w}_i$ above and the convergence of its outer sides imply that
\begin{equation*}
\Big| \sum_{i \in \mathcal C_R} \bar{\mathsf w}_i - \mathsf L_R(f) \Big| \le 2\varpi'
\end{equation*}
with probability tending to one. The number $\varpi' > 0$ was arbitrary, so \eqref{e:reduction-third-comp} follows.

We now combine the three steps. The convergence \eqref{e:reduction-third-comp} and the continuous mapping theorem give
\begin{equation*}
\frac{1}{j!} \Big( \sum_{i \in \mathcal C_R} \bar{\mathsf w}_i \Big)^j \longrightarrow \frac{\mathsf L_R(f)^j}{j!} \qquad \text{in probability}.
\end{equation*}
The estimates \eqref{e:reduction-first-comp} and \eqref{e:reduction-second-comp}, whose right sides tend to zero, hold on $\mathcal A^\star_N \cap \mathcal A_{N, j}$, and $\Prob(\mathcal A^\star_N \cap \mathcal A_{N, j}) \to 1$ by \zcref{l:reduction-structure} and \zcref{p:llt}. Combining these facts with the last display, we obtain \eqref{e:reduction-Pj-limit}. This completes the proof.
\end{proof}

\paragraph{Continuous test functions and the truncation.}
In the next two lemmas and in the proof of \zcref{t:main}, we use the elementary inequality
\begin{equation}\label{e:reduction-exp}
\big| e^{-y_1} - e^{-y_2} \big| \le \min \{1, |y_1 - y_2|\}, \qquad y_1, y_2 \ge 0.
\end{equation}
We first extend \zcref{l:reduction-laplace} to continuous test functions.

\begin{lemma}\label{l:reduction-continuous}
Let $R > 0$ and $x_0 > 0$, and let $f \in C_c^+$ vanish outside $[0, x_0] \times I$. Then
\begin{equation}\label{e:reduction-trunc}
\lim_{N \to \infty} \E e^{-\widehat \Xi_{N, R}(f)} = e^{-\mathsf L_R(f)}.
\end{equation}
\end{lemma}

\begin{proof}
To apply \zcref{l:reduction-laplace}, we approximate $f$ by test functions of the form $h(r^2, E)$. Define $\bar h \colon \R^2 \to [0, \infty)$ by $\bar h(u, E) = 0$ for $E \notin (-2, 2)$ and
\begin{equation*}
\bar h(u, E) = \min\{ 1, (1 + u)_+ \} \, f \big( (\max\{u, 0\})^{1/2}, E \big) \qquad \text{for } u \in \R, \ E \in (-2, 2).
\end{equation*}
Then $\bar h$ is continuous, and
\begin{equation*}
\supp \bar h \subset [-1, x_0^2] \times I, \qquad \bar h(r^2, E) = f(r, E) \quad \text{for } r \ge 0, \ E \in (-2, 2).
\end{equation*}

For $\varpi > 0$, let $h_\varpi$ be the convolution of $\bar h$ with a smooth nonnegative function of integral one supported in the disk of radius $\varpi$. Then $h_\varpi \in C_c^\infty(\R^2)$ is nonnegative, and $\|h_\varpi - \bar h\|_\infty \to 0$ as $\varpi \to 0$, since $\bar h$ is uniformly continuous. For $\varpi < 1$ smaller than the distance from $I$ to $\R \setminus I'$, we have $\supp h_\varpi \subset [-2, x_0^2 + 1] \times \operatorname{int}(I')$. Set
\begin{equation*}
f_\varpi(r, E) = h_\varpi(r^2, E), \qquad (r, E) \in [0, \infty) \times (-2, 2).
\end{equation*}
Then $f$ and $f_\varpi$ vanish outside $[0, x_0 + 1] \times I'$, and
\begin{equation*}
\sup_{(r, E) \in [0, \infty) \times (-2, 2)} |f_\varpi(r, E) - f(r, E)| \le \|h_\varpi - \bar h\|_\infty.
\end{equation*}

We now compare the Laplace functionals of $\widehat \Xi_{N, R}$ at $f$ and at $f_\varpi$. Using \eqref{e:reduction-exp}, we have
\begin{align*}
\big| \E e^{-\widehat \Xi_{N, R}(f)} - \E e^{-\widehat \Xi_{N, R}(f_\varpi)} \big| & \le \E \min \big\{ 1, \|h_\varpi - \bar h\|_\infty \, \widehat \Xi_{N, R}([0, x_0 + 1] \times I') \big\} \\
& \le \Prob \big( (\mathcal A^\star_N)^c \big) + \|h_\varpi - \bar h\|_\infty \, \E \Big[ \1_{\mathcal A^\star_N} \E \big[ \widehat \Xi_N([0, x_0 + 1] \times I') \bigm| H \big] \Big] \\
& \le o(1) + C \|h_\varpi - \bar h\|_\infty.
\end{align*}
In the last inequality, we used \eqref{e:reduction-counts-total} with $x_0 + 1$ in place of $x_0$, and $\widehat \Xi_{N, R} \le \widehat \Xi_N$. Next, \zcref{l:reduction-laplace} and the bounded convergence theorem give
\begin{equation*}
\lim_{N \to \infty} \E e^{-\widehat \Xi_{N, R}(f_\varpi)} = e^{-\mathsf L_R(f_\varpi)}.
\end{equation*}
Moreover,
\begin{equation*}
|\mathsf L_R(f_\varpi) - \mathsf L_R(f)| \le \int |f_\varpi - f| \dd \nu_R \le \|h_\varpi - \bar h\|_\infty \, \nu([0, x_0 + 1] \times I').
\end{equation*}
Combining these estimates, we obtain
\begin{equation*}
\limsup_{N \to \infty} \big| \E e^{-\widehat \Xi_{N, R}(f)} - e^{-\mathsf L_R(f)} \big| \le \big( C + \nu([0, x_0 + 1] \times I') \big) \|h_\varpi - \bar h\|_\infty,
\end{equation*}
and letting $\varpi \to 0$ proves \eqref{e:reduction-trunc}.
\end{proof}

Next, we remove the truncation, using the bound \eqref{e:reduction-counts-tail}, which relies on the second part of \zcref{p:llt}.

\begin{lemma}\label{l:reduction-truncation}
Let $x_0 > 0$, and let $f \in C_c^+$ vanish outside $[0, x_0] \times I$. Then
\begin{equation}\label{e:reduction-lin-limit}
\lim_{N \to \infty} \E e^{-\widehat \Xi_N(f)} = e^{-\mathsf L(f)}.
\end{equation}
\end{lemma}

\begin{proof}
Let $C, c > 0$ be the constants of \zcref{l:reduction-counts} for the window $x_0$, and let $R > 0$. Then
\begin{align*}
\big| \E e^{-\widehat \Xi_N(f)} - \E e^{-\widehat \Xi_{N, R}(f)} \big| & \le \E \min \big\{ 1, \|f\|_\infty (\widehat \Xi_N - \widehat \Xi_{N, R})([0, x_0] \times I') \big\} \\
& \le \Prob \big( (\mathcal A^\star_N)^c \big) + C \|f\|_\infty \sup_{y \ge R} \big( (1 + y^2)^{p_*} e^{-cy} \big),
\end{align*}
where we used \eqref{e:reduction-exp} and the fact that $f$ vanishes outside $[0, x_0] \times I$ in the first inequality, and \eqref{e:reduction-counts-tail} in the second. Further, \eqref{e:reduction-LR} gives
\begin{equation*}
\big| e^{-\mathsf L_R(f)} - e^{-\mathsf L(f)} \big| \le \mathsf L(f) - \mathsf L_R(f) = e^{-R^2/8} \mathsf L(f).
\end{equation*}
Together with \zcref{l:reduction-continuous}, these two bounds yield
\begin{equation*}
\limsup_{N \to \infty} \big| \E e^{-\widehat \Xi_N(f)} - e^{-\mathsf L(f)} \big| \le C \|f\|_\infty \sup_{y \ge R} \big( (1 + y^2)^{p_*} e^{-cy} \big) + e^{-R^2/8} \mathsf L(f).
\end{equation*}
Letting $R \to \infty$, we obtain \eqref{e:reduction-lin-limit}.
\end{proof}

\paragraph{Void probabilities.}
We will deduce \eqref{e:minimum} from the convergence of $\Xi_N$ by means of the following standard consequence of vague convergence.

\begin{lemma}\label{l:reduction-void}
Let $\mathfrak S \subset [0, \infty) \times (-2, 2)$ be a Borel set whose closure in $[0, \infty) \times (-2, 2)$ is compact and whose boundary $\partial \mathfrak S$ in this space satisfies $\nu(\partial \mathfrak S) = 0$. Let $(\mu_N)_{N \ge 2}$ be random locally finite counting measures on $[0, \infty) \times (-2, 2)$ that converge in distribution to $\Xi$ with respect to the vague topology. Then
\begin{equation*}
\lim_{N \to \infty} \Prob \big( \mu_N(\mathfrak S) = 0 \big) = e^{-\nu(\mathfrak S)}.
\end{equation*}
\end{lemma}

\begin{proof}
As $\Xi$ has intensity $\nu$, the assumption $\nu(\partial \mathfrak S) = 0$ implies that $\Xi(\partial \mathfrak S) = 0$ almost surely. Since $\mathfrak S$ is relatively compact, the map $\mu \mapsto \mu(\mathfrak S)$ is continuous, for the vague topology, at every locally finite measure $\mu$ with $\mu(\partial \mathfrak S) = 0$ (see \cite[Chapter 4]{kallenberg2017random}). The continuous mapping theorem therefore shows that $\mu_N(\mathfrak S)$ converges in distribution to $\Xi(\mathfrak S)$, which is a Poisson random variable with mean $\nu(\mathfrak S)$. These random variables take integer values, so
\begin{equation*}
\lim_{N \to \infty} \Prob \big( \mu_N(\mathfrak S) = 0 \big) = \Prob \big( \Xi(\mathfrak S) = 0 \big) = e^{-\nu(\mathfrak S)}.
\end{equation*}
\end{proof}

\begin{proof}[Proof of \zcref{t:main}]
Let $f \in C_c^+$. Since the support of $f$ is compact, $f$ vanishes outside $[0, x_0] \times I$ for some $x_0 > 0$ and some bulk interval $I$. Throughout the proof, $I'$, $\mathcal C$, $\mathcal C_R$, $\widehat \Xi_N$, $\widehat \Xi_{N, R}$, and $\mathcal A^\star_N$ are defined as in the setup above, with this choice of $I$. We extend $f$ by zero to $[0, \infty) \times \R$, and for $\varpi > 0$ we define
\begin{equation}\label{e:reduction-osc}
\operatorname{osc}_f(\varpi) = \sup \big\{ |f(r, E) - f(r', E')| : (r, E), (r', E') \in [0, \infty) \times \R, \ |(r, E) - (r', E')| \le \varpi \big\}.
\end{equation}
We will show that
\begin{equation}\label{e:reduction-goal}
\lim_{N \to \infty} \E e^{-\Xi_N(f)} = e^{-\mathsf L(f)}.
\end{equation}
By \zcref{l:reduction-truncation}, the limit \eqref{e:reduction-lin-limit} holds for the process $\widehat \Xi_N$ of linearized gaps. It remains to pass from $\widehat \Xi_N$ to $\Xi_N$.

\emph{Passage to the resampled matrix and to $H$.} Define
\begin{equation*}
\widetilde \Xi_N = \sum_{i = 1}^{N - 1} \delta_{(d^{-1} \tilde g_i, \, \tilde\lambda_i)},
\end{equation*}
regarded as a measure on $[0, \infty) \times (-2, 2)$ in the same way as $\Xi_N$, so that
\begin{equation*}
\widetilde \Xi_N(f) = \sum_{i} f(d^{-1} \tilde g_i, \tilde\lambda_i).
\end{equation*}
The eigenvalues are continuous functions of the matrix entries, so $\Xi_N(f)$ is a measurable function of $H$, and $\widetilde \Xi_N(f)$ is the same function of $\widetilde H$. Hence \zcref{l:stationary} gives
\begin{equation}\label{e:reduction-stationary-use}
\E e^{-\Xi_N(f)} = \E e^{-\widetilde \Xi_N(f)}.
\end{equation}

We now compare $\widetilde \Xi_N(f)$ with $\widehat \Xi_N(f)$. Note that $\widehat \Xi_N$ records the midpoints $E_i$ in the spectrum of $H$, whereas $\widetilde \Xi_N$ records the left endpoints $\tilde\lambda_i$ in the spectrum of $\widetilde H$. By \eqref{e:reduction-lin-rates}, we have $\delta_N \le d$ for large $N$.

Suppose that $N$ is large and that $\mathcal A^\star_N \cap \mathcal B_N$ holds, so that the conclusions of the second part of \zcref{p:linearization} hold. Let $i \in [N - 1]$ be such that $f(d^{-1} \tilde g_i, \tilde\lambda_i) \neq 0$. Then $\tilde g_i \le x_0 d$ and $\tilde\lambda_i \in I$. We have $x_0 d < s/2$ for large $N$ by \eqref{e:reduction-ratios}, so \eqref{e:reduction-lin-large} implies that $g_i \le s$. Then \eqref{e:reduction-lin-small} gives
\begin{equation*}
|E_i - \tilde\lambda_i| \le |E_i - \widetilde E_i| + \frac{\tilde g_i}{2} \le w_N + \frac{\delta_N}{2} + \frac{x_0 d}{2}, \qquad 2|z_i + S_i| \le d^{-1} \tilde g_i + \frac{\delta_N}{d} \le x_0 + 1.
\end{equation*}
In particular, $E_i \in I'$ for large $N$, so $i \in \mathcal C$.

We have shown that every nonzero term of $\widetilde \Xi_N(f)$ comes from some $i \in \mathcal C$ with $2|z_i + S_i| \le x_0 + 1$. The same holds for the nonzero terms of $\widehat \Xi_N(f)$, so
\begin{equation*}
\widetilde \Xi_N(f) - \widehat \Xi_N(f) = \sum_{\substack{i \in \mathcal C \\ 2|z_i + S_i| \le x_0 + 1}} \Big( f \big( d^{-1} \tilde g_i, \tilde\lambda_i \big) - f \big( 2|z_i + S_i|, E_i \big) \Big).
\end{equation*}
For every $i$ in this sum, \eqref{e:reduction-lin-small} gives
\begin{align*}
\tilde g_i & \le d \Big( 2|z_i + S_i| + \frac{\delta_N}{d} \Big) \le (x_0 + 2) d, \\
\big| d^{-1} \tilde g_i - 2|z_i + S_i| \big| + |\tilde\lambda_i - E_i| & \le \frac{\delta_N}{d} + \frac{\tilde g_i}{2} + |\widetilde E_i - E_i| \le \frac{\delta_N}{d} + \frac{(x_0 + 2) d}{2} + w_N + \frac{\delta_N}{2}.
\end{align*}
We denote the right side of the second line by $\varsigma_N$. It is deterministic, and it tends to zero by \eqref{e:reduction-lin-rates}. Using the definition \eqref{e:reduction-osc}, we conclude that on $\mathcal A^\star_N \cap \mathcal B_N$,
\begin{equation*}
\big| \widetilde \Xi_N(f) - \widehat \Xi_N(f) \big| \le \operatorname{osc}_f(\varsigma_N) \, \widehat \Xi_N([0, x_0 + 1] \times I'),
\end{equation*}
and $\operatorname{osc}_f(\varsigma_N) = o(1)$ because $f$ is uniformly continuous.

We can now bound the difference of the Laplace functionals. Using the first part of \zcref{p:linearization} and \eqref{e:reduction-counts-total} with $x_0 + 1$ in place of $x_0$, we find
\begin{align*}
\big| \E e^{-\widetilde \Xi_N(f)} - \E e^{-\widehat \Xi_N(f)} \big| & \le \Prob \big( (\mathcal A^\star_N)^c \big) + \E \big[ \1_{\mathcal A^\star_N} \Prob( \mathcal B_N^c \mid H ) \big] \\
& \qquad + \operatorname{osc}_f(\varsigma_N) \, \E \Big[ \1_{\mathcal A^\star_N} \E \big[ \widehat \Xi_N([0, x_0 + 1] \times I') \bigm| H \big] \Big] \\
& \le o(1) + o(1) + C \operatorname{osc}_f(\varsigma_N) = o(1).
\end{align*}
Here the conditional probability of $\mathcal B_N^c$ enters only through the bound $|e^{-\widetilde \Xi_N(f)} - e^{-\widehat \Xi_N(f)}| \le 1$. Combining the last estimate with \eqref{e:reduction-lin-limit} and \eqref{e:reduction-stationary-use}, we obtain \eqref{e:reduction-goal}.

\emph{Vague convergence.} The test function $f \in C_c^+$ was arbitrary, so \eqref{e:reduction-goal} holds for all $f \in C_c^+$, and $e^{-\mathsf L(f)} = \E e^{-\Xi(f)}$. The space $[0, \infty) \times (-2, 2)$ is locally compact, separable, and metrizable, and the measures $\Xi_N$ and $\nu$ are locally finite on it. By the Laplace functional criterion for convergence in distribution of random measures (see \cite[Chapter 4]{kallenberg2017random}), the convergence of $\E e^{-\Xi_N(f)}$ to $\E e^{-\Xi(f)}$ for all $f \in C_c^+$ implies that $\Xi_N$ converges in distribution to $\Xi$ for the vague topology. This is the first claim of \zcref{t:main}.

\emph{The smallest gap.} In this step, $I$ denotes a nonempty interval with closure $\bar I \subset (-2, 2)$, as in \zcref{t:main}, and we fix $x \ge 0$. Set $\mathfrak S = [0, x] \times I$. The definition of $\Xi_N$ gives
\begin{equation*}
\Big\{ N^{3/2} \min_{\lambda_i \in I} g_i > x \Big\} = \big\{ \Xi_N(\mathfrak S) = 0 \big\},
\end{equation*}
where the minimum is over $i \in [N - 1]$ such that $\lambda_i \in I$, and an empty minimum is $+\infty$.

The closure of $\mathfrak S$ in $[0, \infty) \times (-2, 2)$ is the compact set $[0, x] \times \bar I$, and the interior of $\mathfrak S$ contains $[0, x) \times \operatorname{int}(I)$. The difference of these two sets is contained in $(\{x\} \times \bar I) \cup ([0, x] \times (\bar I \setminus \operatorname{int}(I)))$, which has Lebesgue measure zero, since $\bar I \setminus \operatorname{int}(I)$ consists of at most two points. The measure $\nu$ has a density with respect to the Lebesgue measure, so $\nu(\partial \mathfrak S) = 0$ and
\begin{equation}\label{e:reduction-boundary}
\nu(\mathfrak S) = \int_I \int_0^x c(E) y \dd y \dd E = \frac{\pi^2 x^2}{12} \int_I \rhosc(E)^3 \dd E,
\end{equation}
where we used the definition of $c(E)$ in the last equality. By the first claim of \zcref{t:main}, we may apply \zcref{l:reduction-void} with $\mu_N = \Xi_N$, which gives
\begin{equation*}
\lim_{N \to \infty} \Prob \big( \Xi_N(\mathfrak S) = 0 \big) = e^{-\nu(\mathfrak S)}.
\end{equation*}
Together with \eqref{e:reduction-boundary}, this proves the second claim of \zcref{t:main}.
\end{proof}