\chapter{The noncompact boundary and Brownian functionals}\label{s:residue}

In this section, we let the damping $z$ of the sigma-model kernels tend to zero. There are two main results. \zcref[S]{t:res-scalar} states that $z K_X(x, y; z)/2 \to \mathcal C_X(x, y)$, and \zcref{l:res-permanent} states that $z^{k - 1} \mathfrak p^{X, z}_k \to k (2k - 3)!! \, M_k^X$ for the permanent pairings $\mathfrak p^{X, z}_k$ of \zcref{d:res-chain}. In \zcref{s:intensity,s:coarse}, these limits identify the eigenvector moments through the extraction theorem \zcref{t:extract-main}, which converts the coefficient of $z^{1 - k}$ at small damping into the density of the diagonal part of a moment measure.

The heuristic is explained in \zcref{s:ov-eigenvectors}, around \eqref{e:ov-gbm}, and we recall it briefly. For small $z$, the killing rate $z d = z(B - F)$ is negligible unless $B$ is of order $1/z$. On this scale, the radial part of the singlet diffusion is a geometric Brownian motion, killed at a rate proportional to its value, and the coordinate $F$ and the angular and anticommuting variables decouple. The identity \eqref{e:ov-laplace} then turns the Laplace transforms of the exponential functional of this geometric Brownian motion into the moment densities $M_k^X$ of \zcref{d:brownian}.

In the physics literature, the same scaling appears in the computation of eigenfunction statistics from the one-dimensional sigma model. As the level broadening tends to zero, the noncompact radial variable is the one that survives, in the scaling form of \cite[(3.9)]{mirlin2000statistics}, and the resulting transfer equation $\partial_\tau W = (z^2 \partial_z^2 - z) W$ of \cite[(3.22)--(3.23)]{mirlin2000statistics} is the backward equation of the killed geometric Brownian motion above; see also \cite{fyodorov1994statistical}. Its stationary solution $2 \sqrt z K_1(2 \sqrt z)$, which goes back to Efetov and Larkin \cite[(55)]{efetov1983kinetics}, is the function $\Upsilon$ of \eqref{e:res-T} below. In this section, we make the passage to this scaling limit rigorous, including the corner of $D$ and, for the permanent pairings, all anticommuting variables. The main difficulty is the second point, and we treat it with a coupling argument in \zcref{s:res-permanent}.

We now outline the section. In \zcref{s:res-diffusions}, we recall the radial and the singlet diffusions of \zcref{s:sigma-scalar}. \zcref[S]{s:res-scaling} introduces the scaled coordinates and the limiting semigroups, and \zcref{s:res-hierarchy} writes all moment densities $M_k^X$ in terms of these semigroups. \zcref[S]{s:res-envelope} proves a tail bound, and \zcref{s:res-scalar} proves \zcref{t:res-scalar}. Finally, \zcref{s:res-chart,s:res-permanent} treat the permanent pairings, for which we need the complete rank-one sigma model in polar coordinates.

\paragraph{Conventions.} Throughout this section, $z > 0$ is real unless stated otherwise, and we set
\[
 \hat z = \frac zX .
\]
We fix $0 < X_- \le X_+ < \infty$ and $\delta \in (0, 1/2]$. Constants $C$ may depend on $X_\pm$ and $\delta$, and on the order $k$ where it appears, and they may change from line to line. We let $D^\circ = (1, \infty) \times (-1, 1)$ denote the interior of $D$.

\section{The radial and the singlet diffusions}\label{s:res-diffusions}

Recall from \zcref{d:kernels} the domain $D$, the function $d = B - F$, the operator $\Delta_D$, the diffusion with generator $X \Delta_D$, which \zcref{s:sigma-scalar} calls the radial diffusion, the semigroup $e^{t (X \Delta_D - z d)}$, and the singlet state $\Psi^{X, z}_t$. Recall from \zcref{s:sigma-scalar} that the coordinates of the radial diffusion are independent diffusions whose finite endpoints are entrance boundaries, so that the radial diffusion started in $D^\circ$ stays in $D^\circ$, and that they solve the stochastic differential equations \eqref{e:res-flat-sde}. Recall also the singlet diffusion of \zcref{l:res-singlet}, which is obtained from the radial diffusion by the Doob transform with the function $1/d$ and is stopped at the corner at its corner time $\tau_{\mathrm c}$. It solves \eqref{e:res-singlet-sde} before $\tau_{\mathrm c}$, and its law is given by the Doob identity \eqref{e:res-doob}. By \eqref{e:res-singlet-fk}, the singlet state is the expectation of its survival factor. All starting points below lie in $D^\circ$.

As in \zcref{s:sigma-scalar}, we let $(\tilde B_t, \tilde F_t)$ and $(B_t, F_t)$ denote the radial and the singlet diffusions, driven by the Brownian motions $\beta^1$, $\beta^2$ and $\beta^{\mathrm s, 1}$, $\beta^{\mathrm s, 2}$ of \eqref{e:res-flat-sde} and \eqref{e:res-singlet-sde}; a pair $(B, F)$ without a time index is a point of $D$. We will use the identity \eqref{e:res-mobius}. The holomorphy and the absolute convergence of the kernels of \zcref{d:kernels} are proved in \zcref{l:sigma-scalar}\ref{i:sigma-scalar-kernels}, where the singlet state is estimated through \zcref{l:res-singlet}(c).

\section{The boundary at infinity}\label{s:res-scaling}

\subsection{Scaled coordinates}

The parameters $X$ and $z$ enter the objects of \zcref{d:kernels} only through $\hat z = z/X$ and the time $Xt$. Indeed, $(\tilde B_{t/X}, \tilde F_{t/X})_{t \ge 0}$ is the radial diffusion with $X = 1$. Using the definition in \zcref{d:kernels}, we find
\[
 e^{t (X \Delta_D - z d)} = e^{Xt (\Delta_D - \hat z d)} , \qquad \Psi^{X, z}_t = \Psi^{1, \hat z}_{Xt} ,
\]
where the second identity follows from the first and \eqref{e:intro-singlet}. From now on, we work with $X = 1$ and the damping $\hat z$, and we use the scaled radial coordinate and the scaled domain
\[
 q = \hat z B , \qquad D_{\hat z} = [\hat z, \infty) \times [-1, 1] ,
\]
and a time $s$ in the scaled variables corresponds to the time $t = s/X$ in \zcref{d:kernels}.

Let $(\tilde B_s, \tilde F_s)$ and $(B_s, F_s)$ be the radial and the singlet diffusions with $X = 1$, started at $(q/\hat z, F)$ with $(q, F) \in (\hat z, \infty) \times (-1, 1)$, and let $\E$ denote the expectation with respect to their laws. We define the \emph{scaled radial diffusion} and the \emph{scaled singlet diffusion} by
\[
 (\tilde q_s, \tilde F_s) = (\hat z \tilde B_s, \tilde F_s) , \qquad
 (q_s, F_s) = (\hat z B_s, F_s) , \qquad s \ge 0 .
\]
Let $\tilde\beta = \beta^1$ and $\beta = \beta^{\mathrm s, 1}$ denote the Brownian motions that drive their first coordinates. We write $\tau_{\mathrm c}$ for the corner time of the singlet. By \eqref{e:res-flat-sde} and \eqref{e:res-singlet-sde},
\begin{equation}\label{e:res-scaled-sde}
\begin{aligned}
 \dd \tilde q_s &= 2 \tilde q_s \dd s + \big( 2 (\tilde q_s^2 - \hat z^2) \big)^{1/2} \dd \tilde\beta_s , \\
 \dd q_s &= 2 \hat z \, \frac{\hat z - q_s F_s}{q_s - \hat z F_s} \dd s + \big( 2 (q_s^2 - \hat z^2) \big)^{1/2} \dd \beta_s , \qquad s < \tau_{\mathrm c} ,
\end{aligned}
\end{equation}
where we used $(1 - BF)/(B - F) = (\hat z - qF)/(q - \hat z F)$. The killing integral becomes
\[
 \hat z \int_0^s d(B_u, F_u) \dd u = \int_0^s (q_u - \hat z F_u) \dd u ,
\]
and the corner corresponds to $(q, F) = (\hat z, 1)$, where $q - \hat z F = 0$. For $s \ge 0$, $(q, F) \in (\hat z, \infty) \times (-1, 1)$, and bounded measurable functions $f$ on $D_{\hat z}$, we define
\begin{equation}\label{e:res-scaled-states}
\begin{gathered}
 \Psi^{(\hat z)}_s(q, F) = \E \exp\Big( -\int_0^s (q_u - \hat z F_u) \dd u \Big) , \\
 \big( \widetilde{\mathcal T}^{(\hat z)}_s f \big)(q, F) = \E \Big[ f(\tilde q_s, \tilde F_s) \exp\Big( -\int_0^s (\tilde q_u - \hat z \tilde F_u) \dd u \Big) \Big] .
\end{gathered}
\end{equation}
Using \zcref{l:res-singlet}(c) and the definition of the radial semigroup, we obtain the exact rescaling
\begin{equation}\label{e:res-rescaling}
 \Psi^{X, z}_t(B, F) = \Psi^{(\hat z)}_{Xt}(\hat z B, F) , \qquad
 \big( e^{t (X \Delta_D - z d)} f \big)(B, F) = \big( \widetilde{\mathcal T}^{(\hat z)}_{Xt} f_{\hat z} \big)(\hat z B, F) ,
\end{equation}
where $f_{\hat z}(q, F) = f(q/\hat z, F)$. In particular, $0 \le \Psi^{(\hat z)}_s \le 1$.

The coefficients of \eqref{e:res-scaled-sde} are close to those of geometric Brownian motion. Indeed, \eqref{e:res-mobius} implies that, for $q \ge \hat z$ and $F \in [-1, 1]$,
\begin{equation}\label{e:res-coefficients}
 \Big| 2 \hat z \, \frac{\hat z - qF}{q - \hat z F} \Big| \le 2 \hat z , \qquad
 0 \le \sqrt 2 \, q - \big( 2 (q^2 - \hat z^2) \big)^{1/2} \le \sqrt 2 \, \hat z ,
\end{equation}
where for the second bound we used
\[
 q - (q^2 - \hat z^2)^{1/2} = \frac{\hat z^2}{q + (q^2 - \hat z^2)^{1/2}} \le \hat z .
\]
Formally setting $\hat z = 0$ in \eqref{e:res-scaled-sde}, we obtain the geometric Brownian motions $\dd q^0_s = \sqrt 2 \, q^0_s \dd \beta_s$ and $\dd \tilde q^0_s = 2 \tilde q^0_s \dd s + \sqrt 2 \, \tilde q^0_s \dd \tilde\beta_s$, killed at the rates $q^0_s$ and $\tilde q^0_s$. We now introduce the corresponding semigroups.

\subsection{The limiting semigroups}

We extend the definition of $g_X$ in \zcref{d:brownian} to all times. For a standard Brownian motion $\beta = (\beta_u)_{u \ge 0}$, we set
\[
 g_1(u) = \sqrt 2 \, \beta_u - u , \qquad A_s = \int_0^s e^{g_1(u)} \dd u , \qquad u, s \ge 0 .
\]
Then $q^0_s = q e^{g_1(s)}$ and $\tilde q^0_s = q e^{g_1(s) + 2s}$ solve the equations above, with $\tilde\beta$ in place of $\beta$ in the second case. For $s \ge 0$, $q > 0$, and measurable functions $f \ge 0$ on $(0, \infty)$, we define
\begin{equation}\label{e:res-T}
 \mathcal T_s f(q) = \E \big[ f(q e^{g_1(s)}) e^{-q A_s} \big] , \qquad
 \Upsilon_s = \mathcal T_s 1 , \qquad
 \Upsilon(q) = 2 \sqrt q \, K_1(2 \sqrt q) ,
\end{equation}
where $K_1$ is the modified Bessel function. The function $\Upsilon_s(q) = \E e^{-q A_s}$ is the Laplace transform of $A_s$, and $\mathcal T_s$ is the semigroup of the geometric Brownian motion $q^0$, killed at the rate $q^0$. We will see in \zcref{l:coarse-side} that $\Upsilon$ is the limit of $\Upsilon_s$ as $s \to \infty$. We also define the differential operator $\mathscr L = -\partial_q (q^2 \partial_q) + q$ on $C_c^\infty((0, \infty))$, and the unitary map $\mathcal U \colon L^2((0, \infty), \dd q) \to L^2(\R, \dd y)$ by $(\mathcal U f)(y) = e^{y/2} f(e^y)$.

\begin{lemma}\label{l:res-limit-semigroups}
\begin{enumerate}[label=(\alph*)]
\item For all $s \ge 0$ and all measurable $f, f' \ge 0$ on $(0, \infty)$,
\[
 \int_0^\infty q^{-2} f'(q) \, \mathcal T_s f(q) \dd q = \int_0^\infty q^{-2} f(q) \, \mathcal T_s f'(q) \dd q .
\]
\item The operator $\mathscr L$ is essentially self-adjoint on $C_c^\infty((0, \infty))$ in $L^2((0, \infty), \dd q)$. We also write $\mathscr L$ for its closure. It satisfies $\mathscr L \ge 1/4$, and $\mathcal U \mathscr L \mathcal U^{-1}$ is the closure of $-\partial_y^2 + e^y + 1/4$ on $C_c^\infty(\R)$.
\item For all $s \ge 0$ and $f \in L^2((0, \infty), \dd q)$, and for almost every $q > 0$,
\begin{equation}\label{e:res-flat-limit}
 (e^{-s \mathscr L} f)(q) = \E \Big[ f(q e^{g_1(s) + 2s}) \exp\Big( -q \int_0^s e^{g_1(u) + 2u} \dd u \Big) \Big] .
\end{equation}
For measurable $f \ge 0$, we use the right side of \eqref{e:res-flat-limit} as the definition of $(e^{-s \mathscr L} f)(q)$ for every $q > 0$.
\item For all $s \ge 0$, $q > 0$, and measurable $f \ge 0$,
\begin{equation}\label{e:res-h-transform}
 \mathcal T_s(q f)(q) = q \, (e^{-s \mathscr L} f)(q) ,
\end{equation}
where $q f$ denotes the function $q \mapsto q f(q)$.
\end{enumerate}
\end{lemma}

The first statement is the symmetry of $\mathcal T_s$ with respect to the speed measure $q^{-2} \dd q$ of $q^0$. The last is the Doob transform of $\mathcal T_s$ by the function $q$, which is harmonic for the generator $q^2 \partial_q^2$ of $q^0$. Under this transform, the driftless geometric Brownian motion $q^0$ becomes the geometric Brownian motion $\tilde q^0$ with generator $q^2 \partial_q^2 + 2q \partial_q$.

\begin{proof}
We begin with (a). Fix a path of $\beta$ on $[0, s]$, and substitute $q = r e^{-g_1(s)}$ in the left side, for which
\[
 q^{-2} \dd q = r^{-2} e^{g_1(s)} \dd r , \qquad q A_s = r \int_0^s e^{g_1(u) - g_1(s)} \dd u .
\]
Tonelli's theorem yields
\[
 \int_0^\infty q^{-2} f'(q) \, \mathcal T_s f(q) \dd q
 = \E \int_0^\infty r^{-2} f(r) \, e^{g_1(s)} f'(r e^{-g_1(s)}) \exp\Big( -r \int_0^s e^{g_1(u) - g_1(s)} \dd u \Big) \dd r .
\]

We now reverse time. The processes
\[
 \hat\beta_u = \beta_s - \beta_{s - u} , \qquad \hat g(u) = \sqrt 2 \, \hat\beta_u - u = g_1(s) - g_1(s - u) , \qquad u \in [0, s] ,
\]
are a standard Brownian motion and a Brownian motion with drift $-1$. In particular,
\[
 g_1(s) = \hat g(s) , \qquad \int_0^s e^{g_1(u) - g_1(s)} \dd u = \int_0^s e^{-\hat g(v)} \dd v .
\]
Next, by the Cameron--Martin theorem \cite{revuz1999continuous}, the variable
\[
 e^{\hat g(s)} = \exp \big( \sqrt 2 \, \hat\beta_s - s \big)
\]
is the density of a measure under which $\hat\beta_u - \sqrt 2 \, u$, $u \le s$, is a standard Brownian motion. Under this measure,
\[
 -\hat g(u) = -\sqrt 2 \big( \hat\beta_u - \sqrt 2 \, u \big) - u
\]
has the law of $g_1$, by the symmetry $\beta \mapsto -\beta$. We obtain
\[
 \int_0^\infty q^{-2} f'(q) \, \mathcal T_s f(q) \dd q = \E \int_0^\infty r^{-2} f(r) f'(r e^{g_1(s)}) e^{-r A_s} \dd r ,
\]
which is the right side of (a).

For (d), we use the Cameron--Martin theorem directly. Under the measure with density $e^{g_1(s)}$, the process
\[
 \beta'_u = \beta_u - \sqrt 2 \, u , \qquad u \le s ,
\]
is a standard Brownian motion, and $g_1(u) = \sqrt 2 \, \beta'_u + u$. This yields
\[
 \mathcal T_s(q f)(q) = q \, \E \big[ e^{g_1(s)} f(q e^{g_1(s)}) e^{-q A_s} \big]
 = q \, \E \Big[ f(q e^{\sqrt 2 \beta'_s + s}) \exp\Big( -q \int_0^s e^{\sqrt 2 \beta'_u + u} \dd u \Big) \Big] ,
\]
and
\[
 \sqrt 2 \, \beta'_u + u = g_1(u) + 2u ,
\]
with $g_1$ built from $\beta'$. This is \eqref{e:res-h-transform}, with the right side of \eqref{e:res-flat-limit} as the definition of $e^{-s \mathscr L} f$.

We now prove (b). The map $\mathcal U$ is unitary, since the substitution $q = e^y$ gives
\[
 \int_\R |e^{y/2} f(e^y)|^2 \dd y = \int_0^\infty |f(q)|^2 \dd q ,
\]
and $\mathcal U$ maps $C_c^\infty((0, \infty))$ onto $C_c^\infty(\R)$. Let $f \in C_c^\infty((0, \infty))$ and $v = \mathcal U f$. The operator $q \partial_q$ corresponds to $\partial_y$, and
\[
 f(e^y) = e^{-y/2} v(y) , \qquad -\partial_q(q^2 \partial_q) = -(q \partial_q)^2 - q \partial_q .
\]
A direct computation gives
\[
 (\mathcal U \mathscr L f)(y) = -v''(y) + \big( e^y + \tfrac14 \big) v(y) .
\]
The operator $-\partial_y^2 + V$ with a nonnegative smooth potential $V$ is essentially self-adjoint on $C_c^\infty(\R)$ \cite{reed1975methods}, and the lower bound follows from
\[
 \int_\R \big( |v'|^2 + (e^y + \tfrac14) |v|^2 \big) \dd y \ge \tfrac14 \|v\|^2 .
\]
Since $\mathcal U$ is unitary, (b) follows.

Finally, we prove (c). Let $\mathscr H$ denote the closure of $-\partial_y^2 + e^y + 1/4$. The potential is continuous and bounded below, and the Feynman--Kac formula \cite{simon2005functional} states that, for $v \in L^2(\R)$ and almost every $y$,
\[
 (e^{-s \mathscr H} v)(y) = e^{-s/4} \, \E \Big[ v(y + \sqrt 2 \, \beta_s) \exp\Big( -\int_0^s e^{y + \sqrt 2 \beta_u} \dd u \Big) \Big] ,
\]
where we used that $\sqrt 2 \, \beta$ has generator $\partial_y^2$. With $v = \mathcal U f$ and $y = \log q$, we find
\[
 (\mathcal U^{-1} e^{-s \mathscr H} \mathcal U f)(q) = \E \Big[ e^{\beta_s/\sqrt 2 - s/4} f(q e^{\sqrt 2 \beta_s}) \exp\Big( -q \int_0^s e^{\sqrt 2 \beta_u} \dd u \Big) \Big] .
\]
Under the measure with density $e^{\beta_s/\sqrt 2 - s/4}$, the Cameron--Martin theorem shows that $\beta'_u = \beta_u - u/\sqrt 2$ is a standard Brownian motion, and
\[
 \sqrt 2 \, \beta_u = \sqrt 2 \, \beta'_u + u = g_1(u) + 2u ,
\]
with $g_1$ built from $\beta'$. By (b), we have
\[
 e^{-s \mathscr L} = \mathcal U^{-1} e^{-s \mathscr H} \mathcal U ,
\]
and we obtain \eqref{e:res-flat-limit}.
\end{proof}

\section{The moment densities as Brownian functionals}\label{s:res-hierarchy}

We now write all moment densities $M_k^X$ of \zcref{d:brownian} in terms of the semigroups of \zcref{l:res-limit-semigroups}. For $k = 2$, this is the identity behind \zcref{t:res-scalar}. For general $k$, it will be used in \zcref{l:res-permanent} and in the long-wire limit of \zcref{s:coarse}.

\begin{lemma}\label{l:res-hierarchy}
Let $k \ge 2$, $X > 0$, and $0 \le x_1 \le \dots \le x_k \le 1$. Set $s_0 = X x_1$, $s_j = X (x_{j + 1} - x_j)$ for $1 \le j \le k - 1$, and $s_k = X (1 - x_k)$. Then
\begin{align}
 \frac{(k - 2)!}{X^{k - 1}} M_k^X(x_1, \dots, x_k)
 &= \int_0^\infty q^{-2} \Upsilon_{s_0}(q) \big[ q \, \mathcal T_{s_1} q \, \mathcal T_{s_2} \cdots q \, \mathcal T_{s_{k - 1}} q \, \Upsilon_{s_k} \big](q) \dd q \notag \\
 &= \int_0^\infty \Upsilon_{s_0}(q) \big[ e^{-s_1 \mathscr L} q \, e^{-s_2 \mathscr L} \cdots q \, e^{-s_{k - 1} \mathscr L} \Upsilon_{s_k} \big](q) \dd q ,
 \label{e:res-hierarchy}
\end{align}
where $q$ denotes multiplication by $q$, there are $k$ such factors in the first line and $k - 2$ in the second, and all operators act on nonnegative functions. All three expressions are finite. In particular, for $k = 2$ and $0 \le x \le y \le 1$,
\begin{equation}\label{e:res-pair}
 \frac{\mathcal C_X(x, y)}{X} = \int_0^\infty \Upsilon_{X x}(q) \big[ e^{-X (y - x) \mathscr L} \Upsilon_{X (1 - y)} \big](q) \dd q .
\end{equation}
\end{lemma}

\begin{proof}
By Brownian scaling, the process $(g_X(t))_{t \in [0, 1]}$ of \zcref{d:brownian} has the law of $(g_1(Xt))_{t \in [0, 1]}$. Under this identification, $Z_X = X^{-1} A_X$. Inserting this into the definition of $M_k^X$, we find
\[
 M_k^X(x_1, \dots, x_k) = X^{k - 1} \, \E \Big[ A_X^{1 - k} \prod_{a = 1}^k e^{g_1(X x_a)} \Big] .
\]
We insert the identity
\[
 A^{1 - k} = \frac{1}{(k - 2)!} \int_0^\infty q^{k - 2} e^{-q A} \dd q , \qquad A > 0 ,
\]
and use Tonelli's theorem. With $q^0_u = q e^{g_1(u)}$, we have
\[
 q A_X = \int_0^X q^0_u \dd u , \qquad \prod_{a = 1}^k e^{g_1(X x_a)} = q^{-k} \prod_{a = 1}^k q^0_{X x_a} ,
\]
and hence
\[
 \frac{(k - 2)!}{X^{k - 1}} M_k^X(x_1, \dots, x_k)
 = \int_0^\infty q^{-2} \, \E \Big[ \prod_{a = 1}^k q^0_{X x_a} \exp\Big( -\int_0^X q^0_u \dd u \Big) \Big] \dd q .
\]

Recall that $q^0$ is a Markov process, killed at the rate $q^0$, with semigroup $\mathcal T_s$. The Markov property at the times $X x_1 \le \dots \le X x_k$ writes the expectation as
\[
 \big[ \mathcal T_{s_0} q \, \mathcal T_{s_1} q \cdots \mathcal T_{s_{k - 1}} q \, \mathcal T_{s_k} 1 \big](q) .
\]
By \zcref{l:res-limit-semigroups}(a), we may move $\mathcal T_{s_0}$ onto the constant function $1$ in the integral against $q^{-2} \dd q$, which turns it into $\Upsilon_{s_0}$. The result is the first line of \eqref{e:res-hierarchy}.

For the second line, we apply \eqref{e:res-h-transform} $k - 1$ times, from the right. The first application gives
\[
 \mathcal T_{s_{k - 1}} (q \Upsilon_{s_k}) = q \, e^{-s_{k - 1} \mathscr L} \Upsilon_{s_k} ,
\]
and each further application moves one factor $q$ to the left through $\mathcal T_{s_j}$ and replaces $\mathcal T_{s_j}$ by $e^{-s_j \mathscr L}$. After $k - 1$ steps, the expression in brackets in the first line becomes
\[
 q^2 \big[ e^{-s_1 \mathscr L} q \cdots q \, e^{-s_{k - 1} \mathscr L} \Upsilon_{s_k} \big] ,
\]
and the factor $q^2$ cancels $q^{-2}$.

The identities hold in $[0, \infty]$, as all expressions are integrals of nonnegative functions. To see that they are finite, we note that Jensen's inequality implies
\[
 Z_X \ge \exp\Big( \int_0^1 g_X(t) \dd t \Big) \ge \exp\Big( -\sup_{t \le 1} |g_X(t)| \Big) .
\]
This bounds the integrand in the definition of $M_k^X$ by
\[
 Z_X^{1 - k} \prod_{a = 1}^k e^{g_X(x_a)} \le \exp\Big( (2k - 1) \sup_{t \le 1} |g_X(t)| \Big) ,
\]
which is integrable because $\sup_{t \le 1} |\beta_t|$ has Gaussian tails.
\end{proof}

\section{A tail envelope}\label{s:res-envelope}

The singlet state is close to one for small $q$ and decays rapidly for large $q$. To integrate the limits of \zcref{s:res-scalar} over $q$, we need a bound for this decay that is uniform in $\hat z$.

\begin{lemma}\label{l:res-envelope}
Let $0 < t_0 \le T$ and $q_* = 4 e^{6T}$. Define $\bar\Upsilon_{t_0, T}(q) = 1$ for $0 < q < q_*$ and
\[
 \bar\Upsilon_{t_0, T}(q) = \exp\big( -t_0 e^{-3T} \sqrt q/2 \big) + \exp\big( -(\log q)^2/(16 T) \big) , \qquad q \ge q_* .
\]
Then for all $0 < \hat z \le 1$, all $(q, F) \in (\hat z, \infty) \times (-1, 1)$ and all $s \in [t_0, T]$,
\begin{equation}\label{e:res-envelope}
 0 \le \Psi^{(\hat z)}_s(q, F) \le \bar\Upsilon_{t_0, T}(q) , \qquad
 0 \le \big( \widetilde{\mathcal T}^{(\hat z)}_s 1 \big)(q, F) \le \bar\Upsilon_{t_0, T}(q) , \qquad
 0 \le \Upsilon_s(q) \le \bar\Upsilon_{t_0, T}(q) .
\end{equation}
Moreover, $\int_0^\infty (1 + q)^r \bar\Upsilon_{t_0, T}(q) \dd q < \infty$ for all $r \ge 0$. In particular, let $\bar\Upsilon = \bar\Upsilon_{X_- \delta, X_+}$. Then for all $X \in [X_-, X_+]$, $0 < z \le X_-$, $t \in [\delta, 1]$, and $(B, F) \in D^\circ$,
\begin{equation}\label{e:res-envelope-original}
 0 \le \Psi^{X, z}_t(B, F) \le \bar\Upsilon(zB/X) , \qquad
 0 \le \big( e^{t (X \Delta_D - z d)} 1 \big)(B, F) \le \bar\Upsilon(zB/X) .
\end{equation}
\end{lemma}

\begin{proof}
The killing rates $q - \hat z F$, $\tilde q - \hat z \tilde F$, and $q^0$ are nonnegative, and all three functions take values in $[0, 1]$. It suffices to consider $q \ge q_*$. We first treat $\Psi^{(\hat z)}_s$. It\^o's formula and \eqref{e:res-scaled-sde} give, on $[0, \tau_{\mathrm c})$,
\begin{gather*}
 \log q_u = \log q + \int_0^u \gamma_v \dd v + M_u , \qquad
 M_u = \int_0^u \Big( 2 - \frac{2 \hat z^2}{q_v^2} \Big)^{1/2} \dd \beta_v , \\
 \gamma_v = 2 \hat z \, \frac{\hat z - q_v F_v}{q_v (q_v - \hat z F_v)} - 1 + \frac{\hat z^2}{q_v^2} .
\end{gather*}
Using \eqref{e:res-coefficients} and $q_v \ge \hat z$, we see that
\[
 -3 \le \gamma_v \le 2 , \qquad \langle M \rangle_u = \int_0^u \Big( 2 - \frac{2 \hat z^2}{q_v^2} \Big) \dd v \le 2u .
\] We extend $M$ to all times by stopping it at $\tau_{\mathrm c}$, and we define the event
\[
 \mathcal A_q = \Big\{ \inf_{u \le T} M_u \ge -\frac12 \log q \Big\} .
\]

We now bound the killing on $\mathcal A_q$. For $u < \tau_{\mathrm c} \wedge T$, the bound $\gamma_v \ge -3$ and $q \ge q_*$ give
\[
 \log q_u \ge \tfrac12 \log q - 3T , \qquad q_u \ge e^{-3T} \sqrt q \ge 2 .
\]
Therefore $\tau_{\mathrm c} > T$ on $\mathcal A_q$, since $q_u \to \hat z \le 1$ as $u \uparrow \tau_{\mathrm c}$. Recalling that $\hat z \le 1$, we obtain on $\mathcal A_q$ that
\[
 q_u - \hat z F_u \ge q_u - 1 \ge \frac{q_u}{2} \ge \frac{e^{-3T} \sqrt q}{2} , \qquad u \le T ,
\]
and the killing integral up to time $s \ge t_0$ is bounded below by $t_0 e^{-3T} \sqrt q/2$.

It remains to bound the probability of $\mathcal A_q^c$. The exponential martingale inequality \cite{revuz1999continuous} and $\langle M \rangle_T \le 2T$ give
\[
 \Prob(\mathcal A_q^c) \le \exp\Big( -\frac{(\frac12 \log q)^2}{2 \cdot 2T} \Big) = \exp\Big( -\frac{(\log q)^2}{16 T} \Big) .
\]
Splitting the expectation in \eqref{e:res-scaled-states} according to $\mathcal A_q$ proves the first bound in \eqref{e:res-envelope}.

The same argument applies to the other two functions. For the radial diffusion, \eqref{e:res-scaled-sde} and It\^o's formula give
\[
 \dd \log \tilde q_v = \Big( 1 + \frac{\hat z^2}{\tilde q_v^2} \Big) \dd v + \Big( 2 - \frac{2 \hat z^2}{\tilde q_v^2} \Big)^{1/2} \dd \tilde\beta_v ,
\]
with drift in $[1, 2]$, and $\tilde q$ never reaches $\hat z$. For $\Upsilon_s$, the logarithm of $q^0_u = q e^{g_1(u)}$ has drift $-1$ and quadratic variation $2u$.

The moments are finite because
\[
 \exp\Big( -\frac{(\log q)^2}{16 T} \Big) = q^{-\log q/(16 T)}
\]
decays faster than every power of $q$. Finally, by \eqref{e:res-rescaling}, we have
\[
 \Psi^{X, z}_t(B, F) = \Psi^{(\hat z)}_{Xt}(zB/X, F) , \qquad \hat z = \frac zX \le 1 , \qquad Xt \in [X_- \delta, X_+] ,
\]
and \eqref{e:res-envelope} implies the first bound in \eqref{e:res-envelope-original}. The second follows in the same way.
\end{proof}

\section{The scalar residue}\label{s:res-scalar}

We now prove the first main result of this section.

\begin{theorem}\label{t:res-scalar}
Let $0 < X_- \le X_+ < \infty$ and $\delta \in (0, 1/2]$. Then
\begin{equation}\label{e:res-scalar}
 \lim_{z \downarrow 0} \frac z2 K_X(x, y; z) = \mathcal C_X(x, y) ,
\end{equation}
uniformly for $X \in [X_-, X_+]$ and $\delta \le x \le y \le 1 - \delta$, where the limit is taken along real $z$.
\end{theorem}

The proof compares the scaled diffusions with the geometric Brownian motions of \zcref{s:res-scaling}, driven by the same Brownian motion. We first record this comparison.

\begin{lemma}\label{l:res-coupling}
For every $T > 0$, there exists $C_T$ such that for all $0 < \hat z \le 1$ and $(q, F) \in (\hat z, \infty) \times (-1, 1)$,
\begin{align}
 \sup_{s \le T} \big| \Psi^{(\hat z)}_s(q, F) - \Upsilon_s(q) \big| &\le C_T \hat z ,
 \label{e:res-side} \\
 \sup_{s, s' \le T} \big| \big( \widetilde{\mathcal T}^{(\hat z)}_s \Psi^{(\hat z)}_{s'} \big)(q, F) - \big( e^{-s \mathscr L} \Upsilon_{s'} \big)(q) \big| &\le C_T \hat z .
 \label{e:res-middle}
\end{align}
\end{lemma}

\begin{proof}
We begin with \eqref{e:res-side}. Let $(q_u, F_u)$ be the scaled singlet diffusion started at $(q, F)$, and recall that $\beta$ drives $q_u$ in \eqref{e:res-scaled-sde}. After the corner time $\tau_{\mathrm c}$, the process $q_u$ is constant. We extend $\beta$ beyond $\tau_{\mathrm c}$ by an independent Brownian motion, on an enlarged probability space, and we compare $q_u$ with the geometric Brownian motion $q^0_u$, where
\[
 q^0_u = q e^{g_1(u)} , \qquad y_u = q_u - q^0_u ,
\]
and $g_1$ is built from $\beta$. Then $y_0 = 0$ and
\[
 \dd y_u = a_u \dd u + (\sqrt 2 \, y_u + e_u) \dd \beta_u , \qquad |a_u| \le 2 \hat z , \quad |e_u| \le \sqrt 2 \, \hat z .
\]
Indeed, for $u < \tau_{\mathrm c}$, the coefficients are
\[
 a_u = 2 \hat z \, \frac{\hat z - q_u F_u}{q_u - \hat z F_u} , \qquad e_u = \big( 2(q_u^2 - \hat z^2) \big)^{1/2} - \sqrt 2 \, q_u ,
\]
and the bounds are \eqref{e:res-coefficients}. For $u \ge \tau_{\mathrm c}$, we have $q_u = \hat z$, $a_u = 0$, and $e_u = -\sqrt 2 \, \hat z$.

By the Burkholder--Davis--Gundy inequality, we have, for $u \le T$,
\[
 \E \sup_{v \le u} |y_v|^2 \le C \Big( T^2 \hat z^2 + T \hat z^2 + \int_0^u \E \sup_{w \le v} |y_w|^2 \dd v \Big) ,
\]
where the left side is finite by the moment bound of \zcref{l:res-singlet}(a). By Gronwall's inequality,
\begin{equation}\label{e:res-coupling}
 \E \sup_{u \le T} |q_u - q^0_u|^2 \le C_T \hat z^2 ,
\end{equation}
with a constant that is uniform in $(q, F)$. Using $|e^{-a} - e^{-a'}| \le |a - a'|$ for $a, a' \ge 0$, we obtain
\[
 \big| \Psi^{(\hat z)}_s(q, F) - \Upsilon_s(q) \big|
 \le \E \Big| \int_0^s (q_u - \hat z F_u) \dd u - \int_0^s q^0_u \dd u \Big|
 \le \int_0^s \E |q_u - q^0_u| \dd u + \hat z s \le C_T \hat z .
\]

We now turn to \eqref{e:res-middle}. The function $\Upsilon_{s'}$ is Lipschitz with constant $s'$, since
\[
 |\partial_q \Upsilon_{s'}(q)| = \E \big[ A_{s'} e^{-q A_{s'}} \big] \le \E A_{s'} = s' .
\]
Let $(\tilde q_u, \tilde F_u)$ be the scaled radial diffusion started at $(q, F)$. We compare it with
\[
 \tilde q^0_u = q e^{g_1(u) + 2u} , \qquad \tilde y_u = \tilde q_u - \tilde q^0_u ,
\]
where $g_1$ is built from $\tilde\beta$. The difference satisfies
\[
 \dd \tilde y_u = 2 \tilde y_u \dd u + (\sqrt 2 \, \tilde y_u + \tilde e_u) \dd \tilde\beta_u , \qquad |\tilde e_u| \le \sqrt 2 \, \hat z ,
\]
and the argument above shows that
\[
 \E \sup_{u \le T} |\tilde y_u|^2 \le C_T \hat z^2 .
\]

We set
\[
 \tilde\kappa = \int_0^s (\tilde q_u - \hat z \tilde F_u) \dd u , \qquad \tilde\kappa^0 = \int_0^s \tilde q^0_u \dd u .
\]
By \eqref{e:res-scaled-states} and \eqref{e:res-flat-limit}, the difference in \eqref{e:res-middle} is
\[
 \E \big[ \Psi^{(\hat z)}_{s'}(\tilde q_s, \tilde F_s) e^{-\tilde\kappa} \big] - \E \big[ \Upsilon_{s'}(\tilde q^0_s) e^{-\tilde\kappa^0} \big] .
\]
All factors lie in $[0, 1]$. Replacing the factors one at a time, we bound the modulus of this difference by
\begin{align*}
 &\E \big| \Psi^{(\hat z)}_{s'}(\tilde q_s, \tilde F_s) - \Upsilon_{s'}(\tilde q_s) \big|
 + \E \big| \Upsilon_{s'}(\tilde q_s) - \Upsilon_{s'}(\tilde q^0_s) \big|
 + \E |\tilde\kappa - \tilde\kappa^0| \\
 &\qquad \le C_T \hat z + T \, \E |\tilde y_s| + \int_0^s \E |\tilde y_u| \dd u + T \hat z \le C_T \hat z ,
\end{align*}
where we used \eqref{e:res-side} for the first term.
\end{proof}

\begin{proof}[Proof of \zcref{t:res-scalar}]
Let $s_0 = X x$, $s_1 = X (y - x)$, and $s_2 = X (1 - y)$. They lie in $[0, X_+]$, and $s_0, s_2 \ge X_- \delta$. In the definition \eqref{e:intro-K} of $K_X$, we use \eqref{e:res-rescaling} and the substitution $q = \hat z B$, for which $\dd B = \dd q/\hat z$ and $z/\hat z = X$. This gives
\begin{equation}\label{e:res-scaled-kernel}
 \frac z2 K_X(x, y; z) = \frac X2 \int_{\hat z}^\infty \int_{-1}^1 \Psi^{(\hat z)}_{s_0}(q, F) \big( \widetilde{\mathcal T}^{(\hat z)}_{s_1} \Psi^{(\hat z)}_{s_2} \big)(q, F) \dd F \dd q .
\end{equation}
By \eqref{e:res-pair}, and since the integrand there is independent of $F$, the target is
\[
 \mathcal C_X(x, y) = \frac X2 \int_0^\infty \int_{-1}^1 \Upsilon_{s_0}(q) \big( e^{-s_1 \mathscr L} \Upsilon_{s_2} \big)(q) \dd F \dd q .
\]
All factors of the two integrands take values in $[0, 1]$.

Let $R \ge 1$, and split the integrals at $q = R$. For $\hat z < q \le R$, \zcref{l:res-coupling} with $T = X_+$ bounds the difference of the integrands by $C \hat z$, uniformly in $F$. The range $q \le \hat z$ contributes at most $X \hat z$ to the target, since its integrand is bounded by one.

For $q > R$, \zcref{l:res-envelope} with $t_0 = X_- \delta$ and $T = X_+$ bounds the first factors of the two integrands by $\bar\Upsilon(q)$. We obtain
\[
 \Big| \frac z2 K_X(x, y; z) - \mathcal C_X(x, y) \Big|
 \le C \Big( R \hat z + \hat z + \int_R^\infty \bar\Upsilon(q) \dd q \Big) ,
\]
uniformly for $X \in [X_-, X_+]$ and $\delta \le x \le y \le 1 - \delta$ (including $x = y$, for which $s_1 = 0$ and the middle factors are the identity). As $z \downarrow 0$, the right side tends to $C \int_R^\infty \bar\Upsilon$, which tends to zero as $R \to \infty$, and we obtain \eqref{e:res-scalar}.
\end{proof}

\section{The rank-one superspace in polar coordinates}\label{s:res-chart}

The permanent pairings of \zcref{d:res-chain} below are built from semigroup products of the complete rank-one sigma model, with all its anticommuting variables, and not only from the radial diffusions of \zcref{s:res-diffusions}. To take their limit, we need the generator in polar coordinates near the noncompact boundary. In this section, we write it down, and we show that on the scale $q = \hat z B$ the terms acting on the hyperbolic angle $\theta_{\mathrm B}$ and the anticommuting variables become negligible. We refer to this as freezing, since these variables do not move in the limit. The spherical diffusion persists and acts trivially on the limiting profiles, which are independent of the spherical variables.

\subsection{The polar chart}

We recall the rank-one objects of \zcref{s:sigma}. The target space $\mathcal M_1$ has four odd generators, and its underlying manifold $\mathcal M_1^0 = H_1 \times S_1$, described in \zcref{s:sigma-body}, is the product of a hyperbolic plane and a sphere. The field $Q$ is a $(2 | 2)$ supermatrix with $Q^2 = \Id$, whose blocks $Q^{\mathrm{BB}}$ and $Q^{\mathrm{FF}}$ are indexed by the retarded and advanced flavors $\mathrm R$ and $\mathrm A$, and $d_1(Q) = \frac12 \Str(\Lambda Q)$. The Casimir operator $\mathsf C_1$ is defined in \zcref{d:sigma-casimir}, the invariant integral and the pairing $\mathcal B_1$ in \zcref{d:sigma-integral}, and the semigroups $e^{-t \mathsf G_{X, z}}$ of $\mathsf G_{X, z} = X \mathsf C_1 + z d_1$ in \zcref{p:sigma-regular}. Since $\mathsf G_{X, z} = X \mathsf G_{1, \hat z}$, we have $e^{-t \mathsf G_{X, z}} = e^{-Xt \mathsf G_{1, \hat z}}$.

We use the polar coordinates $(B, F, \theta_{\mathrm B}, \theta_{\mathrm F}; \xi_1, \dots, \xi_4)$ of \eqref{e:sigma-polar}, in which $Q = e^{-\Theta_\xi} Q_0 e^{\Theta_\xi}$, and we index the rows and columns in the order $(\mathrm{RB}, \mathrm{RF}, \mathrm{AB}, \mathrm{AF})$. Recall that $Q_0 = Q_0(B, F, \theta_{\mathrm B}, \theta_{\mathrm F})$ denotes the point of $\mathcal M_1^0$ with the coordinates of \zcref{s:sigma-body}, whose diagonal is $(B, F, -B, -F)$ and whose entries at $(\mathrm{RB}, \mathrm{AB})$ and $(\mathrm{RF}, \mathrm{AF})$ are $(B^2 - 1)^{1/2} e^{\iu \theta_{\mathrm B}}$ and $\iu (1 - F^2)^{1/2} e^{\iu \theta_{\mathrm F}}$. Recall also that $\Theta_\xi \in \mathfrak k_1$ has the entries $\xi_1$, $\xi_2$, $\xi_3$, $\xi_4$ at $(\mathrm{RB}, \mathrm{RF})$, $(\mathrm{RF}, \mathrm{RB})$, $(\mathrm{AB}, \mathrm{AF})$, $(\mathrm{AF}, \mathrm{AB})$; the blocks of $e^{\pm \Theta_\xi}$ are written out in the proof of \zcref{l:sigma-polar}. Thus $Q^{\mathrm{BB}}$ is, up to nilpotent terms, the point of the hyperboloid with polar coordinates $(B, \theta_{\mathrm B})$, and $Q^{\mathrm{FF}}$ is the point of the sphere with height $F$ and longitude $\theta_{\mathrm F}$. We let $\mathbf e \in S^2$ denote the point with these spherical coordinates. Functions of $(F, \theta_{\mathrm F})$ that are smooth on $S^2$ are then smooth functions of $\mathbf e$, also at the poles $F = \pm 1$. As in \zcref{s:sigma-polar}, we set $n_{\mathrm R} = \xi_1 \xi_2$ and $n_{\mathrm A} = \xi_3 \xi_4$, and $[f]_{\mathrm{top}}$ is the coefficient of $\xi_1 \xi_2 \xi_3 \xi_4 = n_{\mathrm R} n_{\mathrm A}$ in a superfunction $f$ (\zcref{l:sigma-polar}\ref{i:sigma-polar-top}).

By \zcref{l:sigma-polar}\ref{i:sigma-polar-entries}, we have $Q^2 = \Id$ and
\begin{equation}\label{e:res-d-polar}
 d_1(Q) = B - F ,
\end{equation}
without nilpotent terms. By \zcref{l:sigma-polar}\ref{i:sigma-polar-chart}, the polar chart is a coordinate system of $\mathcal M_1$ over the set of body points with $B > 1$ (near $F = \pm 1$, after $(F, \theta_{\mathrm F})$ is replaced by Cartesian coordinates of the sphere). We use two further facts from \zcref{s:sigma-radial}. First, let $f$ be a rapidly decreasing superfunction supported in $\{B \ge 2\}$. Then $f_\emptyset(\Lambda) = 0$, and \zcref{l:sigma-radial-integral} implies that
\begin{equation}\label{e:res-polar-integral}
 \int f = \int_D \frac{1}{(B - F)^2} \int_0^{2\pi} \!\! \int_0^{2\pi} [f]_{\mathrm{top}} \, \frac{\dd \theta_{\mathrm B} \dd \theta_{\mathrm F}}{4 \pi^2} \, \dd B \dd F .
\end{equation}
Second, by \zcref{l:sigma-singlet}, invariant superfunctions are functions of $(B, F)$ in the polar chart, and $e^{-t \mathsf G_{X, z}} 1$ has the radial representative $\Psi^{X, z}_t$.

The source of the permanent pairings is the superfunction
\begin{equation}\label{e:res-source}
 \mathfrak s(Q) = \tfrac12 \big( Q^{\mathrm{BB}}_{\mathrm{RR}} + Q^{\mathrm{BB}}_{\mathrm{RA}} - Q^{\mathrm{BB}}_{\mathrm{AR}} - Q^{\mathrm{BB}}_{\mathrm{AA}} \big) .
\end{equation}
We will show in \zcref{p:coarse-source-limit} that $s_E \mathfrak s$ is the orbit multiplier of \zcref{t:interface}(b) for the local source with $Y^{\mathrm F} = 0$ and $Y^{\mathrm B} = -\frac\iu2 \big( \begin{smallmatrix} 1 & -1 \\ 1 & -1 \end{smallmatrix} \big)$ of \zcref{l:coarse-permanent}(c), which produces permanents of $\Im G$. Multiplying out $Q = e^{-\Theta_\xi} Q_0 e^{\Theta_\xi}$, we find
\begin{align}
 \mathfrak s &= \big( B + (B^2 - 1)^{1/2} \cos\theta_{\mathrm B} \big) \Big( 1 + \frac{n_{\mathrm R} + n_{\mathrm A}}{2} \Big) - \frac F2 (n_{\mathrm R} + n_{\mathrm A}) + \frac{(B^2 - 1)^{1/2} \cos\theta_{\mathrm B}}{4} n_{\mathrm R} n_{\mathrm A} \notag \\
 &\quad - \frac\iu2 (1 - F^2)^{1/2} e^{\iu \theta_{\mathrm F}} \xi_1 \xi_4 + \frac\iu2 (1 - F^2)^{1/2} e^{-\iu \theta_{\mathrm F}} \xi_2 \xi_3 .
 \label{e:res-source-polar}
\end{align}
In particular, $\mathfrak s$ is not invariant. Its body $B + (B^2 - 1)^{1/2} \cos\theta_{\mathrm B}$ is positive, since it is at least $B - (B^2 - 1)^{1/2} = (B + (B^2 - 1)^{1/2})^{-1}$.

\begin{definition}\label{d:res-chain}
Let $X > 0$, $z > 0$, and $k \ge 1$, and let $0 \le x_1 \le \dots \le x_k \le 1$. We define the \emph{permanent pairing}
\[
 \mathfrak p^{X, z}_k(x_1, \dots, x_k) = \mathcal B_1 \big( e^{-x_1 \mathsf G_{X, z}} 1, \, \mathfrak s \, e^{-(x_2 - x_1) \mathsf G_{X, z}} \mathfrak s \cdots e^{-(x_k - x_{k - 1}) \mathsf G_{X, z}} \mathfrak s \, e^{-(1 - x_k) \mathsf G_{X, z}} 1 \big) .
\]
We extend $\mathfrak p^{X, z}_k$ to all $(x_1, \dots, x_k) \in [0, 1]^k$ by symmetry. More generally, for $\hat z > 0$, $n \ge 0$, and $t_0, \dots, t_n \ge 0$, we call the superfunction $e^{-t_0 \mathsf G_{1, \hat z}} \mathfrak s \, e^{-t_1 \mathsf G_{1, \hat z}} \mathfrak s \cdots \mathfrak s \, e^{-t_n \mathsf G_{1, \hat z}} 1$ a \emph{semigroup product with $n$ sources}, and we let $\mathfrak S_k$ denote the set of finite linear combinations of semigroup products with at most $k$ sources.
\end{definition}

Since $\mathfrak s$ is a multiplier of degree one by \zcref{l:sigma-multipliers}, \zcref{p:sigma-regular} shows that the pairing is well defined and finite. The name of $\mathfrak p^{X, z}_k$ refers to \zcref{s:coarse}, where we show, using \zcref{t:interface}(b), that it is the limit of expected permanents of the matrix $\Im G$ at $k$ sites, at the damping $\eps = z/(2\pi)$.

\subsection{Charges}

Semigroup products are not invariant, but they transform in a simple way under phase rotations. For $\gamma = (\gamma_{\mathrm B}, \gamma_{\mathrm F}) \in \R^2$, let
\[
 \mathsf k_\gamma = \diag\big( e^{\iu \gamma_{\mathrm B}/2}, e^{\iu \gamma_{\mathrm F}/2}, e^{-\iu \gamma_{\mathrm B}/2}, e^{-\iu \gamma_{\mathrm F}/2} \big)
\]
in the order of indices above. Recall from \zcref{s:sigma-supermatrices} that $G_0$ is the group of the block diagonal supermatrices $\diag(g_{\mathrm B} | g_{\mathrm F})$ with $g_{\mathrm B} \in U(1, 1)$ and $g_{\mathrm F} \in U(2)$, which acts on $\mathcal M_1$ by conjugation, and that $K_0$ is the subgroup of the elements that commute with $\Lambda$. The matrix $\mathsf k_\gamma$ lies in $K_0$, since its blocks $\diag(e^{\iu \gamma_{\mathrm B}/2}, e^{-\iu \gamma_{\mathrm B}/2})$ and $\diag(e^{\iu \gamma_{\mathrm F}/2}, e^{-\iu \gamma_{\mathrm F}/2})$ are unitary and diagonal. Conjugating $Q_0$ and $e^{\Theta_\xi}$ by $\mathsf k_\gamma$, we see that $\mathsf k_\gamma Q \mathsf k_\gamma^{-1}$ has the same $B$ and $F$ as $Q$, and the remaining polar coordinates
\[
 \theta_{\mathrm B} + \gamma_{\mathrm B}, \quad \theta_{\mathrm F} + \gamma_{\mathrm F}, \quad
 e^{\iu (\gamma_{\mathrm B} - \gamma_{\mathrm F})/2} \xi_1, \quad e^{-\iu (\gamma_{\mathrm B} - \gamma_{\mathrm F})/2} \xi_2, \quad e^{-\iu (\gamma_{\mathrm B} - \gamma_{\mathrm F})/2} \xi_3, \quad e^{\iu (\gamma_{\mathrm B} - \gamma_{\mathrm F})/2} \xi_4 .
\]
For a subset $I \subset \{1, 2, 3, 4\}$, let $\xi^I$ denote the monomial of \zcref{d:grass-algebra}, and set $c_I = \frac12 (|I \cap \{1, 4\}| - |I \cap \{2, 3\}|)$. We say that a superfunction $f$ has \emph{charge} $(m, m')$ if $f(\mathsf k_\gamma Q \mathsf k_\gamma^{-1}) = e^{\iu (m \gamma_{\mathrm B} + m' \gamma_{\mathrm F})} f(Q)$ for all $\gamma$. In the polar chart, this means that
\begin{equation}\label{e:res-charge}
 f = \sum_I e^{\iu (m - c_I) \theta_{\mathrm B}} e^{\iu (m' + c_I) \theta_{\mathrm F}} f_I(B, F) \, \xi^I
\end{equation}
for functions $f_I$ of $(B, F)$. The operators $\mathsf C_1$ and $d_1$ commute with the conjugation by $\mathsf k_\gamma$, by \zcref{l:sigma-casimir}\ref{i:sigma-casimir-commute} and the invariance of $d_1$, and so does the semigroup $e^{-t \mathsf G_{1, \hat z}}$. The entries $Q^{\mathrm{BB}}_{\mathrm{RR}}$ and $Q^{\mathrm{BB}}_{\mathrm{AA}}$ have charge $(0, 0)$, while $Q^{\mathrm{BB}}_{\mathrm{RA}}$ and $Q^{\mathrm{BB}}_{\mathrm{AR}}$ have charges $(1, 0)$ and $(-1, 0)$. It follows that every semigroup product with $n$ sources is a sum of even superfunctions of charges $(m, 0)$ with $|m| \le n$. For even $I$, we have $c_I \in \{-1, 0, 1\}$.

\subsection{The Casimir operator in the polar chart}

Let $\partial_{\xi_a}$ denote the left derivative with respect to $\xi_a$, and set $\partial_+ = \partial_{\theta_{\mathrm B}} + \partial_{\theta_{\mathrm F}}$. We define the odd differential operators
\[
 \mathcal D_1 = \partial_{\xi_1} + \tfrac\iu2 \xi_2 \partial_+ , \qquad
 \mathcal D_2 = \partial_{\xi_2} + \tfrac\iu2 \xi_1 \partial_+ , \qquad
 \mathcal D_3 = \partial_{\xi_3} - \tfrac\iu2 \xi_4 \partial_+ , \qquad
 \mathcal D_4 = \partial_{\xi_4} - \tfrac\iu2 \xi_3 \partial_+ ,
\]
and the antisymmetric matrix $(g^{ab})_{a, b = 1}^4$ with $g^{13} = g^{24} = 0$ and
\begin{gather*}
 g^{12} = g^{34} = \frac{2 (BF - 1)}{(B - F)^2} , \\
 g^{14} = \frac{2 \iu \big( (B^2 - 1)(1 - F^2) \big)^{1/2} e^{\iu (\theta_{\mathrm B} - \theta_{\mathrm F})}}{(B - F)^2} , \qquad
 g^{23} = -\frac{2 \iu \big( (B^2 - 1)(1 - F^2) \big)^{1/2} e^{-\iu (\theta_{\mathrm B} - \theta_{\mathrm F})}}{(B - F)^2} .
\end{gather*}
In the coordinates $(B, F)$, the singlet operator \eqref{e:sigma-singlet-operator} reads
\[
 \Delta^{\mathrm s}_D = \partial_B \big( (B^2 - 1) \partial_B \big) + \partial_F \big( (1 - F^2) \partial_F \big) - \frac{2 (B^2 - 1)}{d} \partial_B + \frac{2 (1 - F^2)}{d} \partial_F .
\]

\begin{lemma}\label{l:res-polar}
For every smooth superfunction $f$ on $\{B > 1\}$, in the polar chart,
\begin{equation}\label{e:res-polar}
 -\mathsf C_1 f = \Delta^{\mathrm s}_D f + \frac{1}{B^2 - 1} \partial_{\theta_{\mathrm B}}^2 f + \frac{1}{1 - F^2} \partial_{\theta_{\mathrm F}}^2 f + \frac12 \sum_{a, b = 1}^4 g^{ab} \mathcal D_b \mathcal D_a f .
\end{equation}
\end{lemma}

We first describe the structure of \eqref{e:res-polar}. There are no mixed terms between the radial coordinates $(B, F)$ and the other coordinates, and the radial part is the singlet operator $\Delta^{\mathrm s}_D$, also on noninvariant superfunctions. The phase terms are those of the polar coordinates of $H_1$ and $S_1$. The entries of the matrix $(g^{ab})$ of the odd part are of order $1/B$ for large $B$, uniformly in $F$, and its Pfaffian $g^{12} g^{34} + g^{14} g^{23} = 4 (B - F)^{-2}$ degenerates only at the corner. The operators $\mathcal D_a$ involve the angles only through $\partial_+$, which shifts $\theta_{\mathrm B}$ and $\theta_{\mathrm F}$ together. Later, \zcref{l:res-freezing} uses three of these facts: the absence of mixed terms, the radial part $\Delta^{\mathrm s}_D$, and the decay of $g^{ab}$.

The proof reduces \eqref{e:res-polar} to a few coefficients that depend only on $(B, F)$, by the symmetry of $\mathsf C_1$ under the stabilizer of $\Lambda$. It computes these coefficients from three sources: the Laplace--Beltrami operator of $\mathcal M_1^0$, the singlet operator of \zcref{l:sigma-radial-part}, and the action of $\mathsf C_1$ on products of two entries of $Q$. We first state the two computations that it needs. We work over the set of body points with $B > 1$ and $|F| < 1$, where the polar coordinates are coordinates, by \zcref{l:sigma-polar}\ref{i:sigma-polar-chart}. The indices $a, b$ run over $\{1, 2, 3, 4\}$. We call $a'$ the partner of $a$, with $1' = 2$, $2' = 1$, $3' = 4$, and $4' = 3$, and we set $\sigma_a = 1$ for $a \in \{1, 2\}$ and $\sigma_a = -1$ for $a \in \{3, 4\}$, so that
\[
 \mathcal D_a = \partial_{\xi_a} + \tfrac\iu2 \sigma_a \xi_{a'} \partial_+ .
\]
We also write $c_a = c_{\{a\}}$ for the numbers of \eqref{e:res-charge}, so that $c_1 = c_4 = \frac12$, $c_2 = c_3 = -\frac12$, and $c_{a'} = -c_a$. At rank one, the even elements of $\mathfrak k$ are the diagonal matrices, and the odd elements are the linear combinations of the four matrix units $E_{\mathrm{RB}, \mathrm{RF}}$, $E_{\mathrm{RF}, \mathrm{RB}}$, $E_{\mathrm{AB}, \mathrm{AF}}$, and $E_{\mathrm{AF}, \mathrm{AB}}$ (\zcref{s:sigma-supermatrices}).

\begin{lemma}\label{l:res-stabilizer}
In the polar chart, the fields $\widehat Y$ of \zcref{l:sigma-action} for $Y \in \mathfrak k$ are as follows. For $Y = \diag(y_1, y_2, y_3, y_4)$,
\begin{equation}\label{e:res-left-even}
 \widehat Y = -\iu (y_1 - y_3) \partial_{\theta_{\mathrm B}} - \iu (y_2 - y_4) \partial_{\theta_{\mathrm F}} + (y_1 - y_2)(\xi_1 \partial_{\xi_1} - \xi_2 \partial_{\xi_2}) + (y_3 - y_4)(\xi_3 \partial_{\xi_3} - \xi_4 \partial_{\xi_4}) ,
\end{equation}
and for the four odd matrix units of $\mathfrak k$,
\begin{equation}\label{e:res-left-odd}
\begin{gathered}
 \widehat{E_{\mathrm{RB}, \mathrm{RF}}} = -\partial_{\xi_1} + \tfrac\iu2 \xi_2 \partial_+ , \qquad \widehat{E_{\mathrm{RF}, \mathrm{RB}}} = -\partial_{\xi_2} + \tfrac\iu2 \xi_1 \partial_+ , \\
 \widehat{E_{\mathrm{AB}, \mathrm{AF}}} = -\partial_{\xi_3} - \tfrac\iu2 \xi_4 \partial_+ , \qquad \widehat{E_{\mathrm{AF}, \mathrm{AB}}} = -\partial_{\xi_4} - \tfrac\iu2 \xi_3 \partial_+ .
\end{gathered}
\end{equation}
In particular, the four odd fields are $-\partial_{\xi_a} + \frac\iu2 \sigma_a \xi_{a'} \partial_+$, none of these fields differentiates in $B$ or $F$, and their coefficients do not depend on $(B, F, \theta_{\mathrm B}, \theta_{\mathrm F})$.
\end{lemma}

\begin{proof}
Let $V$ be a vector field in the polar chart such that $\vartheta V Q = [\vartheta Y, Q]$, where $\vartheta$ is the parameter of \eqref{e:sigma-action}. Then $V = \widehat Y$. Indeed, by \zcref{l:sigma-polar}\ref{i:sigma-polar-entries}, the Jacobian of the eight entries $Q^{\mathrm{BB}}_{\mathrm{RR}}$, $Q^{\mathrm{FF}}_{\mathrm{RR}}$, $Q^{\mathrm{BB}}_{\mathrm{RA}}$, $Q^{\mathrm{FF}}_{\mathrm{RA}}$, $Q^{\mathrm{BF}}_{\mathrm{RR}} = d \xi_1$, $Q^{\mathrm{FB}}_{\mathrm{RR}} = -d \xi_2$, $Q^{\mathrm{BF}}_{\mathrm{AA}} = -d \xi_3$, and $Q^{\mathrm{FB}}_{\mathrm{AA}} = d \xi_4$ with respect to $(B, F, \theta_{\mathrm B}, \theta_{\mathrm F}, \xi_1, \dots, \xi_4)$ has a triangular body with nonzero diagonal entries, so a vector field is determined by its values on these entries.

We begin with $Y = \diag(y_1, y_2, y_3, y_4)$. Conjugation by $e^{sY}$ multiplies the entries of $Q_0$ at $(\mathrm{RB}, \mathrm{AB})$ and $(\mathrm{RF}, \mathrm{AF})$ by $e^{s(y_1 - y_3)}$ and $e^{s(y_2 - y_4)}$, and the entries of $\Theta_\xi$ by the corresponding factors. In the polar coordinates, it therefore replaces $\theta_{\mathrm B}$, $\theta_{\mathrm F}$, $\xi_1$, $\xi_2$, $\xi_3$, and $\xi_4$ by
\[
 \theta_{\mathrm B} - \iu s (y_1 - y_3) , \quad \theta_{\mathrm F} - \iu s (y_2 - y_4) , \quad e^{s(y_1 - y_2)} \xi_1 , \quad e^{s(y_2 - y_1)} \xi_2 , \quad e^{s(y_3 - y_4)} \xi_3 , \quad e^{s(y_4 - y_3)} \xi_4 .
\]
Since
\[
 e^{sY} Q e^{-sY} = e^{-e^{sY} \Theta_\xi e^{-sY}} \big( e^{sY} Q_0 e^{-sY} \big) e^{e^{sY} \Theta_\xi e^{-sY}} ,
\]
the derivative at $s = 0$ gives \eqref{e:res-left-even}.

Next, let $Y = \alpha E_{\mathrm{RB}, \mathrm{RF}} + \beta E_{\mathrm{RF}, \mathrm{RB}}$ with $\alpha, \beta \in \C$. Multiplying the $2 \times 2$ blocks in the flavor block $\mathrm{RR}$, we find $[\vartheta Y, \Theta_\xi] = \vartheta (\alpha \xi_2 + \beta \xi_1) \Id_{\mathrm R}$, which commutes with $\vartheta Y$ and $\Theta_\xi$. Hence
\begin{gather*}
 e^{\vartheta Y} e^{-\Theta_\xi} = e^{-\Theta_{\xi'}} e^{c \Id_{\mathrm R}} , \qquad
 e^{\vartheta Y} Q e^{-\vartheta Y} = e^{-\Theta_{\xi'}} \big( e^{c \Id_{\mathrm R}} Q_0 e^{-c \Id_{\mathrm R}} \big) e^{\Theta_{\xi'}} , \\
 \xi' = (\xi_1 - \vartheta \alpha, \, \xi_2 - \vartheta \beta, \, \xi_3, \, \xi_4) , \qquad c = -\tfrac12 \vartheta (\alpha \xi_2 + \beta \xi_1) .
\end{gather*}
Conjugation by $e^{c \Id_{\mathrm R}}$ multiplies the entries of $Q_0$ at $(\mathrm{RB}, \mathrm{AB})$ and $(\mathrm{RF}, \mathrm{AF})$ by $e^c$, which amounts to replacing $\theta_{\mathrm B}$ and $\theta_{\mathrm F}$ by $\theta_{\mathrm B} - \iu c$ and $\theta_{\mathrm F} - \iu c$. Since $e^{\vartheta Y} Q e^{-\vartheta Y} = Q + [\vartheta Y, Q]$, the terms of first order in $\vartheta$ give the first line of \eqref{e:res-left-odd}. The same computation in the flavor block $\mathrm{AA}$, where conjugation by $e^{c \Id_{\mathrm A}}$ replaces $\theta_{\mathrm B}$ and $\theta_{\mathrm F}$ by $\theta_{\mathrm B} + \iu c$ and $\theta_{\mathrm F} + \iu c$, gives the second line.
\end{proof}

\begin{lemma}\label{l:res-entries}
We have $\mathsf C_1 Q_{\mu\nu} = 0$ for all entries $Q_{\mu\nu}$ of $Q$. Let $Q_{\mu\nu}$ and $Q_{\mu'\nu'}$ be odd entries, so that $|\mu| + |\nu| = |\mu'| + |\nu'| = 1$. Then
\begin{equation}\label{e:res-entries-product}
\begin{split}
 -\mathsf C_1 (Q_{\mu\nu} Q_{\mu'\nu'}) &= (-1)^{|\mu'|} Q_{\mu'\nu} Q_{\mu\nu'} + (-1)^{|\nu|} Q_{\mu\nu'} Q_{\mu'\nu} \\
 &\quad - \delta_{\mu\nu'} \sum_i (-1)^{|i|} Q_{i\nu} Q_{\mu'i} - (-1)^{|\nu|} \delta_{\nu\mu'} (Q^2)_{\mu\nu'} ,
\end{split}
\end{equation}
and the body of this superfunction is
\begin{equation}\label{e:res-carre}
 2 (-1)^{|\nu|} \big( Q_{\mu\nu'} Q_{\mu'\nu} - \delta_{\mu\nu'} \delta_{\nu\mu'} \big)_\emptyset .
\end{equation}
\end{lemma}

\begin{proof}
Recall that $|E_{ij}| = |i| + |j|$ is the parity of $E_{ij}$. By \eqref{e:sigma-action}, for all indices $i, j, \mu, \nu$,
\begin{equation}\label{e:res-action-entries}
 \widehat{E_{ij}} \, Q_{\mu\nu} = \delta_{\mu i} Q_{j\nu} - (-1)^{|E_{ij}| (|\mu| + |i|)} Q_{\mu i} \, \delta_{j\nu} ,
\end{equation}
where the sign comes from moving the odd parameter $\vartheta$ past $Q_{\mu i}$ when $E_{ij}$ is odd. Recall from \eqref{e:sigma-casimir} that
\[
 \mathsf C_1 = - \frac12 \sum_{i, j} (-1)^{|j|} \, \widehat{E_{j i}} \, \widehat{E_{i j}} .
\]
We apply $\widehat{E_{ji}}$ to \eqref{e:res-action-entries}, using \eqref{e:res-action-entries} again, and we sum over $i$ and $j$ with the weights $(-1)^{|j|}$. Since $\sum_i (-1)^{|i|} = 0$ and $\Str Q = 0$ by \eqref{e:sigma-Q-relations}, we find
\[
 -2 \mathsf C_1 Q_{\mu\nu} = \sum_j (-1)^{|j|} Q_{\mu\nu} - 2 \delta_{\mu\nu} \Str Q + \sum_i (-1)^{|i|} Q_{\mu\nu} = 0 .
\]
For homogeneous $f$ and $f'$, the Leibniz rule for the derivations $\widehat{E_{ij}}$, which are odd when $E_{ij}$ is odd, and \eqref{e:sigma-casimir} give
\[
 \mathsf C_1 (f f') = (\mathsf C_1 f) f' + f \, \mathsf C_1 f' - \sum_{i, j} (-1)^{|j| + |E_{ij}| |f|} \, (\widehat{E_{ji}} f)(\widehat{E_{ij}} f') .
\]
For odd entries $f = Q_{\mu\nu}$ and $f' = Q_{\mu'\nu'}$, the first two terms vanish. Inserting \eqref{e:res-action-entries} into the last sum and simplifying the signs with the relations $|\mu| + |\nu| = |\mu'| + |\nu'| = 1$, we obtain \eqref{e:res-entries-product}. Only products of two even entries have a body, and the diagonal blocks of the body $Q_0$ square to the identity. Using also $Q^2 = \Id$, we find \eqref{e:res-carre}.
\end{proof}

\begin{proof}[Proof of \zcref{l:res-polar}]
\emph{Step 1 (a frame that commutes with the fields of $\mathfrak k$).} We consider the two diagonal elements $\mathsf r_{\mathrm B}$ and $\mathsf r_{\mathrm F}$ of $\mathfrak k$ for which $\mathsf k_\gamma = e^{\gamma_{\mathrm B} \mathsf r_{\mathrm B} + \gamma_{\mathrm F} \mathsf r_{\mathrm F}}$, and the operators $\widetilde{\mathcal D}_a$, defined by
\[
 \mathsf r_{\mathrm B} = \tfrac\iu2 \diag(1, 0, -1, 0) , \qquad \mathsf r_{\mathrm F} = \tfrac\iu2 \diag(0, 1, 0, -1) , \qquad \widetilde{\mathcal D}_a = e^{\iu c_a (\theta_{\mathrm B} - \theta_{\mathrm F})} \mathcal D_a .
\]
By \eqref{e:res-left-even}, we have
\[
 \widehat{\mathsf r_{\mathrm B}} = \partial_{\theta_{\mathrm B}} + \iu \sum_a c_a \xi_a \partial_{\xi_a} , \qquad
 \widehat{\mathsf r_{\mathrm F}} = \partial_{\theta_{\mathrm F}} - \iu \sum_a c_a \xi_a \partial_{\xi_a} .
\]
We claim that each of the eight operators $\partial_B$, $\partial_F$, $\partial_{\theta_{\mathrm B}}$, $\partial_{\theta_{\mathrm F}}$, $\widetilde{\mathcal D}_1, \dots, \widetilde{\mathcal D}_4$ supercommutes with each of the six fields \eqref{e:res-left-odd}, $\widehat{\mathsf r_{\mathrm B}}$, and $\widehat{\mathsf r_{\mathrm F}}$. For the first four operators, this holds since the six fields have constant coefficients in $(B, F, \theta_{\mathrm B}, \theta_{\mathrm F})$. The phase of $\widetilde{\mathcal D}_b$ commutes with the odd fields, since $\partial_+$ annihilates $\theta_{\mathrm B} - \theta_{\mathrm F}$. Moreover,
\[
 \big\{ -\partial_{\xi_a} + \tfrac\iu2 \sigma_a \xi_{a'} \partial_+ , \, \mathcal D_b \big\} = \tfrac\iu2 (\sigma_a - \sigma_b) \, \delta_{a' b} \, \partial_+ = 0 , \qquad
 \Big[ \iu \sum_a c_a \xi_a \partial_{\xi_a} , \, \mathcal D_b \Big] = -\iu c_b \mathcal D_b ,
\]
where we used that $a' = b$ implies $\sigma_a = \sigma_b$, and that $c_{b'} = -c_b$. Since $\partial_{\theta_{\mathrm B}} e^{\iu c_b (\theta_{\mathrm B} - \theta_{\mathrm F})} = \iu c_b \, e^{\iu c_b (\theta_{\mathrm B} - \theta_{\mathrm F})}$, the two contributions to $[\widehat{\mathsf r_{\mathrm B}}, \widetilde{\mathcal D}_b]$ cancel, and the same holds for $\widehat{\mathsf r_{\mathrm F}}$.

The eight operators form a frame: their coefficients with respect to the coordinate derivatives form a matrix with invertible body. By induction on the order, every differential operator $P$ of order at most two in the polar chart can therefore be written uniquely as
\[
 P = \sum_w a_w W_w ,
\]
where $W_w$ runs over the identity, the eight operators, and the products $W' W''$ of two of them with $W'$ before $W''$ in the order $\partial_B, \partial_F, \partial_{\theta_{\mathrm B}}, \partial_{\theta_{\mathrm F}}, \widetilde{\mathcal D}_1, \dots, \widetilde{\mathcal D}_4$, and with $W' \ne W''$ if both are odd. Here the $a_w$ are superfunctions, and if $P$ is even, then $a_w$ has the parity of $W_w$.

\emph{Step 2 (reduction to functions of $(B, F)$).} By the last assertion of \zcref{l:sigma-casimir}\ref{i:sigma-casimir-local}, in the coordinates of a section, $\mathsf C_1$ is a sum of finitely many terms, each of which is a superfunction, written on the left, times a product of at most two of the derivatives $\partial_{x_\mu}$ and $\partial_{\chi_a}$. In the polar chart, each of these derivatives becomes a first-order operator in $\partial_B$, $\partial_F$, $\partial_{\theta_{\mathrm B}}$, $\partial_{\theta_{\mathrm F}}$, and $\partial_{\xi_1}, \dots, \partial_{\xi_4}$. Hence $-\mathsf C_1$ is an even differential operator in the polar chart whose order, counting the odd derivatives $\partial_{\xi_a}$ as well as the even ones, is at most two, which is the hypothesis of the expansion of Step 1. By \zcref{l:sigma-casimir}\ref{i:sigma-casimir-commute}, it commutes with all fields $\widehat Y$. We expand $-\mathsf C_1 = \sum_w a_w W_w$ as in Step 1. Let $V$ be one of the six fields of Step 1. Since $V$ supercommutes with each $W_w$, and $a_w$ has the parity of $W_w$, we have
\[
 0 = [-\mathsf C_1, V] = -\sum_w (V a_w) \, W_w ,
\]
and the uniqueness of the expansion gives $V a_w = 0$ for all $w$. The first two fields in \eqref{e:res-left-odd} have the anticommutator $-\iu \partial_+$, and hence $\partial_+ a_w = 0$. The four odd fields then give $\partial_{\xi_a} a_w = 0$ for all $a$, and $\widehat{\mathsf r_{\mathrm B}}$ and $\widehat{\mathsf r_{\mathrm F}}$ show that $a_w$ does not depend on $\theta_{\mathrm B}$ and $\theta_{\mathrm F}$. Thus every $a_w$ is a function of $(B, F)$ without odd terms, and $a_w = 0$ whenever $W_w$ is odd. We conclude that
\begin{equation}\label{e:res-polar-form}
 -\mathsf C_1 = P_{\mathrm{rad}} + \sum_{x, \theta} a_{x \theta} \, \partial_x \partial_\theta + \sum_{\theta \le \theta'} a_{\theta \theta'} \, \partial_\theta \partial_{\theta'} + b_{\mathrm B} \, \partial_{\theta_{\mathrm B}} + b_{\mathrm F} \, \partial_{\theta_{\mathrm F}} + \sum_{a < b} a_{ab} \, \widetilde{\mathcal D}_a \widetilde{\mathcal D}_b ,
\end{equation}
where $x$ runs over $\{B, F\}$ and $\theta, \theta'$ over $\{\theta_{\mathrm B}, \theta_{\mathrm F}\}$, $P_{\mathrm{rad}}$ is a differential operator in $(B, F)$ of order two, and all coefficients are functions of $(B, F)$. In particular, $-\mathsf C_1$ has no terms that mix a derivative in $B$ or $F$ with a derivative in an odd variable.

\emph{Step 3 (the radial and the angular parts).} Let $\phi \in C_c^\infty(D^\circ)$, and let $f$ be the superfunction that equals $\phi(B, F)$ in the polar chart and vanishes elsewhere. It is smooth by \zcref{l:sigma-polar}\ref{i:sigma-polar-chart}, and invariant under $\mathfrak k$ by \zcref{l:res-stabilizer}, since none of the fields there differentiates in $B$ or $F$. Every word in \eqref{e:res-polar-form} outside $P_{\mathrm{rad}}$ ends with $\partial_\theta$ or with some $\widetilde{\mathcal D}_a$, which annihilate $f$. Hence $-\mathsf C_1 f = P_{\mathrm{rad}} \phi$, and \zcref{l:sigma-radial-part} implies that $P_{\mathrm{rad}} \phi = \Delta^{\mathrm s}_D \phi$. As $\phi$ is arbitrary and the coefficients of $P_{\mathrm{rad}}$ are functions of $(B, F)$, we conclude that $P_{\mathrm{rad}} = \Delta^{\mathrm s}_D$.

For the angular part, we use the principal part of $\mathsf C_1$. By \zcref{l:sigma-casimir}\ref{i:sigma-casimir-local}, in the coordinates of a section, the coefficients of the second derivatives of $\mathsf C_1$ in the even coordinates are those of $-\Delta_{\mathcal M_1^0}$ plus superfunctions with nilpotent values. Under a change of coordinates, the bodies of these coefficients transform as for a differential operator on $\mathcal M_1^0$, since the derivatives of the even coordinates in the odd ones, and of the odd coordinates in the even ones, have no body. In the polar chart, the bodies of the coefficients of $\partial_x \partial_y$ for $x, y \in \{B, F, \theta_{\mathrm B}, \theta_{\mathrm F}\}$ are therefore those of the Laplace--Beltrami operator of the metric of \zcref{s:sigma-body},
\[
 \Delta_{\mathcal M_1^0} = \partial_B \big( (B^2 - 1) \partial_B \big) + \frac{1}{B^2 - 1} \partial_{\theta_{\mathrm B}}^2 + \partial_F \big( (1 - F^2) \partial_F \big) + \frac{1}{1 - F^2} \partial_{\theta_{\mathrm F}}^2 .
\]
The products $\widetilde{\mathcal D}_a \widetilde{\mathcal D}_b$ contribute to these coefficients only multiples of $\xi_{a'} \xi_{b'}$, which have no body. Since the coefficients in \eqref{e:res-polar-form} are functions, we conclude that
\[
 a_{x \theta} = a_{\theta_{\mathrm B} \theta_{\mathrm F}} = 0 , \qquad a_{\theta_{\mathrm B} \theta_{\mathrm B}} = \frac{1}{B^2 - 1} , \qquad a_{\theta_{\mathrm F} \theta_{\mathrm F}} = \frac{1}{1 - F^2} .
\]

\emph{Step 4 (the first-order terms).} We write each product $\widetilde{\mathcal D}_a \widetilde{\mathcal D}_b$ with $a < b$ as half the difference $\widetilde{\mathcal D}_a \widetilde{\mathcal D}_b - \widetilde{\mathcal D}_b \widetilde{\mathcal D}_a$ plus half the anticommutator. The phases commute with the operators $\mathcal D_a$, and
\[
 \widetilde{\mathcal D}_a \widetilde{\mathcal D}_b = \tfrac12 \big( \widetilde{\mathcal D}_a \widetilde{\mathcal D}_b - \widetilde{\mathcal D}_b \widetilde{\mathcal D}_a \big) + \tfrac12 e^{\iu (c_a + c_b)(\theta_{\mathrm B} - \theta_{\mathrm F})} \{\mathcal D_a, \mathcal D_b\} , \qquad \{\mathcal D_a, \mathcal D_b\} = \iu \sigma_a \delta_{a' b} \partial_+ .
\]
Since $c_a + c_{a'} = 0$, the anticommutators contribute multiples of $\partial_+$ with coefficients that are functions of $(B, F)$. Combining this with Step 3, we find
\begin{equation}\label{e:res-polar-form2}
 -\mathsf C_1 = \Delta^{\mathrm s}_D + \frac{1}{B^2 - 1} \partial_{\theta_{\mathrm B}}^2 + \frac{1}{1 - F^2} \partial_{\theta_{\mathrm F}}^2 + b'_{\mathrm B} \, \partial_{\theta_{\mathrm B}} + b'_{\mathrm F} \, \partial_{\theta_{\mathrm F}} + \frac12 \sum_{a, b} \tilde g^{ab} \mathcal D_b \mathcal D_a ,
\end{equation}
where $b'_{\mathrm B}$ and $b'_{\mathrm F}$ are functions of $(B, F)$, and $(\tilde g^{ab})$ is an antisymmetric matrix such that $e^{-\iu (c_a + c_b)(\theta_{\mathrm B} - \theta_{\mathrm F})} \tilde g^{ab}$ is a function of $(B, F)$. We show in Step 5 that $\tilde g^{ab} = g^{ab}$.

We now show that $b'_{\mathrm B} = b'_{\mathrm F} = 0$. For superfunctions $f$ and $f'$ with compact support in the polar chart, we set
\[
 \langle f, f' \rangle = \int_D d^{-2} \int_0^{2\pi} \!\! \int_0^{2\pi} [f f']_{\mathrm{top}} \, \dd \theta_{\mathrm B} \dd \theta_{\mathrm F} \, \dd B \dd F .
\]
By \zcref{l:sigma-casimir}\ref{i:sigma-casimir-symmetric} and \eqref{e:sigma-radial-compact}, the operator $-\mathsf C_1$ is symmetric for this form. The first three terms of \eqref{e:res-polar-form2} are symmetric, since
\[
 \Delta^{\mathrm s}_D = d^2 \partial_B \big( d^{-2} (B^2 - 1) \partial_B \big) + d^2 \partial_F \big( d^{-2} (1 - F^2) \partial_F \big) .
\]
Integrating by parts in the odd variables and in the angles, we find that, for homogeneous $f$,
\[
 \langle \mathcal D_a f, f' \rangle = -(-1)^{|f|} \langle f, \mathcal D_a f' \rangle , \qquad
 \langle \mathcal D_b \mathcal D_a f, f' \rangle = -\langle f, \mathcal D_a \mathcal D_b f' \rangle .
\] Since $\mathcal D_a$ annihilates $\tilde g^{ab}$ and $\tilde g^{ab}$ is antisymmetric, the last term of \eqref{e:res-polar-form2} is symmetric. The remaining term $b'_{\mathrm B} \partial_{\theta_{\mathrm B}} + b'_{\mathrm F} \partial_{\theta_{\mathrm F}}$ is antisymmetric, by an integration by parts in the angles. Being also symmetric, it vanishes.

\emph{Step 5 (the odd coefficients).} It remains to show that $\tilde g^{ab} = g^{ab}$. Since $e^{-\iu (c_a + c_b)(\theta_{\mathrm B} - \theta_{\mathrm F})} \tilde g^{ab}$ is a function of $(B, F)$, it suffices to compute bodies. Let $a < b$. We have
\[
 \mathcal D_{a_2} \mathcal D_{a_1} (\xi_a \xi_b) = \delta_{a_1 a} \delta_{a_2 b} - \delta_{a_1 b} \delta_{a_2 a} , \qquad
 \Delta^{\mathrm s}_D (d^2) = d \, \Delta_D d = 2d(B + F) ,
\]
where the second identity follows from \eqref{e:sigma-singlet-operator}. Inserting these identities into \eqref{e:res-polar-form2}, we obtain
\begin{equation}\label{e:res-polar-test}
 -\mathsf C_1 (d^2 \xi_a \xi_b) = 2d(B + F) \, \xi_a \xi_b + d^2 \tilde g^{ab} ,
\end{equation}
whose body is $d^2 \tilde g^{ab}$. By \zcref{l:sigma-polar}\ref{i:sigma-polar-entries}, the superfunction $d^2 \xi_a \xi_b$ is, up to sign, the product of two of the odd entries $Q^{\mathrm{BF}}_{\mathrm{RR}} = d \xi_1$, $Q^{\mathrm{FB}}_{\mathrm{RR}} = -d \xi_2$, $Q^{\mathrm{BF}}_{\mathrm{AA}} = -d \xi_3$, and $Q^{\mathrm{FB}}_{\mathrm{AA}} = d \xi_4$, and \zcref{l:res-entries} computes the body of $-\mathsf C_1$ on such products.

We apply \eqref{e:res-carre} to the six pairs $a < b$, with the bodies of the entries of $Q_0$. For $(a, b) = (1, 2)$, we take $(\mu, \nu, \mu', \nu') = (\mathrm{RB}, \mathrm{RF}, \mathrm{RF}, \mathrm{RB})$, and for $(a, b) = (1, 4)$, we take $(\mu, \nu, \mu', \nu') = (\mathrm{RB}, \mathrm{RF}, \mathrm{AF}, \mathrm{AB})$; the other pairs are analogous. This yields the bodies
\begin{gather*}
 -\mathsf C_1 \big( Q^{\mathrm{BF}}_{\mathrm{RR}} Q^{\mathrm{FB}}_{\mathrm{RR}} \big) \big|_\emptyset = -\mathsf C_1 \big( Q^{\mathrm{BF}}_{\mathrm{AA}} Q^{\mathrm{FB}}_{\mathrm{AA}} \big) \big|_\emptyset = -2 (BF - 1) , \\
 -\mathsf C_1 \big( Q^{\mathrm{BF}}_{\mathrm{RR}} Q^{\mathrm{FB}}_{\mathrm{AA}} \big) \big|_\emptyset = -2 \, Q^{\mathrm{BB}}_{\mathrm{RA}} Q^{\mathrm{FF}}_{\mathrm{AR}} \big|_\emptyset = 2 \iu \big( (B^2 - 1)(1 - F^2) \big)^{1/2} e^{\iu (\theta_{\mathrm B} - \theta_{\mathrm F})} , \\
 -\mathsf C_1 \big( Q^{\mathrm{FB}}_{\mathrm{RR}} Q^{\mathrm{BF}}_{\mathrm{AA}} \big) \big|_\emptyset = 2 \, Q^{\mathrm{FF}}_{\mathrm{RA}} Q^{\mathrm{BB}}_{\mathrm{AR}} \big|_\emptyset = -2 \iu \big( (B^2 - 1)(1 - F^2) \big)^{1/2} e^{-\iu (\theta_{\mathrm B} - \theta_{\mathrm F})} ,
\end{gather*}
for the pairs $(1, 2)$, $(3, 4)$, $(1, 4)$, and $(2, 3)$. For $(1, 3)$ and $(2, 4)$, the entry $Q_{\mu\nu'}$ in \eqref{e:res-carre} is odd, and the body vanishes. We compare these bodies with the body of \eqref{e:res-polar-test}, using
\[
 Q^{\mathrm{BF}}_{\mathrm{RR}} Q^{\mathrm{FB}}_{\mathrm{RR}} = -d^2 \xi_1 \xi_2 , \quad
 Q^{\mathrm{BF}}_{\mathrm{AA}} Q^{\mathrm{FB}}_{\mathrm{AA}} = -d^2 \xi_3 \xi_4 , \quad
 Q^{\mathrm{BF}}_{\mathrm{RR}} Q^{\mathrm{FB}}_{\mathrm{AA}} = d^2 \xi_1 \xi_4 , \quad
 Q^{\mathrm{FB}}_{\mathrm{RR}} Q^{\mathrm{BF}}_{\mathrm{AA}} = d^2 \xi_2 \xi_3 ,
\]
and we find $\tilde g^{ab} = g^{ab}$ for all $a < b$. Together with \eqref{e:res-polar-form2}, this proves \eqref{e:res-polar}.
\end{proof}

\begin{remark}\label{r:res-polar-check}
The formula \eqref{e:res-polar} is also confirmed by exact computer algebra, with the supplementary scripts in the folder \texttt{polar}. The script \texttt{verify\_tangency\_exact.py} computes the sixteen fields $\widehat{E_{ij}}$ in the polar chart and checks \eqref{e:sigma-action} on all entries of $Q$, and \texttt{verify\_op\_exact.py} checks \eqref{e:res-polar} on the superfunctions $\xi^I h$ with $h \in \{1, x_a, x_a x_b\}$, which determine a second-order operator (its output is in \texttt{verify\_op\_exact.log}).
\end{remark}

\subsection{Freezing at the noncompact boundary}

We now pass to the scaled variables. Let $\hat z \in (0, 1]$, and let $f$ be an even superfunction on $\{B > 1\}$ of charge $(m, 0)$. By \eqref{e:res-charge}, we may write
\begin{equation}\label{e:res-components}
 f = \sum_{I \ \mathrm{even}} e^{\iu (m - c_I) \theta_{\mathrm B}} \xi^I f_I(q, \mathbf e) , \qquad q = \hat z B ,
\end{equation}
where $f_I$ is a function of $q$ and of the point $\mathbf e \in S^2$, with charge $c_I$ under the rotations of $S^2$ about its axis. The superfunction $f$ is smooth if and only if the functions $f_I$ are smooth on $(\hat z, \infty) \times S^2$, and we call $(f_I)_I$ the \emph{components} of $f$. Let $\Delta_{S^2} = \partial_F((1 - F^2) \partial_F) + (1 - F^2)^{-1} \partial_{\theta_{\mathrm F}}^2$ denote the Laplace operator of $S^2$. We define
\begin{equation}\label{e:res-scalar-generator}
 \Delta_{\hat z} = (q^2 - \hat z^2) \partial_q^2 + 2 \hat z \, \frac{\hat z - qF}{q - \hat z F} \, \partial_q + \Delta_{S^2} + \frac{2 \hat z (1 - F^2)}{q - \hat z F} \, \partial_F .
\end{equation}
The first two terms form the generator of $q_s$ in \eqref{e:res-scaled-sde}. The last two generate a Brownian motion on $S^2$ with a drift, since the vector field $(1 - F^2) \partial_F$ is the gradient of $F$ on $S^2$, which is smooth.

\begin{lemma}\label{l:res-freezing}
Let $k \ge 1$, $|m| \le k$, and $\hat z \in (0, 1]$. There exist a constant $C_k$ and a matrix $\mathcal V_{\hat z} = \mathcal V_{\hat z, m}$ indexed by the even subsets of $\{1, 2, 3, 4\}$, with entries that are smooth functions of $(q, \mathbf e) \in (\hat z, \infty) \times S^2$, such that the following holds. For every smooth even superfunction $f$ on $\{B > 1\}$ of charge $(m, 0)$, the components of $-\mathsf G_{1, \hat z} f$ are
\begin{equation}\label{e:res-generator-limit}
 \big( -\mathsf G_{1, \hat z} f \big)_I = \big( \Delta_{\hat z} - (q - \hat z F) \big) f_I + \sum_J (\mathcal V_{\hat z})_{IJ} f_J ,
 \qquad
 \|\mathcal V_{\hat z}(q, \mathbf e)\| \le C_k \frac{\hat z}{q} \quad \text{for } q \ge 2 \hat z .
\end{equation}
Moreover, the source satisfies
\begin{equation}\label{e:res-frozen-source}
 \hat z \, \mathfrak s = q \, \mathfrak s_0 + \hat z \, \mathfrak r_{\hat z} , \qquad
 \mathfrak s_0 = (1 + \cos\theta_{\mathrm B}) \Big( 1 + \frac{n_{\mathrm R} + n_{\mathrm A}}{2} \Big) + \frac{\cos\theta_{\mathrm B}}{4} n_{\mathrm R} n_{\mathrm A} ,
\end{equation}
where the coefficients of $\mathfrak r_{\hat z}$ are bounded by an absolute constant on $\{B > 1\}$, uniformly in $\hat z$.
\end{lemma}

In the scaled variables, the generator acts on the components as the scalar operator $\Delta_{\hat z}$ of the singlet diffusion with a spherical angle, plus a matrix potential that vanishes at the noncompact boundary. The source becomes multiplication by $q$ times the element $\mathfrak s_0$, which depends only on $\theta_{\mathrm B}$ and on the odd variables. In the limit, the generator and the source leave the variables $\theta_{\mathrm B}$ and $\xi$ fixed.

\begin{proof}
We use \zcref{l:res-polar} and the substitution $B = q/\hat z$, for which $\partial_B = \hat z \partial_q$. Recall that
\[
 \Delta^{\mathrm s}_D = (B^2 - 1) \partial_B^2 + \frac{2(1 - BF)}{B - F} \partial_B + (1 - F^2) \partial_F^2 - 2F \partial_F + \frac{2(1 - F^2)}{B - F} \partial_F .
\]
In the variable $q$, we have
\[
 (B^2 - 1) \partial_B^2 = (q^2 - \hat z^2) \partial_q^2 , \qquad
 \frac{2(1 - BF)}{B - F} \partial_B = 2 \hat z \, \frac{\hat z - qF}{q - \hat z F} \, \partial_q , \qquad
 \frac{2(1 - F^2)}{B - F} = \frac{2 \hat z (1 - F^2)}{q - \hat z F} ,
\]
Comparing with \eqref{e:res-scalar-generator}, we see that the operator
\[
 \Delta^{\mathrm s}_D + \frac{1}{1 - F^2} \, \partial_{\theta_{\mathrm F}}^2
\]
acts on each component as $\Delta_{\hat z}$. By \eqref{e:res-d-polar}, the multiplication by $\hat z d_1$ is the multiplication by $q - \hat z F$.

We next show that the remaining terms of \eqref{e:res-polar} act on the components as a matrix. By \eqref{e:res-components}, the operator $\partial_{\theta_{\mathrm B}}$ acts on the component of $\xi^I$ as multiplication by $\iu (m - c_I)$, and the term with $\partial_{\theta_{\mathrm B}}^2$ contributes the diagonal entries
\[
 -\frac{(m - c_I)^2}{B^2 - 1} = -\frac{(m - c_I)^2 \hat z^2}{q^2 - \hat z^2} .
\]

For the odd part, note that $\partial_+$ acts on every superfunction of charge $(m, 0)$ as multiplication by $\iu m$, since the coefficient of $\xi^I$ in \eqref{e:res-charge} has the phase charges $m - c_I$ and $c_I$. Multiplication by $\xi_a$ and the derivatives $\partial_{\xi_a}$ preserve the sum of the two charges, so the operators $\mathcal D_a$ act on the coefficients of superfunctions of charge $(m, 0)$ as finite matrices with entries in $\{0, \pm 1, \pm \frac{m}{2}\}$. The last term of \eqref{e:res-polar} acts as a matrix whose entries are linear combinations of the $g^{ab}$, with coefficients bounded in terms of $k$. In the components, the phases $e^{\pm \iu \theta_{\mathrm B}}$ of $g^{14}$ and $g^{23}$ are absorbed by the change of $c_I$, and the factors
\[
 (1 - F^2)^{1/2} e^{\mp \iu \theta_{\mathrm F}}
\]
are smooth functions of $\mathbf e$. We let $\mathcal V_{\hat z}$ be the sum of these two matrices.

Let $B \ge 2$. The bounds $B - F \ge B - 1$ and $(B - 1)^2 \ge B^2/4$ give
\begin{gather*}
 |g^{12}| = |g^{34}| \le \frac{2 (B + 1)}{(B - 1)^2} \le \frac{12}{B} , \qquad
 |g^{14}| = |g^{23}| \le \frac{2B}{(B - 1)^2} \le \frac 8B , \\
 \frac{(m - c_I)^2 \hat z^2}{q^2 - \hat z^2} \le \frac{4 (k + 1)^2}{3 B^2} .
\end{gather*}
Since $1/B = \hat z/q$, these bounds imply the bound on $\mathcal V_{\hat z}$ in \eqref{e:res-generator-limit}.

Finally, \eqref{e:res-source-polar} gives \eqref{e:res-frozen-source} with
\begin{align*}
 \mathfrak r_{\hat z} &= \frac{(q^2 - \hat z^2)^{1/2} - q}{\hat z} \cos\theta_{\mathrm B} \Big( 1 + \frac{n_{\mathrm R} + n_{\mathrm A}}{2} + \frac{n_{\mathrm R} n_{\mathrm A}}{4} \Big) - \frac F2 (n_{\mathrm R} + n_{\mathrm A}) \\
 &\quad - \frac\iu2 (1 - F^2)^{1/2} \big( e^{\iu \theta_{\mathrm F}} \xi_1 \xi_4 - e^{-\iu \theta_{\mathrm F}} \xi_2 \xi_3 \big) ,
\end{align*}
and \eqref{e:res-coefficients} gives
\[
 \big| (q^2 - \hat z^2)^{1/2} - q \big| \le \hat z . \qedhere
\]
\end{proof}

\section{The permanent residue}\label{s:res-permanent}

We now prove the second main result of this section. Its proof extends the coupling argument of \zcref{t:res-scalar} to the matrix potential of \zcref{l:res-freezing}. This is the most technical step of the section.

\begin{lemma}\label{l:res-permanent}
Let $k \ge 2$, $0 < X_- \le X_+ < \infty$, and $\delta \in (0, 1/2]$. Then
\begin{equation}\label{e:res-permanent}
 \lim_{z \downarrow 0} z^{k - 1} \mathfrak p^{X, z}_k(x_1, \dots, x_k) = k (2k - 3)!! \, M_k^X(x_1, \dots, x_k) ,
\end{equation}
uniformly for $X \in [X_-, X_+]$ and $(x_1, \dots, x_k)$ such that $\delta \le x_1$, $x_k \le 1 - \delta$, and $x_{j + 1} - x_j \ge \delta$ for $1 \le j \le k - 1$.
\end{lemma}

The constant $k(2k - 3)!!$ is the constant $\varkappa_{k, k!}$ of \eqref{e:extract-constant}, which the extraction theorem requires for the moment measures of \zcref{s:coarse} whose weights are squared permanents of eigenvector entries (\zcref{d:coarse-permanent}).

We fix the notation for the proof. We use the scaled times $s_0 = X x_1$, $s_j = X (x_{j + 1} - x_j)$ for $1 \le j \le k - 1$, and $s_k = X (1 - x_k)$, which lie in $[s_*, T]$ with $s_* = X_- \delta$ and $T = X_+$. We let $0 < \hat z \le \hat z_0 = \min(1, X_-)$. Since $e^{-t \mathsf G_{X, z}} = e^{-Xt \mathsf G_{1, \hat z}}$, we have
\[
 \mathfrak p^{X, z}_k = \hat z^{-k} \mathcal B_1 \big( \Psi^{(\hat z)}_{s_0}, \mathfrak c_1 \big) .
\]
Here $\Psi^{(\hat z)}_{s_0}$ stands for the invariant superfunction $e^{-s_0 \mathsf G_{1, \hat z}} 1$, whose radial representative, in the variables $(q, F)$, is the function $\Psi^{(\hat z)}_{s_0}$ of \eqref{e:res-rescaling}. The \emph{partial products} $\mathfrak c_j$ are defined by
\begin{equation}\label{e:res-chain-pieces}
 \mathfrak c_k = \hat z \, \mathfrak s \, e^{-s_k \mathsf G_{1, \hat z}} 1 , \qquad
 \mathfrak c_j = \hat z \, \mathfrak s \, e^{-s_j \mathsf G_{1, \hat z}} \mathfrak c_{j + 1} , \quad 1 \le j \le k - 1 .
\end{equation}
We define the \emph{frozen partial products} by
\begin{equation}\label{e:res-frozen-chain}
 \mathfrak c^0_j = \mathfrak s_0^{k - j + 1} \varphi_j , \qquad
 \varphi_k = q \, \Upsilon_{s_k} , \qquad
 \varphi_j = q \, \mathcal T_{s_j} \varphi_{j + 1} , \quad 1 \le j \le k - 1 ,
\end{equation}
where $\mathfrak s_0$ is the element of \eqref{e:res-frozen-source} and $\varphi_j$ is a function of $q$. The proof has four steps. We first prove a Feynman--Kac formula for semigroup products away from the corner (Step 1) and a polynomial bound (Step 2). We then show that $\mathfrak c_j \to \mathfrak c^0_j$ (Step 3), and we integrate over $q$ (Step 4). In the first three steps, the norm $\| \cdot \|$ of a semigroup product at a point is the maximum of the moduli of its components \eqref{e:res-components}, over the finitely many charges $|m| \le k$ and even $I$.

We also use the diffusion of the scalar part of the generator. For $q > 2 \hat z$ and $\mathbf e \in S^2$, let $(q_u, \mathbf e_u)$ denote the diffusion on $(\hat z, \infty) \times S^2$ with generator $\Delta_{\hat z}$, started at $(q, \mathbf e)$. Its radial part $q_u$ is driven by a Brownian motion $\beta$, and the motion on $S^2$ by an independent one. The coefficients of $\Delta_{\hat z}$ are smooth for $q > \hat z$, and the diffusion is well defined up to the hitting time $\tau_2$ of $\{q = 2 \hat z\}$, which corresponds to $B = 2$. By \eqref{e:res-scalar-generator}, the process $(q_u, F_u)$, with $F_u$ the height of $\mathbf e_u$, solves the equation of the scaled singlet diffusion in \eqref{e:res-scaled-sde} up to $\tau_2$. We need the following bounds for it.

\begin{lemma}\label{l:res-radial}
Let $T > 0$ and $n \ge 1$. There exists $C$ such that for all $0 < \hat z \le 1$, $q > 2 \hat z$, and $\mathbf e \in S^2$,
\begin{equation}\label{e:res-radial-moments}
 \E \sup_{u \le T} q_{u \wedge \tau_2}^n \le C q^n , \qquad
 \E q_{u \wedge \tau_2}^{-2} \le e^{10 u} q^{-2} , \quad u \ge 0 ,
\end{equation}
and
\begin{equation}\label{e:res-exit}
 \Prob(\tau_2 \le T) \le \exp\Big( -\frac{(\log(q/(2 \hat z)) - 3T)^2}{4T} \Big) , \qquad q \ge 2 \hat z e^{3T} .
\end{equation}
Moreover, for every compact set $\mathcal K \subset (0, \infty)$,
\begin{equation}\label{e:res-tight}
 \lim_{R \to \infty} \sup_{\hat z \le 1, \, q \in \mathcal K, \, \mathbf e \in S^2} \Prob \big( q_u \notin [R^{-1}, R] \text{ for some } u \le T \wedge \tau_2 \big) = 0 .
\end{equation}
\end{lemma}

\begin{proof}
For $u < \tau_2$, we have $q_u > 2 \hat z$, and \eqref{e:res-coefficients} bounds the drift of $q_u$ by $2 \hat z \le q_u$ and its diffusion coefficient by $\sqrt 2 \, q_u$. It\^o's formula and the Burkholder--Davis--Gundy inequality give the first bound in \eqref{e:res-radial-moments}. Further, the drift of $q_u^{-2}$ satisfies
\[
 -2 q_u^{-3} \cdot 2 \hat z \, \frac{\hat z - q_u F_u}{q_u - \hat z F_u} + 6 (q_u^2 - \hat z^2) q_u^{-4} \le 10 q_u^{-2} ,
\]
and $q_u^{-2} \le (2 \hat z)^{-2}$ before $\tau_2$. Taking expectations of the stopped process and applying Gronwall's inequality, we obtain the second bound.

As in the proof of \zcref{l:res-envelope}, the process $\log q_u - \log q + 3u$ is bounded below by a continuous martingale with quadratic variation at most $2u$, for $u < \tau_2$. The exponential martingale inequality implies \eqref{e:res-exit} and the lower half of \eqref{e:res-tight}. The upper half follows from the first bound in \eqref{e:res-radial-moments} and Markov's inequality.
\end{proof}

The proof also needs the regularity of semigroup products for small values of the last time $s_n$ in \eqref{e:sigma-chain}, since the Feynman--Kac formula below evaluates them there. This regularity is proved in \zcref{l:sigma-chains}. The semigroup products of \zcref{d:res-chain} are semigroup products \eqref{e:sigma-chain} with the multipliers $\mathfrak s$ (\zcref{l:sigma-multipliers}), and we apply \zcref{l:sigma-chains} with $X = 1$, $z_0 = 1$, $z = \hat z \in (0, 1]$, and times in $[0, T]$. We use part \ref{i:sigma-chains-compact} of \zcref{l:sigma-chains}, which is the last assertion of \zcref{p:sigma-regular}\ref{i:sigma-regular-uniform}, to bound semigroup products on the compact set $\{B \le 3\}$ of $\mathcal M_1^0$, which contains the corner, in Step 2 below, uniformly in $\hat z$ and in the times, and its part \ref{i:sigma-chains-time} for the regularity up to time zero, at fixed $\hat z$. These parts bound coefficients in the odd generators $\chi_a$. Recall that these are the coefficients in the coordinates of a section (the regular coordinates of \zcref{d:sigma-target}), and that the translated charts of \zcref{s:sigma-weighted} are the charts $x \mapsto g' x g'^{-1}$ obtained from one fixed chart at $\Lambda$ by the action of an element $g' \in G_0$. In polar coordinates, part \ref{i:sigma-chains-time} implies the following corollary.

\begin{corollary}\label{c:res-chains-polar}
Let $0 < \hat z \le 1$, let $f \in \mathfrak S_k$ be a semigroup product of charge $(m, 0)$, and let $r \ge 0$. The components \eqref{e:res-components} of $e^{-v \mathsf G_{1, \hat z}} f$ are smooth functions of $(v, q, \mathbf e) \in [0, \infty) \times (\hat z, \infty) \times S^2$, and they solve
\begin{equation}\label{e:res-component-equation}
 \partial_v u = \big( \Delta_{\hat z} - (q - \hat z F) + \mathcal V_{\hat z} \big) u \qquad \text{on } \{q > 2 \hat z\} .
\end{equation}
Moreover, there exist $N$ and $C$ such that these components and their derivatives of order at most two are bounded by $C (1 + B)^N$ on $\{B \ge 2\}$, for all $v \in [0, r]$.
\end{corollary}

\begin{proof}
By \zcref{l:sigma-chains}\ref{i:sigma-chains-time}, with $X = 1$, $z = \hat z$, and the multipliers $\mathfrak s$, the map
\[
 (v, Q) \mapsto \big( e^{-v \mathsf G_{1, \hat z}} f \big)(Q) , \qquad (v, Q) \in [0, \infty) \times \mathcal M_1^0 ,
\]
is smooth, also at $v = 0$. It solves $\partial_v u = -\mathsf G_{1, \hat z} u$, and \zcref{l:res-freezing} turns this equation into \eqref{e:res-component-equation}. The weighted bound in \zcref{l:sigma-chains}\ref{i:sigma-chains-time} and the Sobolev embedding theorem on the balls of the translated charts show that, for some $N$, the coefficients of $e^{-v \mathsf G_{1, \hat z}} f$ in the translated charts, and their derivatives of order at most two, are bounded by $C (1 + B)^N$, uniformly for $v \in [0, r]$.

It remains to pass to the polar chart on $\{B \ge 2\}$. We write $Q_0 = g \Lambda g^{-1}$ with a body element $g = \diag(g_{\mathrm B} | g_{\mathrm F})$. We may choose $g_{\mathrm B}$ as a smooth function of $(B, \theta_{\mathrm B})$ whose entries are bounded, with their derivatives, by $C B^{1/2}$ on $\{B \ge 2\}$ (a boost of rapidity $\operatorname{arccosh} B$), and $g_{\mathrm F}$ as a smooth function of $\mathbf e$ on each of two open sets that cover $S^2$. Then, by \eqref{e:sigma-polar}, we have
\[
 Q = g \, e^{-g^{-1} \Theta_\xi g} \Lambda e^{g^{-1} \Theta_\xi g} \, g^{-1} .
\]
In the translated chart centered at the body point $Q_0$, the coordinates of $Q$ are therefore a fixed smooth function of the odd matrix $g^{-1} \Theta_\xi g$. This matrix is linear in $\xi$, with coefficients bounded by $C B$. The transition between the two charts and its derivatives are thus bounded by powers of $B$, and the chain rule transfers the bounds to the components.
\end{proof}

\begin{proof}[Proof of \zcref{l:res-permanent}]
\emph{Step 1 (a Feynman--Kac formula).} Let $(q_u, \mathbf e_u)$ and $\tau_2$ be as in \zcref{l:res-radial}, and let $\mathcal E_u$ be the matrix solution of
\[
 \dd \mathcal E_u = \mathcal E_u \, \mathcal V_{\hat z}(q_u, \mathbf e_u) \dd u , \qquad \mathcal E_0 = \Id , \qquad
 \|\mathcal V_{\hat z}(q_u, \mathbf e_u)\| \le \frac{C_k \hat z}{q_u} \le C_k \quad (u < \tau_2) ,
\]
where the bound is \eqref{e:res-generator-limit}. Gronwall's inequality implies
\begin{equation}\label{e:res-gronwall}
 \|\mathcal E_u\| \le e^{C_k u} , \qquad
 \|\mathcal E_u - \Id\| \le C_k e^{C_k u} \hat z \int_0^u q_v^{-1} \dd v , \qquad u \le \tau_2 .
\end{equation}

Let $f \in \mathfrak S_k$ be a semigroup product of charge $(m, 0)$ and $r \ge 0$. By \zcref{c:res-chains-polar}, the components of $e^{-v \mathsf G_{1, \hat z}} f$ are smooth in $(v, q, \mathbf e)$, also at $v = 0$, they solve \eqref{e:res-component-equation}, and they grow at most polynomially in $B$, uniformly for $v \in [0, r]$. We apply It\^o's formula to the process
\[
 \mathcal E_v \exp\Big( -\int_0^v (q_w - \hat z F_w) \dd w \Big) \big( e^{-(r - v) \mathsf G_{1, \hat z}} f \big)(q_v, \mathbf e_v) , \qquad v \le \tau_2 \wedge r .
\]
The drift of this process vanishes. The polynomial bound on the components, \eqref{e:res-gronwall}, and \eqref{e:res-radial-moments} bound it by an integrable random variable. Hence its martingale part is a true martingale, and taking expectations, we obtain
\begin{equation}\label{e:res-stopped-fk}
 \big( e^{-r \mathsf G_{1, \hat z}} f \big)(q, \mathbf e)
 = \E \Big[ \mathcal E_{\tau_2 \wedge r} \exp\Big( -\int_0^{\tau_2 \wedge r} (q_v - \hat z F_v) \dd v \Big) \big( e^{-(r - \tau_2 \wedge r) \mathsf G_{1, \hat z}} f \big)(q_{\tau_2 \wedge r}, \mathbf e_{\tau_2 \wedge r}) \Big] ,
\end{equation}
as an identity between vectors of components, at points with $q > 2 \hat z$.

\emph{Step 2 (a polynomial bound).} We claim that for $0 \le n \le k$, there exists $C$ such that
\begin{equation}\label{e:res-polynomial-bound}
 \big\| e^{-r \mathsf G_{1, \hat z}} \mathfrak s \, e^{-r_1 \mathsf G_{1, \hat z}} \mathfrak s \cdots \mathfrak s \, e^{-r_n \mathsf G_{1, \hat z}} 1 \big\| \le C (1 + B)^n
\end{equation}
on $\{B > 1\}$ (at points with $B \le 3$, the norm is the maximum of the coefficients in the regular coordinates), for all $0 < \hat z \le \hat z_0$, $r \in [0, T]$, and $r_1, \dots, r_n \in [s_*, T]$. We argue by induction on $n$. Let $f = 1$ if $n = 0$, and
\[
 f = \mathfrak s \, e^{-r_1 \mathsf G_{1, \hat z}} \cdots e^{-r_n \mathsf G_{1, \hat z}} 1
\]
if $n \ge 1$. The entries of $Q^{\mathrm{BB}}$ are bounded by $CB$. Together with the induction hypothesis, this implies
\[
 \|f\| \le C (1 + B)^n \qquad \text{on } \{B > 1\} .
\]

We first bound $e^{-r \mathsf G_{1, \hat z}} f$ on the compact set $\{B \le 3\}$. This superfunction is a semigroup product \eqref{e:sigma-chain} with the multipliers $\mathfrak s$ and times in $[0, T]$. We apply \zcref{l:sigma-chains}\ref{i:sigma-chains-compact} with $X = 1$, $z_0 = 1$, $z = \hat z$, and the compact set $\{B \le 3\}$ of $\mathcal M_1^0$. It bounds the coefficients of $e^{-r \mathsf G_{1, \hat z}} f$ on $\{B \le 3\}$, uniformly in $r \in [0, T]$ and $0 < \hat z \le \hat z_0$; this includes times $r$ close to zero.

We now insert these bounds on $\{B \le 3\}$ into \eqref{e:res-stopped-fk}, at points with $B > 2$. The last factor in \eqref{e:res-stopped-fk} is bounded by a constant on the event $\{\tau_2 \le r\}$, where it is evaluated at $B = 2$. On the complementary event, it reduces to $f(q_r, \mathbf e_r)$, and
\[
 \|f(q_r, \mathbf e_r)\| \le C (1 + B_r)^n = C \hat z^{-n} (\hat z + q_r)^n , \qquad B_r = \frac{q_r}{\hat z} .
\]
Using \eqref{e:res-gronwall}, the nonnegativity of the killing and the moment bound
\[
 \E (\hat z + q_{r \wedge \tau_2})^n \le C (\hat z + q)^n = C \hat z^n (1 + B)^n ,
\]
we obtain \eqref{e:res-polynomial-bound}. Since $\mathfrak c_j$ is $\hat z^{k - j + 1}$ times a semigroup product with $k - j + 1$ sources, \eqref{e:res-polynomial-bound} with $r = 0$ gives
\begin{equation}\label{e:res-chain-bound}
 \|\mathfrak c_j\| \le C \hat z^{k - j + 1} (1 + B)^{k - j + 1} = C (\hat z + q)^{k - j + 1} , \qquad 1 \le j \le k .
\end{equation}
For the frozen partial products, we differentiate
\[
 \mathcal T_s \varphi(q) = \E \big[ \varphi(q e^{g_1(s)}) e^{-q A_s} \big]
\]
in $q$ and argue by induction, using that $e^{g_1(s)}$ and $A_s$ have moments of all orders. We find
\[
 |\varphi_j(q)| + |\varphi_j'(q)| \le C (1 + q)^{k - j + 1} , \qquad 1 \le j \le k .
\]

\emph{Step 3 (the partial products freeze).} We show by downward induction on $j$ that $\mathfrak c_j \to \mathfrak c^0_j$ as $\hat z \to 0$, componentwise, uniformly for $q$ in compact subsets of $(0, \infty)$, for $\mathbf e \in S^2$, and for the times in $[s_*, T]$. For $j = k$, \eqref{e:res-frozen-source} gives
\[
 \mathfrak c_k = (q \, \mathfrak s_0 + \hat z \, \mathfrak r_{\hat z}) \Psi^{(\hat z)}_{s_k} , \qquad \mathfrak c^0_k = q \, \mathfrak s_0 \Upsilon_{s_k} ,
\]
and \eqref{e:res-side} gives the claim.

For the induction step, let $j < k$, $s = s_j$, and $q > 0$, and take $\hat z$ so small that $q \ge 2 \hat z e^{3T}$. By \eqref{e:res-frozen-source} and \eqref{e:res-chain-bound}, it suffices to show that
\[
 \lim_{\hat z \to 0} e^{-s \mathsf G_{1, \hat z}} \mathfrak c_{j + 1} = \mathfrak s_0^{k - j} \mathcal T_s \varphi_{j + 1} .
\]
Let $q^0_u = q e^{g_1(u)}$, with $g_1$ built from the Brownian motion $\beta$ of Step 1, so that
\[
 \mathfrak s_0^{k - j} \mathcal T_s \varphi_{j + 1}(q) = \E \big[ e^{-\kappa^0} \mathfrak c^0_{j + 1}(q^0_s) \big] , \qquad \kappa^0 = \int_0^s q^0_u \dd u .
\]
We compare this with \eqref{e:res-stopped-fk} for $f = \mathfrak c_{j + 1}$ and $r = s$, and we set
\[
 \kappa = \int_0^{\tau_2 \wedge s} (q_u - \hat z F_u) \dd u .
\]

We first consider the event $\{\tau_2 \le s\}$. There, the integrand of the frozen expectation is bounded by $C (1 + q^0_s)^k$, and the integrand of \eqref{e:res-stopped-fk} by a constant, by \eqref{e:res-polynomial-bound} at $B = 2$ and \eqref{e:res-gronwall}. The probability of this event tends to zero by \eqref{e:res-exit}, so the Cauchy--Schwarz inequality shows that its contribution tends to zero.

We now turn to the event $\{s < \tau_2\}$, on which the difference of the integrands is bounded by
\[
 \big( \|\mathcal E_s - \Id\| + |\kappa - \kappa^0| \big) \|\mathfrak c_{j + 1}(q_s, \mathbf e_s)\|
 + \|\mathfrak c_{j + 1}(q_s, \mathbf e_s) - \mathfrak c^0_{j + 1}(q_s)\|
 + \|\mathfrak c^0_{j + 1}(q_s) - \mathfrak c^0_{j + 1}(q^0_s)\| .
\]
We bound the three terms in turn. The proof of \eqref{e:res-coupling} rests on the equation of $q_u$ in \eqref{e:res-scaled-sde} and on the bounds \eqref{e:res-coefficients}. Both hold for the present process up to $\tau_2$, since its pair $(q_u, F_u)$ solves \eqref{e:res-scaled-sde} there, and the same proof shows that
\[
 \E \sup_{u \le s \wedge \tau_2} |q_u - q^0_u|^2 \le C \hat z^2 , \qquad \E \big[ |\kappa - \kappa^0|^2 ; s < \tau_2 \big] \le C \hat z^2 .
\]
By \eqref{e:res-gronwall}, the Cauchy--Schwarz inequality and \eqref{e:res-radial-moments}, we have
\[
 \E \big[ \|\mathcal E_s - \Id\|^2 ; s < \tau_2 \big] \le C \hat z^2 \int_0^s \E q_{u \wedge \tau_2}^{-2} \dd u \le C \hat z^2 q^{-2} .
\]
Together with \eqref{e:res-chain-bound}, the moment bounds of \zcref{l:res-radial} and the Cauchy--Schwarz inequality, these bounds show that the expectation of the first term tends to zero.

The expectation of the second term tends to zero by the induction hypothesis on the event $\{q_s \in [R^{-1}, R]\}$. On the complementary event, we use \eqref{e:res-chain-bound}, the bound on $\varphi_{j + 1}$ and the tightness \eqref{e:res-tight}, and we let $R \to \infty$ after $\hat z \to 0$.

The third term is at most
\[
 \|\mathfrak s_0\|^{k - j} \sup |\varphi'_{j + 1}| \, |q_s - q^0_s| ,
\]
with the supremum over the segment between $q_s$ and $q^0_s$. Its expectation is $O(\hat z)$, by the bound on $\varphi'_{j + 1}$, the moment bounds, and the coupling. All these bounds are uniform in the stated parameters. We multiply by $\hat z \mathfrak s$ and use \eqref{e:res-frozen-source}. This completes the induction step, since
\[
 q \mathfrak s_0 \cdot \mathfrak s_0^{k - j} \mathcal T_s \varphi_{j + 1} = \mathfrak s_0^{k - j + 1} \varphi_j = \mathfrak c^0_j .
\]

\emph{Step 4 (integration).} Let $\chi \in C^\infty(\R)$ satisfy $\chi = 1$ on $(-\infty, 3]$ and $\chi = 0$ on $[4, \infty)$. The superfunction $\chi(d_1)$ is invariant, with radial representative $\chi(d)$, by \eqref{e:res-d-polar}. Since $B \ge d - 1$, the function $1 - \chi(d)$ vanishes on $\{B < 2\}$, and $\chi(d)$ vanishes outside the compact set $\{d \le 4\}$. We split
\[
 \mathcal B_1 \big( \Psi^{(\hat z)}_{s_0}, \mathfrak c_1 \big) = \int \chi(d_1) \Psi^{(\hat z)}_{s_0} \mathfrak c_1 + \int \big( 1 - \chi(d_1) \big) \Psi^{(\hat z)}_{s_0} \mathfrak c_1 .
\]
\zcref[S]{p:sigma-regular}\ref{i:sigma-regular-uniform} and \eqref{e:res-chain-pieces} bound the coefficients of the first integrand in regular coordinates by $C \hat z^k$, on its compact support. Hence, by \eqref{e:sigma-integral}, the first integral is $O(\hat z^k)$, which is $o(\hat z)$ since $k \ge 2$.

The second integrand is supported in $\{B \ge 2\}$, and we use \eqref{e:res-polar-integral}. The functions $\Psi^{(\hat z)}_{s_0}$ and $\chi(d)$ have no odd variables in the polar chart, and
\[
 \big[ (1 - \chi(d)) \Psi^{(\hat z)}_{s_0} \mathfrak c_1 \big]_{\mathrm{top}} = (1 - \chi(d)) \Psi^{(\hat z)}_{s_0} [\mathfrak c_1]_{\mathrm{top}} .
\]
We substitute $q = \hat z B$, for which
\[
 (B - F)^{-2} \dd B = \hat z (q - \hat z F)^{-2} \dd q , \qquad d = \frac{q - \hat z F}{\hat z} .
\]
We obtain
\[
 \hat z^{k - 1} \mathfrak p^{X, z}_k + o(1) = \int_{2 \hat z}^\infty \int_{-1}^1 \Big( 1 - \chi\Big( \frac{q - \hat z F}{\hat z} \Big) \Big) \frac{\Psi^{(\hat z)}_{s_0}(q, F)}{(q - \hat z F)^2} \big\langle [\mathfrak c_1]_{\mathrm{top}} \big\rangle (q, F) \dd F \dd q ,
\]
where $\langle \cdot \rangle$ denotes the average over $\theta_{\mathrm B}$ and $\theta_{\mathrm F}$.

For $q \ge 2 \hat z$ and $F \in [-1, 1]$, we have
\[
 q - \hat z F \ge q - \hat z \ge \frac q2 .
\]
Therefore \zcref{l:res-envelope} with $t_0 = s_*$ and \eqref{e:res-chain-bound} bound the integrand by
\[
 C q^{-2} (\hat z + q)^k \, \bar\Upsilon_{s_*, T}(q) \le C q^{k - 2} \bar\Upsilon_{s_*, T}(q) ,
\]
which is integrable since $k \ge 2$. By Step 3 and \eqref{e:res-side}, the integrand converges to
\[
 q^{-2} \Upsilon_{s_0}(q) \big\langle [\mathfrak s_0^k]_{\mathrm{top}} \big\rangle \varphi_1(q) ,
\]
uniformly for $q$ in compact subsets of $(0, \infty)$, $F \in [-1, 1]$, and the parameters. Splitting the integral over $q$ at $R^{-1}$ and $R$, as in the proof of \zcref{t:res-scalar}, we conclude that, uniformly in the parameters,
\begin{equation}\label{e:res-frozen-limit}
 \lim_{\hat z \downarrow 0} \hat z^{k - 1} \mathfrak p^{X, z}_k
 = 2 \, \big\langle [\mathfrak s_0^k]_{\mathrm{top}} \big\rangle \int_0^\infty q^{-2} \Upsilon_{s_0}(q) \varphi_1(q) \dd q ,
\end{equation}
where the factor two is the integral over $F$.

It remains to compute the constant. Let $t = 1 + \cos\theta_{\mathrm B}$. Since $n_{\mathrm R}^2 = n_{\mathrm A}^2 = 0$, the binomial expansion of
\[
 \mathfrak s_0^k = \Big( t + \frac t2 (n_{\mathrm R} + n_{\mathrm A}) + \frac{t - 1}{4} n_{\mathrm R} n_{\mathrm A} \Big)^k
\]
gives
\[
 [\mathfrak s_0^k]_{\mathrm{top}} = k t^{k - 1} \frac{t - 1}{4} + \binom k2 t^{k - 2} \frac{t^2}{2} = \frac k4 t^{k - 1} (kt - 1) .
\]
The averages of the powers of $t$ are
\[
 \frac{1}{2\pi} \int_0^{2\pi} (1 + \cos\theta)^n \dd \theta = 2^{-n} \binom{2n}{n} , \qquad n \ge 0 ,
\]
and we find
\[
 \big\langle [\mathfrak s_0^k]_{\mathrm{top}} \big\rangle = \frac k4 \Big( k \, 2^{-k} \binom{2k}{k} - 2^{1 - k} \binom{2k - 2}{k - 1} \Big) = \frac{k (2k - 3)!!}{2 (k - 2)!} .
\]
In the last step, we used the identities
\[
 k \binom{2k}{k} = 2(2k - 1) \binom{2k - 2}{k - 1} , \qquad \binom{2k - 2}{k - 1} = \frac{2^{k - 1} (2k - 3)!!}{(k - 1)!} .
\]
Finally, unwinding \eqref{e:res-frozen-chain}, we find
\[
 \varphi_1 = q \, \mathcal T_{s_1} q \cdots q \, \mathcal T_{s_{k - 1}} q \, \Upsilon_{s_k} ,
\]
and the first line of \eqref{e:res-hierarchy} identifies the integral in \eqref{e:res-frozen-limit} with $(k - 2)! \, X^{1 - k} M_k^X$. Thus
\[
 \lim_{z \downarrow 0} z^{k - 1} \mathfrak p^{X, z}_k = X^{k - 1} \lim_{\hat z \downarrow 0} \hat z^{k - 1} \mathfrak p^{X, z}_k = X^{k - 1} \cdot 2 \cdot \frac{k (2k - 3)!!}{2 (k - 2)!} \cdot \frac{(k - 2)!}{X^{k - 1}} M_k^X = k (2k - 3)!! \, M_k^X ,
\]
which is \eqref{e:res-permanent}.
\end{proof}