)$; and that $C$ lives at the opposite ends of the $H_2$ flow segments, so $C^* C$ is the term we wish to control.


Unfortunately, the decay of $L$ at the front face is two orders worse than that of $C^* C$.  To overcome this issue, we present the following lemma, to be followed by the end of the proof of the theorem.

\begin{lemma}[Decomposition of $L$]
$L\in\Psi^{2m-1,2l+2}_{2,h}(\mathcal{C})$ in Equation \eqref{remainder 2} may be decomposed as $L = L_1 + L_2$, where $L_1\in\Psi^{2m-1,2l}_{2,h}(\mathcal{C})$ and $L_2\in\Re^{2l+2}$.
\end{lemma}

\begin{proof}
We make full use of the assumption that $\mathrm{sub}(P)\equiv 0$.  We expand the total symbol $p$ (of $P = {}^h \mathrm{Op_W}(p)$) in powers of $h$:
\[
p(\widetilde{x},\widetilde{\xi}) = p_0(\widetilde{\xi}) + O\left(h^2\right) = p_0 + p_1,
\]
where $p_1$ signifies everything but the principal part of $P$.

Next, we again make use of the Weyl formula, in conjunction with equation \eqref{remainder 2}.  If we replace $p$ by $p_0$ in the asymptotic sum, and replace $a$ by $\widetilde{a}_1$, the term $\alpha=\beta=0$ directly cancels, and the term $|\alpha+\beta| = 1$ is quantized to give ${}^h \mathrm{Op_W}(\{p_0,\widetilde{a}_1\})\in\Psi^{2m,2l}_{2,h}(\mathcal{C})$.


This differs from $\int^\delta_0 B_s^* B_s \ ds + C^* C - D^* D$ by a remainder $\tilde{L}\in\Psi^{2m-1,2l}_{2,h}(\mathcal{C})$, since ${}^h \mathrm{Op_W}(\{p_0,\widetilde{a}_1\})$ and $\int^\delta_0 B_s^* B_s \ ds + C^* C - D^* D$ both have symbol $\mathbf{H}(\widetilde{a}_1) = \rho_\mathrm{ff} \widetilde{H}_2(\widetilde{a}_1)$.


Therefore, $L\in\Psi^{2m-1,2l+2}_{2,h}(\mathcal{C})$ is generated by quantizing all terms involving $p_1$, plus all terms with $|\alpha+\beta| > 1$ with $p_0$ in place of $p$, plus $\tilde{L}$ plus the residual operator $L_2 := R\in\Re^{2l+2}$.

First, we carefully study the terms in the Weyl expansion arising from $p=p_0 + p_1$ from all ``slices'' $|\alpha+\beta| > 1$.  To do this, we introduce local coordinates in one coordinate patch of $S_\mathrm{tot}$: $\widetilde{x}$, $\widetilde{\xi}_1$, $H := h / {\widetilde{\xi}_1}$, $\Xi_2 := \widetilde{\xi}_2 / \widetilde{\xi}_1,\ldots,\Xi_d := \widetilde{\xi}_d / \widetilde{\xi}_1$, and $\widetilde{\xi}''$.  As usual, $H$ is a defining function for the side face and $\widetilde{\xi}_1$ for the front face.  At this point, we use our earlier observation that $p$ is smooth on $T^*\mathbb{T}^n \times [0,1)$ (in particular $p$ is smooth in $\widetilde{\xi}'$, unlike $a$), so for all multi-indices $\gamma$ we have $\partial^\gamma_{\widetilde{\xi}} p = O(1)$ and $\partial^\gamma_{\widetilde{x}} p = O(1)$.  ``Slices'' for which $|\alpha+\beta| > 1$ satisfy (1) $|\alpha|\geq 2$ or (2) $|\beta|\geq 1$, $|\alpha|\leq |\beta|$ (these cases are not exclusive).  Example \ref{ex:symbol class} gives
\[
h^{|\alpha+\beta| - 1}\left(\partial^\alpha_{\widetilde{x}} \partial^\beta_{\widetilde{\xi}} a\right)\left(\partial^\alpha_{\widetilde{\xi}} \partial^\beta_{\widetilde{x}} p\right)\in S^{2m + 1-|\alpha+\beta|,2l+2-|\alpha|-(\beta_{d+1}+\ldots+\beta_n)}(S_\mathrm{tot}).
\]
If $\gamma\neq 0$, we can improve on $\partial^\gamma_{\widetilde{x}} p = O(1)$.  We have:
\begin{align}
p(\widetilde{x},\widetilde{\xi}) &= p_0(\widetilde{\xi}) + O\left(h^2\right) \nonumber \\
&= {p_0}\restriction_\mathcal{C}(\widetilde{\xi}'') + \rho_\mathrm{ff} \left[\frac{\partial p_0}{\partial\widetilde{\xi}'}\right]_\mathcal{C}(\widetilde{\xi}'')\cdot\Gamma' + \frac{\rho_\mathrm{ff}^2}{2} \left[\frac{\partial^2 p_0}{{\partial\widetilde{\xi}'}^2}\right]_\mathcal{C}(\widetilde{\xi}'')\cdot (\Gamma')^2 + O\left(\rho_\mathrm{ff}^3\right) + O\left(\rho_\mathrm{sf}^2 \rho_\mathrm{ff}^2\right), \label{eqn:Taylor expansion}
\end{align}
where $SN^+(\mathcal{C}\times\{0\}) = \{\rho_\mathrm{ff} = 0\}$.


The $O\left(\rho_\mathrm{ff}^3\right)$ term is the remainder in the Taylor expansion of the principal symbol, so it is independent of $\widetilde{x}$.  Then we differentiate \eqref{eqn:Taylor expansion} to get $\partial^\gamma_{\widetilde{x}} p = O\left(\rho^2_\mathrm{sf} \rho^2_\mathrm{ff}\right) = O\left(H^2 \widetilde{\xi}_1^2\right)$.  This implies, in case (2), that
\[
h^{|\alpha+\beta| - 1}\left(\partial^\alpha_{\widetilde{x}} \partial^\beta_{\widetilde{\xi}} a\right)\left(\partial^\alpha_{\widetilde{\xi}} \partial^\beta_{\widetilde{x}} p\right)\in S^{2m-1-|\alpha+\beta|,2l-|\alpha|-(\beta_{d+1}+\ldots+\beta_n)}(S_\mathrm{tot}).
\]
We are thus able to conclude that all terms of the form $h^{|\alpha+\beta| - 1}\left(\partial^\alpha_{\widetilde{x}} \partial^\beta_{\widetilde{\xi}} a\right)\left(\partial^\alpha_{\widetilde{\xi}} \partial^\beta_{\widetilde{x}} p\right)$ for which $|\alpha+\beta| > 1$ are $O\left(\rho_\mathrm{sf}^{-2m+1} \rho_\mathrm{ff}^{-2l}\right)$.

Finally, we consider the terms arising when $p = p_1 = O\left(\rho_\mathrm{sf}^2 \rho_\mathrm{ff}^2\right)$ and $|\alpha+\beta|\leq 1$.  If $\alpha=\beta=0$, it is easily seen that $h^{-1} a p_1\in S^{2m-1,2l}(S_\mathrm{tot})$.  Next suppose $\alpha=0$, $|\beta|=1$.  We assume the worst: one of the first $d$ components of $\beta$ is equal to one.  Then, since $\partial^\beta_{\widetilde{x}} p_1 = O\left(\rho_\mathrm{sf}^2 \rho_\mathrm{ff}^2\right)$ and $h^{-1}(h \partial_{\widetilde{\xi}})^\beta a\in S^{2m,2l+2}(S_\mathrm{tot})$, we have
\[
h^{-1} \left(\partial^\beta_{\widetilde{x}} p_1\right) \left(h\partial_{\widetilde{\xi}}\right)^\beta a\in S^{2m-2,2l}(S_\mathrm{tot}).
\]
The remaining case is $|\alpha|=1$, $\beta=0$.  We again assume the worst: $\alpha_1 = 1$.  Since $\partial_{\widetilde{\xi}_1} p_1 = O\left(H^2 \widetilde{\xi}_1\right)$ and $\partial_{\widetilde{x}_1} a\in S^{2m,2l+1}(S_\mathrm{tot})$, we have
\[
(\partial_{\widetilde{x}_1}