\chapter{The first moment}\label{s:firstmoment}

In this section, we compute the integral of the one-interval hole probability over the starting point of the interval, at the threshold relevant for the largest gaps. This calculation determines the centering, the scaling, and the constant $\Lambda_I$ in \zcref{t:main}. We work in eigenvalue coordinates, with the notation of \zcref{s:mainresults}. Recall the quantities $\ell_N(x)$ and $\chi_N$ from \eqref{e:overviewell}. For $x \in \R$, we also define $s_N(x) = N \sigma_I \ell_N(x)$, so that
\begin{equation}\label{e:sNdef}
s_N(x) = b_N + \frac{x - \frac98 \log b_N}{a_N} , \qquad \ell_N(x) = \frac{s_N(x)}{N \sigma_I} , \qquad \chi_N = a_N N \sigma_I .
\end{equation}
These definitions are chosen so that, for $i \in \mathcal G_N(I)$,
\begin{equation}\label{e:taushift}
\tau_{i, N} - x = \chi_N \big( \lambda_{i + 1} - \lambda_i - \ell_N(x) \big) .
\end{equation}
Note that $s_N(x) = b_N + O_x(\log \log N / \sqrt{\log N})$, where $O_x$ indicates that the implicit constant depends on $x$.

\begin{proposition}\label{p:firstmoment}
For every $x \in \R$, we have
\begin{equation}\label{e:firstmoment}
\lim_{N \to \infty} \chi_N \int_{y_-}^{y_+ - \ell_N(x)} E_{1, N} \big( (y, y + \ell_N(x)) \big) \dd y = \Lambda_I e^{-x} .
\end{equation}
Moreover, there exist $C_x, N_x > 0$ such that for all $N \ge N_x$,
\begin{equation}\label{e:firstsup}
\sup_{y \in [y_-, y_+ - \ell_N(x)]} \chi_N \, E_{1, N} \big( (y, y + \ell_N(x)) \big) \le C_x \log N .
\end{equation}
\end{proposition}

\begin{proof}
Throughout the proof, $x$ is fixed, and we abbreviate $s_N = s_N(x)$ and $\ell_N = \ell_N(x)$. We begin with the value of $q_1$ at $s_N$. Recall that $F(b_N) = \log N$, $F'(b_N) = a_N$, and $F'' = 1/32$. Taylor's theorem then gives
\begin{equation}\label{e:FsN}
F(s_N) = \log N + x - \frac98 \log b_N + \frac{1}{64} \Big( \frac{x - \frac98 \log b_N}{a_N} \Big)^2 = \log N + x - \frac98 \log b_N + o(1) ,
\end{equation}
where we used $a_N \ge c \sqrt{\log N}$. Using \zcref{l:q1}(a), \eqref{e:FsN}, and $s_N / b_N \to 1$, we have
\begin{equation}\label{e:q1sN}
q_1(s_N) = \kappa_1 s_N^{-1/8} e^{-F(s_N)} (1 + o(1)) = \frac{\kappa_1 b_N e^{-x}}{N} (1 + o(1)) .
\end{equation}

For $y \in I$, we set
\begin{equation}
\Psi(y) = \sigma(y) / \sigma_I \ge 1 ,
\end{equation}
and for $y \in [y_-, y_+ - \ell_N]$, we set
\begin{equation}
s(y) = N \sigma(y) \ell_N = s_N \Psi(y) ,
\end{equation}
so that
\begin{equation}
(y, y + \ell_N) = (y, y + s(y)/(N \sigma(y))) .
\end{equation}
Since $I \subset (-2, 2)$ is compact and $s(y) \le C \sqrt{\log N}$, \zcref{c:onehole} gives
\begin{equation}\label{e:Ereplace}
E_{1, N}\big( (y, y + \ell_N) \big) = q_1\big( s_N \Psi(y) \big) (1 + o(1))
\end{equation}
uniformly in $y \in [y_-, y_+ - \ell_N]$.

Next, \eqref{e:Ereplace} and \eqref{e:q1sN} imply that
\begin{equation}
\chi_N E_{1, N}((y, y + \ell_N)) \le C \chi_N q_1(s_N) \le C_x a_N b_N ,
\end{equation}
where we used that $q_1$ is nonincreasing and $\Psi \ge 1$. This implies \eqref{e:firstsup}, since $a_N b_N \le 2 \log N$ by \eqref{e:abdef}.

We now prove \eqref{e:firstmoment}. Let $d_0 \in (0, |I|/2)$ be a small constant, which will be chosen below depending only on $I$, and let $\mathcal U$ denote the set of $y \in I$ with $\dist(y, \partial_* I) < d_0$. Since $\Psi$ is continuous on $I$ and attains the value $1$ only on $\partial_* I$, there exists $c_0 > 0$ such that $\Psi \ge 1 + c_0$ on $I \setminus \mathcal U$. It follows from \zcref{l:q1}(a) and \eqref{e:FsN} that
\begin{equation}
q_1\big( (1 + c_0) s_N \big) \le C e^{- (1 + c_0)^2 s_N^2 / 64} \le N^{-(1 + c_0)^2 + o(1)} ,
\end{equation}
so \eqref{e:Ereplace} and the monotonicity of $q_1$ imply that the contribution of $I \setminus \mathcal U$ to the left side of \eqref{e:firstmoment} is bounded by $C \chi_N N^{-(1 + c_0)^2 + o(1)} \to 0$.

Fix $\xi \in \partial_* I$. Recall that
\begin{equation}
|\xi| = \max(|y_-|, |y_+|) > 0 ,
\end{equation}
since $I$ is nondegenerate. We parametrize the points of $\mathcal U$ near $\xi$ as $y = \xi - \sgn(\xi) t$ with $t \in [0, d_0)$, so that $t$ is the distance from $y$ to $\xi$, and $y$ moves into the interior of $I$ as $t$ increases. Note that $\sigma'(\xi) = - \xi / \sigma_I$, which gives
\begin{equation}
\Psi\big( \xi - \sgn(\xi) t \big) = 1 + \frac{|\xi|}{\sigma_I^2} \, t + O(t^2) .
\end{equation}
We choose $d_0$ so small that the function $t \mapsto \Psi(\xi - \sgn(\xi) t)$ is increasing on $[0, d_0)$, with derivative between $|\xi| / (2 \sigma_I^2)$ and $2 |\xi| / \sigma_I^2$.

We now change variables by setting
\begin{equation}\label{e:zdef}
z = (\log N) \big( \Psi(y) - 1 \big) ,
\end{equation}
so that $s(y) = s_N (1 + z / \log N)$, and $\dd t = \sigma_I^2 (|\xi| \log N)^{-1} (1 + O(t)) \dd z$. In the integral, $t$ ranges over $[0, d_0)$ if $\xi = y_-$, and over $[\ell_N, d_0)$ if $\xi = y_+$. In both cases, we let $[\underline z_N, \overline z_N)$ denote the corresponding range of $z$. We have
\begin{equation}
0 \le \underline z_N \le C \ell_N \log N \to 0
\end{equation}
and $\overline z_N \ge c \log N$.

Setting $w = z / \log N \ge 0$, we have
\begin{equation}\label{e:Fincrement}
F\big( s_N (1 + w) \big) - F(s_N) = \frac{s_N^2}{64} (2 w + w^2) + \frac{s_N}{8} w .
\end{equation}
By \eqref{e:FsN}, we have $s_N^2 / 64 = \log N + O(\sqrt{\log N})$. Therefore, the right side of \eqref{e:Fincrement} equals $2z + o(1)$ uniformly for $z$ in compact sets. Further, it is at least $z$ for all $z \ge 0$ when $N$ is large. Combining this with \zcref{l:q1}(a), which shows that the ratio $q_1(s) / (\kappa_1 s^{-1/8} e^{-F(s)})$ takes values in $[1/2, 2]$ for $s \ge s_N$ and $N$ large, we find
\begin{equation}\label{e:qratio}
\lim_{N \to \infty} \frac{q_1(s_N(1 + z/\log N))}{q_1(s_N)} = e^{-2z} , \qquad \frac{q_1(s_N(1 + z/\log N))}{q_1(s_N)} \le 4 e^{-z} ,
\end{equation}
where the limit is locally uniform in $z \ge 0$, and the bound holds for all $z \ge 0$ and $N$ large.

It follows from \eqref{e:Ereplace} that the contribution of the part of $\mathcal U$ near $\xi$ to the left side of \eqref{e:firstmoment} is
\begin{equation}
(1 + o(1)) \, \chi_N \, q_1(s_N) \, \frac{\sigma_I^2}{|\xi| \log N} \int_{\underline z_N}^{\overline z_N} \frac{q_1(s_N(1 + z/\log N))}{q_1(s_N)} \big( 1 + O(t) \big) \dd z .
\end{equation}
By \eqref{e:qratio} and dominated convergence, the integral converges to
\begin{equation}
\int_0^\infty e^{-2z} \dd z = 1/2 .
\end{equation}
Using \eqref{e:q1sN} and $\chi_N = a_N N \sigma_I$, we find that the contribution converges to
\begin{equation}
\lim_{N \to \infty} \frac{a_N b_N}{2 \log N} \cdot \kappa_1 \frac{\sigma_I^3}{|\xi|} e^{-x} = \kappa_1 \frac{\sigma_I^3}{|\xi|} e^{-x} ,
\end{equation}
where we used that $a_N b_N / (2 \log N) \to 1$, which follows from \eqref{e:abdef}. Summing over $\xi \in \partial_* I$, we obtain \eqref{e:firstmoment}.
\end{proof}

\begin{remark}\label{r:heuristic}
The proof of \zcref{p:firstmoment} explains the form of the centering in \zcref{t:main}. Heuristically, the integral of $E_{1, N}((y, y + \ell))$ over $y$ is the expected value of $\sum_i (\lambda_{i + 1} - \lambda_i - \ell)_+$, so the expected number of gaps longer than $\ell$ is minus its derivative in $\ell$. With $\ell = s / (N \sigma_I)$, the proof shows that the integral is approximately $q_1(s) \sigma_I^2 / (2 |\xi| \log N)$ for each $\xi \in \partial_* I$, and differentiating $q_1(s) \approx \kappa_1 s^{-1/8} e^{-F(s)}$ produces a factor $F'(s) \approx s/32$. So the expected number of gaps longer than $s/(N \sigma_I)$ is of order $N s \, q_1(s) / \log N$, which is of order $N q_1(s) / s$ since $s^2 \approx 64 \log N$. This number is of order one when $N q_1(s) \approx s \approx b_N$, or equivalently when $F(s) \approx \log N - (9/8) \log b_N$. The linear term $s/8$ in $F$, which comes from the term $\mp t/2$ in \eqref{e:ehrhardtintro}, produces the constant $-4$ in \eqref{e:firstorder}. By contrast, for the GUE the corresponding exponent $s^2/32$ has no linear term.
\end{remark}