\chapter{The metallic expansion}\label{s:metallic}

In this section we prove \zcref{p:metallic}, which gives the first corrections to the eigenvector correlation $\mathcal C_X$ and to the function $I_X$ as $X \downarrow 0$. We also prove that the kernel $K_X(x, y; z)$ tends to $(1 - e^{-2z})/z^2 = 2 I_{\GUE}(z)$ as $X \downarrow 0$, uniformly in $0 < x \le y < 1$. This limit is the metallic endpoint of the last row of \zcref{p:endpoints}, which is proved in \zcref{s:scales}.

The two statements of \zcref{p:metallic} have different proofs. The correlation $\mathcal C_X$ is a Brownian functional, and we expand it by Taylor's theorem in the parameter $\sqrt{2X}$ of a Gaussian integral (\zcref{s:met-profile}). The function $I_X$ is defined in \zcref{d:kernels} through the diffusion $(\tilde B_t, \tilde F_t)$ with generator $X \Delta_D$. Heuristically, the semigroups in \eqref{e:intro-K} act almost as multiplication operators for small $X$, since the diffusion moves only by $O(\sqrt X)$ during the times $x$, $y - x$, and $1 - y$. In the limiting case $X = 0$, the semigroups would be the multiplications by $e^{-tzd}$, the singlet state would be $e^{-tzd}$, and the kernel would be $\int_D e^{-zd} \dd B \dd F$. We expand the semigroups in powers of $X$ by Duhamel's formula, with remainders in closed form (\zcref{s:met-singlet}). The main point is to bound these remainders uniformly over the closed time simplex $0 \le x \le y \le 1$. The difficulty is that some of the three times vanish on the faces of the simplex, where the corresponding factors give no decay in $B$. We overcome this with an envelope for $\log(1 + \tilde d_t)$, where $\tilde d_t = \tilde B_t - \tilde F_t$ is the function $d = B - F$ of \zcref{d:kernels} along the diffusion. This is the quantity $\log(1 + \gamma \tilde d_t)$ of \zcref{l:met-ito,l:met-envelope} with $\gamma = 1$, the value used in this section; \zcref{s:scales} uses other values of $\gamma$. The envelope shows that the killing accumulated over all the time intervals decays faster than every power of $B$. The coefficients are then elementary integrals, which we evaluate in \zcref{s:met-scalar}. In \zcref{s:met-modes}, we express them through the Neumann Laplacian and compare them, heuristically, with the corrections of Kravtsov and Mirlin \cite{kravtsov1994level}.

\section{The eigenvector correlation}\label{s:met-profile}

Recall the Brownian motion $\beta$ and the objects $g_X$, $Z_X$, and $\mathcal C_X$ of \zcref{d:brownian}, and the Neumann Green kernel $G_{\mathrm N}$ of \zcref{p:metallic}.

\begin{lemma}\label{l:met-profile}
Let $X_0 > 0$. There exists $C > 0$ such that for all $X \in (0, X_0]$,
\begin{equation}\label{e:met-profile}
\sup_{x, y \in [0, 1]} \big| \mathcal C_X(x, y) - 1 - 2 X G_{\mathrm N}(x, y) \big| \le C X^2 .
\end{equation}
\end{lemma}

\begin{proof}
For $\tau \in \R$ and $x, y \in [0, 1]$, we set
\[
g^\tau(u) = \tau \beta_u - \frac{\tau^2 u}{2}, \qquad Z^\tau = \int_0^1 e^{g^\tau(u)} \dd u, \qquad R^\tau(x, y) = \frac{e^{g^\tau(x) + g^\tau(y)}}{Z^\tau} .
\]
Since $g_X = g^{\sqrt{2X}}$, we have $\mathcal C_X(x, y) = \E R^{\sqrt{2X}}(x, y)$. We will show that $\tau \mapsto \E R^\tau(x, y)$ is an even function of class $C^4$ whose fourth derivative is bounded uniformly in $x$, $y$, and $|\tau| \le \sqrt{2 X_0}$, and we will compute its second derivative at $\tau = 0$.

We begin by justifying differentiation under the expectation. Let $\tau_0 = \sqrt{2X_0}$, and set
\[
\beta^* = \sup_{u \in [0, 1]} |\beta_u| .
\]
Differentiating $e^{g^\tau(u)}$ $j$ times in $\tau$ produces $e^{g^\tau(u)}$ times a polynomial of degree at most $j$ in $\beta_u - \tau u$ and $u$. For $|\tau| \le \tau_0$ and $u \in [0, 1]$, we have
\[
\big| \partial_\tau^j e^{g^\tau(u)} \big| \le C_j (1 + \beta^* + \tau_0)^j e^{g^\tau(u)}, \qquad \big| \partial_\tau^j Z^\tau \big| \le C_j (1 + \beta^* + \tau_0)^j Z^\tau,
\]
and the same bound holds for $e^{g^\tau(x) + g^\tau(y)}$. Expanding the derivatives of $R^\tau = e^{g^\tau(x) + g^\tau(y)} (Z^\tau)^{-1}$ by the Leibniz rule, we find that for $0 \le j \le 4$
\begin{equation}\label{e:met-profile-domination}
\sup_{x, y \in [0, 1]} \big| \partial_\tau^j R^\tau(x, y) \big| \le C_j (1 + \beta^* + \tau_0)^j R^\tau(x, y) \le C_j (1 + \beta^* + \tau_0)^j e^{3 \tau_0 \beta^* + \tau_0^2/2},
\end{equation}
where we used $e^{g^\tau(x) + g^\tau(y)} \le e^{2\tau_0 \beta^*}$ and $Z^\tau \ge e^{-\tau_0 \beta^* - \tau_0^2/2}$ in the last step. The reflection principle implies that $\beta^*$ has Gaussian tails, and the right side of \eqref{e:met-profile-domination} is therefore integrable.

Therefore $\tau \mapsto \E R^\tau(x, y)$ is of class $C^4$ on $[-\tau_0, \tau_0]$, with derivatives $\E \partial_\tau^j R^\tau(x, y)$. Further, its fourth derivative is bounded uniformly in $x$ and $y$. Since $-\beta$ is a Brownian motion, $R^{-\tau}(x, y)$ has the same law as $R^\tau(x, y)$. It follows that $\E R^\tau(x, y)$ is an even function of $\tau$, whose odd derivatives vanish at $\tau = 0$. Taylor's theorem and $R^0 = 1$ then yield
\begin{equation}\label{e:met-profile-taylor}
\E R^\tau(x, y) = 1 + \frac{\tau^2}{2} \E \partial_\tau^2 R^\tau(x, y) \big|_{\tau = 0} + O(\tau^4),
\end{equation}
uniformly in $x, y \in [0, 1]$ and $|\tau| \le \tau_0$.

We now compute the second derivative. Let
\[
\beta^{(1)} = \int_0^1 \beta_u \dd u, \qquad \beta^{(2)} = \int_0^1 (\beta_u^2 - u) \dd u .
\]
At $\tau = 0$, we have
\[
\begin{gathered}
\partial_\tau e^{g^\tau(x) + g^\tau(y)} = \beta_x + \beta_y, \qquad \partial_\tau^2 e^{g^\tau(x) + g^\tau(y)} = (\beta_x + \beta_y)^2 - (x + y), \\
\partial_\tau Z^\tau = \beta^{(1)}, \qquad \partial_\tau^2 Z^\tau = \beta^{(2)} .
\end{gathered}
\]
Differentiating the quotient twice, we find
\[
\partial_\tau^2 R^\tau(x, y) \big|_{\tau = 0} = (\beta_x + \beta_y)^2 - (x + y) - 2 (\beta_x + \beta_y) \beta^{(1)} - \beta^{(2)} + 2 \big( \beta^{(1)} \big)^2 .
\]
We have $\E \beta^{(2)} = 0$, $\E (\beta^{(1)})^2 = 1/3$, $\E \beta_x \beta_y = \min(x, y)$, and
\[
\E \beta_x \beta^{(1)} = \int_0^1 \min(x, u) \dd u = x - \frac{x^2}{2} .
\]
Using these identities and $\min(x, y) - x - y = -\max(x, y)$, we conclude that
\[
\E \partial_\tau^2 R^\tau(x, y) \big|_{\tau = 0} = 2 \min(x, y) - 2x - 2y + x^2 + y^2 + \frac23 = 2 G_{\mathrm N}(x, y) .
\]
Inserting this into \eqref{e:met-profile-taylor} with $\tau = \sqrt{2X}$ proves \eqref{e:met-profile}.
\end{proof}

\section{Feynman--Kac preliminaries}\label{s:met-fk}

We now collect the facts about the diffusions of \zcref{d:kernels} and \zcref{r:intro-singlet} that the expansion of $I_X$ uses. Since \zcref{s:scales} uses them with a different time scale and a different size of the killing, we state them with general parameters.

We say that a function on $D$ is smooth if it is the restriction to $D$ of a smooth function on an open neighborhood of $D$ in $\R^2$, and we let $C^\infty(D)$ denote the space of these functions. A smooth function of $(B, F)$ is a smooth function on $\mathbb H^2 \times S^2$ (the hyperbolic cosine $B$ and the cosine $F$ of the two distances to the base points are smooth functions of the squares of these distances). For $f, g \in C^\infty(D)$, we set
\[
\langle \nabla_D f, \nabla_D g \rangle = (B^2 - 1) \, \partial_B f \, \overline{\partial_B g} + (1 - F^2) \, \partial_F f \, \overline{\partial_F g}, \qquad |\nabla_D f|^2 = \langle \nabla_D f, \nabla_D f \rangle,
\]
which is the inner product of the Riemannian gradients on $\mathbb H^2 \times S^2$, linear in the first argument and conjugate linear in the second. We say that a function $f$ on $D$ has \emph{polynomial growth} if there exist $C, q \ge 0$ such that $|f| \le C (1 + d)^q$ on $D$. Note that $1 \le B \le 1 + d$, since $|F| \le 1$. In particular, every polynomial in $B$ and $F$ has polynomial growth. We will often use the identities
\begin{equation}\label{e:met-d-identities}
\Delta_D d = 2(B + F), \qquad |\nabla_D d|^2 = B^2 - F^2 = d (B + F), \qquad 0 \le B + F \le d + 2 .
\end{equation}

Recall that the radial diffusion $(\tilde B_t, \tilde F_t)$ of \zcref{d:kernels} has generator $X \Delta_D$, and that the singlet diffusion $(B_t, F_t)$ of \zcref{r:intro-singlet} has generator $X \Delta^{\mathrm s}_D$, where $\Delta^{\mathrm s}_D = d \Delta_D d^{-1}$, and is stopped at the corner $(1, 1)$. We write $\tilde d_t = \tilde B_t - \tilde F_t$ and $d_t = B_t - F_t$. For a measurable function $f$ of polynomial growth, $t \ge 0$, and $\Re z > 0$, we define $e^{t(X \Delta_D - z d)} f$ by the Feynman--Kac formula of \zcref{d:kernels}, whose expectation is finite by \eqref{e:met-moments} below. By the Markov property of $(\tilde B_t, \tilde F_t)$, these operators form a semigroup on the functions of polynomial growth, and
\begin{equation}\label{e:met-fk-domination}
\big| e^{t(X \Delta_D - z d)} f \big| \le e^{t(X \Delta_D - (\Re z) d)} |f| .
\end{equation}

\begin{lemma}\label{l:met-ito}
Let $X$, $\gamma$, and $T$ be positive, and let $\Theta \ge X (1 + 2\gamma) T$.
\begin{enumerate}[label=(\alph*)]
\item For $f \in C^\infty(D)$ and $p \in D$, the process
\[
f(\tilde B_t, \tilde F_t) - f(p) - X \int_0^t \Delta_D f(\tilde B_s, \tilde F_s) \dd s, \qquad t \ge 0,
\]
is a continuous local martingale under $\E_p$, with quadratic variation
\[
2X \int_0^t |\nabla_D f|^2 (\tilde B_s, \tilde F_s) \dd s .
\]
For a complex martingale $M$, quadratic variation here means the positive
process $[\Re M]+[\Im M]$.
\item Let $(R_t)_{t \ge 0}$ denote either $\gamma \tilde d_t$ under $\E_p$ or $\gamma d_t$ under $\E^{\mathrm s}_p$, where in the second case $p$ lies in the interior of $D$, and set $\ell_t = \log(1 + R_t)$. There exists a continuous local martingale $N$ with $N_0 = 0$ and $\langle N \rangle_T \le 2\Theta$ such that $\ell_t - \ell_0 - N_t$ is a continuous process of finite variation with values in $[-\Theta, 2\Theta]$ for $t \le T$.
\item With $R_t$ as in (b), for all $m \ge 0$,
\begin{equation}\label{e:met-moments}
\sup_{t \le T} \E (1 + R_t)^m \le e^{m(m + 2) \Theta} (1 + R_0)^m .
\end{equation}
\item With $R_t$ as in (b), and with $\Prob$ denoting the law under which $R$ is defined,
\begin{equation}\label{e:met-log-tail}
\Prob\Big( \inf_{t \le T} (1 + R_t) < (1 + R_0)^{1/2} \Big) \le \exp\Big( - \frac{\big( \frac12 \log(1 + R_0) - \Theta \big)_+^2}{4 \Theta} \Big) .
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
We begin with (a), and first let $p \in D^\circ$. By the boundary classification that precedes \eqref{e:res-flat-sde}, the radial diffusion started at $p$ stays in $D^\circ$. Its coordinates solve the stochastic differential equations \eqref{e:res-flat-sde}, driven by independent standard Brownian motions $\beta^1$ and $\beta^2$. In particular,
\[
\dd \langle \tilde B \rangle_t = 2X (\tilde B_t^2 - 1) \dd t, \qquad \dd \langle \tilde F \rangle_t = 2X (1 - \tilde F_t^2) \dd t, \qquad \langle \tilde B, \tilde F \rangle_t = 0 .
\]
Since $f$ is smooth on a neighborhood of $D$, Itô's formula applies to $f(\tilde B_t, \tilde F_t)$, and we find
\[
\begin{aligned}
\dd f(\tilde B_t, \tilde F_t) = {} & X \Delta_D f(\tilde B_t, \tilde F_t) \dd t + \big( 2X (\tilde B_t^2 - 1) \big)^{1/2} \partial_B f(\tilde B_t, \tilde F_t) \dd \beta^1_t \\
& + \big( 2X (1 - \tilde F_t^2) \big)^{1/2} \partial_F f(\tilde B_t, \tilde F_t) \dd \beta^2_t ,
\end{aligned}
\]
where we used $\Delta_D = (B^2 - 1) \partial_B^2 + 2B \partial_B + (1 - F^2) \partial_F^2 - 2F \partial_F$. The local martingale part has the quadratic variation $2X \int_0^t |\nabla_D f|^2 (\tilde B_s, \tilde F_s) \dd s$, which proves (a) for $p \in D^\circ$.

Now let $p$ lie on the boundary of $D$. We write $N_t$ for the process in (a), and we set
\[
A_t = 2X \int_0^t |\nabla_D f|^2 (\tilde B_s, \tilde F_s) \dd s .
\]
The diffusion does not reach the boundary of $D$ at positive times (see the discussion after \zcref{d:kernels}). For $\eps > 0$, the Markov property at time $\eps$ and the first case show that $N_t - N_\eps$ and $|N_t - N_\eps|^2 - (A_t - A_\eps)$, $t \ge \eps$, are continuous local martingales for the filtration of the diffusion. For $n \ge 1$, let $T_n$ be the first time $t$ with $|N_t| + A_t \ge n$. The stopped increments
\[
N_{t\wedge T_n}-N_{\eps\wedge T_n},\qquad
|N_{t\wedge T_n}-N_{\eps\wedge T_n}|^2-(A_{t\wedge T_n}-A_{\eps\wedge T_n}),\qquad t\ge\eps,
\]
are bounded martingales: they vanish on $\{T_n\le\eps\}$, and on $\{T_n>\eps\}$ the preceding local martingales are stopped at $T_n$. We let $\eps \downarrow 0$, using $N_0 = A_0 = 0$ and bounded convergence, and find that $N_{t \wedge T_n}$ and $|N_{t \wedge T_n}|^2 - A_{t \wedge T_n}$ are martingales on $[0, \infty)$. Since $T_n \to \infty$, (a) follows for $p$.

We next prove (b) for the radial diffusion. Let $\ell(r) = \log(1 + \gamma r)$ for $r \ge 0$. Applying (a) to $\ell(d)$, we see that $\ell_t - \ell_0 - X \int_0^t \Delta_D \ell(d)(\tilde B_s, \tilde F_s) \dd s$ is a local martingale with quadratic variation $2X \int_0^t |\nabla_D \ell(d)|^2 (\tilde B_s, \tilde F_s) \dd s$. The identities \eqref{e:met-d-identities} give
\[
\Delta_D \ell(d) = \frac{2 \gamma (B + F)}{1 + \gamma d} - \frac{\gamma^2 d (B + F)}{(1 + \gamma d)^2}, \qquad |\nabla_D \ell(d)|^2 = \frac{\gamma^2 d (B + F)}{(1 + \gamma d)^2} .
\]
We write $R = \gamma d$. The last inequality in \eqref{e:met-d-identities} implies that $0 \le \gamma (B + F) \le R + 2\gamma$, and we find
\[
0 \le \frac{2 \gamma (B + F)}{1 + R} \le \frac{2 (R + 2 \gamma)}{1 + R} \le 2 (1 + 2\gamma), \qquad 0 \le \frac{\gamma^2 d (B + F)}{(1 + R)^2} \le \frac{R (R + 2\gamma)}{(1 + R)^2} \le 1 + 2\gamma .
\]
Multiplying by $X$, we obtain
\[
-X(1 + 2\gamma) \le X \Delta_D \ell(d) \le 2 X (1 + 2\gamma), \qquad 2 X |\nabla_D \ell(d)|^2 \le 2X(1 + 2\gamma) .
\]
These bounds, integrated over $[0, t]$ with $t \le T$, imply (b).

For the singlet diffusion, the drift $-2 X d^{-1} \langle \nabla_D \, \cdot \, , \nabla_D d \rangle$ of $X \Delta^{\mathrm s}_D$ cancels the first-order term of $\Delta_D \ell(d)$. Indeed, by \eqref{e:met-d-identities},
\[
\Delta^{\mathrm s}_D \ell(d) = \Delta_D \ell(d) - \frac{2 \ell'(d) |\nabla_D d|^2}{d} = \ell''(d) |\nabla_D d|^2 = - \frac{\gamma^2 d (B + F)}{(1 + \gamma d)^2},
\]
which is smooth on $D$ and vanishes at the corner. By the stochastic differential equation for the singlet diffusion in \zcref{l:res-singlet}, Itô's formula applies to $\ell(d)$ up to the time at which the corner is reached, and $\ell_t$ is constant after this time. The bounds above then imply (b), with $\ell_t - \ell_0 - N_t$ nonincreasing.

For (c), we apply Itô's formula to $e^{m \ell_t} = (1 + R_t)^m$. By the bounds in the proof of (b), its drift is bounded by $m (2 + m) X (1 + 2\gamma) e^{m \ell_t}$. We stop at the exit time of $\{R \le n\}$, take expectations, and apply Gronwall's lemma. Letting $n \to \infty$ and using Fatou's lemma, we obtain \eqref{e:met-moments}.

We now prove (d). We may assume that $\delta = \frac12 \log(1 + R_0) - \Theta$ is positive. Using (b), we see that on the event in \eqref{e:met-log-tail},
\[
\inf_{t \le T} N_t \le -\frac12 \log(1 + R_0) + \Theta = -\delta .
\]
For $\sigma > 0$, the process $\exp(-\sigma N_t - \sigma^2 \langle N \rangle_t / 2)$ is a nonnegative local martingale, hence a supermartingale, and $\langle N \rangle_T \le 2 \Theta$. Doob's maximal inequality implies that
\[
\Prob\Big( \inf_{t \le T} N_t \le -\delta \Big) \le \Prob\Big( \sup_{t \le T} e^{-\sigma N_t - \sigma^2 \langle N \rangle_t / 2} \ge e^{\sigma \delta - \sigma^2 \Theta} \Big) \le e^{-\sigma \delta + \sigma^2 \Theta},
\]
and the choice $\sigma = \delta / (2\Theta)$ gives \eqref{e:met-log-tail}.
\end{proof}

We will use \zcref{l:met-ito} through the following bound on the Feynman--Kac weight of a path along which polynomial factors and killing factors are collected at finitely many times. For $\lambda > 0$, $\Theta > 0$, and $\tau \ge 0$, we define
\begin{equation}\label{e:met-envelope-function}
\mathcal E_\tau(R) = \exp\big( -\lambda \tau \big( (1 + R)^{1/2} - 1 \big) \big) + \exp\Big( - \frac{\big( \frac12 \log(1 + R) - \Theta \big)_+^2}{8 \Theta} \Big), \qquad R \ge 0,
\end{equation}
and we suppress the dependence on $\lambda$ and $\Theta$ in the notation. The function $\mathcal E_\tau$ is nonincreasing in $\tau$ and bounded by two. Observe that the second term in \eqref{e:met-envelope-function} decays faster than every power of $1 + R$, since its logarithm behaves like $-(\log(1 + R))^2/(32 \Theta)$ as $R \to \infty$. In particular, $(1 + R)^q \mathcal E_\tau(R)$ is integrable over $[0, \infty)$ for all $q \ge 0$ and $\tau > 0$.

\begin{lemma}\label{l:met-envelope}
Let $X$, $\gamma$, $T$, $\Theta$, and $R_t$ be as in \zcref{l:met-ito}, and let $\lambda > 0$, $n \ge 1$, and $q \ge 0$. Let $0 \le t_1 \le \dots \le t_n \le T$ and $r_1, \dots, r_n \ge 0$, and set $\tau = t_n + r_1 + \dots + r_n$. Then
\begin{equation}\label{e:met-envelope}
\E \bigg[ \prod_{i = 1}^n (1 + R_{t_i})^q e^{-\lambda r_i R_{t_i}} \exp\Big( -\lambda \int_0^{t_n} R_s \dd s \Big) \bigg] \le e^{2 n q (n q + 1) \Theta} (1 + R_0)^{n q} \, \mathcal E_\tau(R_0) .
\end{equation}
\end{lemma}

\begin{proof}
Let $G$ denote the event that $1 + R_t \ge (1 + R_0)^{1/2}$ for all $t \le T$. The exponential factors in \eqref{e:met-envelope} satisfy, on $G$,
\[
\prod_{i = 1}^n e^{-\lambda r_i R_{t_i}} \exp\Big( -\lambda \int_0^{t_n} R_s \dd s \Big) \le \exp\big( -\lambda \tau \big( (1 + R_0)^{1/2} - 1 \big) \big) .
\]
Using Hölder's inequality and \eqref{e:met-moments} with $m = nq$, we find
\[
\E \prod_{i = 1}^n (1 + R_{t_i})^q \le \prod_{i = 1}^n \Big( \E (1 + R_{t_i})^{n q} \Big)^{1/n} \le e^{n q (n q + 2) \Theta} (1 + R_0)^{n q} .
\]
We bound the exponential factors by one on the complement of $G$. By the Cauchy--Schwarz inequality, Hölder's inequality, \eqref{e:met-moments} with $m = 2nq$, and \eqref{e:met-log-tail}, we have
\[
\E \Big[ \prod_{i = 1}^n (1 + R_{t_i})^q \1_{G^c} \Big] \le e^{n q (2 n q + 2) \Theta} (1 + R_0)^{n q} \exp\Big( - \frac{\big( \frac12 \log(1 + R_0) - \Theta \big)_+^2}{8 \Theta} \Big) .
\]
Adding the two bounds yields \eqref{e:met-envelope}.
\end{proof}

The expansions of the next section rest on the following form of Duhamel's formula.

\begin{lemma}\label{l:met-duhamel}
Let $X > 0$, $\Re z > 0$, and $T > 0$. Let $u \colon [0, T] \times D \to \C$ be such that $u(t, \cdot) \in C^\infty(D)$ for all $t$, the functions $u$, $\partial_t u$, $\partial_B u$, $\partial_F u$, and $\Delta_D u$ are continuous on $[0, T] \times D$, and there exist $C, q \ge 0$ such that
\[
|u| + |\partial_t u| + |\nabla_D u| + |\Delta_D u| \le C (1 + d)^q \qquad \text{on } [0, T] \times D .
\]
Let $g = \partial_t u - (X \Delta_D - z d) u$. Then
\begin{equation}\label{e:met-duhamel}
e^{T(X \Delta_D - z d)} u(0) = u(T) - \int_0^T e^{(T - r)(X \Delta_D - z d)} g(r) \dd r .
\end{equation}
\end{lemma}

\begin{proof}
Fix $p \in D$, and consider under $\E_p$ the process
\[
N_r = \exp\Big( -z \int_0^r \tilde d_s \dd s \Big) u(T - r, \tilde B_r, \tilde F_r), \qquad 0 \le r \le T .
\]
By \zcref{l:met-ito}(a) and Itô's formula for its product with the factor of finite variation, the process
\[
N_r - N_0 + \int_0^r e^{-z \int_0^s \tilde d_v \dd v} g(T - s, \tilde B_s, \tilde F_s) \dd s, \qquad 0 \le r \le T,
\]
is a local martingale, whose quadratic variation at time $T$ has expectation at most $2X \int_0^T \E_p |\nabla_D u|^2 (T - s, \tilde B_s, \tilde F_s) \dd s$. This expectation is finite by the growth assumption and \eqref{e:met-moments} with $\gamma = 1$, and a local martingale with this property is a martingale with mean zero.

Taking expectations at $r = T$, and using Fubini's theorem, which applies since $g$ has polynomial growth uniformly in $t$, we obtain
\[
\E_p N_T - u(T, p) = - \int_0^T \Big( e^{s(X \Delta_D - z d)} g(T - s) \Big)(p) \dd s .
\]
Since $\E_p N_T = (e^{T(X \Delta_D - z d)} u(0))(p)$, the change of variables $r = T - s$ gives \eqref{e:met-duhamel}.
\end{proof}

\section{Expansion of the singlet state}\label{s:met-singlet}

We now expand the singlet state and the right branch $e^{b(X \Delta_D - zd)} \Psi^{X, z}_c$ of the kernel \eqref{e:intro-K} in powers of $X$. Throughout this section, $\Re z > 0$. Recall the singlet operator $\Delta^{\mathrm s}_D = d \Delta_D d^{-1}$, and let $\C + d \, C^\infty(D)$ denote the space of functions $c + d g$ with $c \in \C$ and $g \in C^\infty(D)$. Since $1/d$ is harmonic for $\Delta_D$ away from the corner, we have
\begin{equation}\label{e:met-singlet-operator}
\Delta^{\mathrm s}_D (c + d g) = d \, \Delta_D g, \qquad c \in \C, \ g \in C^\infty(D),
\end{equation}
away from the corner, and we use the right side to define $\Delta^{\mathrm s}_D$ on $\C + d \, C^\infty(D)$. For $k \ge 0$ and $t \ge 0$, we define functions $\Psi^{[k]}_t$ on $D$ by
\begin{equation}\label{e:met-singlet-recursion}
\Psi^{[0]}_t = e^{-t z d}, \qquad \Psi^{[k]}_t = \int_0^t e^{-(t - r) z d} \, \Delta^{\mathrm s}_D \Psi^{[k - 1]}_r \dd r \quad (k \ge 1).
\end{equation}
We will see that $\Psi^{[k]}_t$ is the coefficient of $X^k$ in the singlet state $\Psi^{X, z}_t$. For $\tau \ge 0$, we call a function on $D$ an \emph{exponential polynomial of order $\tau$} if it is the product of $e^{-\tau z d}$ and a polynomial in $B$ and $F$.

\begin{lemma}\label{l:met-singlet}
The following statements hold.
\begin{enumerate}[label=(\alph*)]
\item For all $k \ge 0$ and $t \ge 0$, we have $\Psi^{[k]}_t \in \C + d \, C^\infty(D)$, and $\Psi^{[k]}_t$ is an exponential polynomial of order $t$, whose polynomial factor has coefficients that are polynomials in $t$ and $z$ and is divisible by $d$ if $k \ge 1$. Moreover, $d^{-1} \Delta^{\mathrm s}_D \Psi^{[k]}_t$ is an exponential polynomial of order $t$, and
\begin{equation}\label{e:met-first-coefficient}
\Delta^{\mathrm s}_D \Psi^{[0]}_t = z^2 t^2 d (B + F) e^{-tzd}, \qquad \Psi^{[1]}_t = \frac{z^2 t^3}{3} \, d (B + F) e^{-tzd} .
\end{equation}
\item Let $X > 0$ and $M \ge 0$. For all $t \ge 0$,
\begin{equation}\label{e:met-singlet-expansion}
\Psi^{X, z}_t = \sum_{k = 0}^M X^k \Psi^{[k]}_t + X^{M + 1} d \, \mathcal R_M(t), \qquad \mathcal R_M(t) = \int_0^t e^{(t - r)(X \Delta_D - z d)} \big[ d^{-1} \Delta^{\mathrm s}_D \Psi^{[M]}_r \big] \dd r .
\end{equation}
\end{enumerate}
\end{lemma}

\begin{proof}
We begin with (a), and first treat $k = 0$. We have $\Psi^{[0]}_t = 1 + d g_t$, where $g_t = (e^{-tzd} - 1)/d$ is an entire function of $d$. Let $\psi$ be a smooth function on $[0, \infty)$. By \eqref{e:met-d-identities}, we have
\[
\Delta_D \psi(d) = (B + F) \big( 2\psi'(d) + d \psi''(d) \big) = (B + F) \, \frac{\dd^2}{\dd r^2} \big( r \psi(r) \big) \Big|_{r = d} .
\]
With $r g_t(r) = e^{-tzr} - 1$, this identity yields $\Delta_D g_t = z^2 t^2 (B + F) e^{-tzd}$, and the first identity in \eqref{e:met-first-coefficient} follows from \eqref{e:met-singlet-operator}. Inserting it into \eqref{e:met-singlet-recursion} gives the second.

Now let $k \ge 2$, and suppose that $\Psi^{[k - 1]}_r = d e^{-rzd} \rho_r$, where $\rho_r$ is a polynomial in $B$ and $F$ whose coefficients are polynomials in $r$ and $z$. The product rule and \eqref{e:met-d-identities} show that $e^{rzd} \Delta_D(e^{-rzd} \rho_r)$ is a polynomial of the same kind. By \eqref{e:met-singlet-operator}, we have $\Delta^{\mathrm s}_D \Psi^{[k - 1]}_r = d \Delta_D(e^{-rzd} \rho_r)$, and inserting this into \eqref{e:met-singlet-recursion}, we see that $\Psi^{[k]}_t$ has the same form, with $t$ in place of $r$ and another polynomial of the same kind. By induction, (a) follows.

For (b), we compare the difference quotients by $d$. By \eqref{e:intro-singlet}, we have
\[
\frac{\Psi^{X, z}_t - 1}{d} = -z \int_0^t e^{v(X \Delta_D - z d)} 1 \dd v .
\]
We set
\[
u(t) = \frac1d \Big( \sum_{k = 0}^M X^k \Psi^{[k]}_t - 1 \Big),
\]
which belongs to $C^\infty(D)$ by (a), and which satisfies $u(0) = 0$, since $\Psi^{[0]}_0 = 1$ and $\Psi^{[k]}_0 = 0$ for $k \ge 1$. Differentiating \eqref{e:met-singlet-recursion} in $t$, we find $\partial_t \Psi^{[0]}_t = -zd \Psi^{[0]}_t$, and $\partial_t \Psi^{[k]}_t = -zd \Psi^{[k]}_t + \Delta^{\mathrm s}_D \Psi^{[k - 1]}_t$ for $k \ge 1$. We sum these identities with the weights $X^k$ and use $\Delta^{\mathrm s}_D (1 + d u(t)) = d \Delta_D u(t)$, which follows from \eqref{e:met-singlet-operator}. This gives
\[
d \, \partial_t u(t) = -z d \big( 1 + d u(t) \big) + X d \Delta_D u(t) - X^{M + 1} \Delta^{\mathrm s}_D \Psi^{[M]}_t ,
\]
and after division by $d$,
\[
\partial_t u(t) - (X \Delta_D - z d) u(t) = -z - X^{M + 1} d^{-1} \Delta^{\mathrm s}_D \Psi^{[M]}_t .
\]

The function $u$ satisfies the assumptions of \zcref{l:met-duhamel} on $[0, t]$, by (a). We apply \zcref{l:met-duhamel} with $T = t$. Using \eqref{e:met-duhamel} and $u(0) = 0$, we find
\[
u(t) = -z \int_0^t e^{(t - r)(X \Delta_D - z d)} 1 \dd r - X^{M + 1} \mathcal R_M(t) = \frac{\Psi^{X, z}_t - 1}{d} - X^{M + 1} \mathcal R_M(t) .
\]
Multiplying by $d$ proves \eqref{e:met-singlet-expansion}.
\end{proof}

We next treat the right branch. For $n \ge 0$ and $b, c \ge 0$, we define
\begin{equation}\label{e:met-branch-recursion}
\Psi^{[0]}_{b, c} = e^{-(b + c) z d}, \qquad \Psi^{[n]}_{b, c} = e^{-bzd} \Psi^{[n]}_c + \int_0^b e^{-(b - r) z d} \Delta_D \Psi^{[n - 1]}_{r, c} \dd r \quad (n \ge 1),
\end{equation}
so that $\Psi^{[n]}_{0, c} = \Psi^{[n]}_c$. By \zcref{l:met-singlet}(a) and the product rule, $\Psi^{[n]}_{b, c}$ is an exponential polynomial of order $b + c$, whose polynomial factor has coefficients that are polynomials in $b$, $c$, and $z$.

\begin{lemma}\label{l:met-branch}
Let $X > 0$, $M \ge 0$, and $b, c \ge 0$. Then
\begin{equation}\label{e:met-branch-expansion}
e^{b(X \Delta_D - z d)} \Psi^{X, z}_c = \sum_{n = 0}^M X^n \Psi^{[n]}_{b, c} + X^{M + 1} \mathcal R_M(b, c),
\end{equation}
where
\[
\mathcal R_M(b, c) = e^{b(X \Delta_D - z d)} \big[ d \, \mathcal R_M(c) \big] + \int_0^b e^{(b - r)(X \Delta_D - z d)} \Delta_D \Psi^{[M]}_{r, c} \dd r .
\]
\end{lemma}

\begin{proof}
We set
\[
u(b) = \sum_{n = 0}^M X^n \Psi^{[n]}_{b, c} .
\]
Differentiating \eqref{e:met-branch-recursion} in $b$, we find
\[
\partial_b u(b) = (X \Delta_D - z d) u(b) - X^{M + 1} \Delta_D \Psi^{[M]}_{b, c},
\]
and by \eqref{e:met-singlet-expansion}, we have $u(0) = \Psi^{X, z}_c - X^{M + 1} d \mathcal R_M(c)$. The function $u$ satisfies the assumptions of \zcref{l:met-duhamel}, and \eqref{e:met-duhamel} implies that
\[
e^{b(X \Delta_D - z d)} u(0) = u(b) + X^{M + 1} \int_0^b e^{(b - r)(X \Delta_D - z d)} \Delta_D \Psi^{[M]}_{r, c} \dd r .
\]
The function $d \mathcal R_M(c)$ has polynomial growth by \eqref{e:met-fk-domination}, \zcref{l:met-singlet}(a), and \eqref{e:met-moments}, and the semigroup may be applied to it. Inserting the expression for $u(0)$ into this identity gives \eqref{e:met-branch-expansion}.
\end{proof}

\section{The kernel and the function $I_X$}\label{s:met-scalar}

We now insert the expansions of \zcref{s:met-singlet} into the definition \eqref{e:intro-K} of $K_X$. For $0 < x \le y < 1$, we write $a = x$, $b = y - x$, and $c = 1 - y$, so that $(a, b, c)$ ranges over the time simplex $a, b, c \ge 0$, $a + b + c = 1$, with $a, c > 0$. For $m \ge 0$, we define
\begin{equation}\label{e:met-kernel-coefficients}
K^{[m]}(x, y; z) = \int_D \sum_{i + n = m} \Psi^{[i]}_a \Psi^{[n]}_{b, c} \dd B \dd F, \qquad I^{[m]}(z) = \int_0^1 \int_x^1 K^{[m]}(x, y; z) \dd y \dd x .
\end{equation}
The integrand of $K^{[m]}$ is an exponential polynomial of order $a + b + c = 1$, and these integrals converge absolutely (see \eqref{e:met-radial} below), and $K^{[m]}$ extends continuously to the closed simplex.

\begin{proposition}\label{p:met-kernel}
Let $X_0 > 0$ and $M \ge 0$, and let $\mathcal K \subset \{z \in \C : \Re z > 0\}$ be compact. There exists $C > 0$ such that for all $X \in (0, X_0]$, $z \in \mathcal K$, and $0 < x \le y < 1$,
\begin{equation}\label{e:met-kernel-expansion}
\Big| K_X(x, y; z) - \sum_{m = 0}^M X^m K^{[m]}(x, y; z) \Big| \le C X^{M + 1} .
\end{equation}
Consequently,
\begin{equation}\label{e:met-scalar-expansion}
\Big| I_X(z) - \sum_{m = 0}^M X^m I^{[m]}(z) \Big| \le C X^{M + 1} .
\end{equation}
\end{proposition}

\begin{proof}
Let $\alpha = \min_{\mathcal K} \Re z$, which is positive. By \eqref{e:met-singlet-expansion} with $t = a$ and \eqref{e:met-branch-expansion}, we have
\[
\Psi^{X, z}_a \, e^{b(X \Delta_D - z d)} \Psi^{X, z}_c = \sum_{m = 0}^M X^m \sum_{i + n = m} \Psi^{[i]}_a \Psi^{[n]}_{b, c} + X^{M + 1} \mathcal R ,
\]
where
\begin{equation}\label{e:met-all-remainders}
\mathcal R = d \, \mathcal R_M(a) \, e^{b(X \Delta_D - z d)} \Psi^{X, z}_c + \Big( \sum_{i = 0}^M X^i \Psi^{[i]}_a \Big) \mathcal R_M(b, c) + \sum_{\substack{0 \le i, n \le M \\ i + n \ge M + 1}} X^{i + n - M - 1} \Psi^{[i]}_a \Psi^{[n]}_{b, c} .
\end{equation}
It suffices to show that $\int_D |\mathcal R| \dd B \dd F$ is bounded uniformly in $X$, $z$, and the simplex. We bound each factor in \eqref{e:met-all-remainders} pointwise with \zcref{l:met-envelope}, applied to the radial diffusion with $\gamma = 1$, $T = 1$, $\Theta = 3 X_0$, and $\lambda = \alpha$. Below, $C$ and $q$ denote constants depending only on $X_0$, $M$, and $\mathcal K$, whose values may change from line to line.

We first describe the form in which the semigroups appear. Let $f_1, \dots, f_n$ be exponential polynomials of orders $r_1, \dots, r_n$ whose polynomial factors have coefficients bounded for $z \in \mathcal K$, and let $s_1, \dots, s_n \ge 0$ with $s_1 + \dots + s_n \le 1$. Consider the function
\[
e^{s_1 (X \Delta_D - z d)} \Big[ f_1 \, e^{s_2 (X \Delta_D - z d)} \big[ f_2 \cdots e^{s_n (X \Delta_D - z d)} f_n \big] \Big] .
\]
The Markov property and \eqref{e:met-fk-domination} bound its absolute value at $p$ by a constant times the left side of \eqref{e:met-envelope}, with $t_i = s_1 + \dots + s_i$. \zcref[S]{l:met-envelope} therefore bounds it by $C (1 + d(p))^q \mathcal E_\tau(d(p))$, where
\[
\tau = s_1 + \dots + s_n + r_1 + \dots + r_n .
\]
We call $\tau$ the total time of such an expression.

We now bound the factors of the first two terms of \eqref{e:met-all-remainders}. Recall from \zcref{l:met-singlet}(a) that $\Psi^{[i]}_a$ is an exponential polynomial of order $a$. Using $(1 + d)^{1/2} - 1 \le d$, we obtain
\[
|\Psi^{[i]}_a| \le C (1 + d)^q e^{-\alpha a d} \le C (1 + d)^q \mathcal E_a(d) .
\]
The function $\mathcal R_M(a)$ in \eqref{e:met-singlet-expansion} is an integral over $r \in [0, a]$ of expressions of total time $(a - r) + r = a$. It follows that
\[
|d \mathcal R_M(a)| \le C (1 + d)^q \mathcal E_a(d) .
\]

For the right branch, we recall from \eqref{e:met-singlet-expansion} with $M = 0$ that $\Psi^{X, z}_c = e^{-czd} + X d \mathcal R_0(c)$, and therefore
\[
e^{b(X \Delta_D - z d)} \Psi^{X, z}_c = e^{b(X \Delta_D - z d)} e^{-czd} + X \int_0^c e^{b(X \Delta_D - z d)} \Big[ d \, e^{(c - r)(X \Delta_D - z d)} \big[ d^{-1} \Delta^{\mathrm s}_D \Psi^{[0]}_r \big] \Big] \dd r .
\]
Both terms are expressions, or integrals of expressions, of total time $b + c$, so that
\[
\big| e^{b(X \Delta_D - zd)} \Psi^{X, z}_c \big| \le C (1 + d)^q \mathcal E_{b + c}(d) .
\]
The same argument applies to the two terms of $\mathcal R_M(b, c)$ in \zcref{l:met-branch}. In the first term, the times add up to $b + (c - r) + r = b + c$. In the second, $\Delta_D \Psi^{[M]}_{r, c}$ is an exponential polynomial of order $r + c$, and the times add up to $(b - r) + (r + c) = b + c$. We conclude that
\[
|\mathcal R_M(b, c)| \le C (1 + d)^q \mathcal E_{b + c}(d) .
\]

Finally, each term of the last sum in \eqref{e:met-all-remainders} is an exponential polynomial of order $a + b + c = 1$. Combining these bounds, we find
\[
|\mathcal R| \le C (1 + d)^{2q} \big( \mathcal E_a(d) \, \mathcal E_{b + c}(d) + \mathcal E_1(d) \big) \le 3C (1 + d)^{2q} \mathcal E_{1/2}(d),
\]
where we used that $\max(a, b + c) \ge 1/2$, that $\mathcal E_\tau$ is nonincreasing in $\tau$, and that $\mathcal E_\tau \le 2$. (When $a$ is small, the left factor decays slowly, but then $b + c \ge 1/2$, and the right factor provides the decay.)

It remains to integrate over $D$. For a nonnegative measurable function $\psi$ on $[0, \infty)$, the change of variables to $d = B - F$ and $B + F$ gives
\begin{equation}\label{e:met-radial}
\int_D \psi(d) \dd B \dd F = \int_0^\infty \psi(d) \min(d, 2) \dd d .
\end{equation}
Applying \eqref{e:met-radial} with $\psi = (1 + d)^{2q} \mathcal E_{1/2}(d)$, which is integrable, we find that $\int_D |\mathcal R| \dd B \dd F \le C$. We have proved \eqref{e:met-kernel-expansion}. Integrating it over the simplex, we find \eqref{e:met-scalar-expansion}.
\end{proof}

We now evaluate the coefficients, using the variables $d = B - F$ and $B + F$ on $D$. For fixed $d \ge 0$, the variable $B + F$ ranges over $[|d - 2|, d + 2]$, and $\dd B \dd F = \frac12 \dd d \, \dd (B + F)$. It follows that for $j \ge 0$ and a measurable function $\psi$ on $[0, \infty)$ such that $|\psi(d)| (1 + d)^{j + 1}$ is integrable,
\begin{equation}\label{e:met-fiber}
\int_D \psi(d) (B + F)^j \dd B \dd F = \int_0^\infty \psi(d) \, \frac{(d + 2)^{j + 1} - |d - 2|^{j + 1}}{2 (j + 1)} \dd d .
\end{equation}
For $j = 0$, $1$, and $2$, the fraction in \eqref{e:met-fiber} takes the values
\[
\min(d, 2), \qquad 2d, \qquad
\begin{cases}
d^3/3 + 4d, & d \le 2, \\
2d^2 + 8/3, & d \ge 2,
\end{cases}
\]
respectively. For $m \ge 0$, we also define
\[
K^{[m]}_\triangle(B, F; z) = e^{zd} \int_0^1 \int_0^{1 - a} \sum_{i + n = m} \Psi^{[i]}_a \Psi^{[n]}_{b, 1 - a - b} \dd b \dd a ,
\]
which is a polynomial in $B$ and $F$, since the integrand is an exponential polynomial of order one. By Fubini's theorem, we have
\begin{equation}\label{e:met-scalar-coefficient}
I^{[m]}(z) = \int_D e^{-zd} K^{[m]}_\triangle \dd B \dd F .
\end{equation}

\begin{lemma}\label{l:met-coefficients}
For $\Re z > 0$ and $0 < x \le y < 1$,
\begin{equation}\label{e:met-coefficients}
K^{[0]}(x, y; z) = \frac{1 - e^{-2z}}{z^2}, \qquad I^{[0]}(z) = I_{\GUE}(z), \qquad I^{[1]}(z) = 0, \qquad I^{[2]}(z) = \frac{1 - e^{-2z}}{45} .
\end{equation}
\end{lemma}

\begin{proof}
By the definitions \eqref{e:met-singlet-recursion} and \eqref{e:met-branch-recursion}, we have
\[
\Psi^{[0]}_a \Psi^{[0]}_{b, c} = e^{-azd} e^{-(b + c)zd} = e^{-zd},
\]
and \eqref{e:met-radial} gives
\[
K^{[0]}(x, y; z) = \int_0^\infty e^{-zd} \min(d, 2) \dd d = \frac{1 - e^{-2z}}{z^2} .
\]
Since the simplex has area $1/2$, this identity implies that $I^{[0]} = I_{\GUE}$.

We next compute $K^{[1]}_\triangle$. The sum in its definition has the three terms $\Psi^{[1]}_a \Psi^{[0]}_{b, c}$, $\Psi^{[0]}_a e^{-bzd} \Psi^{[1]}_c$, and $\Psi^{[0]}_a$ times the integral in \eqref{e:met-branch-recursion} with $n = 1$. By \eqref{e:met-first-coefficient}, the first two terms are given by
\[
\Psi^{[1]}_a \Psi^{[0]}_{b, c} = \frac{z^2 a^3}{3} \, d (B + F) e^{-zd}, \qquad \Psi^{[0]}_a e^{-bzd} \Psi^{[1]}_c = \frac{z^2 c^3}{3} \, d (B + F) e^{-zd} .
\]

For the third term, we use the identity
\[
\Delta_D e^{-\tau zd} = \big( \tau^2 z^2 d (B + F) - 2 \tau z (B + F) \big) e^{-\tau zd}, \qquad \tau \ge 0,
\]
which follows from \eqref{e:met-d-identities}, and we find
\[
\int_0^b e^{-(b - r) zd} \Delta_D e^{-(r + c) zd} \dd r = e^{-(b + c) zd} (B + F) \Big( \frac{z^2 d}{3} \big( (b + c)^3 - c^3 \big) - z \big( (b + c)^2 - c^2 \big) \Big) .
\]
Adding the three terms and using $b + c = 1 - a$, we find
\[
K^{[1]}_\triangle = (B + F) \int_0^1 \int_0^{1 - a} \Big( \frac{z^2 d}{3} \big( a^3 + (1 - a)^3 \big) - z \big( (1 - a)^2 - c^2 \big) \Big) \dd b \dd a, \qquad c = 1 - a - b .
\]
Integrating componentwise over the simplex, we have
\[
\int_0^1 \int_0^{1 - a} \big( a^3, (1 - a)^3, (1 - a)^2, c^2 \big) \dd b \dd a = \Big( \frac{1}{20}, \frac15, \frac14, \frac{1}{12} \Big),
\]
where for $c^2$ we used that $a$, $b$, and $c$ play symmetric roles. Inserting these values, we find
\[
K^{[1]}_\triangle = (B + F) \Big( \frac{d z^2}{12} - \frac{z}{6} \Big) .
\]
Inserting this into \eqref{e:met-scalar-coefficient}, and using \eqref{e:met-fiber} with $j = 1$ and the integrals
\begin{equation}\label{e:met-laplace-powers}
\int_0^\infty e^{-zd} d^n \dd d = \frac{n!}{z^{n + 1}}, \qquad n \ge 0,
\end{equation}
we obtain
\[
I^{[1]}(z) = \int_0^\infty e^{-zd} \Big( \frac{d z^2}{12} - \frac{z}{6} \Big) 2d \dd d = \frac{z^2}{6} \cdot \frac{2}{z^3} - \frac{z}{3} \cdot \frac{1}{z^2} = 0 .
\]

The second-order term is computed in the same way, and we give the steps. The sum in the definition of $K^{[2]}_\triangle$ has the three terms $\Psi^{[2]}_a \Psi^{[0]}_{b, c}$, $\Psi^{[1]}_a \Psi^{[1]}_{b, c}$, and $\Psi^{[0]}_a \Psi^{[2]}_{b, c}$. Besides \eqref{e:met-d-identities}, we use the identities
\[
\begin{gathered}
\Delta_D (B + F) = 2d, \qquad |\nabla_D (B + F)|^2 = d (B + F), \\
2 \langle \nabla_D d, \nabla_D (B + F) \rangle = d^2 + (B + F)^2 - 4,
\end{gathered}
\]
which follow from $\Delta_D B = 2B$, $\Delta_D F = -2F$, and the definition of $\langle \nabla_D \cdot, \nabla_D \cdot \rangle$. Inserting them into the complex product rule $\Delta_D(fg) = f \Delta_D g + g \Delta_D f + 2 \langle \nabla_D f, \nabla_D \overline g \rangle$, we find, for $w \in \C$,
\begin{equation}\label{e:met-laplace-moments}
\begin{aligned}
e^{w d} \Delta_D \big( (B + F) e^{-w d} \big) &= (w^2 d - 3w) (B + F)^2 - w d^2 + 2d + 4w, \\
e^{w d} \Delta_D \big( d (B + F) e^{-w d} \big) &= (w^2 d^2 - 5 w d + 3) (B + F)^2 + 3d^2 - w d^3 + 4 w d - 4 .
\end{aligned}
\end{equation}
Below, all integrals over the simplex are over $a \in [0, 1]$ and $b \in [0, 1 - a]$, with $c = 1 - a - b$, and we will use the identities
\begin{equation}\label{e:met-simplex-moments}
\int_0^1 \int_0^{1 - a} a^n \dd b \dd a = \int_0^1 \tau^n (1 - \tau) \dd \tau = \frac{1}{(n + 1)(n + 2)}, \qquad n \ge 0 .
\end{equation}

We begin with the first term. Using \eqref{e:met-first-coefficient} and \eqref{e:met-singlet-operator}, we find
\[
\Delta^{\mathrm s}_D \Psi^{[1]}_r = \frac{z^2 r^3}{3} \, d \, \Delta_D \big( (B + F) e^{-rzd} \big) .
\]
We insert the first identity in \eqref{e:met-laplace-moments} with $w = rz$ into \eqref{e:met-singlet-recursion} with $k = 2$, and we integrate over $r \in [0, t]$, which gives
\[
e^{tzd} \Psi^{[2]}_t = \Big( \frac{z^4 t^6 d^2}{18} - \frac{z^3 t^5 d}{5} \Big) (B + F)^2 + \frac{z^2 t^4 d^2}{6} + \frac{z^3 t^5 (4d - d^3)}{15} .
\]
We have $e^{zd} \Psi^{[0]}_{b, c} = e^{azd}$, and integrating this function with $t = a$ over the simplex with \eqref{e:met-simplex-moments} gives the first line of \eqref{e:met-second-terms} below.

For the second term, we add the two terms of $\Psi^{[1]}_{b, c}$ computed above, which gives
\begin{equation}\label{e:met-branch-first}
e^{(b + c) zd} \Psi^{[1]}_{b, c} = (B + F) \Big( \frac{z^2 d}{3} (b + c)^3 - z \big( (b + c)^2 - c^2 \big) \Big) .
\end{equation}
We multiply by $e^{azd} \Psi^{[1]}_a = (z^2 a^3/3) \, d (B + F)$, use $b + c = 1 - a$, and integrate over the simplex with the values
\[
\int_0^1 \int_0^{1 - a} \big( a^3 (1 - a)^3, a^3 (1 - a)^2, a^3 c^2 \big) \dd b \dd a = \Big( \frac{1}{280}, \frac{1}{140}, \frac{1}{420} \Big),
\]
which follow from the Beta integral. This gives the second line of \eqref{e:met-second-terms}.

For the third term, we recall from \eqref{e:met-branch-recursion} that
\[
\Psi^{[2]}_{b, c} = e^{-bzd} \Psi^{[2]}_c + \int_0^b e^{-(b - r) zd} \Delta_D \Psi^{[1]}_{r, c} \dd r .
\]
The part $\Psi^{[0]}_a e^{-bzd} \Psi^{[2]}_c = e^{-zd} e^{czd} \Psi^{[2]}_c$ has the same simplex integral as the first term, since $a$ and $c$ play symmetric roles.

In the second part, we substitute $\tau = r + c$, and we integrate over $(b, c)$ instead of $(a, b)$, which preserves the Lebesgue measure. For $0 \le c \le \tau \le 1$, the variable $b$ then ranges over $[\tau - c, 1 - c]$, an interval of length $1 - \tau$. Multiplied by $e^{zd}$, the second part therefore has the simplex integral
\[
\int_0^1 (1 - \tau) \int_0^\tau G(\tau, c) \dd c \dd \tau, \qquad G(\tau, c) = e^{\tau zd} \Delta_D \Big( (B + F) e^{-\tau zd} \Big( \frac{z^2 d}{3} \tau^3 - z (\tau^2 - c^2) \Big) \Big),
\]
where we used \eqref{e:met-branch-first} with $(r, c)$ in place of $(b, c)$. Inserting \eqref{e:met-laplace-moments} with $w = \tau z$ and using $\int_0^\tau (\tau^2 - c^2) \dd c = 2\tau^3/3$, we obtain
\[
\begin{aligned}
\int_0^\tau G(\tau, c) \dd c = {} & \frac{z^2 \tau^4}{3} \big( z^2 \tau^2 d^2 - 7 z \tau d + 9 \big) (B + F)^2 \\
& + \frac{z^2 \tau^4}{3} (5 d^2 - 12) + \frac{z^3 \tau^5}{3} (4d - d^3) - \frac{4 z \tau^3 d}{3} .
\end{aligned}
\]
Integrating against $1 - \tau$ with \eqref{e:met-simplex-moments} and adding the first part, we obtain the third line of
\begin{equation}\label{e:met-second-terms}
\begin{aligned}
\int_0^1 \int_0^{1 - a} e^{zd} \Psi^{[2]}_a \Psi^{[0]}_{b, c} \dd b \dd a &= \frac{z^2 d^2}{180} + \frac{z^3 (4d - d^3)}{630} + (B + F)^2 \Big( \frac{z^4 d^2}{1008} - \frac{z^3 d}{210} \Big), \\
\int_0^1 \int_0^{1 - a} e^{zd} \Psi^{[1]}_a \Psi^{[1]}_{b, c} \dd b \dd a &= (B + F)^2 \Big( \frac{z^4 d^2}{2520} - \frac{z^3 d}{630} \Big), \\
\int_0^1 \int_0^{1 - a} e^{zd} \Psi^{[0]}_a \Psi^{[2]}_{b, c} \dd b \dd a &= -\frac{zd}{15} - \frac{2 z^2}{15} + \frac{11 z^2 d^2}{180} + \frac{z^3 (4d - d^3)}{105} \\
&\qquad + (B + F)^2 \Big( \frac{z^2}{10} - \frac{19 z^3 d}{315} + \frac{z^4 d^2}{144} \Big) .
\end{aligned}
\end{equation}
Adding the three lines, we find
\begin{equation}\label{e:met-second-polynomial}
K^{[2]}_\triangle = -\frac{dz}{15} + \frac{(d^2 - 2) z^2}{15} + \frac{(4d - d^3) z^3}{90} + (B + F)^2 \Big( \frac{z^2}{10} - \frac{d z^3}{15} + \frac{d^2 z^4}{120} \Big) .
\end{equation}

It remains to integrate over $D$. We write the right side of \eqref{e:met-second-polynomial} as $\psi_0(d) + \psi_2(d) (B + F)^2$, and we insert it into \eqref{e:met-scalar-coefficient} with $m = 2$. For $j = 0$ and $j = 2$, we write the fractions in \eqref{e:met-fiber} as
\[
\min(d, 2) = d - (d - 2)_+, \qquad \frac{(d + 2)^3 - |d - 2|^3}{6} = \frac{d^3}{3} + 4d - \frac{(d - 2)_+^3}{3} .
\]
Substituting $d = 2 + u$ in the integrals over $\{d > 2\}$, we find
\[
\begin{aligned}
I^{[2]}(z) = {} & \int_0^\infty e^{-zd} \Big( \psi_0(d) \, d + \psi_2(d) \Big( \frac{d^3}{3} + 4d \Big) \Big) \dd d \\
& - e^{-2z} \int_0^\infty e^{-zu} \Big( \psi_0(2 + u) \, u + \psi_2(2 + u) \frac{u^3}{3} \Big) \dd u .
\end{aligned}
\]
Expanding the polynomial factors of the two integrands, we find
\[
\begin{aligned}
\psi_0(d) \, d + \psi_2(d) \Big( \frac{d^3}{3} + 4d \Big) &= \frac{4z^2}{15} d - \Big( \frac{z}{15} + \frac{2z^3}{9} \Big) d^2 + \Big( \frac{z^2}{10} + \frac{z^4}{30} \Big) d^3 - \frac{z^3}{30} d^4 + \frac{z^4}{360} d^5, \\
\psi_0(2 + u) \, u + \psi_2(2 + u) \frac{u^3}{3} &= \frac{2(z^2 - z)}{15} u + \Big( \frac{4z^2}{15} - \frac{z}{15} - \frac{4z^3}{45} \Big) u^2 + \Big( \frac{z^2}{10} - \frac{z^3}{9} + \frac{z^4}{90} \Big) u^3 \\
&\qquad + \Big( \frac{z^4}{90} - \frac{z^3}{30} \Big) u^4 + \frac{z^4}{360} u^5 .
\end{aligned}
\]
By \eqref{e:met-laplace-powers}, the terms of order $z^{-2}$ and $z^{-1}$ cancel in both integrals, and the constant terms add up to $1/45$. We conclude that $I^{[2]}(z) = (1 - e^{-2z})/45$.
\end{proof}

\begin{remark}\label{r:met-script}
The supplementary script \texttt{metallic\_coefficients.py} confirms the computations in the proof of \zcref{l:met-coefficients} in exact arithmetic. It implements \eqref{e:met-singlet-recursion}, \eqref{e:met-branch-recursion}, the integrals over the simplex, and the integrals over $D$, the latter both through \eqref{e:met-fiber} and directly in the coordinates $(B, F)$. It confirms \eqref{e:met-second-polynomial} and the four values in \eqref{e:met-coefficients}. (All the functions involved are holomorphic in $z$, and the script evaluates them for real $z > 0$.)
\end{remark}

\begin{proof}[Proof of \zcref{p:metallic}]
The expansion \eqref{e:intro-metallic-C} is \zcref{l:met-profile}. We apply \zcref{p:met-kernel} with $M = 2$ and $X_0 = 1$. Using \zcref{l:met-coefficients}, we obtain
\[
I_X(z) = I_{\GUE}(z) + \frac{X^2}{45} \big( 1 - e^{-2z} \big) + O(X^3),
\]
uniformly for $z$ in compact subsets of $\{\Re z > 0\}$, which is \eqref{e:intro-metallic-I}.
\end{proof}

The case $M = 0$ of \zcref{p:met-kernel} will be used in \zcref{s:scales}.

\begin{corollary}\label{c:met-kernel}
Let $\mathcal K \subset \{z \in \C : \Re z > 0\}$ be compact. There exists $C > 0$ such that for all $X \in (0, 1]$, $z \in \mathcal K$, and $0 < x \le y < 1$,
\[
\Big| K_X(x, y; z) - \frac{1 - e^{-2z}}{z^2} \Big| \le C X .
\]
\end{corollary}

\begin{proof}
This is \eqref{e:met-kernel-expansion} with $M = 0$ and $X_0 = 1$, together with the first identity in \eqref{e:met-coefficients}.
\end{proof}

\section{The coefficients and the diffusion modes}\label{s:met-modes}

We now express the coefficients through the Neumann Laplacian on $[0, 1]$, with eigenvalues $0$ and $(\pi n)^2$ for $n \ge 1$ and normalized eigenfunctions $1$ and $\sqrt 2 \cos(\pi n t)$. Recall from \zcref{s:intro-results} that
\[
G_{\mathrm N}(x, y) = \sum_{n \ge 1} \frac{2 \cos(\pi n x) \cos(\pi n y)}{(\pi n)^2},
\]
so that the integral operator with kernel $G_{\mathrm N}$ satisfies $\Tr G_{\mathrm N}^p = \sum_{n \ge 1} (\pi n)^{-2p}$ for $p \ge 1$.

\begin{remark}\label{r:met-traces}
The coefficients in \zcref{p:metallic} are spectral quantities of $G_{\mathrm N}$. The first correction to $\mathcal C_X$ is the kernel of $2X G_{\mathrm N}$. In $I_X$, the coefficient $I^{[1]}$ of $X$ vanishes, and the coefficient of $X^2$ satisfies $I^{[2]}(z) = (1 - e^{-2z})/45$ by \zcref{l:met-coefficients}, with
\[
\frac{1}{45} = 2 \sum_{n \ge 1} \frac{1}{(\pi n)^4} = 2 \Tr G_{\mathrm N}^2 .
\]
Carrying the recursion one order further, the supplementary script of \zcref{r:met-script} gives $I^{[3]}(z) = -8z/945 = -8 z \Tr G_{\mathrm N}^3$.

In the metallic perturbation theory of the physics literature, $2X G_{\mathrm N}$ has the form of the diffusion propagator of a wire with hard-wall ends. In the normalization of Mirlin's review \cite{mirlin2000statistics}, in which the dimensionless conductance is $g_{\rm M}$, the propagator on its diagonal is $(2/g_{\rm M}) G_{\mathrm N}(x, x)$ \cite[(3.55)]{mirlin2000statistics}, and the correction to the correlation of one eigenfunction at two distinct points is the propagator itself \cite[(3.74)]{mirlin2000statistics}. This agrees with \eqref{e:intro-metallic-C} under the identification $g_{\rm M} = 1/X$ of the conductance, which we make in \zcref{r:met-km}.

The quadratic term has the following meaning for the pair measure $\nu_X$ of \zcref{c:intro-pair}. For $z = \pi \eps - \iu \pi v$, we have
\[
\Re \big( 1 - e^{-2z} \big) = 1 - e^{-2\pi \eps} \cos(2 \pi v) = \int_\R P_\eps(v - u) \, 2 \sin^2(\pi u) \dd u,
\]
where we used $\int_\R P_\eps(v - u) \cos(2\pi u) \dd u = e^{-2\pi\eps} \cos(2\pi v)$. The measure $\nu_{\GUE}(\dd s) = (1 - (\sin \pi s/(\pi s))^2) \dd s$ satisfies
\[
(P_\eps * \nu_{\GUE})(v) = 1 + \Re I_{\GUE}(\pi \eps - \iu \pi v) - P_\eps(v)
\]
by \zcref{l:scales-gue}, which \zcref{s:scales} proves by a direct computation. Together with \eqref{e:intro-pair-transform} and \zcref{p:metallic}, this implies that the Poisson transform $(P_\eps * \nu_X)(v)$ is the sum of the Poisson transforms of the densities $1 - (\sin \pi u/(\pi u))^2$ and $4 X^2 \Tr(G_{\mathrm N}^2) \sin^2(\pi u)$, up to an error $O(X^3)$, uniformly for $\eps$ and $v$ in compact subsets of $(0, \infty)$ and $\R$. This is a statement about the Poisson transforms at positive $\eps$, and we do not prove a corresponding expansion of $\nu_X$ itself.
\end{remark}

\begin{remark}\label{r:met-km}
The quadratic term in \zcref{r:met-traces} has the form of the correction of order $1/g_{\rm KM}^2$ that Kravtsov and Mirlin \cite[(15)]{kravtsov1994level} obtained for the pair correlation of a metallic sample in the unitary class, whose dimensionless conductance $g_{\rm KM}$ is the ratio of the Thouless energy to the mean level spacing. In dimension one, their correction has the coefficient $a_1/(\pi^2 g_{\rm KM}^2)$ with $a_1 = \Tr G_{\mathrm N}^2$ \cite[(13)]{kravtsov1994level}. In their notation, the diffusion modes appear through the denominators $-\iu s + \pi^2 g_{\rm KM} n^2$ \cite[(16)]{kravtsov1994level}, so that their correction is $\pi^{-2} \sum_{n \ge 1} (\pi^2 g_{\rm KM} n^2)^{-2} \sin^2(\pi s)$. The two corrections agree if $g_{\rm KM} = 1/(2\pi X)$. We use this heuristic identification of the conductance in our comparisons with the physics literature. In the normalization of \cite[(2.27)]{mirlin2000statistics}, where the conductance is $g_{\rm M} = 2\pi g_{\rm KM}$, it reads $g_{\rm M} = 1/X$.

This identification can also be read off from our generator by the following heuristic computation, which we do not justify. Let $\varrho \in \R^4$ denote geodesic normal coordinates on $\mathbb H^2 \times S^2$ at the point that corresponds to the corner, so that $d = |\varrho|^2/2 + O(|\varrho|^4)$, and $\Delta_D$ is the Euclidean Laplacian to leading order. In the Gaussian approximation of $X \Delta_D - zd$, a path $\varrho(t)$, $0 \le t \le 1$, with coefficients $\varrho_n$ in the Neumann basis $1, \sqrt 2 \cos(\pi n t)$, has the Feynman--Kac action
\[
\sum_{n \ge 0} \Big( z + \frac{(\pi n)^2}{2X} \Big) \frac{|\varrho_n|^2}{2} .
\]
The spectral variable corresponds to $z = -\iu \pi s$, so that $\pi^{-1}(z + (\pi n)^2/(2X)) = -\iu s + (\pi n)^2/(2\pi X)$, and the comparison with the denominators $-\iu s + \pi^2 g_{\rm KM} n^2$ gives $g_{\rm KM} = 1/(2\pi X)$. Then $\pi^{-2} \sum_n (\pi^2 g_{\rm KM} n^2)^{-2} = 4 X^2 \sum_n (\pi n)^{-4}$, in agreement with the coefficient $2X^2/45$ of $\sin^2(\pi s)$ in \zcref{r:met-traces}. This agreement is a consistency check, not part of our proofs.
\end{remark}