\chapter{Factorial moments and the proof of \texorpdfstring{\zcref{t:main}}{Theorem 1.1}}\label{s:factorial}

In this section, we complete the proofs of \zcref{t:main} and \zcref{c:ordered}. We follow the approach of Feng and Wei \cite{FengWei2025}, who reduced the Poisson convergence to the convergence of the moments in \eqref{e:positivemoments} below; we state and prove this reduction as \zcref{l:poisson} in \zcref{s:poisson}. The new input is the use of the estimates for several intervals from \zcref{s:severalintervals} in place of the negative association of determinantal processes used in \cite{FengWei2025}.

\section{The positive-part moments}

Fix $k \ge 1$ and $\mathbf x = (x_1, \dots, x_k) \in \R^k$, and set $\ell_j = \ell_N(x_j)$, with $\ell_N(x)$ as in \eqref{e:sNdef}. We work with $N$ sufficiently large that $0 < \ell_j < |I|$ for every $j$. We define
\begin{equation}\label{e:positivemoments}
M_{N, k}(\mathbf x) = \E \sum_{\substack{i_1, \dots, i_k \in \mathcal G_N(I) \\ \text{distinct}}} \prod_{j = 1}^k \big( \tau_{i_j, N} - x_j \big)_+ ,
\end{equation}
where we write $(t)_+ = \max(t, 0)$. For a realization of the eigenvalues, we let $\mathcal Y_N$ denote the set of $\mathbf y = (y_1, \dots, y_k) \in \prod_{j = 1}^k [y_-, y_+ - \ell_j]$ for which there exist distinct $i_1, \dots, i_k \in \mathcal G_N(I)$ such that
\begin{equation}\label{e:inside}
[y_j, y_j + \ell_j] \subset (\lambda_{i_j}, \lambda_{i_j + 1}) , \qquad 1 \le j \le k .
\end{equation}

\begin{lemma}\label{l:volume}
We have
\begin{equation}\label{e:volume}
M_{N, k}(\mathbf x) = \chi_N^k \int_{\prod_j [y_-, y_+ - \ell_j]} \Prob(\mathbf y \in \mathcal Y_N) \dd \mathbf y .
\end{equation}
\end{lemma}

\begin{proof}
Consider a gap $(\lambda_i, \lambda_{i + 1})$ and $\ell > 0$. The set of $y$ such that $[y, y + \ell] \subset (\lambda_i, \lambda_{i + 1})$ is an interval of length $(\lambda_{i + 1} - \lambda_i - \ell)_+$. When $i \in \mathcal G_N(I)$, this interval is a subset of $[y_-, y_+ - \ell]$. Given $\mathbf y \in \mathcal Y_N$, the indices $i_1, \dots, i_k$ in \eqref{e:inside} are determined by $\mathbf y$, since the gaps are disjoint. Therefore, $\mathcal Y_N$ is the disjoint union, over ordered $k$-tuples of distinct indices in $\mathcal G_N(I)$, of products of such intervals, and
\begin{equation}
|\mathcal Y_N| = \sum_{\substack{i_1, \dots, i_k \in \mathcal G_N(I) \\ \text{distinct}}} \prod_{j = 1}^k \big( \lambda_{i_j + 1} - \lambda_{i_j} - \ell_j \big)_+ .
\end{equation}
By \eqref{e:taushift}, the right side equals $\chi_N^{-k}$ times the sum in \eqref{e:positivemoments}. Taking expectations and using Fubini's theorem, we obtain \eqref{e:volume}.
\end{proof}

Note that the event $\{ \mathbf y \in \mathcal Y_N \}$ implies that the set $\bigcup_j (y_j, y_j + \ell_j)$ contains no eigenvalue. The converse holds when the intervals are separated, provided that every interval of moderate length in $I$ contains an eigenvalue. To make this precise, we set
\begin{equation}\label{e:etaN}
\eta_N = N^{-1/3} ,
\end{equation}
and we let $\mathcal R_N$ denote the event that every closed interval $B \subset I$ with $|B| \ge \eta_N$ contains an eigenvalue of $H_N$.

\begin{lemma}\label{l:bridges}
For every $D > 0$, there exists $N_0 > 0$ such that $\Prob(\mathcal R_N^c) \le N^{-D}$ for $N \ge N_0$.
\end{lemma}

\begin{proof}
Let $\gamma > 0$ be a parameter, which will be chosen at the end of the proof. Let $\delta > 0$ be such that $I \subset [-2 + \delta, 2 - \delta]$, and set $s = \gamma \sqrt{\log N}$. We define
\begin{equation}
B_m = [y_- + m \eta_N / 2, y_- + (m + 1) \eta_N/2]
\end{equation}
for the integers $m \ge 0$ with $B_m \subset I$; there are at most $2 |I| N^{1/3}$ of them. Every closed interval $B \subset I$ with $|B| \ge \eta_N$ contains one of the $B_m$.

Let $y$ denote the left endpoint of $B_m$. Since $s / (N \sigma(y)) \le \eta_N / 2$ for $N \ge N_0$, the interval $(y, y + s/(N \sigma(y)))$ contains no eigenvalue whenever $B_m$ contains none. Using \zcref{c:onehole} and \zcref{l:q1}(a), we find that the hole probability of this interval is bounded by
\begin{equation}
2 q_1(s) \le C e^{-F(s)} \le C N^{-\gamma^2/64} .
\end{equation}
A union bound over $m$ gives $\Prob(\mathcal R_N^c) \le C N^{1/3 - \gamma^2/64}$, and we choose $\gamma$ so that $\gamma^2 / 64 > D + 1$.
\end{proof}

\begin{remark}
\zcref{l:bridges} also follows from the rigidity of eigenvalues proved in \cite[Theorem~2.2]{ErdosYauYin2012}, after a rescaling of the GOE by the factor $(1 + 1/N)^{-1/2}$ that makes the row sums of the variance matrix equal to one. However, the argument above requires only the one-interval estimate.
\end{remark}

\begin{proposition}\label{p:moments}
For every $k \ge 1$ and $\mathbf x \in \R^k$, we have
\begin{equation}\label{e:momentslimit}
\lim_{N \to \infty} M_{N, k}(\mathbf x) = \prod_{j = 1}^k \Lambda_I e^{-x_j} .
\end{equation}
\end{proposition}

\begin{proof}
We define
\begin{equation}
\mathcal H_j(y) = E_{1, N}\big( (y, y + \ell_j) \big) , \qquad \mathcal H(\mathbf y) = E_{1, N} \Big( \bigcup_{j = 1}^k (y_j, y_j + \ell_j) \Big) .
\end{equation}
By \zcref{p:firstmoment}, we have
\begin{equation}\label{e:EjAj}
\lim_{N \to \infty} \chi_N \int_{y_-}^{y_+ - \ell_j} \mathcal H_j(y) \dd y = \Lambda_I e^{-x_j} , \qquad \sup_{y \in [y_-, y_+ - \ell_j]} \chi_N \mathcal H_j(y) \le C \log N .
\end{equation}
Here, and for the rest of this proof, the constants $C$ may depend on $k$, $\mathbf x$, and $I$. Let $\mathcal S_N$ denote the set of $\mathbf y \in \prod_j [y_-, y_+ - \ell_j]$ such that
\begin{equation}\label{e:separated}
|y_i - y_j| \ge 4 \eta_N \quad (i \neq j) , \qquad y_j - y_- \ge 4 \eta_N , \qquad y_+ - (y_j + \ell_j) \ge 4 \eta_N ,
\end{equation}
and let $\mathcal S_N^c$ denote its complement in $\prod_j [y_-, y_+ - \ell_j]$. We may assume that $\ell_j \le \eta_N$ for all $j$, since $\ell_j \le C \sqrt{\log N} / N$.

We first bound the integral of the product of the $\mathcal H_j$ over $\mathcal S_N^c$. The set $\mathcal S_N^c$ is covered by the sets where $|y_i - y_j| < 4 \eta_N$ for some $i \neq j$, and the sets where $y_j$ lies within $4 \eta_N$ of $y_-$ or of $y_+ - \ell_j$ for some $j$. To bound the integral over the set where $|y_i - y_j| < 4 \eta_N$, we integrate first in $y_i$, bound $\mathcal H_i$ by its supremum, and integrate the remaining factors over their full ranges. Using \eqref{e:EjAj}, we see that the contribution of this set to $\chi_N^k \int_{\mathcal S_N^c} \prod_j \mathcal H_j(y_j) \dd \mathbf y$ is bounded by $C \eta_N \log N$. The boundary sets are treated in the same way. Summing these contributions, we have
\begin{equation}\label{e:badmass}
\chi_N^k \int_{\mathcal S_N^c} \prod_{j = 1}^k \mathcal H_j(y_j) \dd \mathbf y \le C \eta_N \log N .
\end{equation}

We now prove the upper bound in \eqref{e:momentslimit}. Suppose first that two of the intervals $[y_j, y_j + \ell_j]$ intersect. Then they cannot be contained in two disjoint gaps, so $\Prob(\mathbf y \in \mathcal Y_N) = 0$. Otherwise, we have
\begin{equation}
\Prob(\mathbf y \in \mathcal Y_N) \le \mathcal H(\mathbf y) ,
\end{equation}
and the intervals $(y_j, y_j + \ell_j)$ have disjoint closures and lie in $I$. Set
\begin{equation}
L = N \sum_j \ell_j \le C \sqrt{\log N} .
\end{equation}
By \eqref{e:quasilambda}, we have
\begin{equation}
\mathcal H(\mathbf y) \le C e^{C \sqrt{\log N}} \prod_j \mathcal H_j(y_j) .
\end{equation}
When moreover $\mathbf y \in \mathcal S_N$, the intervals are separated by at least $3 \eta_N \ge N^{-1/2}$, and \eqref{e:factorlambda} gives
\begin{equation}
\mathcal H(\mathbf y) = (1 + O(e^{C \sqrt{\log N}} N^{-1/4})) \prod_j \mathcal H_j(y_j) .
\end{equation}

Inserting these bounds into \eqref{e:volume} and using \eqref{e:badmass}, we find
\begin{equation}
M_{N, k}(\mathbf x) \le \big( 1 + O(e^{C \sqrt{\log N}} N^{-1/4}) \big) \prod_{j = 1}^k \Big( \chi_N \int_{y_-}^{y_+ - \ell_j} \mathcal H_j \Big) + C e^{C \sqrt{\log N}} \eta_N \log N .
\end{equation}
By \eqref{e:EjAj}, the right side converges to $\prod_j \Lambda_I e^{-x_j}$.

To prove the lower bound, suppose that $\mathbf y \in \mathcal S_N$, that the set $\bigcup_j (y_j, y_j + \ell_j)$ contains no eigenvalue, and that $\mathcal R_N$ holds. We order the intervals from left to right. The interval between $y_-$ and the first prescribed interval, the intervals between consecutive prescribed intervals, and the interval between the last prescribed interval and $y_+$ all have length at least $3 \eta_N$ and lie in $I$, so each contains an eigenvalue. Further, the points $y_j$ and $y_j + \ell_j$ are almost surely not eigenvalues. Hence each prescribed interval $[y_j, y_j + \ell_j]$ is contained in a gap $(\lambda_i, \lambda_{i + 1})$ with $\lambda_i, \lambda_{i + 1} \in I$, and different prescribed intervals belong to different gaps. This shows that $\mathbf y \in \mathcal Y_N$, and therefore
\begin{equation}
\Prob(\mathbf y \in \mathcal Y_N) \ge \mathcal H(\mathbf y) - \Prob(\mathcal R_N^c) , \qquad \mathbf y \in \mathcal S_N .
\end{equation}

Using \eqref{e:volume}, \eqref{e:factorlambda}, \zcref{l:bridges} with $D = k + 1$, and $\chi_N \le C N \sqrt{\log N}$, we have
\begin{equation}
M_{N, k}(\mathbf x) \ge \big( 1 - O(e^{C \sqrt{\log N}} N^{-1/4}) \big) \chi_N^k \int_{\mathcal S_N} \prod_{j = 1}^k \mathcal H_j(y_j) \dd \mathbf y - C (N \sqrt{\log N})^k N^{-k - 1} .
\end{equation}
Finally, \eqref{e:badmass} implies that $\chi_N^k$ times the difference between the integrals of $\prod_j \mathcal H_j(y_j)$ over $\mathcal S_N$ and over $\prod_j [y_-, y_+ - \ell_j]$ is $o(1)$. By \eqref{e:EjAj}, the latter integral, multiplied by $\chi_N^k$, converges to $\prod_j \Lambda_I e^{-x_j}$. This completes the proof.
\end{proof}

\section{Proofs of the main results}

\begin{proof}[Proof of \zcref{t:main}]
\zcref{p:moments} shows that the point processes $\mathcal X_N$ satisfy the hypothesis of \zcref{l:poisson} with $\Lambda = \Lambda_I$. Hence \zcref{t:main} follows from \zcref{l:poisson}.
\end{proof}

\begin{proof}[Proof of \zcref{c:ordered}]
For $x \in \R$, let $\mathcal X_N((x, \infty))$ denote the number of $i \in \mathcal G_N(I)$ with $\tau_{i, N} > x$. Then
\begin{equation}
\{ \tau^\downarrow_{k, N} \le x \} = \{ \mathcal X_N((x, \infty)) \le k - 1 \} ,
\end{equation}
with the convention that $\tau^{\downarrow}_{k, N} = - \infty$ if $|\mathcal G_N(I)| < k$. By \zcref{t:main}, $\mathcal X_N((x, \infty))$ converges in distribution to a Poisson random variable with mean $\Lambda_I e^{-x}$, which gives \eqref{e:ordered}. Taking $k = 1$, we obtain \eqref{e:gumbel}.

We now prove \eqref{e:firstorder}. Let $\mathcal E_N = \{ \mathcal G_N(I) \neq \varnothing \}$. By \eqref{e:ordered} with $k = 1$, we have $\limsup_{N \to \infty} \Prob(\mathcal E_N^c) \le \exp(- \Lambda_I e^{-x})$ for every $x \in \R$. Letting $x \to -\infty$ shows that $\Prob(\mathcal E_N) \to 1$. Let $\widetilde \tau_{1, N}$ equal $\tau^\downarrow_{1, N}$ on $\mathcal E_N$ and $0$ on its complement. Then \eqref{e:gumbel} implies that $\widetilde \tau_{1, N}$ is tight. On $\mathcal E_N$, the definition \eqref{e:taudef} gives
\begin{equation}
N \sigma_I m_{1, N} = b_N + a_N^{-1} \big( \widetilde \tau_{1, N} - (9/8) \log b_N \big) .
\end{equation}
From \eqref{e:abdef}, we have
\begin{equation}
\begin{gathered}
b_N = 8 \sqrt{\log N} - 4 + O\big( (\log N)^{-1/2} \big) , \\
\frac{1}{a_N} = \frac{4}{\sqrt{\log N + 1/4}} = \frac{4}{\sqrt{\log N}} + O\big( (\log N)^{-3/2} \big) ,
\end{gathered}
\end{equation}
and $\log b_N = (1/2) \log \log N + \log 8 + o(1)$. Inserting these expansions on $\mathcal E_N$ and using $\Prob(\mathcal E_N) \to 1$, we obtain \eqref{e:firstorder}.
\end{proof}

\begin{proof}[Proof of the claims in \zcref{r:universality}]
Given a real symmetric $N \times N$ matrix $M$ with eigenvalues $\mu_1 \le \dots \le \mu_N$, a set $I' \subset \R$, and $k \ge 1$, we let $T_k(M; I')$ denote the $k$-th largest of the gaps $\mu_{i + 1} - \mu_i$ with $\mu_i \in I'$, with $T_k(M; I') = 0$ if there are fewer than $k$ such gaps. We also set
\begin{equation}\label{e:tauhat}
\widehat \tau_k(M; I') = a_N \big( N \sigma_I \, T_k(M; I') - b_N \big) + \frac98 \log b_N ,
\end{equation}
so that the quantity $\widehat \tau_{k, N}$ in \zcref{r:universality} is $\widehat \tau_k(W_N; I)$. Throughout the proof, we fix $k \ge 1$, and we let $F_k(x)$ denote the right side of \eqref{e:ordered}.

We first compare $W_N$ with a Gaussian matrix. Set
\begin{equation}
\theta_N = (1 + 1/N)^{-1/2}
\end{equation}
and $G_N = \theta_N H_N$. The entries $(G_N)_{ij}$, $i \le j$, are independent centered Gaussian random variables with variance $(1 + \1_{i = j})/(N + 1)$. Hence every row of the variance matrix of $G_N$ sums to $1$, and $G_N$ is the GOE matrix in the normalization of \cite[Definition~1.1]{LandonLopattoMarcinek2020}.

Let $\Phi \in C_c^\infty(\R)$, and set $\nu = N / \sqrt{\log N}$ and
\begin{equation}
S_N(t) = \Phi\Big( a_N \sigma_I \sqrt{\log N} \, t - a_N b_N + \frac98 \log b_N \Big) , \qquad t \in \R ,
\end{equation}
so that
\begin{equation}
\Phi(\widehat \tau_k(M; I)) = S_N(\nu \, T_k(M; I)) .
\end{equation}
The derivatives of $S_N$ of order at most five are bounded by $C (\log N)^5$, since $a_N \sigma_I \sqrt{\log N} \le C \log N$. Using \cite[Corollary~1.6]{LandonLopattoMarcinek2020}, whose error term involves derivatives of the test function of order at most five, we find
\begin{equation}\label{e:LLMcompare}
\big| \E \Phi\big( \widehat \tau_k(W_N; I) \big) - \E \Phi\big( \widehat \tau_k(G_N; I) \big) \big| \le C N^{-c} (\log N)^5 .
\end{equation}

Next, we show that $\Prob(\widehat \tau_k(G_N; I) \le x) \to F_k(x)$ for all $x \in \R$. Set $I_N = \theta_N^{-1} I$. The matrix $G_N$ has eigenvalues $\theta_N \lambda_i$, so $T_k(G_N; I) = \theta_N T_k(H_N; I_N)$. Fix $x \in \R$ and $\eta > 0$. By \eqref{e:tauhat} and \eqref{e:sNdef}, the events $\{ \widehat \tau_k(G_N; I) > x \}$ and $\{ T_k(H_N; I_N) > \ell_N(x) / \theta_N \}$ coincide for $N$ large. Note that $\ell_N(x + \eta) - \ell_N(x) = \eta / \chi_N$ is much larger than $\ell_N(x) (\theta_N^{-1} - 1) \le C \sqrt{\log N} / N^2$. Hence, for $N$ large, we have
\begin{equation}
\ell_N(x) \le \ell_N(x) / \theta_N \le \ell_N(x + \eta) ,
\end{equation}
and therefore
\begin{equation}\label{e:tausandwich}
\Prob\big( \widehat \tau_k(H_N; I_N) > x + \eta \big) \le \Prob\big( \widehat \tau_k(G_N; I) > x \big) \le \Prob\big( \widehat \tau_k(H_N; I_N) > x \big) .
\end{equation}

It remains to compare $\widehat \tau_k(H_N; I_N)$ with $\tau^\downarrow_{k, N}$. Let $x' \in \R$. The events $\{ \widehat \tau_k(H_N; I_N) > x' \}$ and $\{ \tau^\downarrow_{k, N} > x' \}$ can differ only if there exists a gap $(\lambda_i, \lambda_{i + 1})$ of length at least $\ell_N(x')$ such that exactly one of the conditions $\lambda_i \in I_N$ and $i \in \mathcal G_N(I)$ holds. The closure of such a gap contains a point within distance $2/N$ of $y_-$ or of $y_+$, since $|\theta_N^{-1} y_\pm - y_\pm| \le 2/N$. A gap of length at least $\ell_N(x')$ whose closure contains a point within distance $2/N$ of $y_\pm$ contains one of the intervals
\begin{equation}
\Big( y_\pm - \frac{\ell_N(x')}{2} + \frac2N , \, y_\pm - \frac2N \Big) , \qquad \Big( y_\pm + \frac2N , \, y_\pm + \frac{\ell_N(x')}{2} - \frac2N \Big) .
\end{equation}

These four intervals lie in $[-2 + \delta, 2 - \delta]$ for some $\delta > 0$ depending on $I$. Further, $\sigma(y) \ge \sigma_I - C \ell_N(x')$ on them, so each has the form $(y, y + s/(N \sigma(y)))$ with $s \ge s_N(x')/2 - C$, where we recall that
\begin{equation}
s_N(x') = N \sigma_I \ell_N(x') .
\end{equation}
Using \zcref{c:onehole}, \zcref{l:q1}(a), and the monotonicity of $q_1$, we see that the probability that one of them contains no eigenvalue is bounded by $C q_1(s_N(x')/2 - C) \le N^{-1/4 + o(1)}$. Therefore,
\begin{equation}
|\Prob(\widehat \tau_k(H_N; I_N) > x') - \Prob(\tau^\downarrow_{k, N} > x')| \le N^{-1/4 + o(1)}
\end{equation}
for every fixed $x'$. Inserting this with $x' = x$ and $x' = x + \eta$ into \eqref{e:tausandwich} and using \eqref{e:ordered}, we find that every limit point of $\Prob(\widehat \tau_k(G_N; I) \le x)$ belongs to $[F_k(x), F_k(x + \eta)]$. Letting $\eta \downarrow 0$ and using the continuity of $F_k$, we have
\begin{equation}
\Prob(\widehat \tau_k(G_N; I) \le x) \to F_k(x) .
\end{equation}

Since $F_k$ is the distribution function of a probability measure on $\R$, it follows that
\begin{equation}
\E \Phi(\widehat \tau_k(G_N; I)) \to \int \Phi \dd F_k
\end{equation}
for all $\Phi \in C_c^\infty(\R)$. Then \eqref{e:LLMcompare} implies that the same holds for $W_N$ in place of $G_N$. Convergence of $\E \Phi$ for all $\Phi \in C_c^\infty(\R)$ to the integral against a probability measure implies convergence in distribution, so
\begin{equation}
\Prob(\widehat \tau_k(W_N; I) \le x) \to F_k(x)
\end{equation}
for all $x \in \R$.
\end{proof}