\chapter{Stability of the transfer kernel}\label{s:stab}

In this section we prove that the powers of the adapted transfer
operator $A_h$ of \zcref{s:tube-gauge} stay bounded over $h^{-4}$ steps,
on arbitrary inputs and uniformly in $h$. This is the stability
hypothesis of the approximation theorem used in \zcref{s:cons}, and
\zcref{s:end,s:assembly} use the underlying four-term bound for one step.
The main result is \zcref{t:stab-main}.

We compare $A_h$ with the product $C_h$ of the cutoff Mehler operator of
\zcref{s:tube-cutoff} and the heat semigroup of the saddle manifold over
the time of one link. Since $A_h - C_h$ is not small in norm, we measure
it relative to the defects of $C_h$, in the form \eqref{e:tube-four-term},
which \zcref{c:tools-placements} turns into the hypothesis of
\zcref{l:tools-defect}. The comparison is made through the kernel of
$A_h$, written in coordinates centered at the midpoint of the two
endpoints of a link, and it rests on three structural facts. First, the
chordal Gaussian of a link is exactly even under the reversal of the
tangent increment (\zcref{l:stab-chord}). Second, once the action of the
stabilizer on the normal variables is factored out, the remaining kernel
is a scalar superfunction of the unscaled normal coordinates. The odd
normal variables, integrated against the odd Mehler factor, therefore
produce a multiple of the identity up to $O(h)$
(\zcref{l:stab-frame,l:stab-moment}). Third, we control differences
across a link by the heat semigroup of the saddle manifold, through a
difference energy along geodesics (\zcref{l:stab-energy}), and the
marginals of kernels that are odd under the reversal gain a factor $h^2$
(\zcref{l:stab-parity}). Recall that $h^2 = W^{-1}$ is the size of the
increment of one link and $\delta_W = h^4$ its variance. By \zcref{r:stab-fock}, an energy estimate in a metric of product form
cannot control the odd normal variables, and the proof below
integrates them out against the full Mehler factor of the step, half-weight
included, before any norm is taken.

\section{The reference contraction and the main theorem}
\label{s:stab-main}

\subsection{The underlying manifold, the coefficient bundle, and the body heat}

We let $|\mathcal M|$ denote the underlying manifold of the saddle manifold through
$R_*$, with the invariant Riemannian metric that
$g_{R_*}(Y, Z) = 4s_E^2\Str(YZ)$ induces on the real form of the even
tangent directions. (On the fermion--fermion block, the sign of the
supertrace is compensated by the real form, as in \zcref{s:sigma}.) We
let $\mu$ denote its Riemannian volume and $n_b$ its dimension. In the
balanced signature, $|\mathcal M|$ is the product of the hyperbolic and
the compact symmetric spaces that form the underlying manifold of $\mathcal M_k$ (see
\zcref{s:sigma}). Recall from \zcref{s:tube-prelim} that $U(k, k|2k)$
denotes the pair of the group
$G_0 = \{\diag(g_{\rm B} \,|\, g_{\rm F}) : g_{\rm B} \in U(k, k), \ g_{\rm F} \in U(2k)\}$
and the Lie superalgebra $\mathfrak{gl}(2k|2k)$, so that $G_0$ is the
underlying manifold of this supergroup. In every signature, $|\mathcal M|$
is a complete Riemannian symmetric space on which $G_0$ acts transitively
by isometries, with compact stabilizer $K_0$. We fix $\varrho_0 > 0$ smaller
than half the injectivity radius of $|\mathcal M|$. For
$d(x, y) < 2\varrho_0$ there exists a unique minimal geodesic from $y$ to
$x$, and we let $m(x, y)$ denote its midpoint.

We let $\Lambda_T$ denote the exterior algebra of the odd tangent
coordinates of the saddle manifold. We measure its elements in the
coefficient norm $| \cdot |_{2; \omega}$ of \zcref{d:grass-norm}, taken in an
orthonormal basis of the odd tangent coordinates, with a scalar weight
$0 < \omega \le 1$:
\[
 |c|_{2; \omega}^2 = \sum_{j \ge 0}\omega^{2j}|c_j|^2 ,
\]
where $c_j$ is the component of $c$ of odd degree $j$ and $|c_j|$ its
invariant norm. This is the norm of the continuum model in
\zcref{s:sigma-weighted}. By \zcref{l:grass-norm}, multiplication by a
nilpotent even element has norm $O(\omega^2)$ in this norm, and an odd
derivative has norm $O(\omega^{-1})$. The compact group $K_0$ acts
unitarily on $\mathcal H_{\rm nor} \otimes \Lambda_T$, by rotations of the
normal and odd tangent coordinates, and this action preserves
$\mathbf N$. We let $\mathcal E$ denote the associated Hilbert bundle over
$|\mathcal M|$ with fiber $\mathcal H_{\rm nor} \otimes \Lambda_T$, and
$\nabla$ its canonical connection, induced by the
$\mathfrak k_0$-component of the Maurer--Cartan form. For
$d(x, y) < 2\varrho_0$ we let $\tau_{xy}$ denote the parallel transport of
$\mathcal E$ along the minimal geodesic from $y$ to $x$. This transport is unitary, it commutes with $\mathbf N$, and
\[
 \tau_{yx} = \tau_{xy}^{-1} .
\]

We let $\Delta_\nabla$ denote the Friedrichs extension of
$-\nabla^*\nabla$ from the smooth compactly supported sections. On each
summand of $\mathcal E$ of finite rank (an eigenspace of $\mathbf N$
tensored with $\Lambda_T$), $\Delta_\nabla$ is essentially self-adjoint on
these sections, since $|\mathcal M|$ is complete \cite{chernoff1973essential}.
For $q \in \R$ we set
\[
 \mathcal H^{\rm tube}_q = L^2\bigl(|\mathcal M|, v^{2q}\dd\mu; \mathcal E\bigr) ,
 \qquad
 \mathbf H_h = e^{\delta_W\Delta_\nabla/2} ,
\]
where $v$ is the weight \eqref{e:sigma-weight}. Since $v$ is a function
of the BB block of the point, and the BB factor of $|\mathcal M|$ is the
hyperbolic space $H_k$ of \zcref{s:sigma} in every signature, $v$ is
defined on $|\mathcal M|$. By \zcref{l:sigma-body}\ref{i:sigma-body-derivatives},
$\log v$ is Lipschitz with bounded second derivatives. We call $\mathbf H_h$ the body heat. In terms of a Brownian
motion on $|\mathcal M|$ with generator $\frac12\Delta$, it is the
semigroup at time $\delta_W$. With the generator $\Delta$ it is the
semigroup at time $\tau_{\rm phys} = \delta_W/2$, which is the time of one
link: by \eqref{e:tube-heat-kernel} and the discussion after
\eqref{e:tube-decomposition}, the tangent part of one link is the heat
semigroup of $L_H(0)$ at time $\tau_{\rm phys}$.

\subsection{The reference contraction}

We now define the frozen source kick. In the normalization of
\zcref{s:hist}, \eqref{e:hist-normalized} gives the bosonic part
$e^{-\iu\eta\Tr(S^{\rm BB}\Gamma_{BB})}$ of the source factor $s_\Gamma$
at a point $S$, where $\eta = \eps\eta_0$ is of order $W^{-2} = \delta_W$.
Since the BB factor of $|\mathcal M|$ is the hyperbolic space $H_k$, the body of the BB block
of the point $R(x, 0)$ of the saddle manifold has the form
$\kappa_E\Id - \iu s_EQ_B$ with $Q_B \in H_k$, and we write
$Q_B = Q_B(x)$. The body of the bosonic part of $s_\Gamma$ at this point is
then
\[
 e^{-\iu\eta\Tr((\kappa_E\Id - \iu s_EQ_B)\Gamma_{BB})}
 = c_\Gamma \, e^{-\eta s_E\Tr(Q_B\Gamma_{BB})} ,
 \qquad
 c_\Gamma = e^{-\iu\eta\kappa_E\Tr\Gamma_{BB}} .
\]
We define the \emph{frozen source kick} $\mathbf S_h$ to be the
multiplication by $e^{-\eta s_E\Tr(Q_B\Gamma_{BB})}$. By the last display,
it is the multiplication by the body of the bosonic part of
$c_\Gamma^{-1}s_\Gamma$ at $U = 0$. We keep the constant $c_\Gamma$ out of the kick,
since $\Tr\Gamma_{BB}$ is not real in general on the parameter ball
$\mathcal U_2$ of \eqref{e:end-damping}, and $|c_\Gamma|$ may then exceed
one. Since $\Tr\Lambda = 0$ and $\|\Gamma_{BB} - \Lambda\|_{\rm op} \le 1/4$
on $\mathcal U_2$, we have $|\Tr\Gamma_{BB}| \le k/2$. Since $\eta$ is of
order $\delta_W$, this implies
\begin{equation}\label{e:stab-scalar}
 |c_\Gamma - 1| \le C\delta_W .
\end{equation}
We account for $c_\Gamma$ in the bound for the source,
\zcref{l:stab-src}.

When the damping is real, we have $\Gamma_{BB} = \Lambda$ and
$c_\Gamma = 1$, and \zcref{l:rep-kick} gives $0 < \mathbf S_h \le 1$. For
complex $\Gamma \in \mathcal U_2$, the kick is complex, and we have
\begin{equation}\label{e:stab-kick}
 |\mathbf S_h| \le e^{-c\delta_Wv} \le 1 ,
 \qquad
 |\nabla\mathbf S_h| + |\nabla^2\mathbf S_h| \le C\delta_Wv\,e^{-c\delta_Wv} .
\end{equation}
Indeed, by \eqref{e:sigma-weight} and
\zcref{l:sigma-body}\ref{i:sigma-body-trace}, we have $\Lambda Q_B \ge 0$ and
$\Tr(\Lambda Q_B) = 2(v - 1) \ge 2k$. Together with these relations,
\eqref{e:end-damping} implies that
\[
 \Re\Tr(Q_B\Gamma_{BB}) = \Re\Tr\bigl(\Lambda Q_B \, \Gamma_{BB}\Lambda\bigr)
 \ge c_\eps\Tr(\Lambda Q_B) = 2c_\eps(v - 1) \ge c_\eps v .
\]
The derivatives of $\Tr(Q_B\Gamma_{BB})$ along the orbit are bounded by
$Cv$, since a unit tangent generator acts on $Q_B$ by a commutator of
size $O(v)$ and $\|\Gamma\|$ is bounded, and the second bound in
\eqref{e:stab-kick} follows as in the proof of \zcref{l:rep-kick}. We
define the reference step
\[
 C_h = \mathbf M^\chi \otimes \mathbf S_h\mathbf H_h ,
\]
where $\mathbf M^\chi$ is the cutoff Mehler operator of
\zcref{s:tube-cutoff} with $\mathbf M = \mathbf M_h^{\rm q}$. Thus
$\mathbf M^\chi$ already contains the full scalar $\lambda$ of
\zcref{l:tube-quadratic}, including the physical root phase $\zeta_p$,
exactly once. Since the scalar cutoff
$\chi_{r/h}$ is invariant under $K_0$, it commutes with the parallel
transport, with $\mathbf S_h$, and with $\mathbf H_h$. Since $\|\mathbf M^\chi\| \le 1$,
$|\mathbf S_h| \le 1$ and $\|\mathbf H_h\| \le 1$ on
$\mathcal H^{\rm tube}_0$, the operator $C_h$ is a contraction of
$\mathcal H^{\rm tube}_0$, and we let
\[
 D_R = \Id - C_h^*C_h , \qquad D_L = \Id - C_hC_h^*
\]
denote its defects. We also set
\[
 D_{\rm body, R} = \Id - \mathbf H_h^2 + \mathbf H_h(\Id - |\mathbf S_h|^2)\mathbf H_h ,
 \qquad
 D_{\rm body, L} = \Id - |\mathbf S_h|^2 + \mathbf S_h(\Id - \mathbf H_h^2)\mathbf S_h^* ,
\]
so that $D_{\rm body, R} = \Id - \mathbf H_h|\mathbf S_h|^2\mathbf H_h$ and
$D_{\rm body, L} = \Id - \mathbf S_h\mathbf H_h^2\mathbf S_h^*$.

We now compare these defects. The normal factor
$(\mathbf M^\chi)^*\mathbf M^\chi$ and the body factor
$\mathbf H_h|\mathbf S_h|^2\mathbf H_h$ act on different tensor factors,
so they commute. For commuting operators $a$, $b$ with $0 \le a, b \le \Id$
we have
\[
 \Id - ab - \tfrac12(\Id - a) - \tfrac12(\Id - b) = \tfrac12\bigl(a(\Id - b) + b(\Id - a)\bigr) \ge 0 .
\]
Combining this inequality with $D_{\rm nor} \le 2\bar D_R$ from
\zcref{l:tube-slow}, and arguing in the same way on the left, we obtain
\begin{equation}\label{e:stab-product-defects}
 D_{\rm nor} + D_{\rm body, R} \le 4D_R ,
 \qquad
 D_{\rm nor} + D_{\rm body, L} \le 4D_L .
\end{equation}

The weighted spaces need a conjugated reference. Multiplication by $v^q$
is a unitary map from $\mathcal H^{\rm tube}_q$ onto
$\mathcal H^{\rm tube}_0$. We set
\[
 C_{h, q} = v^{-q}C_hv^{q} .
\]
Since $C_h$ is a contraction of $\mathcal H^{\rm tube}_0$, the operator
$C_{h, q}$ is a contraction of $\mathcal H^{\rm tube}_q$, with the defects
$v^{-q}D_Rv^q$ and $v^{-q}D_Lv^q$. Note that $C_h$ itself is in general
not a contraction of $\mathcal H^{\rm tube}_q$ for $q \ne 0$, since the
conjugate $v^q\mathbf H_hv^{-q}$ has norm $1 + O(q^2\delta_W)$, as the model
case $e^{qx}e^{t\partial_x^2}e^{-qx} = e^{t(\partial_x - q)^2}$ on $\R$
shows.

\subsection{The main theorem}

We consider the adapted transfer operator $A_h$ of \eqref{e:tube-table},
with the invariant cutoff $\chi_{\rm inv}$ at radius $r$ on both sides and
all odd coefficients retained. It acts on functions on the real tube
of \zcref{d:tube-cycle}. The Sch\"afer--Wegner cycle of
\zcref{s:rep} is different, and the two are connected in
\zcref{s:end,s:assembly}.

\begin{theorem}\label{t:stab-main}
Let $T > 0$ and $q \in \R$. There exist $\omega \in (0, 1]$, then
$r \in (0, r_*]$, then $h_0 > 0$ and constants $A_1$ and $B$, depending on
$k$, $q$, $T$, the compact sets of energies, of $\eps$ and of $X$, and the
damping ball $\mathcal U_2$, with the following property.
For $0 < h \le h_0$, in every signature, and uniformly for the damping
$\Gamma$ in the parameter ball $\mathcal U_2$ of \eqref{e:end-damping},
the operator $A_h$ on $\mathcal H^{\rm tube}_q$ satisfies the four-term
bound
\begin{multline}\label{e:stab-four-term}
 |\langle g, (A_h - C_{h, q})f\rangle|
 \le a\|D_{L, q}^{1/2}g\|\|D_{R, q}^{1/2}f\| \\
 + A_1\sqrt{\delta_W}\bigl(\|g\|\|D_{R, q}^{1/2}f\| + \|D_{L, q}^{1/2}g\|\|f\|\bigr)
 + A_1\delta_W\|g\|\|f\|
\end{multline}
relative to the contraction $C_{h, q} = v^{-q}C_hv^q$, with its defects
$D_{R, q} = v^{-q}D_Rv^q$ and $D_{L, q} = v^{-q}D_Lv^q$ and with
$a \le 1/16$. Consequently
\[
\begin{gathered}
 A_h^*A_h \le (1 + B\delta_W/2)\Id - \tfrac12D_{R, q} , \\
 A_hA_h^* \le (1 + B\delta_W/2)\Id - \tfrac12D_{L, q} , \\
 \sup_{n\delta_W \le T}\|A_h^n\|_{\mathcal H^{\rm tube}_q \to \mathcal H^{\rm tube}_q} \le e^{BT/4} ,
\end{gathered}
\]
with adjoints in $\mathcal H^{\rm tube}_q$. The same holds for every step
of the same form with a smaller damping in $\mathcal U_2$. The conclusion
remains true, with the same $r$, if $\omega$ is replaced by any
$\omega' \in (0, \omega]$, with $h_0$, $A_1$ and $B$ depending also on
$\omega'$.
\end{theorem}

The second part of the theorem follows from the first by the reduction
to \zcref{c:tools-placements} in \zcref{r:tube-four-term-reduction},
applied on $\mathcal H^{\rm tube}_q$ with $\mathbf T = A_h$,
$C = C_{h, q}$, $\delta_0 = \delta_W$, and $t = T$. Recall that this
reduction takes $B = \max\{1, 2304A_1^2\}$ and requires $a \le 1/16$ and
$B\delta_0 \le 1$. We therefore choose $h_0$ so small that
$Bh_0^4 \le 1$.

Since multiplication by $v^q$ is unitary from $\mathcal H^{\rm tube}_q$
onto $\mathcal H^{\rm tube}_0$, \eqref{e:stab-four-term} is equivalent to
the four-term bound for $A_{h, q} = v^qA_hv^{-q}$ on $\mathcal H^{\rm tube}_0$
relative to $C_h$. This is the form in which we prove it. The constants
$\omega$ and $r$ are fixed before $h \to 0$. We write $A_h - C_h$ as a sum
of terms in \zcref{s:stab-kernel,s:stab-frame}, prove the three tools of
\zcref{s:stab-energy,s:stab-parity}, bound the terms in
\zcref{s:stab-uncharged,s:stab-words}, and collect the bounds, together
with the weights, in \zcref{s:stab-proof}.

\section{The kernel in midpoint coordinates}\label{s:stab-kernel}

By \eqref{e:tube-table} and \eqref{e:tube-raw-kernel}, the kernel of $A_h$
with respect to $\dd U'\dd\nu(y)$ is
\begin{equation}\label{e:stab-kernel}
 A_h(x, U; y, U')
 = s_\Gamma\bigl(R(x, U)\bigr)\,w_{\rm out}(U)\,
 K_{\tau_{\rm phys}}\bigl(R(x, U), R(y, U')\bigr)\,w_{\rm in}(U') ,
\end{equation}
where we set
\[
 w_{\rm out} = \chi_{\rm inv}\varpi w_0^{1/2}\gamma_h^{-1} ,
 \qquad
 w_{\rm in} = \chi_{\rm inv}\varpi w_0^{1/2}\gamma_h .
\]
Both normal multipliers are invariant under conjugation by $K$, since
$\varpi$, $w_0$, $\gamma_h$, and $\chi_{\rm inv}$ are. This is an identity
of all Grassmann coefficients. The kernel decays like a Gaussian at scale
$h^2$ in the distance between the endpoints, so that up to an error of
size $e^{-c/\delta_W}$ in every norm used below we may restrict to
$d(x, y) < \varrho_0$.

\subsection{Midpoint frames}

We now introduce coordinates in which the reversal of a link is
explicit. For $d(x, y) < \varrho_0$, with all odd coordinates retained by
finite nilpotent expansion around the body, let $m = m(x, y)$ be the
midpoint and let $Y = Y(x, y) \in \mathfrak m$ be such that
\[
 x = T(m)e^{h^2Y/2} \cdot R_* , \qquad y = T(m)e^{-h^2Y/2} \cdot R_* ,
\]
where $g \cdot R_*$ denotes the point $gR_*g^{-1}$ of the saddle
manifold. Here $Y$ is the full tangent increment, with its even part
$Y_0$ and its odd part. These coordinates exist and are unique near the
diagonal by the inverse function theorem for supermanifolds
\cite{leites1980introduction,varadarajan2004supersymmetry}, applied to
the map $(m, Y) \mapsto (x, y)$. The body of this map is a
diffeomorphism near the diagonal, by the geodesic midpoint of the
symmetric space for $d(x, y) < 2\varrho_0$. Its odd differential at
$Y = 0$ is the invertible map
\[
 (\delta m, \delta Y) \mapsto (\delta m + h^2\delta Y/2, \delta m - h^2\delta Y/2) ,
\]
and the differential stays invertible for $|h^2Y|$ small. We define
the stabilizer elements
\begin{equation}\label{e:stab-midframe}
 k_+(x, y) = e^{-h^2Y/2}T(m)^{-1}T(x) , \qquad
 k_-(x, y) = e^{h^2Y/2}T(m)^{-1}T(y) ,
\end{equation}
which lie in $K$ by the choice of $Y$. Exchanging $x$ and $y$ fixes $m$
and replaces $Y$ by $-Y$. By \eqref{e:stab-midframe}, this implies
\begin{equation}\label{e:stab-swap}
 k_+(y, x) = k_-(x, y) , \qquad k_-(y, x) = k_+(x, y) ,
\end{equation}
and $k_\pm(x, x) = \Id$. We call $(m, Y)$ the midpoint coordinates of the
pair $(x, y)$. By homogeneity, all bounds below on these coordinates hold
on uniformly sized charts.

\begin{lemma}\label{l:stab-chord}
Let $U$ and $U''$ be normal, and write $U = U_m + h^2V/2$ and
$U'' = U_m - h^2V/2$, with $S = R_* + U_m$. Then
\begin{equation}\label{e:stab-midpoint}
\begin{aligned}
 &e^{h^2\operatorname{ad}_Y/2}(R_* + U) - e^{-h^2\operatorname{ad}_Y/2}(R_* + U'') \\
 &\qquad = 2\sinh(h^2\operatorname{ad}_Y/2)S + h^2\cosh(h^2\operatorname{ad}_Y/2)V \\
 &\qquad = h^2(V + [Y, S]) + h^6\bigl([Y, [Y, V]]/8 + [Y, [Y, [Y, S]]]/24\bigr) + O(h^{10})
\end{aligned}
\end{equation}
as a convergent series, where $O(h^{10})$ is bounded by
$Ch^{10}(|V| + |Y|)^5$ on the tube. The function
\[
 \zeta(U, U''; Y) = \Str\bigl(e^{h^2\operatorname{ad}_Y/2}(R_* + U) - e^{-h^2\operatorname{ad}_Y/2}(R_* + U'')\bigr)^2
\]
satisfies $\zeta(U, U''; -Y) = \zeta(U, U''; Y)$ exactly, and
\[
 \zeta = h^4\bigl(\Str V^2 + \Str[Y, S]^2\bigr) + O\bigl(h^8(|V| + |Y|)^4\bigr) .
\]
\end{lemma}

\begin{proof}
The identity \eqref{e:stab-midpoint} follows from
$e^{\pm h^2\operatorname{ad}_Y/2}(S \pm h^2V/2)$ and the definitions of
$\sinh$ and $\cosh$, and the expansion is the Taylor series. Since $\Sigma$
commutes with the block-diagonal $S$ and $V$, and $\Sigma Y\Sigma^{-1} = -Y$
for off-diagonal $Y$, we have
\[
 \Sigma(\operatorname{ad}_Y^jX)\Sigma^{-1} = (-1)^j\operatorname{ad}_Y^jX
\]
for block-diagonal $X$. Conjugating the left side of
\eqref{e:stab-midpoint} by $\Sigma$ replaces $Y$ by $-Y$, and
$\zeta$ is invariant under this conjugation, since $\Str$ is. Finally,
$V$ is block diagonal and $[Y, S]$ is off diagonal, and the cross term in
$\Str(V + [Y, S])^2$ vanishes.
\end{proof}

The leading Gaussian of a link is thus the flat normal increment
Gaussian times the conditional tangent Gaussian with the quadratic form
$\Str[Y, S]^2$, whose body is uniformly elliptic on the tube by root
separation. By the end of \zcref{s:tube-operator}, its normalized integral
over $\mathfrak m$ is $\varpi(S)^{-2}$ times the flat one.

\section{The frame factorization and scalar amplitudes}\label{s:stab-frame}

We now factor out the way in which the stabilizer acts on the normal variables.
For $k \in K$ and a function $f$ of $U$, we set
\[
 (\pi(k)f)(U) = f(k^{-1}Uk) .
\]
For the body $k_0$ of $k$, the operator $\pi(k_0)$ is unitary on
$\mathcal H_{\rm nor}$ and preserves $\mathbf N$. We write $k = k_0k_{\rm rem}$,
where $k_{\rm rem}$ has positive Grassmann degree. Then
$\pi(k_{\rm rem}) = \exp(\mathrm d\pi(\log k_{\rm rem}))$ is a finite sum,
since $\log k_{\rm rem}$ is nilpotent, and for $\mathfrak e \in \mathfrak k$ the
operator $\mathrm d\pi(\mathfrak e)$ is the odd or compact charge of $\mathfrak e$ of
\zcref{s:tube-operator}.

\begin{lemma}\label{l:stab-frame}
For $d(x, y) < \varrho_0$, let $(m, Y)$ be the midpoint coordinates and
$k_\pm = k_\pm(x, y)$. Let $\widetilde\Pi(Y)$ be the operator on functions
of the normal variable with the kernel
\[
 \Pi(U, U''; Y) = w_{\rm out}(U)\exp\Bigl\{-\frac{\zeta(U, U''; Y)}{2h^4}\Bigr\}w_{\rm in}(U'')
\]
with respect to $\dd U''$. Then the kernel \eqref{e:stab-kernel}, as an
operator from functions of $U'$ at $y$ to functions of $U$ at $x$, is
\begin{equation}\label{e:stab-frame}
 s_\Gamma\bigl(R(x, \cdot)\bigr)\,\pi(k_+)^{-1}\,\widetilde\Pi(Y)\,\pi(k_-) .
\end{equation}
The kernel $\Pi$ is a scalar superfunction, in the following sense. It
is a function of the normal coordinates $U$ and $U''$ and of $Y$ with
values in the Grassmann algebra generated by their odd coordinates, and
its coefficients are smooth in the even coordinates on the tube and
analytic on regions where the cutoffs are locally constant. It
does not depend on $m$, and $\widetilde\Pi(-Y) = \widetilde\Pi(Y)$.
\end{lemma}

\begin{proof}
By \eqref{e:stab-midframe}, we have $T(x) = T(m)e^{h^2Y/2}k_+$ and
$T(y) = T(m)e^{-h^2Y/2}k_-$. The elements $k_\pm$ stabilize $R_*$, and hence
\[
 R(x, U) = T(m)e^{h^2Y/2}(R_* + k_+Uk_+^{-1})e^{-h^2Y/2}T(m)^{-1} ,
\]
and similarly for $R(y, U')$ with $k_-$ and $e^{-h^2Y/2}$. Using the
invariance of the supertrace under conjugation by $T(m)$, we obtain
\[
 \Str\bigl(R(x, U) - R(y, U')\bigr)^2 = \zeta(k_+Uk_+^{-1}, k_-U'k_-^{-1}; Y) .
\]

We now substitute $U'' = k_-U'k_-^{-1}$ in the integral over $U'$. The
multipliers $w_{\rm out}$ and $w_{\rm in}$ are invariant under conjugation
by $K$. On each block of $\mathfrak k$, we have
$k_-U'k_-^{-1} = \lambda_{k_-}\rho_{k_-^{-1}}U'$, with the left and right
multiplications $\lambda$ and $\rho$. Using $\Sdet = \exp\Str\log$,
\eqref{e:tube-tensor-trace}, and the fact that $k_-$ is close to the
identity, we find
\[
 \Sdet\lambda_{k_-} = \Sdet(k_-)^{\operatorname{sdim}} ,
 \qquad
 \Sdet\rho_{k_-^{-1}} = \Sdet(k_-)^{-\operatorname{sdim}} ,
\]
where $\operatorname{sdim}$ is the superdimension of the block. Hence the
Berezinian of the substitution is one, and we obtain \eqref{e:stab-frame}.

The geometric factors of $\Pi$ are exponentials, Berezinians, and analytic
functions of the supermatrices $U$, $U''$, and $Y$. The invariant cutoffs
are smooth, so their finite nilpotent expansions give smooth
coefficients, which are analytic where the cutoffs are locally constant.
None of these factors involves $m$. This proves the second assertion. The evenness of
$\widetilde\Pi$ in $Y$ follows from \zcref{l:stab-chord}.
\end{proof}

The next lemma rests on \zcref{l:stab-frame}. All non-scalar structure of the kernel in the odd
normal variables comes from the frame actions $\pi(k_\pm)$, which are
finite sums of words in the normal charges. Without the factorization, a
constant matrix on $\mathcal H_{\rm odd}$, such as an odd creation
operator, could not be excluded from the amplitude, and the conclusion of
\zcref{l:stab-moment} would fail for it.

\begin{lemma}\label{l:stab-moment}
In the standardized coordinates $U = hZ$ and $U'' = hZ''$ and the positive
Wick realization, let $V = (Z - Z'')/h$ and $U_m = h(Z + Z'')/2$, and let
$\Pi(U, U''; Y) = k_z(Z, Z'')\,a(U_m, V; Y)$ with the phase-free even Mehler kernel $k_z$ of
\eqref{e:tube-mehler}; the scalar $\lambda$ of $\mathbf M_h^{\rm q}$ is
included in $a$. Let $b$ be a polynomial in the normalized normal
increments, with coefficients that are scalar superfunctions of the same
kind as $\Pi$. After the finite integration over the odd normal
variables, the operator with the kernel $b\,\Pi$ is $\mathbf M[c]$, in the
notation of \zcref{l:tube-amplitude}, with
\[
 c(U_m, V; Y) = c^{\rm sc}(U_m, V; Y)\,\Id_{\mathcal H_{\rm odd}} + h\,c^{\rm rem}(U_m, V; Y) ,
\]
where $c^{\rm sc}$ is scalar, and $c^{\rm sc}$ and $c^{\rm rem}$ are
amplitudes with bounded seminorms, times a polynomial envelope in $Y$.
Consequently its vacuum column and the vacuum column of its adjoint are
scalar multiples of $\Omega_0$ up to $O(h)$ in every graph norm.
\end{lemma}

\begin{proof}
We first express the odd normal dependence. Let $\theta$ and $\theta''$
be the copies of a standardized odd normal coordinate at the two
endpoints. The corresponding odd coordinates of the unscaled midpoint and
of the unscaled increment, normalized by $h^2$, are
\[
 \tfrac h2(\theta + \theta'') , \qquad h^{-1}(\theta - \theta'') .
\]
Since $\Pi$ and $b$ are scalar superfunctions of
the unscaled coordinates, $a$ and $b$ are polynomials in
$h(\theta + \theta'')/2$ and $(\theta - \theta'')/h$, with coefficients that
are smooth functions of the even variables with the bounds of the
amplitude class. For $a$, this follows from \zcref{l:stab-chord} after
division by the quadratic exponent of $k_z$, restricted to
$|V| + |Y| \le h^{-\alpha}$ for a small fixed $\alpha > 0$, with a
Gaussian-small complement.

Next, we integrate the odd variables. Multiplying the Berezin kernel of
an operator $X$ of parity $|X|$ by $(\theta + \theta'')/2$ corresponds to
the bounded map $X \mapsto \frac12(\theta X + (-1)^{|X|}X\theta)$, and
multiplying by $\theta - \theta''$ corresponds to the graded commutator
with $\theta$. Every nonconstant monomial in the midpoint odd variables
carries an explicit factor $h$. By \zcref{l:tube-odd-gaussian}
with $\kappa_{\rm o} = h$, applied to the odd Mehler factor
$e^{-z\mathbf N_F}$, every product of normalized increments acting on this
factor is a number times the identity plus $O(h)$, and
$e^{-z\mathbf N_F} = \Id + O(h^2)$. We conclude that at every point
$(U_m, V)$ the coefficient is a scalar times the identity plus $O(h)$. The
same holds for every derivative in $(U_m, V)$, since the odd computation
does not involve these variables. This proves the decomposition.

Finally, applying part (iii) of \zcref{l:tube-amplitude} to $c^{\rm sc}$
and to $c^{\rm rem}$, we obtain $\mathbf M[c]\Omega_0 = c_0^{\rm sc}\Omega_0 + O(h)$
in every graph norm, with a scalar $c_0^{\rm sc}$. The same argument
applies to $\mathbf M[c]^* = \mathbf M[c^\sharp]$.
\end{proof}

In particular, for an odd charge
$\mathfrak q_a = \sum_j(F_jx_j + G_j\partial_j)$, the matrix commutators in
\eqref{e:tube-odd-commutator} satisfy $[F_j, c] = h[F_j, c^{\rm rem}]$ and
$[G_j, c] = h[G_j, c^{\rm rem}]$. Together with
\eqref{e:tube-odd-commutator}, this implies that
$[\mathfrak q_a, \mathbf M[c]]$ is $h$ times a sum of operators
$\mathbf M[c']$ with at most one letter on either side.
By \eqref{e:tube-compact-commutator}, the same holds for the even part of
a compact charge, and its finite matrix part gives
$[\mathfrak j_{a, F}, \mathbf M[c]] = h\mathbf M[[\mathfrak j_{a, F}, c^{\rm rem}]]$.

\subsection{Words in the charges}

\begin{lemma}\label{l:stab-words}
For $d(x, y) < \varrho_0$, write $k_\pm = k_{\pm, 0}k_{\pm, \rm rem}$.
Then
\[
 \log k_{\pm, \rm rem} = h^2\sum_\alpha c_{\pm, \alpha}\mathfrak e_\alpha ,
\]
where the $\mathfrak e_\alpha$ run over a basis of the even and odd generators of
$\mathfrak k$, and the $c_{\pm, \alpha}$ are analytic superfunctions of the
midpoint coordinates and of the odd coordinates of $m$, with every fixed
covariant derivative bounded on uniformly sized charts, times a
polynomial in $Y$. By \eqref{e:stab-swap},
$c_{+, \alpha}(y, x) = c_{-, \alpha}(x, y)$. The expansion
\[
 \pi(k_{\pm, \rm rem})
 = \sum_{n = 0}^D\frac{h^{2n}}{n!}\Bigl(\sum_\alpha c_{\pm, \alpha}\,\mathrm d\pi(\mathfrak e_\alpha)\Bigr)^n
\]
terminates at an order $D$ depending only on $k$. It is a finite sum of
words in the odd and compact charges in which a word of degree $n$
carries the factor $h^{2n}$, and after the integration over the odd
tangent increments, normalized by $h^2$, no negative power of $h$ arises.
\end{lemma}

\begin{proof}
By the construction of the midpoint coordinates above, $k_\pm$ are
analytic functions of the pair $(x, y)$, with all derivatives of fixed order
bounded on uniformly sized translated charts, and $k_\pm = \Id$ at
coincident endpoints. The factor $k_{\pm, \rm rem} = k_{\pm, 0}^{-1}k_\pm$
has positive Grassmann degree, and its logarithm is a finite Grassmann
polynomial. Since every nonzero coefficient of this polynomial vanishes
at coincident endpoints, it contains at least one even or odd endpoint
difference. Such a difference is of size $h^2$, and this gives the
factor $h^2$. The odd
coordinates of $m$ are not small; they occur only in the bounded
coefficients.

The induced action of $\mathfrak e_\alpha$ on the normal variable is linear with
Berezinian one, and its differential $\mathrm d\pi(\mathfrak e_\alpha)$ is an odd
charge $\mathfrak q_a$ or a compact charge $\mathfrak j_a$. The
exponential series terminates since each coefficient has positive degree
in a finite Grassmann algebra, and the noncommuting products of charges
are kept in every term. Finally, the odd tangent endpoint differences,
normalized by $h^2$, are integrated against the odd tangent Gaussian,
which is the odd heat over the time $\delta_W$. By
\zcref{l:tube-odd-gaussian} with $\kappa_{\rm o} = h^2$, pairs of them
have bounded covariance and an unpaired one gains $h^2$, so that the
explicit factor of a word survives the integration.
\end{proof}

\subsection{The decomposition of the step}

We define, for $d(x, y) < \varrho_0$,
\[
 \tau_{xy}' = \pi(k_{+, 0})^{-1}\pi(k_{-, 0}) ,
 \qquad
 \Pi_x(x, y) = \pi(k_{+, 0})^{-1}\widetilde\Pi(Y)\pi(k_{+, 0}) ,
 \qquad
 L_\pm = \mathrm d\pi(\log k_{\pm, \rm rem}) ,
\]
where $L_+$ acts on the fiber at $x$ and $L_-$ on the fiber at $y$. By
\eqref{e:stab-midframe}, the transvection $T(m)e^{h^2Y}T(m)^{-1}$ (the
isometry of the saddle manifold that translates along the geodesic
through $m$ and transports frames in parallel along it) carries the frame
$T(y)$ to the frame $T(x)k_+^{-1}k_-$ at $x$. The body of $\tau_{xy}'$ is therefore the parallel
transport along the minimal geodesic in the frames $T$, and we identify
it with $\tau_{xy}$ on the normal factor. Note also that, by the
conjugation invariance of $\Str$ and of the multipliers,
$\pi(k)^{-1}\widetilde\Pi(Y)\pi(k) = \widetilde\Pi(k^{-1}Yk)$ for
$k \in K_0$. Thus
\begin{equation}\label{e:stab-Pi-x}
 \Pi_x(x, y) = \widetilde\Pi(Y_x) ,
\end{equation}
where $Y_x = k_{+, 0}^{-1}Yk_{+, 0}$ is the increment written in the frame
$T(x)$. By \zcref{l:stab-frame}, the kernel of $A_h$ at $(x, y)$ is
\[
 s_\Gamma\,\pi(k_{+, \rm rem})^{-1}\Pi_x(x, y)\tau_{xy}\pi(k_{-, \rm rem})
 = s_\Gamma\Bigl(\sum_{n \le D}\frac{(-L_+)^n}{n!}\Bigr)\Pi_x(x, y)\tau_{xy}\Bigl(\sum_{n \le D}\frac{L_-^n}{n!}\Bigr) .
\]
We call the term with $n = 0$ on both sides the uncharged part, the two
terms with one charge factor the degree-one part, and the rest the
higher words. We set $L_-^{\rm out} = \tau_{xy}L_-\tau_{xy}^{-1}$, which
acts on the fiber at $x$. The degree-one part is then
\begin{equation}\label{e:stab-degree-one}
 \Pi_x\tau_{xy}L_- - L_+\Pi_x\tau_{xy}
 = \tfrac12\bigl\{\Pi_x, L_-^{\rm out} - L_+\bigr\}\tau_{xy}
 + \tfrac12\bigl[\Pi_x, L_-^{\rm out} + L_+\bigr]\tau_{xy} ,
\end{equation}
with the anticommutator $\{X, X'\} = XX' + X'X$ and the commutator.

The reference $C_h$ has the kernel $\mathbf S_h(x)\,\mathbf M^\chi\,p^\nabla_h(x, y)$,
where $p^\nabla_h(x, y)$ is the kernel of $\mathbf H_h$, an operator from
the fiber at $y$ to the fiber at $x$. We let $\mathbf F_h$ be the frozen
normal transfer with cutoffs of \zcref{s:tube-powers}, and we let $p_h$
be the scalar heat kernel of $e^{\delta_W\Delta/2}$ on $|\mathcal M|$. We
write
\[
 A_h - C_h = \mathbf E_{\rm src} + \mathbf E_{\rm nor} + \mathbf E_0 + \mathbf E_{\rm hol} + \mathbf E_1 + \mathbf E_{\ge 2} ,
\]
where, with kernels on $d(x, y) < \varrho_0$ and with the Gaussian-small
remainder included in $\mathbf E_0$, we set
\begin{gather*}
 \mathbf E_{\rm src} = (s_\Gamma - \mathbf S_h)\,\mathcal K_h ,
 \qquad
 \mathbf E_{\rm nor} = (\mathbf F_h - \mathbf M^\chi) \otimes \mathbf S_h\mathbf H_h ,
 \\
 \mathbf E_0(x, y) = \mathbf S_h(x)\bigl(\Pi_x(x, y) - \mathbf F_hp_h(x, y)\bigr)\tau_{xy} ,
 \\
 \mathbf E_{\rm hol}(x, y) = \mathbf S_h(x)\mathbf F_h\bigl(p_h(x, y)\tau_{xy} - p^\nabla_h(x, y)\bigr) ,
\end{gather*}
where $\mathcal K_h$ denotes $A_h$ without the source factor, and where
$\mathbf E_1$ and $\mathbf E_{\ge 2}$ are $\mathbf S_h$ times the degree-one
part and the higher words. The sum telescopes, since $\mathbf F_h$
commutes with the transports and with $\mathbf S_h$. Each of
$\mathbf E_{\rm nor}$, $\mathbf E_0$, $\mathbf E_{\rm hol}$, $\mathbf E_1$, and
$\mathbf E_{\ge 2}$ has the form $\mathbf S_hX$ with an operator $X$ that does
not depend on the damping, and these five operators $X$ add up to
$\mathcal K_h - \mathbf M^\chi \otimes \mathbf H_h$.

We bound each term in the following form. For an operator $X$ on
$\mathcal H^{\rm tube}_0$ and numbers $a', A' \ge 0$, we say that $X$
\emph{satisfies the four-term bound with parameters} $a'$ and $A'$ if
$C_h + X$ satisfies \eqref{e:tube-four-term} relative to $C_h$ with
$\delta_0 = \delta_W$, $a = a'$, and $A_1 = A'$, that is, if
\eqref{e:stab-four-term} holds with $q = 0$, with $X$ in place of
$A_h - C_h$, $a'$ in place of $a$, and $A'$ in place of $A_1$. If two
operators satisfy such bounds, then their sum satisfies the four-term
bound whose parameters are the sums of theirs. For an operator $X$ on
$\mathcal H^{\rm tube}_0$, recall from \zcref{s:tube-oscillator} its fast
block $\mathbf P_0^\perp X\mathbf P_0^\perp$, its vacuum row $\mathbf P_0X\mathbf P_0^\perp$ and
vacuum column $\mathbf P_0^\perp X\mathbf P_0$, and its vacuum scalar $\mathbf P_0X\mathbf P_0$,
where $\mathbf P_0$ acts on the normal factor. We will need the following
block criterion for $\mathbf E_1$ and $\mathbf E_{\ge 2}$.

\begin{lemma}\label{l:stab-blocks}
Let $X$ be an operator on $\mathcal H^{\rm tube}_0$, and let
$b_0, b, b_1, b_2 \ge 0$. Suppose that the fast block of $X$ satisfies
\eqref{e:tube-four-term}, with $\mathbf P_0^\perp X\mathbf P_0^\perp$ in place of
$\mathbf T - C$, with $D_{\rm nor} + D_{\rm body, L}$ and $D_{\rm nor} + D_{\rm body, R}$
in place of $D_L$ and $D_R$, with $\delta_0 = \delta_W$, $a = b_0$, and
$A_1 = b$, and that
\begin{equation}\label{e:stab-blocks}
 \|\mathbf P_0X\mathbf P_0^\perp\| + \|\mathbf P_0^\perp X\mathbf P_0\| \le b_1h^3 ,
 \qquad
 \|\mathbf P_0X\mathbf P_0\| \le b_2h^4 .
\end{equation}
Then $X$ satisfies the four-term bound with parameters $4b_0$ and
$C(b + b_1 + b_2)$.
\end{lemma}

\begin{proof}
We write $X$ as the sum of its fast block, its vacuum row and column, and
its vacuum scalar. By \eqref{e:stab-product-defects}, we have
$\|(D_{\rm nor} + D_{\rm body, R})^{1/2}f\| \le 2\|D_R^{1/2}f\|$, and similarly
on the left, so that the fast block satisfies the four-term bound with
parameters $4b_0$ and $2b$. Since $\Re z_\sigma \ge c_0h^2$ for all
$\sigma$, we have $D_{\rm nor} \ge c_0h^2\mathbf P_0^\perp$ for $h$ small. This
implies
\[
 \|\mathbf P_0^\perp f\| \le Ch^{-1}\|D_{\rm nor}^{1/2}f\| \le 2Ch^{-1}\|D_R^{1/2}f\| .
\]
Using \eqref{e:stab-blocks} and $h^2 = \sqrt{\delta_W}$, we obtain
\[
 |\langle g, \mathbf P_0X\mathbf P_0^\perp f\rangle| \le b_1h^3\|g\|\|\mathbf P_0^\perp f\|
 \le 2Cb_1\sqrt{\delta_W}\|g\|\|D_R^{1/2}f\| ,
\]
the same bound with the roles of $f$ and $g$ exchanged for the vacuum
column, and
$|\langle g, \mathbf P_0X\mathbf P_0f\rangle| \le b_2\delta_W\|g\|\|f\|$. Adding these
bounds proves the lemma.
\end{proof}

\section{The geodesic difference energy}\label{s:stab-energy}

We measure the difference across one link of a section of $\mathcal E$
by the following energy. For $c > 0$ and $0 < t \le 1$, we set
\[
 \mathfrak D_{c, t}(f) = \iint_{d(x, y) < \varrho_0}t^{-n_b/2}e^{-c\,d(x, y)^2/t}
 \|f(x) - \tau_{xy}f(y)\|^2\dd\mu(x)\dd\mu(y) ,
\]
and we write $\mathfrak D_c = \mathfrak D_{c, \delta_W}$. The next lemma
bounds this energy by the defect of the body heat. It holds for every
fiber, uniformly, and in particular for the normal fiber, on which the
connection acts through the compact charges, which are unbounded.

\begin{lemma}\label{l:stab-energy}
Let $(\mathcal Y, g)$ be a Riemannian manifold of dimension $n$ on which a
group of isometries acts transitively, let $\varrho_0$ be smaller than its
injectivity radius, and let $\mathcal E = \bigoplus_j\mathcal E_j$ be an
orthogonal sum of Hermitian vector bundles of finite rank over
$\mathcal Y$ with a metric connection $\nabla$ that preserves the
summands. Let $\tau_{xy}$ be the parallel transport along the minimal
geodesic from $y$ to $x$, let $\Delta_\nabla$ be the Friedrichs extension of
$-\nabla^*\nabla$, and define $\mathfrak D_{c, t}$ as above with $n$ in place
of $n_b$. Then for every $c > 0$ there exists $C$, depending only on $c$,
$n$, $\varrho_0$, and the geometry of $\mathcal Y$, such that for all
$0 < t \le 1$ and all $f \in L^2(\mathcal Y; \mathcal E)$,
\begin{equation}\label{e:stab-energy}
 \mathfrak D_{c, t}(f) \le C\langle f, (\Id - e^{t\Delta_\nabla})f\rangle .
\end{equation}
\end{lemma}

With $\mathcal Y = |\mathcal M|$, the bundle of \zcref{s:stab-main}
decomposed into the eigenspaces of $\mathbf N$ and $t = \delta_W$, we have
$e^{t\Delta_\nabla} = \mathbf H_h^2$, so that the right side of
\eqref{e:stab-energy} becomes $C\langle f, (\Id - \mathbf H_h^2)f\rangle$.
In the proof, we bound the energy of a smooth section by its gradient
along geodesics, extend this bound to the form domain, and pass to all
of $L^2$ by the spectral theorem.

\begin{proof}
We first prove that for every smooth compactly supported section $f$,
\begin{equation}\label{e:stab-energy-gradient}
 \mathfrak D_{c, t}(f) \le Ct\|\nabla f\|^2 .
\end{equation}
Let $d(x, y) < \varrho_0$, let $u = \exp_x^{-1}(y)$, so that
$|u| = d(x, y)$, and let $\gamma(s) = \exp_x(su)$ for $s \in [0, 1]$. The
section $\sigma(s) = \tau_{x\gamma(s)}f(\gamma(s))$ of the fiber at $x$
satisfies $\sigma(0) = f(x)$ and $\sigma(1) = \tau_{xy}f(y)$, and
$\sigma'(s) = \tau_{x\gamma(s)}(\nabla_{\gamma'(s)}f)(\gamma(s))$, since
$\tau$ is the parallel transport of $\nabla$ along $\gamma$. By the
Cauchy--Schwarz inequality,
\begin{equation}\label{e:stab-energy-pointwise}
 \|f(x) - \tau_{xy}f(y)\|^2 \le d(x, y)^2\int_0^1\|\nabla f(\gamma(s))\|^2\dd s .
\end{equation}

We now insert \eqref{e:stab-energy-pointwise} into $\mathfrak D_{c, t}(f)$,
and we set $G = \|\nabla f\|^2$. Let $J_x(u)$ be the Jacobian of $\exp_x$,
so that $\dd\mu(\exp_xu) = J_x(u)\dd u$. The isometries act transitively,
so $J_x$ is the same function for all $x$ in isometric frames; it is
continuous and positive on $|u| \le \varrho_0$, and hence bounded above
and below by positive constants there.

For $s \in [1/2, 1]$, we fix $x$ and substitute
$z = \exp_x(su)$ for $y = \exp_x(u)$. Then
\[
 \dd\mu(y) = \frac{J_x(u)}{J_x(su)s^n}\dd\mu(z) \le C\dd\mu(z) ,
 \qquad
 d(x, z) \le d(x, y) = d(x, z)/s \le 2d(x, z) .
\]
These bounds imply
\begin{multline*}
 \int_{d(x, y) < \varrho_0}t^{-n/2}e^{-c\,d(x, y)^2/t}d(x, y)^2G(\exp_x(su))\dd\mu(y) \\
 \le C\int_{d(x, z) < \varrho_0}t^{-n/2}e^{-c\,d(x, z)^2/t}d(x, z)^2G(z)\dd\mu(z) .
\end{multline*}
Integrating over $x$ and exchanging the integrals, we bound the
contribution of $s \in [1/2, 1]$ by
\[
 C\int G(z)\int_{d(x, z) < \varrho_0}t^{-n/2}e^{-c\,d(x, z)^2/t}d(x, z)^2\dd\mu(x)\dd\mu(z)
 \le Ct\int G\dd\mu ,
\]
where we used the bound on $J$ and the identity
\[
 \int_{\R^n}t^{-n/2}e^{-c|u|^2/t}|u|^2\dd u = C_ct .
\]
For $s \in [0, 1/2]$
we use $\gamma(s) = \exp_y((1 - s)\exp_y^{-1}x)$ and exchange the roles of
$x$ and $y$, which is allowed since the weight is symmetric. This proves
\eqref{e:stab-energy-gradient}.

Next, we extend \eqref{e:stab-energy-gradient} to the form domain. Using
$\|a - b\|^2 \le 2\|a\|^2 + 2\|b\|^2$ and the bound on $J$, we have
$\mathfrak D_{c, t}(f) \le C_0\|f\|^2$, with $C_0$ depending on $c$, $n$,
$\varrho_0$, and the geometry. By the definition of the Friedrichs
extension, the smooth compactly supported sections are a core for its
form $\|\nabla f\|^2$. Approximating in the form norm and using the
continuity of $\mathfrak D_{c, t}$ on $L^2$, we obtain
\eqref{e:stab-energy-gradient} for all $f$ in the form domain of
$-\Delta_\nabla$, with $\|\nabla f\|^2 = \langle f, -\Delta_\nabla f\rangle$.

Finally, let $f \in L^2$, and set
\[
 f_1 = \mathbf 1_{[0, 1/t]}(-\Delta_\nabla)f , \qquad f_2 = f - f_1 .
\]
Since $\lambda \mapsto 1 - e^{-\lambda}$ is concave and vanishes at zero,
we have $\lambda \le (1 - e^{-1})^{-1}(1 - e^{-\lambda})$ for
$0 \le \lambda \le 1$, and $1 \le (1 - e^{-1})^{-1}(1 - e^{-\lambda})$ for
$\lambda \ge 1$. By the spectral theorem,
\[
 t\|\nabla f_1\|^2 \le (1 - e^{-1})^{-1}\langle f_1, (\Id - e^{t\Delta_\nabla})f_1\rangle ,
 \qquad
 \|f_2\|^2 \le (1 - e^{-1})^{-1}\langle f_2, (\Id - e^{t\Delta_\nabla})f_2\rangle .
\]
We use $\mathfrak D_{c, t}(f) \le 2\mathfrak D_{c, t}(f_1) + 2\mathfrak D_{c, t}(f_2)$,
\eqref{e:stab-energy-gradient} for $f_1$, the bound
$\mathfrak D_{c, t}(f_2) \le C_0\|f_2\|^2$, and the orthogonality of $f_1$
and $f_2$ for the spectral measure of $\Delta_\nabla$, and we obtain
\eqref{e:stab-energy}. The constants do not depend on the summands
$\mathcal E_j$, and both sides of \eqref{e:stab-energy} are sums over $j$,
since $\tau$ and $\Delta_\nabla$ preserve the decomposition.
\end{proof}

We use \zcref{l:stab-energy} through the symmetrization identity in the
proof of \zcref{l:tools-dirichlet}(b). Let $R(x, y)$, for
$d(x, y) < \varrho_0$, be operators on the fiber at $x$ such that
\[
 R(y, x) = \tau_{yx}R(x, y)\tau_{xy} ,
 \qquad
 \|R(x, y)\| \le b\,\delta_W^{-n_b/2}e^{-c\,d(x, y)^2/\delta_W} ,
\]
let $T$ have the kernel $R(x, y)\tau_{xy}$, and let
$R_{\rm marg}(x) = \int R(x, y)\dd\mu(y)$ be its marginal. Exchanging $x$ and $y$ and
averaging, as in that proof, we have
\[
 \langle g, (T - R_{\rm marg})f\rangle
 = -\frac12\iint\bigl\langle g(x) - \tau_{xy}g(y), R(x, y)\bigl(f(x) - \tau_{xy}f(y)\bigr)\bigr\rangle\dd\mu(x)\dd\mu(y) .
\]
The Cauchy--Schwarz inequality and \eqref{e:stab-energy} now imply
\begin{equation}\label{e:stab-dirichlet}
 |\langle g, (T - R_{\rm marg})f\rangle|
 \le Cb\,\|(\Id - \mathbf H_h^2)^{1/2}g\|\,\|(\Id - \mathbf H_h^2)^{1/2}f\| .
\end{equation}
This identity uses the symmetry of the kernel, and \eqref{e:stab-energy}
takes the place of the positivity hypothesis of \zcref{l:tools-dirichlet}(b).
Without the symmetry, we write
\[
 \langle g, (T - R_{\rm marg})f\rangle
 = \iint\bigl\langle R(x, y)^*g(x), \tau_{xy}f(y) - f(x)\bigr\rangle\dd\mu(x)\dd\mu(y) ,
\]
and applying the Cauchy--Schwarz inequality with the weight
$\delta_W^{\pm n_b/2}e^{\mp c'd(x, y)^2/\delta_W}$, where $c' = c/2$, we
obtain the one-sided version
\begin{multline}\label{e:stab-dirichlet-one}
 |\langle g, (T - R_{\rm marg})f\rangle| \\
 \le \Bigl(\iint\delta_W^{n_b/2}e^{c'd(x, y)^2/\delta_W}\|R(x, y)^*g(x)\|^2\dd\mu(x)\dd\mu(y)\Bigr)^{1/2}
 \mathfrak D_{c'}(f)^{1/2} .
\end{multline}
By \eqref{e:stab-energy}, its last factor satisfies
$\mathfrak D_{c'}(f)^{1/2} \le C\|(\Id - \mathbf H_h^2)^{1/2}f\|$.

\section{Marginals and midpoint parity}\label{s:stab-parity}

The marginal of a kernel of size one is of size one. The next lemma shows that
the marginal of a kernel that is odd under the reversal of the link gains a
factor $h^2$. This is the source of the factor $\delta_W$ on the marginals of
the degree-one words.

\begin{lemma}\label{l:stab-parity}
Let $\mathcal X$ be a Banach space, let $x \in |\mathcal M|$, and let
$F(w, u) \in \mathcal X$ be defined for $w, u$ in the tangent space at $x$
with $|w| < \varrho_0$. Suppose that $s \mapsto F(su, u)$ is continuously
differentiable and that
\[
 F(0, -u) = -F(0, u) ,
 \qquad
 \|F(su, u)\| + \Bigl\|\frac{\dd}{\dd s}F(su, u)\Bigr\| \le Ae^{-c|u|^2} .
\]
Then for $0 < t \le 1$,
\[
 \Bigl\|\int_{|u| < \varrho_0/t}F(tu/2, u)J_x(tu)\dd u\Bigr\| \le CAt ,
\]
where $J_x$ is the Jacobian of $\exp_x$, and $C$ depends on $c$, $n_b$, and
the geometry.
\end{lemma}

\begin{proof}
Since the geodesic symmetry at $x$ is an isometry with differential
$-\Id$ at $x$, we have $J_x(-w) = J_x(w)$. Substituting $u \mapsto -u$ in half
of the integral, we can replace $F(tu/2, u)$ by
$\frac12(F(tu/2, u) + F(-tu/2, -u))$. Since $F(0, u) + F(0, -u) = 0$,
\[
 F(tu/2, u) + F(-tu/2, -u)
 = \int_0^{t/2}\Bigl(\frac{\dd}{\dd s}F(su, u) + \frac{\dd}{\dd s}F(-su, -u)\Bigr)\dd s ,
\]
and by the hypothesis, the norm of the right side is bounded by
$tAe^{-c|u|^2}$. Since $J_x$ is bounded, integrating over $u$ proves the
lemma.
\end{proof}

\begin{corollary}\label{c:stab-parity}
Let $\mathcal X_x$ be a Banach space of operators on the fiber at $x$
whose norm is invariant under conjugation by the transports, as is the
case for the operator norm and for $X \mapsto \|D_+^{-1/2}XD_+^{-1/2}\|$.
Let $R(x, y)$, for $d(x, y) < \varrho_0$, be of the form
$R(x, y) = h^{-2n_b}\tau_{xm}\widetilde R(m, Y)\tau_{mx}$ in the midpoint
coordinates $(m, Y)$, with $\widetilde R(m, -Y) = -\widetilde R(m, Y)$
and
\[
 \|\widetilde R(m, Y)\| + \bigl\|(\nabla_Y\widetilde R)(m, Y)\bigr\| \le Ae^{-c|Y|^2} ,
\]
where $(\nabla_Y\widetilde R)(m, Y)$ is the covariant derivative of
$\widetilde R$ in $m$ along the geodesic through $m$ in the direction $Y$,
with $Y$ parallel. Then $\|\int R(x, y)\dd\mu(y)\| \le CAh^2$ for all $x$.
\end{corollary}

\begin{proof}
We write $y = \exp_x(h^2u)$, so that
\[
 \dd\mu(y) = h^{2n_b}J_x(h^2u)\dd u , \qquad m = \exp_x(h^2u/2) ,
\]
and $Y = -P_{x \to m}u$, where $P_{x \to m}$ is the parallel transport of
tangent vectors along the radial geodesic. Indeed, $x = \exp_m(h^2Y/2)$,
and the geodesic from $x$ to $y$ passes through $m$ with direction
$P_{x \to m}u$. Thus $h^{2n_b}R(x, y) = F(h^2u/2, u)$ with
\[
 F(w, u) = \tau_{x, \exp_xw}\widetilde R\bigl(\exp_xw, -P_{x \to \exp_xw}u\bigr)\tau_{\exp_xw, x} .
\]
Since $\widetilde R$ is odd in $Y$, the function $F$ satisfies
\[
 F(0, -u) = \widetilde R(x, u) = -F(0, u) .
\]
Along $w = su$, set $z(s) = \exp_x(su)$ and
$Y(s) = -P_{x \to z(s)}u$. Then $z'(s) = -Y(s)$, while $Y(s)$ and the
fiber transports are parallel along this geodesic. Hence
$\frac{\dd}{\dd s}F(su, u)$ is the negative of the transported derivative
$(\nabla_Y\widetilde R)(z(s), Y(s))$. Its norm has the bound required in
\zcref{l:stab-parity}. Applying
\zcref{l:stab-parity} with $t = h^2$, we obtain the claim.
\end{proof}

We now apply \zcref{c:stab-parity} to the degree-one part of the step.
We let $W^{\rm a}$ denote the anticommutator part of
\eqref{e:stab-degree-one}, the operator with the kernel
$W^{\rm a}(x, y)\tau_{xy}$, where
\[
 W^{\rm a}(x, y) = \tfrac12\bigl\{\Pi_x(x, y), (L_-^{\rm out} - L_+)(x, y)\bigr\} ,
\]
and we let $W^{\rm a}_{\rm marg}(x) = \int W^{\rm a}(x, y)\dd\mu(y)$ denote its
marginal. The next lemma is the source of
the factor $\delta_W$ on the marginal of the degree-one part. It measures
operators on the fiber at $x$ in the norm
\[
 \|X\|_{\rm f} = \bigl\|D_{\rm nor}^{-1/2}\mathbf P_0^\perp X\mathbf P_0^\perp D_{\rm nor}^{-1/2}\bigr\| ,
\]
which bounds the fast block of $X$ bilaterally in the normal defect, and
in which \zcref{l:tube-word} measures words.

\begin{lemma}\label{l:stab-degree-one-parity}
Before the odd tangent integration, the full superkernel satisfies
\begin{equation}\label{e:stab-anti-odd}
 W^{\rm a}(y, x) = -\tau_{yx}W^{\rm a}(x, y)\tau_{xy} ,
\end{equation}
where the reversal changes the signs of both the even and the odd
tangent increments. For the resulting operator kernel on the body,
for all $x$,
\begin{equation}\label{e:stab-anti-marg}
 \|W^{\rm a}_{\rm marg}(x)\|_{\rm f} \le C(rh + h^2) .
\end{equation}
\end{lemma}

\begin{proof}
We begin with \eqref{e:stab-anti-odd}. By \eqref{e:stab-swap}, we have
$L_-(y, x) = L_+(x, y)$ and $L_+(y, x) = L_-(x, y)$, as operators on the
fibers at $x$ and $y$, respectively. Recall that
$L_-^{\rm out} = \tau_{xy}L_-\tau_{xy}^{-1}$ and $\tau_{yx} = \tau_{xy}^{-1}$.
Therefore
\[
 (L_-^{\rm out} - L_+)(y, x)
 = \tau_{yx}L_+(x, y)\tau_{xy} - L_-(x, y)
 = -\tau_{yx}(L_-^{\rm out} - L_+)(x, y)\tau_{xy} .
\]
On the other hand, before the odd tangent integration,
$\Pi_x(y, x) = \tau_{yx}\Pi_x(x, y)\tau_{xy}$. Indeed,
\eqref{e:stab-Pi-x} gives $\Pi_x(x, y) = \widetilde\Pi(Y_x)$, with the
increment $Y_x$ in the frame $T(x)$. The transport changes this frame to
$T(y)$, the reversal changes the sign of the increment, and
$\widetilde\Pi(-Y) = \widetilde\Pi(Y)$ by \zcref{l:stab-frame}. Inserting the
two superkernel identities into the definition of $W^{\rm a}$, we obtain
\eqref{e:stab-anti-odd}. This is not a reversal identity for the
operator-valued body kernel: an odd tangent increment becomes an odd
derivative after integration, and that derivative does not change sign
when only the body endpoints are exchanged.

For \eqref{e:stab-anti-marg}, we write the kernel in the midpoint
coordinates of $(x, y)$, as in \zcref{c:stab-parity}:
\[
 W^{\rm a}(x, y) = h^{-2n_b}\tau_{xm}\widetilde W(m, Y)\tau_{mx} .
\]
The pair $(y, x)$ has the midpoint $m$ and the increment $-Y$, and the
transports along one geodesic compose. By \eqref{e:stab-anti-odd}, we
have $\widetilde W(m, -Y) = -\widetilde W(m, Y)$, where $Y$ is the full
tangent increment, and the reversal changes the sign of its even part
$Y_0$ and of its odd part. We expand $\widetilde W$ in the normalized odd
tangent increments. A term of even degree in these increments is then
odd in $Y_0$, and a term of odd degree is even in $Y_0$.

We next bound the size of these terms. The norm $\| \cdot \|_{\rm f}$ is
invariant under conjugation by the transports, since they commute with
$\mathbf N$. By \zcref{l:stab-words}, $h^{-2}L_\pm$ is a sum of odd and
compact charges, with coefficients that are bounded, with bounded
covariant derivatives, times a polynomial in $Y$. By
\zcref{l:stab-moment}, $\Pi_x$ is an operator $\mathbf M[c]$ whose
amplitude carries the tangent Gaussian of the link, which absorbs these
polynomials. The single-charge bounds at the end of
\zcref{l:tube-word}, which part (iv) of \zcref{l:tube-amplitude} extends
from $\mathbf M$ to $\mathbf M[c]$, give
$\|h^4\mathfrak q_a\mathbf M[c]\|_{\rm f} \le Ch^2$ and
$\|h^4\mathfrak j_a\mathbf M[c]\|_{\rm f} \le C(rh + h^2)$, and the adjoint
version bounds the products with the charge on the right. Since
$\widetilde\Pi$ does not depend on $m$, by \zcref{l:stab-frame}, the same
bounds hold for the covariant derivative along the geodesic. Since
$\widetilde W$ is a sum of such products of a charge with $\Pi_x$, with
the factor $h^2$ of $L_\pm$, every coefficient of its expansion, and its
covariant derivative, has norm $\| \cdot \|_{\rm f}$ at most
$Ae^{-c|Y_0|^2}$, with
\[
 A = Ch^{-2}(rh + h^2) = C(r/h + 1) .
\]

Finally, we integrate. For a term of even degree, which is odd in $Y_0$,
the full-pairing part of the finite odd Gaussian computation is a
scalar on $\Lambda_T$, and its coefficient remains odd in $Y_0$.
\zcref[S]{c:stab-parity}, applied in the norm $\| \cdot \|_{\rm f}$,
bounds its marginal by $CAh^2$. The remaining even-degree terms are
$O(Ah^4)$: every non-full pairing leaves at least two factors of size
$h^2$, and the odd heat differs from the identity by $O(h^4)$.
A term of odd degree contains an odd number of
normalized odd tangent increments. As in the proof of
\zcref{l:stab-words}, its integration over the odd tangent coordinates is
the computation of \zcref{l:tube-odd-gaussian} with $\kappa_{\rm o} = h^2$,
and it gives an extra factor $h^2$. The marginal of such a term is again
bounded by $CAh^2$. Since $Ah^2 = C(rh + h^2)$, we obtain \eqref{e:stab-anti-marg}.
Odd midpoint coordinates are retained as bounded midpoint
multiplications throughout this computation; the precise finite
operator rules, including their covariance and derivative bounds, are
given in \zcref{l:stab-body-parity} below.
\end{proof}

The proof of \eqref{e:stab-anti-marg} also shows that the marginal of
the first variation of a weight or of the source factor across a link,
which is antisymmetric under the reversal and carries the factor $h^2$,
is of order $\delta_W$.

\section{The uncharged part}\label{s:stab-uncharged}

\begin{lemma}\label{l:stab-schur}
Let $a(U_m, V; Y)$ be a normal amplitude in the unscaled midpoint and the
normalized increment, as in \zcref{l:stab-moment}, with
$\|a(U_m, V; Y)\| \le e_0(1 + |V| + |Y|)^l$ on the tube. After the
integration over the odd normal variables, the operator with the kernel
$k_z\,a$ has norm at most $Ce_0(1 + |Y|)^l$ on $\mathcal H_{\rm nor}$, and
so does its adjoint. With the tangent Gaussian included, the bound is
$Ce_0(1 + |Y|)^le^{-c|Y|^2}$.
\end{lemma}

\begin{proof}
By \zcref{l:tube-odd-gaussian} with $\kappa_{\rm o} = h$, every word of fixed length
in the normalized odd increments acting on the odd Mehler factor is a
bounded matrix, and the odd midpoint variables carry their factor $h$.
The odd integration thus costs no power of $h$, and we obtain an operator $\mathbf M[c]$
with $\|c(U_m, V; Y)\| \le Ce_0(1 + |V| + |Y|)^l$ at fixed $Y$. By part (i)
of \zcref{l:tube-amplitude}, the norm of this operator and that of its
adjoint are bounded by $Ce_0(1 + |Y|)^l$. The anharmonic normal multipliers in $w_{\rm out}$ and
$w_{\rm in}$ are bounded on the tube.
\end{proof}

\begin{lemma}\label{l:stab-body-parity}
Consider a scalar midpoint superkernel of the kind in
\zcref{l:stab-frame}, with its complete normal Mehler factor, and factor
out its body Gaussian at scale $h^2$. Suppose that its amplitude and
the finitely many derivatives used below, including two midpoint
covariant derivatives, have polynomial Gaussian envelopes bounded by
$A$. The same hypothesis may be imposed in the
fast-block norm $\|\cdot\|_{\rm f}$ when normal charges are present and
their bounds have already been applied. Suppose that full link reversal
has sign $\sigma\in\{1,-1\}$. After the normal and odd tangent
integrations, the operator-valued body kernel has a decomposition
\begin{equation}\label{e:stab-body-parity}
 \begin{gathered}
 R=R^{(0)}+R^{(1)}+R^{(2)},\\
 R^{(j)}(y,x)=\sigma(-1)^j\tau_{yx}R^{(j)}(x,y)\tau_{xy}
 \quad(j=0,1),
 \end{gathered}
\end{equation}
where, for a slightly widened body Gaussian
$w_h(x,y)=\delta_W^{-n_b/2}e^{-c d(x,y)^2/\delta_W}$,
\begin{equation}\label{e:stab-body-parity-bounds}
 \|R^{(j)}(x,y)\|\le CAh^{2j}w_h(x,y)\quad(j=0,1,2).
\end{equation}
The corresponding midpoint amplitudes have the same bounds after up to
two covariant derivatives in $m$ in the direction $Y_0$, with $Y_0$ parallel.
Here and below the constants may depend on the fixed degree metric
$\omega$. In particular, when $\sigma=1$,
\begin{equation}\label{e:stab-body-odd-marg}
 \sup_x\Bigl\|\int R^{(1)}(x,y)\dd\mu(y)\Bigr\|\le CAh^4 .
\end{equation}
\end{lemma}

\begin{proof}
Use body midpoint frames, transporting the two endpoint odd tangent
coordinates to $m$. Denote their odd midpoint by $\vartheta$ and their
difference divided by $h^2$ by $\eta$. The change from the full
supermidpoint coordinates to these endpoint coordinates has bounded
scaled coefficients. Indeed, if $\Phi(m,\Delta)=(x,y)$ is the analytic
midpoint map, then $x-y=0$ at $\Delta=0$, and
\[
 \frac{x-y}{h^2}
 =\int_0^1D_\Delta(x-y)(m,t h^2Y)[Y]\dd t .
\]
Together with the endpoint average, this map has an invertible
differential at $h=0$. Its analytic inverse, expanded in the finitely
many odd coordinates, therefore has bounded coefficients with polynomial
increment envelopes. Retain the full finite Berezinian of the endpoint
measure in the kernel coefficients. After removing the Gaussian scale
factors, it too has bounded analytic coefficients and inherits the full
reversal covariance, with the fixed Berezin orientation retained.
For the kernels here this uses the invariant measure $\nu$ and the
even exponential density $j$ of \eqref{e:tube-exp-density}; geodesic
symmetry preserves the endpoint measure.
Nilpotent even coordinate shifts are Taylor
expanded in the kernel coefficients in the fixed endpoint body
coordinates, before applying the kernel to a vector. Equivalently,
each unscaled endpoint difference supplies $h^2$, cancelling the
$h^{-2}$ of a derivative of the normalized body Gaussian. No derivative
of the input, and no negative power of $h$, is introduced by this change.
The change intertwines endpoint interchange with
$(Y_0,\eta)\mapsto(-Y_0,-\eta)$ and fixes $\vartheta$. Normal frame
actions remain in the charged factors of
\eqref{e:stab-degree-one} and \eqref{e:stab-higher-term}; no normal
change of variables or charge is inserted in this coefficient conversion.

Factor out the frozen odd tangent heat $H_F$, whose quadratic generator
and its required coefficient derivatives are $O(h^4)$. Expand the
remaining amplitude in $\vartheta$ and $\eta$. The coefficient of
$\vartheta^J\eta^I$ has reversal parity $\sigma(-1)^{|I|}$ in $Y_0$.
First integrate the full normal Mehler factor, retaining the order of
all coefficient operators. By \zcref{l:stab-schur}, this costs no power
of $h$: an odd normal midpoint carries $h$, and normalized odd normal
increments are treated with $\kappa_{\rm o}=h$ in
\zcref{l:tube-odd-gaussian}. The same applies to normal-amplitude
derivatives, which differentiate bounded unscaled midpoint coefficients,
and to the stated fast-block bounds. Thus the coefficient operators and
their midpoint covariant derivatives have the assumed envelopes.

For the tangent integration, write
$\Delta_IH_F=h^{-2|I|}\operatorname{ad}_{\theta_{i_1}}\cdots
\operatorname{ad}_{\theta_{i_{|I|}}}H_F$, with graded commutators.
The partial-pairing formula of \zcref{l:tube-odd-gaussian}, now with
$\kappa_{\rm o}=h^2$, gives the sharper even-degree assertion
\[
 \Delta_IH_F=
 \begin{cases}
  c_I\Id_{\Lambda_T}+O(h^4),& |I|\text{ even},\\
  O(h^2),& |I|\text{ odd}.
 \end{cases}
\]
For even $|I|$, every non-full pairing leaves at least two linear factors
of size $h^2$, and $H_F-\Id=O(h^4)$; this also covers $I=\emptyset$.
For odd $|I|$, at least one such factor remains. Differentiating the
finite formula preserves these orders. Tangent midpoint coordinates
act by the bounded maps
\[
 J_\theta(X)=\tfrac12\bigl(\theta X+(-1)^{|X|}X\theta\bigr).
\]
In particular, applying the midpoint maps to $c_I\Id$ gives multiplication
by the corresponding midpoint monomial. These maps preserve both error
orders. They act in the graded tensor product with the normal coefficient
operators in their original order; only fixed signs arise, and none
depends on $Y_0$.

Let $R^{(0)}$ consist of the full-pairing terms of even $|I|$, let
$R^{(1)}$ consist of all odd-$|I|$ terms, and put the even-degree
$O(h^4)$ terms in $R^{(2)}$. The moment and midpoint maps are independent
of the sign of $Y_0$ and covariant under $K_0$. The coefficient parities
therefore give the genuine body-kernel identities
\eqref{e:stab-body-parity}, while the Gaussian envelope bounds give
\eqref{e:stab-body-parity-bounds}, including the covariant derivatives.
For $\sigma=1$, the midpoint amplitude of $R^{(1)}$ is odd in $Y_0$ and
has size and derivative $CAh^2$. Applying \zcref{c:stab-parity} proves
\eqref{e:stab-body-odd-marg}.
\end{proof}

We now bound $\mathbf E_0$, which has the kernel
$\mathbf S_h(x)R_0(x, y)\tau_{xy}$ with $R_0 = \Pi_x - \mathbf F_hp_h$. Recall that $w_{\rm out}$ and
$w_{\rm in}$ contain the same half-weights and cutoffs as $\mathbf F_h$,
so that $R_0$ is the difference of two normal operators with the same
outer multipliers. We compare their exponents. By \eqref{e:stab-Pi-x},
$\Pi_x(x, y) = \widetilde\Pi(Y_x)$, and by \zcref{l:stab-chord},
\begin{equation}\label{e:stab-exponent}
\begin{gathered}
 \frac{\zeta}{2h^4}
 = \frac12\Str V^2 + \frac12\Str[Y, R_*]^2 + \mathcal Q_\zeta(U_m; Y) + h^4\mathcal R_h(V, Y; U_m) ,
 \\
 \mathcal Q_\zeta = \Str\bigl([Y, R_*][Y, U_m]\bigr) + \frac12\Str[Y, U_m]^2 ,
\end{gathered}
\end{equation}
where $\mathcal R_h$ retains the full analytic remainder obtained by
squaring the convergent series in \eqref{e:stab-midpoint}; its leading
term is quartic in $(V, Y)$. Write $V_0$ and $Y_0$ for the even increments.
On the local increment domain $|h^2Y_0| < \varrho_0$, with the normal
endpoints in the tube, every Grassmann coefficient of
$\mathcal R_h$, and every fixed derivative of that coefficient in the
even variables, is bounded by a fixed polynomial envelope in $(V_0, Y_0)$,
uniformly for small $h$. Indeed, after the leading quadratic terms are
removed, the convergent series has an overall factor $h^4$; the remaining
analytic functions of $h^2Y_0$ and their derivatives are bounded on this
domain. Finite expansion in the odd variables preserves polynomial
envelopes. The $\cosh$ term in \eqref{e:stab-midpoint} is block diagonal
and the $\sinh$ term is off diagonal, so their cross term has zero
supertrace. The body remainder therefore satisfies the sharper bound
$|\mathcal R_{h,\emptyset}| \le C|Y_0|^2(|V_0|^2 + |Y_0|^2)$.
\begin{enumerate}[label=(\alph*)]
\item The body part of $\mathcal Q_\zeta$ is $O(r|Y|^2)$ on the tube,
 where the body of $U_m$ has norm less than $2r$. Its odd parts come from the
 odd normal midpoint coordinates in $[Y_0, U_m]$, which carry the factor
 $h$, and from $[Y_{\rm odd}, U_{m, \rm body}]$, which carries $r$.
 Differentiating the normalized Gaussian in $\mathcal Q_\zeta$ and using
 \zcref{l:stab-schur}, we find that replacing $\mathcal Q_\zeta$ by zero
 changes the normal operator by at most $C(r + h)$ times a slightly
 widened Gaussian in $Y$. The normalized Gaussians with and without
 $\mathcal Q_\zeta$ have the same integral in $Y$.
\item The odd tangent Gaussian $e^{-\frac12\Str[Y_{\rm odd}, R_*]^2}$ is the
 odd heat over the time $\delta_W$. In the norm $| \cdot |_{2; \omega}$
 of $\Lambda_T$, which is the degree metric of \zcref{s:sigma-weighted},
 odd derivatives cost $\omega^{-1}$, and by \zcref{l:tube-odd-gaussian} with
 $\kappa_{\rm o} = h^2$ this factor differs from the identity of
 $\Lambda_T$ by $O(\delta_W\omega^{-2})$, which gives a scalar term with
 constant $C(\omega)$.
\item The factor $\varpi(U_m + h^2V/2)\varpi(U_m - h^2V/2)\varpi(S)^{-2}$,
 which combines the half-density with the normalization of the tangent
 Gaussian of \zcref{s:tube-operator}, differs from one by
 $O(\delta_W|V|^2)$, since it is even in $V$ and equals one at $V = 0$.
\item The term $h^4\mathcal R_h$, the density $j(h^2Y) = 1 + O(\delta_W|Y|^2)$ of the
 measure in midpoint coordinates, and the difference between $p_h$ and
 the Gaussian $(2\pi\delta_W)^{-n_b/2}e^{-d(x, y)^2/(2\delta_W)}$ with
 $d(x, y) = h^2|Y|_{g_{R_*}}$ change the kernel by $\delta_W$ times a
 polynomial envelope against a slightly widened Gaussian in $(V_0, Y_0)$.
 For the remainder in the exponent, decrease $\varrho_0$ if necessary:
 its body bound gives
 $h^4|\mathcal R_{h,\emptyset}| \le C\varrho_0^2(|V_0|^2 + |Y_0|^2)$,
 so the interpolated exponents remain uniformly coercive. The identity
 $e^{-A-B} - e^{-A} = -B\int_0^1e^{-A-tB}\dd t$ and the finite nilpotent
 expansion then give this bound, with the coefficient and derivative
 envelopes just established.
\end{enumerate}
Apply \zcref{l:stab-body-parity} with $\sigma=1$ to the uncharged
superkernel, and subtract the symmetric body kernel $\mathbf F_hp_h$
from its leading part. This gives body operator kernels
\begin{equation}\label{e:stab-R0}
\begin{gathered}
 R_0=R_{\rm ev}+R_{\rm odd}+R_{\rm rem},\\
 R_{\rm ev}(y,x)=\tau_{yx}R_{\rm ev}(x,y)\tau_{xy},\\
 R_{\rm odd}(y,x)=-\tau_{yx}R_{\rm odd}(x,y)\tau_{xy},\\
 \|R_{\rm ev}(x,y)\|\le C(r+\omega^2+h)w_h(x,y),\\
 \|R_{\rm odd}(x,y)\|\le C(r,\omega)h^2w_h(x,y),\\
 \|R_{\rm rem}(x,y)\|\le C(r,\omega)\delta_Ww_h(x,y),\\
 \sup_x\Bigl\|\int R_{\rm odd}(x,y)\dd\mu(y)\Bigr\|
 \le C(r,\omega)\delta_W .
\end{gathered}
\end{equation}
The midpoint covariant derivatives through order two have the same
Gaussian envelopes, by the bounded fixed-order derivatives of the
geometric factors and the smooth cutoffs in the unscaled normal variables,
and by the homogeneous midpoint frames. To see the small coefficient in the
leading even part, items (a), (c), and (d) give $C(r+h)$ before the
endpoint coefficient conversion. Its body is unchanged by that
conversion. In the leading even normal contractions, a new tangent
midpoint soul has even degree at least two, so its multiplication norm
is $O(\omega^2)$. A term with an unscaled odd normal midpoint factor,
or with odd normal parity after contraction, is $O(h)$ by
\zcref{l:stab-moment}. Paired normalized odd normal increments remain
in the bounded even coefficient; no small factor is attributed to such
a pair. Terms with an
unpaired odd tangent increment belong to $R_{\rm odd}$, and the
even-degree moment remainders belong to $R_{\rm rem}$. Fixed positive
powers of $h^2$ from the analytic coefficient conversion are absorbed in
$Ch$ by decreasing $h_0$ after $r$ and $\omega$ are fixed. Thus $C$ in
the leading bound may be chosen independently of these small parameters;
the remaining constants are allowed to depend on $r$ and $\omega$.

We also need the marginal as an operator on all of $\Lambda_T$.
Freeze the tangent midpoint when taking its leading even contractions.
They act by multiplication in the odd midpoint, and the normalized
tangent super-Gaussian has integral $\varpi(S)^{-2}$ times the flat
one, independently of that midpoint. Its leading mass is therefore
$\varpi(S)^{-2}\Id_{\Lambda_T}$, not merely its value on the constant
tangent coefficient. The half-density cancellation in item (c) then
gives $\mathbf F_h$. The variation of an even midpoint amplitude has
first term $h^2\nabla_{Y_0}$, which is odd in $Y_0$ and integrates to
zero when the midpoint is frozen; its second-order Taylor remainder is
$O(h^4)$ with a polynomial Gaussian envelope. For the odd-degree terms,
\eqref{e:stab-body-odd-marg} supplies the same order. The even-degree
moment remainders, the odd heat minus the identity, the density
$j-1$, and item (d) are already $O(\delta_W)$. Applying
\zcref{l:stab-schur} to these errors and using
$\int p_h(x,y)\dd\mu(y)=1$, we obtain
\begin{equation}\label{e:stab-R0-row}
 \Bigl\|\int R_0(x, y)\dd\mu(y)\Bigr\| \le C(r, \omega)\delta_W .
\end{equation}
Subtracting the last two parts of \eqref{e:stab-R0} shows that the
marginal of $R_{\rm ev}$ is also $O(\delta_W)$. The full body operator
kernel $R_0$ need not be symmetric: full superkernel reversal and
ordinary body-kernel reversal differ on the odd-degree moment terms.

\begin{lemma}\label{l:stab-uncharged}
Let $h$ be small enough, depending on $r$ and $\omega$. Then for all $f$ and
$g$ in $\mathcal H^{\rm tube}_0$,
\begin{multline*}
 |\langle g, \mathbf E_0f\rangle|
 \le C(r+\omega^2+h)\|D_{\rm body, L}^{1/2}g\|\|D_{\rm body, R}^{1/2}f\|\\
 +C(r,\omega)\sqrt{\delta_W}\|g\|\|D_{\rm body,R}^{1/2}f\|
 + C(r, \omega)\delta_W\|g\|\|f\| .
\end{multline*}
\end{lemma}

\begin{proof}
We have $\langle g, \mathbf E_0f\rangle = \langle\mathbf S_h^*g, T_0f\rangle$,
where $T_0$ has the kernel $R_0(x,y)\tau_{xy}$. Decompose it by
\eqref{e:stab-R0}. The kernel $R_{\rm ev}$ is genuinely symmetric on
the body, so \eqref{e:stab-dirichlet}, applied after subtracting its
marginal, gives
\begin{multline*}
 |\langle\mathbf S_h^*g,(T_{\rm ev}-R_{{\rm ev},\rm marg})f\rangle|
 \\
 \le C(r+\omega^2+h)
 \|(\Id-\mathbf H_h^2)^{1/2}\mathbf S_h^*g\|
 \|(\Id-\mathbf H_h^2)^{1/2}f\| .
\end{multline*}
For $R_{\rm odd}$ use \eqref{e:stab-dirichlet-one} instead. Its
Gaussian bound in \eqref{e:stab-R0} makes the first factor there at
most $C(r,\omega)h^2\|\mathbf S_h^*g\|$, while
\zcref{l:stab-energy} bounds the second by
$C\|(\Id-\mathbf H_h^2)^{1/2}f\|$. Its marginal is
$O(\delta_W)$ by \eqref{e:stab-R0}. The remainder has operator norm
$C(r,\omega)\delta_W$ by Schur's test, and the even marginal has the
same bound by \eqref{e:stab-R0-row} and the paragraph following it.
Finally,
$\mathbf S_h(\Id - \mathbf H_h^2)\mathbf S_h^* \le D_{\rm body, L}$ and
$\Id - \mathbf H_h^2 \le D_{\rm body, R}$, while
$\|\mathbf S_h^*g\|\le\|g\|$ and $h^2=\sqrt{\delta_W}$.
These observations give the stated three terms.
\end{proof}

\section{Holonomy, charged words, the frozen normal transfer and the source}
\label{s:stab-words}

\subsection{Holonomy}

We now compare the body heat $\mathbf H_h$ with the operator with kernel
$p_h(x, y)\tau_{xy}$, which uses the geodesic transport. Recall that $p_h$
is the scalar heat kernel of $e^{\delta_W\Delta/2}$ on $|\mathcal M|$. Let $(X_t)$ be a
Brownian motion on $|\mathcal M|$ with generator $\frac12\Delta$, started at
$x$, and let $/\!/_t$ be the stochastic parallel transport of $\mathcal E$
along $X$, from the fiber at $x$ to the fiber at $X_t$. On each summand
of $\mathcal E$ of finite rank, $\Delta_\nabla$ is essentially self-adjoint,
as noted in \zcref{s:stab-main}, and the Feynman--Kac--It\^o formula for
the Bochner Laplacian \cite{guneysu2017covariant} gives
\begin{equation}\label{e:stab-fki}
 (\mathbf H_hf)(x) = \E_x\bigl[/\!/_{\delta_W}^{-1}f(X_{\delta_W})\bigr] .
\end{equation}
The law of $X_{\delta_W}$ has the density $p_h(x, \cdot)$, and the operator
with kernel $p_h(x, y)\tau_{xy}$ acts by $f \mapsto \E_x[\tau_{xX_{\delta_W}}f(X_{\delta_W})]$,
up to the Gaussian-small part of the kernel with $d(x, y) \ge \varrho_0$.
We define the holonomy
\[
 \gamma_x = /\!/_{\delta_W}^{-1}\tau_{X_{\delta_W}x} \in K_0 ,
\]
of the loop that goes from $x$ to $X_{\delta_W}$ along the geodesic and
returns along the Brownian path. Then
\begin{equation}\label{e:stab-hol-formula}
 (\mathbf E_{\rm hol}f)(x) = -\mathbf S_h(x)\mathbf F_h\,\E_x\bigl[(\pi(\gamma_x) - \Id)\tau_{xX_{\delta_W}}f(X_{\delta_W})\bigr]
\end{equation}
up to a Gaussian-small error.

\begin{lemma}\label{l:stab-holonomy}
Let $\tau_x$ be the exit time of $X$ from the ball $B(x, \varrho_0/2)$, and
let $| \cdot |$ be the norm in a fixed faithful unitary representation of
$K_0$. For every $j \ge 1$ there exists $C_j$, and there exist $C, c > 0$,
such that for all $x$ and $0 < t \le 1$,
\[
 \E_x\bigl[|\gamma_x(t) - \Id|^j ; \tau_x > t\bigr] \le C_jt^j ,
 \qquad
 \Prob_x(\tau_x \le t) \le Ce^{-c/t} ,
\]
where $\gamma_x(t)$ is the holonomy defined above with $\delta_W$ replaced by
$t$.
\end{lemma}

\begin{proof}
On $B(x, \varrho_0)$ we use the radial gauge: the frame of the principal
$K_0$-bundle obtained by parallel transport along the geodesics from
$x$. By homogeneity, it is uniform in $x$. The connection form $A$ of this
frame, with values in $\mathfrak k_0$, satisfies $A_x = 0$ and $|A(z)| \le Cd(x, z)$,
with all derivatives bounded. Since the transport along radial geodesics
is trivial in this gauge, $\gamma_x(t) = g_t^{-1}$, where $g_t$ solves the
Stratonovich equation $\dd g = -A(X)\circ\dd X\,g$ with $g_0 = \Id$.

In normal coordinates at $x$, we have
$\dd X^i = \sigma^i_a(X)\dd\beta^a + \frac12b^i(X)\dd s$, with $\sigma$ and $b$
bounded uniformly in $x$, where $\beta$ is a Brownian motion in
$\R^{n_b}$. In It\^o form,
\[
 g_t - \Id = -\int_0^tA_i\sigma^i_a(X_s)g_s\dd\beta^a_s + \int_0^t\widetilde B(X_s)g_s\dd s ,
\]
with $|\widetilde B| \le C$ on the ball, since $A$, $\partial A$, and $\sigma$
are bounded there and $|g_s| = 1$. By the
Burkholder--Davis--Gundy inequality,
\[
 \E_x\sup_{s \le t \wedge \tau_x}\Bigl|\int_0^sA\sigma g\dd\beta\Bigr|^j
 \le C_j\E_x\Bigl(\int_0^{t \wedge \tau_x}|A(X_s)|^2\dd s\Bigr)^{j/2}
 \le C_jt^{j/2}\E_x\sup_{s \le t \wedge \tau_x}d(x, X_s)^j
 \le C_jt^j ,
\]
where the last step is the same inequality for the coordinate
martingale. The drift term is bounded by $Ct$. The exit bound follows from
the exponential martingale inequality for the coordinate martingale,
whose quadratic variation grows at a bounded rate.
\end{proof}

We now bound $\mathbf E_{\rm hol}$ through \eqref{e:stab-hol-formula}.

\begin{lemma}\label{l:stab-hol}
For all $f$ and $g$ in $\mathcal H^{\rm tube}_0$,
\begin{equation}\label{e:stab-hol}
 |\langle g, \mathbf E_{\rm hol}f\rangle|
 \le C(r + h)\sqrt{\delta_W}\,\|g\|\|D_{\rm nor}^{1/2}f\| + C\delta_W\|g\|\|f\| .
\end{equation}
In particular, $\mathbf E_{\rm hol}$ satisfies the four-term bound with
parameters $0$ and $C$.
\end{lemma}

\begin{proof}
We first consider the event
\[
 \mathcal A_x = \{\tau_x > \delta_W, \ |\gamma_x - \Id| \le 1/2\} .
\]
On $\mathcal A_x$, we set $\mathfrak e = \log\gamma_x \in \mathfrak k_0$, so that
$|\mathfrak e| \le 2|\gamma_x - \Id|$ and
\[
 \pi(\gamma_x) - \Id = \int_0^1\pi(e^{s\mathfrak e})\,\mathrm d\pi(\mathfrak e)\dd s ,
 \qquad
 \mathrm d\pi(\mathfrak e) = \sum_a\mathfrak e^a(\mathfrak j_a \otimes \Id + \Id \otimes m_a) ,
\]
where the $m_a$ are bounded matrices on $\Lambda_T$. The operator
$\pi(e^{s\mathfrak e})$ is unitary and commutes with $\mathbf F_h$ and with $\mathbf N$,
and $\mathfrak j_a\mathbf P_0 = 0$. By part (iv) of \zcref{l:tube-amplitude},
in which the cutoff bounds the coordinate factor of a compact charge by
$r/h$, we have
\[
 \|\mathbf F_h\mathfrak j_aD_{\rm nor}^{-1/2}\mathbf P_0^\perp\| \le C(r/h + 1)h^{-1} .
\]
Therefore, on $\mathcal A_x$ and for every vector $u$ in the fiber at $x$,
\[
 \bigl\|\mathbf F_h(\pi(\gamma_x) - \Id)u\bigr\|
 \le C|\gamma_x - \Id|\bigl((r/h + 1)h^{-1}\|D_{\rm nor}^{1/2}u\| + \|u\|\bigr) ,
\]
where the second term comes from the matrices $m_a$ on $\Lambda_T$. On the
complementary event, we use $\|\pi(\gamma_x) - \Id\| \le 2$. By
\zcref{l:stab-holonomy} with $j = 2$ and Chebyshev's inequality, this
event has probability at most $C\delta_W^2 + Ce^{-c/\delta_W}$, and
\zcref{l:stab-holonomy} with $j = 2$ also gives
\[
 \E_x\bigl[|\gamma_x - \Id|^2 ; \tau_x > \delta_W\bigr] \le C\delta_W^2 .
\]

We insert these bounds into \eqref{e:stab-hol-formula} with
$u = \tau_{xX_{\delta_W}}f(X_{\delta_W})$, and we apply the Cauchy--Schwarz
inequality in $\E_x \otimes \dd\mu(x)$. The invariance of $\mu$ for $X$,
\[
 \int\E_x\varphi(X_{\delta_W})\dd\mu(x) = \int\varphi\dd\mu
 \qquad \text{for all measurable } \varphi \ge 0 ,
\]
and the fact that the transports commute with $D_{\rm nor}$ then give
\[
 |\langle g, \mathbf E_{\rm hol}f\rangle|
 \le C\delta_W(r/h + 1)h^{-1}\|g\|\|D_{\rm nor}^{1/2}f\| + C\delta_W\|g\|\|f\| + C\delta_W\|g\|\|f\| ,
\]
where the last term comes from the complementary event, on which we take
the square root of its probability in the Cauchy--Schwarz inequality. Since
$\delta_W(r/h + 1)h^{-1} = \sqrt{\delta_W}(r + h)$, this proves
\eqref{e:stab-hol}. The last assertion follows from
$D_{\rm nor} \le 4D_R$, which is \eqref{e:stab-product-defects}, and from
$r + h \le 2$.
\end{proof}

\subsection{The degree-one part}

\begin{lemma}\label{l:stab-degree-one}
The operator $\mathbf E_1$ satisfies the four-term bound with parameters
$C(r + h)$ and $C$.
\end{lemma}

\begin{proof}
We have $\mathbf E_1 = \mathbf S_hX$, where $X$ is the degree-one part, the
operator with the kernel \eqref{e:stab-degree-one} on $d(x, y) < \varrho_0$.
Since $\mathbf S_h$ is a function on $|\mathcal M|$ with $|\mathbf S_h| \le 1$,
it commutes with $D_{\rm nor}$ and $\mathbf P_0$, and
$\|D_{\rm nor}^{1/2}\mathbf S_h^*g\| \le \|D_{\rm nor}^{1/2}g\|$. We verify the
hypotheses of \zcref{l:stab-blocks} for $\mathbf E_1$, and to this end we
bound $\langle g', Xf\rangle$ for a vector $g'$ in the place of
$\mathbf S_h^*g$. Throughout, $w(x, y) = C\delta_W^{-n_b/2}e^{-c\,d(x, y)^2/\delta_W}$
denotes a slightly widened body Gaussian. It absorbs the polynomials in
$Y$, and $\int w(x, y)\dd\mu(y)$ and $\int w(x, y)\dd\mu(x)$ are bounded.

We begin with the fast block of the commutator part
$X^{\rm c} = \frac12[\Pi_x, L_-^{\rm out} + L_+]\tau_{xy}$ of
\eqref{e:stab-degree-one}. \zcref[S]{l:stab-words} shows that
$h^{-2}(L_-^{\rm out} + L_+)$ is a sum of charges, with coefficients bounded by a polynomial in $Y$.
The remark after \zcref{l:stab-moment} shows that the commutator of a
charge with $\Pi_x$ is $h$ times a sum of operators $\mathbf M[c']$ with at most
one letter on either side. By part (iv) of \zcref{l:tube-amplitude} with
$s \le 1$, and by its adjoint for a letter on the right, each of these
operators has norm $\| \cdot \|_{\rm f}$ at most $Ch^{-2}$, where
$\| \cdot \|_{\rm f}$ is the norm of \zcref{s:stab-parity}; here we used that
$D_+$ and $D_{\rm nor}$ are comparable on the range of $\mathbf P_0^\perp$. The
kernel of the fast block of $X^{\rm c}$ has norm $\| \cdot \|_{\rm f}$ at
most $Ch^2h \cdot h^{-2}w(x, y) = Ch\,w(x, y)$. Since the transports
commute with $D_{\rm nor}$, the Cauchy--Schwarz inequality implies
\[
 |\langle g', \mathbf P_0^\perp X^{\rm c}\mathbf P_0^\perp f\rangle|
 \le Ch\|D_{\rm nor}^{1/2}g'\|\|D_{\rm nor}^{1/2}f\| .
\]

Next, we treat the fast block of the anticommutator part $W^{\rm a}$, with
the one-sided form \eqref{e:stab-dirichlet-one} of the energy, applied to
$R(x, y) = \mathbf P_0^\perp W^{\rm a}(x, y)\mathbf P_0^\perp$, whose marginal is
$\mathbf P_0^\perp W^{\rm a}_{\rm marg}\mathbf P_0^\perp$. Using part (iv) of
\zcref{l:tube-amplitude} with one inverse half-defect, we find
\[
 \|W^{\rm a}(x, y)^*\mathbf P_0^\perp u\| \le C(r + h)w(x, y)\|D_{\rm nor}^{1/2}u\|
\]
for vectors $u$ in the fiber at $x$. Indeed, for an odd charge, part (iv) with
$s = 1$ gives the factor $h^2h^{-1} = h$, and for a compact charge the
coordinate factor, cut off at $r/h$, and one derivative give
$h^2(r/h)h^{-1} = r$.

With $c' = c/2$ in \eqref{e:stab-dirichlet-one},
where $c$ is the constant in $w$, the first factor there is therefore at
most $C(r + h)\|D_{\rm nor}^{1/2}g'\|$. Since $\mathbf P_0^\perp$ commutes with
$\mathbf H_h$, \eqref{e:stab-energy} implies that the second factor
satisfies
$\mathfrak D_{c'}(\mathbf P_0^\perp f)^{1/2} \le C\|(\Id - \mathbf H_h^2)^{1/2}f\|$.
We obtain
\[
 |\langle g', \mathbf P_0^\perp(W^{\rm a} - W^{\rm a}_{\rm marg})\mathbf P_0^\perp f\rangle|
 \le C(r + h)\|D_{\rm nor}^{1/2}g'\|\|(\Id - \mathbf H_h^2)^{1/2}f\| .
\]
By \eqref{e:stab-anti-marg} of \zcref{l:stab-degree-one-parity}, the
marginal satisfies
\[
 |\langle g', \mathbf P_0^\perp W^{\rm a}_{\rm marg}\mathbf P_0^\perp f\rangle|
 \le C(rh + h^2)\|D_{\rm nor}^{1/2}g'\|\|D_{\rm nor}^{1/2}f\| .
\]
We add the three bounds, take $g' = \mathbf S_h^*g$, and use
$\Id - \mathbf H_h^2 \le D_{\rm body, R}$. The fast block of $\mathbf E_1$
then satisfies the hypothesis of \zcref{l:stab-blocks} with
$b_0 = C(r + h)$ and $b = 0$.

It remains to bound the vacuum blocks. The degree-one part has the kernel
$\Pi_x\tau_{xy}L_- - L_+\Pi_x\tau_{xy}$, the left side of
\eqref{e:stab-degree-one}. The
charges commute with $\mathbf P_0$ and annihilate $\Omega_0$, so that
$\mathbf P_0L_\pm = L_\pm\mathbf P_0 = 0$. Hence the vacuum scalar of $X$
vanishes, and its vacuum row has the kernel
$\mathbf P_0(\Pi_x - c_\Pi)\tau_{xy}L_-$, where $c_\Pi$ is the scalar of the
vacuum row of $\Pi_x$ given by \zcref{l:stab-moment}. That lemma also shows
that $\mathbf P_0(\Pi_x - c_\Pi)$ is $O(h)$ in every graph norm. Since $h^{-2}L_-$ is
a sum of charges, which are bounded relative to $\mathbf N + 1$, the norm of
this kernel is bounded by $Ch^3w(x, y)$. The vacuum column is bounded in the same
way, and Schur's test gives
$\|\mathbf P_0X\mathbf P_0^\perp\| + \|\mathbf P_0^\perp X\mathbf P_0\| \le Ch^3$. The same bounds
hold for $\mathbf E_1 = \mathbf S_hX$, and \zcref{l:stab-blocks} with $b_1 = C$
and $b_2 = 0$ proves the lemma.
\end{proof}

\subsection{Higher words}

\begin{lemma}\label{l:stab-higher}
The operator $\mathbf E_{\ge 2}$ satisfies the four-term bound with
parameters $Cr^2 + C(r)h$ and $C$.
\end{lemma}

\begin{proof}
We have $\mathbf E_{\ge 2} = \mathbf S_hX$, where $X$ is the sum of the higher
words. As in the proof of \zcref{l:stab-degree-one}, we bound
$\langle g', Xf\rangle$ for a vector $g'$ in the place of $\mathbf S_h^*g$, and
$w(x, y)$ denotes the widened body Gaussian of that proof.

We first write out the terms. By the expansion of the kernel of $A_h$
after \eqref{e:stab-Pi-x}, the kernel of $X$ on $d(x, y) < \varrho_0$ is
the sum, over $0 \le n_1, n_2 \le D$ with $n = n_1 + n_2 \ge 2$, of
\begin{equation}\label{e:stab-higher-term}
 \frac{(-1)^{n_1}}{n_1!\,n_2!}L_+^{n_1}\,\Pi_x(x, y)\,\tau_{xy}L_-^{n_2}
 = \frac{(-1)^{n_1}}{n_1!\,n_2!}L_+^{n_1}\,\Pi_x(x, y)\,(L_-^{\rm out})^{n_2}\tau_{xy} .
\end{equation}
The transported charges are again charges, since the transport acts on
the normal variables through $K_0$. Then \zcref{l:stab-words} shows that
$h^{-2n}L_+^{n_1}(L_-^{\rm out})^{n_2}$ is a finite sum of products
$\mathbf w_+\mathbf w_-$, with coefficients bounded by polynomials in $Y$,
where $\mathbf w_+$ is a word of degree $n_1$ and $\mathbf w_-$ a word of
degree $n_2$. The number of terms depends only on $k$, and we fix one of
them. We let $a$ and $b$ denote the numbers of odd and compact charges in
$\mathbf w_+\mathbf w_-$, so that $a + b = n$.

Next, we bound the vacuum blocks. The charges commute with $\mathbf P_0$ and
annihilate $\Omega_0$, so that $\mathbf P_0\mathbf w = \mathbf w\mathbf P_0 = 0$ for every
word $\mathbf w$ of positive degree. All vacuum blocks of the term thus
vanish if $n_1 \ge 1$ and $n_2 \ge 1$. If $n_2 = 0$, its vacuum row
and vacuum scalar vanish, and its vacuum column has the kernel
$h^{2n}\mathbf w_+(\Pi_x - c_\Pi')\mathbf P_0\tau_{xy}$ times a coefficient, where
$c_\Pi'$ is the scalar of the vacuum column of $\Pi_x$ given by
\zcref{l:stab-moment}. That lemma also shows that $(\Pi_x - c_\Pi')\Omega_0$ is $O(h)$ in
every graph norm, and a word of degree $n$ is bounded relative to
$(\mathbf N + 1)^n$. The norm of this kernel is therefore bounded by
$Ch^{2n + 1}w(x, y) \le Ch^5w(x, y)$. The case $n_1 = 0$ is symmetric. Schur's test gives \eqref{e:stab-blocks} with $b_1 = Ch^2$ and
$b_2 = 0$.

We now turn to the fast block, in the norm $\| \cdot \|_{\rm f}$ of
\zcref{s:stab-parity}. We move the charges of $\mathbf w_-$ to the left of
$\Pi_x$. Writing $\mathbf w_- = \mathfrak c_1\cdots\mathfrak c_{n_2}$ with charges
$\mathfrak c_i$, we have
\begin{equation}\label{e:stab-higher-move}
 \mathbf w_+\Pi_x\mathbf w_-
 = \mathbf w_+\mathbf w_-\Pi_x
 + \sum_{i = 1}^{n_2} \pm \,\mathbf w_+\mathfrak c_1\cdots\mathfrak c_{i - 1}\,[\Pi_x, \mathfrak c_i]\,\mathfrak c_{i + 1}\cdots\mathfrak c_{n_2} ,
\end{equation}
with graded commutators. Recall from \zcref{l:stab-moment} that $\Pi_x$ is an operator
$\mathbf M[c]$ between the invariant cutoffs contained in $w_{\rm out}$ and
$w_{\rm in}$, which commute with the charges, and the tangent Gaussian in
its amplitude absorbs the polynomials in $Y$. For the first term of
\eqref{e:stab-higher-move}, \zcref{l:tube-word}, which part (iv) of
\zcref{l:tube-amplitude} extends to $\mathbf M[c]$ on the range of
$\mathbf P_0^\perp$, gives
\[
 \|h^{2n}\mathbf w_+\mathbf w_-\Pi_x\|_{\rm f} \le \bigl(Ch^ar^b + C(r)h\bigr)w(x, y) \le \bigl(Cr^2 + C(r)h\bigr)w(x, y) ,
\]
where we used $h \le r \le 1$ and $a + b \ge 2$ in the second
inequality.

For the other terms of \eqref{e:stab-higher-move}, we count the letters.
The word $\mathbf w_+\mathbf w_-$ is a sum of products of at most $a + 2b$
letters, with fixed matrices as coefficients. The commutator
$[\Pi_x, \mathfrak c_i]$ is $h$ times a sum of operators $\mathbf M[c']$ with
at most one letter. For the terms of $\mathfrak c_i$ that contain letters,
this follows from the remark after \zcref{l:stab-moment} and from
\eqref{e:tube-compact-commutator}. For a finite matrix $F$, it follows from
$[F, \mathbf M[c]] = h\mathbf M[[F, c^{\rm rem}]]$, which holds by
\zcref{l:stab-moment}. Replacing $\mathfrak c_i$ by its commutator with
$\Pi_x$ therefore produces the factor $h$ and does not increase the
number of letters.

We then move the remaining charges and letters to the left
of the amplitude operator, one at a time. By part (ii) of
\zcref{l:tube-amplitude} and by \eqref{e:tube-odd-commutator} and
\eqref{e:tube-compact-commutator}, each such move produces commutators
that either carry a factor $h$ and remove a letter or keep the number of
letters. Every resulting term is $h^{2n + 1}$ times an oscillator
polynomial of degree $s \le a + 2b$ applied to an operator $\mathbf M[c'']$,
and part (iv) of \zcref{l:tube-amplitude} bounds it in the norm
$\| \cdot \|_{\rm f}$ by
\[
 Ch^{2n + 1}h^{-\max(2, s)}w(x, y) \le Ch^{a + 1}w(x, y) \le Ch\,w(x, y) ,
\]
since $2n - (a + 2b) = a$.

Combining the bounds for the fast block and using the Cauchy--Schwarz
inequality as in the proof of \zcref{l:stab-degree-one}, we obtain
\[
 |\langle g', \mathbf P_0^\perp X\mathbf P_0^\perp f\rangle|
 \le \bigl(Cr^2 + C(r)h\bigr)\|D_{\rm nor}^{1/2}g'\|\|D_{\rm nor}^{1/2}f\| .
\]
With $g' = \mathbf S_h^*g$, this is the hypothesis of \zcref{l:stab-blocks}
for $\mathbf E_{\ge 2}$ with $b_0 = Cr^2 + C(r)h$ and $b = 0$, and the lemma
follows from \zcref{l:stab-blocks}.
\end{proof}

\subsection{The frozen normal transfer}

\begin{lemma}\label{l:stab-nor}
The operator $\mathbf E_{\rm nor}$ satisfies the four-term bound with
parameters $Cr + C(r)h$ and $C(r)$.
\end{lemma}

\begin{proof}
By \zcref{l:tube-powers} with $m = 0$, $\mathbf F_h - \mathbf M^\chi$
satisfies \eqref{e:tube-four-term} relative to $\mathbf M^\chi$, with the
defects $\bar D_L$ and $\bar D_R$, the parameter $\delta_0 = \delta_W$, the
bilateral coefficient $Cr + C(r)h$ and $A_1 = C(r)$, including the
nilpotent terms of the invariant cutoff. This is a bound on the normal
factor. It extends to the operator $(\mathbf F_h - \mathbf M^\chi) \otimes \Id$,
which acts on each fiber, with $\bar D_L \otimes \Id$ and
$\bar D_R \otimes \Id$ in place of $\bar D_L$ and $\bar D_R$. To see this,
we apply the bound at each point of $|\mathcal M|$ to the components of
$f$ and $g$ in an orthonormal basis of $\Lambda_T$, and we sum with the
Cauchy--Schwarz inequality. Let $C_{\rm body} = \mathbf S_h\mathbf H_h$
denote the body factor of $C_h$, a contraction that commutes with the
normal operators. Then
\[
\begin{gathered}
 \langle g, \mathbf E_{\rm nor}f\rangle
 = \bigl\langle(\Id \otimes C_{\rm body}^*)g, \bigl((\mathbf F_h - \mathbf M^\chi) \otimes \Id\bigr)f\bigr\rangle ,
 \\
 \|(\bar D_L \otimes \Id)^{1/2}(\Id \otimes C_{\rm body}^*)g\| \le \|(\bar D_L \otimes \Id)^{1/2}g\| .
\end{gathered}
\]
Moreover,
$(\mathbf M^\chi)^*\mathbf M^\chi \otimes \mathbf H_h|\mathbf S_h|^2\mathbf H_h
\le (\mathbf M^\chi)^*\mathbf M^\chi \otimes \Id$ gives $\bar D_R \otimes \Id \le D_R$,
since these positive operators commute, and likewise
$\bar D_L \otimes \Id \le D_L$. Inserting these bounds into the four-term
bound for $(\mathbf F_h - \mathbf M^\chi) \otimes \Id$ proves the lemma.
\end{proof}

\subsection{The source}

Let $s_\Gamma^{\rm BB}$ and $s_\Gamma^{\rm FF}$ denote the bosonic and the
fermionic part of the source factor. Since $\Gamma$ is block diagonal,
\eqref{e:hist-normalized} gives, at a point $S$ of the tube,
\begin{equation}\label{e:stab-src-parts}
 s_\Gamma = s_\Gamma^{\rm BB}s_\Gamma^{\rm FF} ,
 \qquad
 s_\Gamma^{\rm BB}(S) = e^{-\iu\eta\Tr(S^{\rm BB}\Gamma_{BB})} ,
 \qquad
 s_\Gamma^{\rm FF}(S) = e^{\iu\eta\Tr(S^{\rm FF}\Gamma_{FF})} .
\end{equation}
We let $D_S = \Id - |\mathbf S_h|^2$ denote the defect of the frozen source
kick. It is the multiplication by a function on $|\mathcal M|$, and
\eqref{e:stab-kick} gives $D_S \ge 1 - e^{-2c\delta_Wv}$, where $c$ is the
constant of \eqref{e:stab-kick}. The soul of the BB block in the odd
tangent coordinates contributes the term $\omega^2$ to the bilateral
parameter in the next lemma. For this reason $\omega$ must be small in
\zcref{t:stab-main}, as in the continuum model, where the soul of the
potential is bounded in \eqref{e:sigma-potential-bounds}.

\begin{lemma}\label{l:stab-src}
There exists $C_1$ such that the following holds if
$C_1(r + \omega^2) \le c/4$, where $c$ is the constant of
\eqref{e:stab-kick}. For $h$ small enough, depending on $r$ and $\omega$,
the operator $\mathbf E_{\rm src}$ satisfies the four-term bound with
parameters $C(r + \omega^2) + C(r, \omega)h$ and $C(r, \omega)$.
\end{lemma}

\begin{proof}
We first prove that, on the tube and in the operator norm on the fiber,
\begin{equation}\label{e:stab-src-multiplier}
 \bigl\|c_\Gamma^{-1}s_\Gamma^{\rm BB} - \mathbf S_h\bigr\|
 \le C_1(r + \omega^2 + h)\,\delta_Wv\,e^{-c\delta_Wv/2} .
\end{equation}
Let $S = R(x, U)$ be a point of the tube. As in \zcref{d:sigma-target} and
the proof of \zcref{l:sigma-multipliers}, we write the frame as
$T(x) = ge^{\Theta}$, where $g = \diag(g_{\rm B} \,|\, g_{\rm F})$ is a body
element with $g_{\rm B}$ positive definite and
$\|g_{\rm B}\|\|g_{\rm B}^{-1}\| \le 2v$, and where $\Theta$ is an odd tangent
supermatrix whose entries are the odd tangent coordinates. Then
$S^{\rm BB} = g_{\rm B}\bigl(e^{\Theta}(R_* + U)e^{-\Theta}\bigr)^{\rm BB}g_{\rm B}^{-1}$,
and we split
\begin{equation}\label{e:stab-src-split}
 S^{\rm BB} = \kappa_E\Id - \iu s_EQ_B + \mathcal N_T + \mathcal N_U ,
 \qquad
 \mathcal N_U = g_{\rm B}\bigl(e^{\Theta}Ue^{-\Theta}\bigr)^{\rm BB}g_{\rm B}^{-1} ,
\end{equation}
where $\kappa_E\Id - \iu s_EQ_B$ is the body of the BB block of $R(x, 0)$,
and $\mathcal N_T$ is its soul.

We bound the two parts in turn. Since $g$ is block diagonal and the odd
powers of $\Theta$ have vanishing diagonal blocks, the entries of
$\mathcal N_T$ are polynomials of even degree at least two in the odd
tangent coordinates, with coefficients bounded by
$C\|g_{\rm B}\|\|g_{\rm B}^{-1}\| \le Cv$. Multiplication by $\mathcal N_T$
therefore has norm at most $C\omega^2v$, by \zcref{l:grass-norm}(a) and (d). The body of
$\mathcal N_U$ is $g_{\rm B}U^{\rm BB}g_{\rm B}^{-1}$, and on the tube
\[
 \|g_{\rm B}U^{\rm BB}g_{\rm B}^{-1}\| \le 2v|U^{\rm BB}| \le 4rv .
\]
Every term of its soul contains two
odd tangent coordinates and an even entry of $U$, or one odd tangent
coordinate and one odd entry of $U$. In the standardized coordinates
$U = hZ$, an odd entry of $U$ is $h$ times a bounded operator on
$\mathcal H_{\rm odd}$, and the soul of $\mathcal N_U$ therefore has norm at
most $C(\omega^2r + \omega h)v \le C(\omega^2 + h)v$.

By the definitions of $c_\Gamma$ and $\mathbf S_h$ in \zcref{s:stab-main}, the
body of the BB block of $R(x, 0)$ contributes $c_\Gamma\mathbf S_h$ to
$s_\Gamma^{\rm BB}$. All these terms are even and commute, and we obtain
\[
 c_\Gamma^{-1}s_\Gamma^{\rm BB} = \mathbf S_he^{\psi} ,
 \qquad
 \psi = -\iu\eta\Tr\bigl((\mathcal N_T + \mathcal N_U)\Gamma_{BB}\bigr) .
\]
Using $\eta \le C\delta_W$, $|\Tr A| \le 2k\|A\|$, and $\|\Gamma_{BB}\| \le 2$ on
$\mathcal U_2$, we find from the bounds above that
$\|\psi\| \le C_1(r + \omega^2 + h)\delta_Wv$. Together with
$\|e^{\psi} - \Id\| \le \|\psi\|e^{\|\psi\|}$ and \eqref{e:stab-kick}, this gives
\[
 \bigl\|c_\Gamma^{-1}s_\Gamma^{\rm BB} - \mathbf S_h\bigr\|
 = \|\mathbf S_h(e^{\psi} - \Id)\|
 \le C_1(r + \omega^2 + h)\,\delta_Wv\,e^{(C_1(r + \omega^2 + h) - c)\delta_Wv} .
\]
By the hypothesis of the lemma, and for $h$ so small that
$C_1h \le c/4$, the exponent is at most $-c\delta_Wv/2$, which proves
\eqref{e:stab-src-multiplier}.

Next, we factor the difference through $D_S$. For $y \ge 0$, we have
$ye^{-cy/2} \le C(1 - e^{-2cy})$, since the quotient of the two sides is
continuous on $(0, \infty)$ and tends to $(2c)^{-1}$ as $y \to 0$ and to
zero as $y \to \infty$. Since $D_S \ge 1 - e^{-2c\delta_Wv}$, the bound
\eqref{e:stab-src-multiplier} implies that
\begin{equation}\label{e:stab-src-factor}
 c_\Gamma^{-1}s_\Gamma^{\rm BB} - \mathbf S_h = D_S^{1/2}b_SD_S^{1/2} ,
 \qquad
 b_S = D_S^{-1}\bigl(c_\Gamma^{-1}s_\Gamma^{\rm BB} - \mathbf S_h\bigr) ,
 \qquad
 \|b_S\| \le C(r + \omega^2 + h) ,
\end{equation}
and $b_S$ commutes with $D_S$. The FF block of $S$ is bounded on the tube, with all its
coefficients, since $g_{\rm F}$ is unitary, and therefore
$\|s_\Gamma^{\rm FF} - \Id\| \le C\delta_W$. Using \eqref{e:stab-src-parts},
\eqref{e:stab-src-factor}, and \eqref{e:stab-scalar}, we obtain
\begin{equation}\label{e:stab-src-decomposition}
 s_\Gamma - \mathbf S_h = D_S^{1/2}b_S'D_S^{1/2} + \mathbf S_h\bigl(c_\Gamma s_\Gamma^{\rm FF} - \Id\bigr) ,
 \qquad
 b_S' = c_\Gamma s_\Gamma^{\rm FF}b_S ,
\end{equation}
with $\|b_S'\| \le 2\|b_S\|$ and $\|c_\Gamma s_\Gamma^{\rm FF} - \Id\| \le C\delta_W$
for $h$ small.

We also need a bound for the step without the frozen kick. Recall that
$\mathcal K_h - \mathbf M^\chi \otimes \mathbf H_h$ is the sum of the five
operators $X$ with $\mathbf S_hX \in \{\mathbf E_{\rm nor}, \mathbf E_0, \mathbf E_{\rm hol}, \mathbf E_1, \mathbf E_{\ge 2}\}$.
In the proofs of \zcref{l:stab-uncharged}, \zcref{l:stab-hol},
\zcref{l:stab-degree-one}, \zcref{l:stab-higher}, and \zcref{l:stab-nor},
the vector $g$ enters only through $\mathbf S_h^*g$, or through
$(\Id \otimes C_{\rm body}^*)g$ for $\mathbf E_{\rm nor}$, and only through the norm of
this vector and its norms with $D_{\rm nor}$, $\Id - \mathbf H_h^2$, and
$\bar D_L \otimes \Id$, which are at most its norm. These proofs
therefore apply with this vector replaced by an arbitrary vector of norm
one. Adding the resulting bounds, and using
\eqref{e:stab-product-defects} on the right, we find that for $h$ small
enough, depending on $r$ and $\omega$,
\begin{equation}\label{e:stab-right}
 \bigl\|(\mathcal K_h - \mathbf M^\chi \otimes \mathbf H_h)f\bigr\|
 \le C\|D_R^{1/2}f\| + C(r, \omega)\sqrt{\delta_W}\|f\| .
\end{equation}
In particular, $\|\mathcal K_h\| \le C$.

We now bound $\mathbf E_{\rm src} = (s_\Gamma - \mathbf S_h)\mathcal K_h$. The
second term of \eqref{e:stab-src-decomposition} has norm at most
$C\delta_W$, and together with $\|\mathcal K_h\| \le C$ it contributes a
scalar term $C\delta_W\|g\|\|f\|$. For the first term, we have
\[
 |\langle g, D_S^{1/2}b_S'D_S^{1/2}\mathcal K_hf\rangle|
 \le \|b_S'\|\,\|D_S^{1/2}g\|\,\|D_S^{1/2}\mathcal K_hf\| .
\]
We have $\|D_S^{1/2}g\| \le 2\|D_L^{1/2}g\|$, since $D_S \le D_{\rm body, L}$
and by \eqref{e:stab-product-defects}. The multiplier $D_S^{1/2}$ commutes
with $\mathbf M^\chi$, and $\mathbf H_hD_S\mathbf H_h \le D_{\rm body, R}$. Therefore
\[
 \|D_S^{1/2}(\mathbf M^\chi \otimes \mathbf H_h)f\|
 \le \langle f, \mathbf H_hD_S\mathbf H_hf\rangle^{1/2}
 \le \|D_{\rm body, R}^{1/2}f\|
 \le 2\|D_R^{1/2}f\| ,
\]
and \eqref{e:stab-right} bounds the remaining part of
$\|D_S^{1/2}\mathcal K_hf\|$. Combining these bounds with
\eqref{e:stab-src-factor}, we obtain
\[
 |\langle g, \mathbf E_{\rm src}f\rangle|
 \le C(r + \omega^2 + h)\|D_L^{1/2}g\|\bigl(\|D_R^{1/2}f\| + C(r, \omega)\sqrt{\delta_W}\|f\|\bigr)
 + C\delta_W\|g\|\|f\| ,
\]
which is the four-term bound with the stated parameters.
\end{proof}

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

We first treat the weights. Let $q \in \R$, and set
\[
 \Delta_v(x, y) = \log v(x) - \log v(y) ,
\]
which is exactly antisymmetric under the reversal and satisfies
$|\Delta_v(x, y)| \le Cd(x, y)$, since $\log v$ is Lipschitz with bounded
second derivatives. The kernel of $A_{h, q} = v^qA_hv^{-q}$ is
$e^{q\Delta_v(x, y)}$ times the kernel of $A_h$, and we write
\begin{equation}\label{e:stab-weight-split}
 A_{h, q} - C_h = v^q(A_h - C_h)v^{-q} + \bigl(v^qC_hv^{-q} - C_h\bigr) .
\end{equation}
The second term comes from the conjugation of the reference:
\[
 v^qC_hv^{-q} - C_h = \mathbf M^\chi \otimes \mathbf S_h\bigl(v^q\mathbf H_hv^{-q} - \mathbf H_h\bigr) ,
\]
with the kernel $\mathbf S_h(x)\mathbf M^\chi(e^{q\Delta_v(x, y)} - 1)p_h^\nabla(x, y)$.
We expand
\[
 e^{q\Delta_v} - 1 = q\Delta_v + O\bigl(q^2d(x, y)^2e^{C|q|d(x, y)}\bigr) ,
\]
and we write $p_h^\nabla = p_h\tau + (p_h^\nabla - p_h\tau)$. The term
$q\Delta_vp_h\tau$ is antisymmetric under the reversal and of size
$C|q|\sqrt{\delta_W}$ times a widened Gaussian. By
\eqref{e:stab-dirichlet-one}, its part without its marginal gives a one-sided
term with coefficient $C|q|$, and its marginal is $O(\delta_W)$ by
\zcref{c:stab-parity}. The remainder is $O(q^2\delta_W)$ by Schur's test,
and the part with $p_h^\nabla - p_h\tau$ is bounded as $\mathbf E_{\rm hol}$,
with the factor $\E_x[e^{2q\Delta_v(x, X_{\delta_W})}] \le C(q)$. For the
first term of \eqref{e:stab-weight-split}, we conjugate each of the terms
$\mathbf E_{\rm src}$, $\mathbf E_{\rm nor}$, $\mathbf E_0$, $\mathbf E_{\rm hol}$,
$\mathbf E_1$, $\mathbf E_{\ge 2}$. The factor $e^{q\Delta_v}$ is absorbed by
widening the Gaussians, and its first variation $q\Delta_v$, which carries
$h^2$ and is antisymmetric, produces one-sided terms through
\eqref{e:stab-dirichlet-one} and marginals of order $\delta_W$ through
\zcref{c:stab-parity}. In $\mathbf E_{\rm nor}$, the operator
$v^qC_{\rm body}v^{-q}$, with $C_{\rm body} = \mathbf S_h\mathbf H_h$ as in the proof of
\zcref{l:stab-nor}, has norm at most $1 + C(q)\delta_W$ by Schur's test, since
$e^{q\Delta_v} > 0$ and its marginal is $1 + O(\delta_W)$ by the same parity
argument. The argument for $\mathbf E_{\rm nor}$ then applies with this
constant.

\begin{proof}[Proof of \zcref{t:stab-main}]
It suffices, by the reduction after the theorem, to prove the four-term
bound for $A_{h, q}$ on $\mathcal H^{\rm tube}_0$ relative to $C_h$. The
reference $C_h$ is a contraction for every $\Gamma \in \mathcal U_2$, by
\eqref{e:stab-kick}, as \zcref{c:tools-placements} requires. We first
collect the bounds for the terms of $A_h - C_h$. Using
\zcref{l:stab-uncharged}, for $h$ small enough depending on $r$ and
$\omega$, and \eqref{e:stab-product-defects}, we see that
the term $\mathbf E_0$, which contains the Gaussian-small remainder from
$d(x, y) \ge \varrho_0$, satisfies the four-term bound with parameters
$C(r+\omega^2+h)$ and $C(r,\omega)$. The operators $\mathbf E_{\rm hol}$,
$\mathbf E_1$, $\mathbf E_{\ge 2}$, $\mathbf E_{\rm nor}$, and $\mathbf E_{\rm src}$
satisfy the four-term bound with the parameters stated in
\zcref{l:stab-hol}, \zcref{l:stab-degree-one}, \zcref{l:stab-higher},
\zcref{l:stab-nor}, and \zcref{l:stab-src}, for $h$ small enough,
depending on $r$ and $\omega$, if $C_1(r + \omega^2) \le c/8$. For
$q \ne 0$, the paragraph above adds the weight terms of
\eqref{e:stab-weight-split}, with constants that depend also on $q$. All
of these bounds hold uniformly for $\Gamma \in \mathcal U_2$, since they use
the damping only through \eqref{e:stab-kick} and \eqref{e:stab-scalar}.
Adding them, we find that $A_{h, q}$ satisfies \eqref{e:stab-four-term}
relative to $C_h$ with
\[
 a = C_k(r + \omega^2) + C_k(r, \omega)h , \qquad A_1 = C_k(r, \omega, q) .
\]

We choose $\omega$ so small that $C_k\omega^2 \le 1/64$ and
$C_1\omega^2 \le c/16$, and then $r$ so small that $C_kr \le 1/64$ and
$C_1r \le c/16$. We let $h_0$ be so small that
$C_k(r, \omega)h_0 \le 1/32$, that $Bh_0^4 \le 1$ for the constant $B$
recalled after the theorem, and that the lemmas above apply for
$h \le h_0$. Then $a \le 1/16$ and $B\delta_W \le 1$ for $h \le h_0$. By
\zcref{r:tube-four-term-reduction}, applied on
$\mathcal H^{\rm tube}_0$ with $\mathbf T = A_{h, q}$, $C = C_h$,
$\delta_0 = \delta_W$, and $t = T$, we obtain the two operator
inequalities of the theorem for $A_{h, q}$, with the defects of $C_h$,
and the power bound $\|A_{h, q}^n\| \le e^{BT/4}$ on
$\mathcal H^{\rm tube}_0$ for $n\delta_W \le T$. Since multiplication by
$v^q$ is unitary, these statements are equivalent to those on
$\mathcal H^{\rm tube}_q$. The constant $B$
depends on $k$, $q$, $r$, and $\omega$, which are fixed before $h \to 0$.
The constant $c$ of \eqref{e:stab-kick}, which enters the choices of
$\omega$ and $r$, can be taken to be $s_Ec_\eps\eta/\delta_W$, by the
derivation of \eqref{e:stab-kick}, where
$\eta/\delta_W = \eps/((4 - E^2)X_N(E)\rho(E))$ and $c_\eps$ is the
constant of \eqref{e:end-damping} on $\mathcal U_2$. It therefore depends on
the compact sets of energies, of $\eps$ and of $X$, and on
$\mathcal U_2$, as stated.

The two conditions on $\omega$ above hold for every
$\omega' \in (0, \omega]$ as well, and the conditions on $r$ do not involve
$\omega$. The argument above with $\omega'$ in place of $\omega$, and with
$h_0$ chosen depending on $\omega'$, therefore proves the theorem for
$\omega'$ with the same $r$.

In a signature $p \ne k$, the reference contains the full factor
$\lambda=\zeta_p(\mu_+\mu_-)^{(p-k)^2/2}$ through $\mathbf M^\chi$,
as defined in \zcref{l:tube-quadratic}. The bounds above are unchanged, and the defect
of the reference contains $(1 - |\lambda|^2)\Id$, which by
\eqref{e:tube-unbalanced-defect} is bounded below by a positive multiple
of $h^2\Id$. A smaller damping changes only
$\mathbf S_h$ and $c_\Gamma$. It preserves \eqref{e:stab-scalar}, and it
preserves \eqref{e:stab-kick}, possibly with $c/2$ in place of $c$. Since
we chose $C_1(r + \omega^2) \le c/8$, the hypothesis of
\zcref{l:stab-src} holds with $c/2$ in place of $c$, and the proof applies
without change.
\end{proof}

\section{Gain of body weights}\label{s:stab-weight-gain}

We now deduce from \zcref{t:stab-main} that the adapted step improves polynomial body weights; \zcref{s:end,s:assembly} use this. Recall from \zcref{s:stab-main} that the weighted space $\mathcal H^{\rm tube}_p$ carries the weight $v^{2p}$, where $v$ is the weight \eqref{e:sigma-weight}, so that
\begin{equation}\label{e:asm-weights}
 \|v^{1/2} f\|_{\mathcal H^{\rm tube}_p} = \|f\|_{\mathcal H^{\rm tube}_{p + 1/2}} .
\end{equation}

The continuum semigroup improves polynomial body weights by \zcref{p:sigma-regular}(b), because its potential contains a positive BB damping term, which grows like the weight $v$ of \eqref{e:sigma-weight}. The following lemma is the discrete analogue for the adapted chain. Its proof is an energy estimate followed by the pigeonhole principle. The idea is to set aside part of the BB damping of the source factor, and the defect sums of \zcref{l:tools-defect} then show that this part dissipates the weight $\delta_W v$ on average over the steps.

\begin{lemma}\label{l:asm-weight-gain}
Let $p \in \R$, let $m \ge 1$ be an integer and let $T > 0$. There exists $C = C_{p, m, T}$ such that for all integers $1 \le n \le T/\delta_W$,
\begin{equation}\label{e:asm-weight-gain}
 \|A_h^n\|_{\mathcal H^{\rm tube}_p \to \mathcal H^{\rm tube}_{p + m/2}} \le C (n \delta_W)^{-m/2} .
\end{equation}
The constant may depend on the tube.
\end{lemma}

\begin{proof}
We begin by setting aside part of the BB damping. Let $c_1$ denote the constant $c$ of \eqref{e:stab-kick}, so that the frozen kick $\mathbf S_h$ of $A_h$ has modulus at most $e^{-c_1 \delta_W v}$. Recall that $\mathbf S_h$ is the frozen kick of $c_\Gamma^{-1} s_\Gamma$, with the constant $c_\Gamma$ of \eqref{e:stab-scalar}. Let $0 < c \le c_1/2$, and set
\[
 \mathbf S_\flat = e^{-c \delta_W v/2} ,
 \qquad
 A_h^\flat = \mathbf S_\flat^{-2} A_h ,
 \qquad
 A_h^\sharp = \mathbf S_\flat A_h^\flat \mathbf S_\flat .
\]
Then $A_h^\flat$ differs from $A_h$ only through its frozen source kick. Indeed, $\mathbf S_\flat$ depends only on the tangent variables, so that the normal part of the source factor of $A_h^\flat$ is that of $A_h$, and its frozen kick is $\mathbf S_\flat^{-2} \mathbf S_h$, where $\mathbf S_h$ is the frozen kick of $A_h$ in \eqref{e:stab-kick}. Since $c \le c_1/2$, the kick $\mathbf S_\flat^{-2} \mathbf S_h$ has modulus at most $e^{-(c_1 - c) \delta_W v} \le e^{-c_1 \delta_W v/2}$, and it satisfies \eqref{e:stab-kick} with $c_1/2$ in place of $c_1$. Moreover, $A_h^n = \mathbf S_\flat (A_h^\sharp)^{n - 1} \mathbf S_\flat A_h^\flat$ for all $n \ge 1$.

We apply the stability theorem \zcref{t:stab-main} in $\mathcal H^{\rm tube}_p$ to the step $A_h^\flat$. This is allowed by the last paragraph of its proof, since $A_h^\flat$ differs from $A_h$ only through its frozen kick, which satisfies \eqref{e:stab-kick} with half the constant. By the theorem, $A_h^\flat$ satisfies the four-term bound \eqref{e:stab-four-term} relative to a contraction
\[
 C_\flat = v^{-p} C_{h, \flat} v^p
\]
of $\mathcal H^{\rm tube}_p$, where $C_{h, \flat}$ is the reference of $A_h^\flat$ on $\mathcal H^{\rm tube}_0$, with $a \le 1/16$. Since multiplication by $v^p$ is unitary from $\mathcal H^{\rm tube}_p$ onto $\mathcal H^{\rm tube}_0$, the operator $C_\flat$ is a contraction of $\mathcal H^{\rm tube}_p$. The first part of the proof of \zcref{c:tools-placements}, applied with $A_0$ equal to the constant $A_1$ of \eqref{e:stab-four-term}, shows that the four-term bound implies \eqref{e:tools-form-bound} relative to $C_\flat$, with $\delta_0 = \delta_W$, $\theta = 1/8$, and $A = A_\flat = \max\{1, 2304 A_1^2\}$, since $A_\flat \delta_W \le 1$ for $h \le h_0$. Let $D_{R, \flat}$ and $D_{L, \flat}$ denote the defects of $C_\flat$ in $\mathcal H^{\rm tube}_p$.

We first show that for all $f \in \mathcal H^{\rm tube}_p$ and $n \le T/\delta_W$,
\begin{equation}\label{e:asm-killing-sum}
 \sum_{j = 0}^{n - 1} \|(\Id - \mathbf S_\flat^2)^{1/2} (A_h^\sharp)^j f\|_{\mathcal H^{\rm tube}_p}^2 \le C_{p, T} \|f\|_{\mathcal H^{\rm tube}_p}^2 .
\end{equation}
We use $C' = \mathbf S_\flat C_\flat \mathbf S_\flat$, which is a contraction since $0 < \mathbf S_\flat \le 1$, as the reference for $A_h^\sharp$. Its defects satisfy
\begin{equation}\label{e:asm-sandwich}
\begin{aligned}
 D_R' = \Id - C'^* C'
 &= (\Id - \mathbf S_\flat^2) + \mathbf S_\flat D_{R, \flat} \mathbf S_\flat + \mathbf S_\flat C_\flat^* (\Id - \mathbf S_\flat^2) C_\flat \mathbf S_\flat 
 \ge (\Id - \mathbf S_\flat^2) + \mathbf S_\flat D_{R, \flat} \mathbf S_\flat, \\
 D_L' = \Id - C' C'^*
 &= (\Id - \mathbf S_\flat^2) + \mathbf S_\flat D_{L, \flat} \mathbf S_\flat + \mathbf S_\flat C_\flat (\Id - \mathbf S_\flat^2) C_\flat^* \mathbf S_\flat 
 \ge (\Id - \mathbf S_\flat^2) + \mathbf S_\flat D_{L, \flat} \mathbf S_\flat .
\end{aligned}
\end{equation}


Observe that $A_h^\sharp - C' = \mathbf S_\flat (A_h^\flat - C_\flat) \mathbf S_\flat$. Since $\mathbf S_\flat$ is self-adjoint, the form bound for $A_h^\flat$ implies
\[
 |\langle g, (A_h^\sharp - C') f \rangle|
 \le \tfrac18 \, \|(D_{L, \flat} + A_\flat \delta_W)^{1/2} \mathbf S_\flat g\| \, \|(D_{R, \flat} + A_\flat \delta_W)^{1/2} \mathbf S_\flat f\| .
\]
Using \eqref{e:asm-sandwich} and $\mathbf S_\flat^2 \le \Id$, we find
\[
 \|(D_{R, \flat} + A_\flat \delta_W)^{1/2} \mathbf S_\flat f\|^2
 = \langle f, (\mathbf S_\flat D_{R, \flat} \mathbf S_\flat + A_\flat \delta_W \mathbf S_\flat^2) f \rangle
 \le \|(D_R' + A_\flat \delta_W)^{1/2} f\|^2 ,
\]
and similarly on the left. Therefore $A_h^\sharp$ satisfies \eqref{e:tools-form-bound} relative to $C'$, with the same constants. Since $D_R' \ge \Id - \mathbf S_\flat^2$, the defect sums \eqref{e:tools-defect-sums} of \zcref{l:tools-defect} imply \eqref{e:asm-killing-sum}.

We now convert \eqref{e:asm-killing-sum} into a gain of weight. By the elementary inequality $x e^{-c x} \le c^{-1} (1 - e^{-c x})$ for $x \ge 0$, we have $\delta_W v \mathbf S_\flat^2 \le c^{-1} (\Id - \mathbf S_\flat^2)$ pointwise. Using $A_h^{j + 1} = \mathbf S_\flat (A_h^\sharp)^j \mathbf S_\flat A_h^\flat$, \eqref{e:asm-weights}, and the fact that $\mathbf S_\flat$ commutes with $v$, we obtain
\[
 \sum_{j = 1}^{n} \delta_W \|A_h^j f\|_{\mathcal H^{\rm tube}_{p + 1/2}}^2
 \le c^{-1} \sum_{j = 0}^{n - 1} \|(\Id - \mathbf S_\flat^2)^{1/2} (A_h^\sharp)^j \mathbf S_\flat A_h^\flat f\|_{\mathcal H^{\rm tube}_p}^2
 \le C_{p, T}' \|f\|_{\mathcal H^{\rm tube}_p}^2 ,
\]
where in the last inequality we used \eqref{e:asm-killing-sum} and the boundedness of $\mathbf S_\flat A_h^\flat$ on $\mathcal H^{\rm tube}_p$.

The interval $[\lceil n/2 \rceil, n]$ contains at least $n/2$ integers. By the pigeonhole principle, there exists $j$ in this interval such that
\[
 \|A_h^j f\|_{\mathcal H^{\rm tube}_{p + 1/2}}^2 \le 2 C_{p, T}' (n \delta_W)^{-1} \|f\|_{\mathcal H^{\rm tube}_p}^2 .
\]
Since $(n - j) \delta_W \le T$, the power bound of \zcref{t:stab-main} in $\mathcal H^{\rm tube}_{p + 1/2}$ bounds $A_h^{n - j}$ on that space, and we conclude that
\[
 \|A_h^n f\|_{\mathcal H^{\rm tube}_{p + 1/2}} \le C (n \delta_W)^{-1/2} \|f\|_{\mathcal H^{\rm tube}_p} .
\]
This is \eqref{e:asm-weight-gain} for $m = 1$.

Let $m \ge 2$, and suppose first that $n \ge m$. We split the $n$ steps into $m$ consecutive blocks of lengths at least $n/(2m)$, and we apply the case $m = 1$ on the $i$-th block, from $\mathcal H^{\rm tube}_{p + (i - 1)/2}$ to $\mathcal H^{\rm tube}_{p + i/2}$. The product of the $m$ bounds is bounded by $C (n \delta_W)^{-m/2}$.

Now suppose that $1 \le n < m$. We write $A_h = \mathbf S_\flat^2 A_h^\flat$, where $\mathbf S_\flat^2 = e^{-c \delta_W v}$ and $A_h^\flat$ is bounded on $\mathcal H^{\rm tube}_p$ by \zcref{t:stab-main}. Then
\[
\begin{aligned}
 \|A_h^n f\|_{\mathcal H^{\rm tube}_{p + m/2}}
 &= \|v^{m/2} \mathbf S_\flat^2 A_h^\flat A_h^{n - 1} f\|_{\mathcal H^{\rm tube}_p} \\
 &\le \Bigl( \sup_{x > 0} x^{m/2} e^{-c \delta_W x} \Bigr) \|A_h^\flat A_h^{n - 1} f\|_{\mathcal H^{\rm tube}_p}
 \le C_m \delta_W^{-m/2} \|f\|_{\mathcal H^{\rm tube}_p} ,
\end{aligned}
\]
and $\delta_W^{-m/2} \le m^{m/2} (n \delta_W)^{-m/2}$ since $n < m$.
\end{proof}

The last step of the proof applies to all $n \ge 1$ and gives
\begin{equation}\label{e:asm-weight-crude}
 \|A_h^n\|_{\mathcal H^{\rm tube}_p \to \mathcal H^{\rm tube}_{p + m/2}} \le C_{p, m, T} W^m ,
 \qquad 1 \le n \le T/\delta_W .
\end{equation}
In \zcref{s:assembly}, this bound is used for the first steps after the right block of the chain, where a cost polynomial in $W$ is acceptable.