If $\e$ is small enough and $\la\leq\e$, then there exists a unique solution \[\pth{F^{(1)}_\A,F^{(2,1)}_\A,F^{(2,2)}_\A,\F_\A,\h_\la}\] to the system \eqref{eq F1}-\eqref{eq h} on $[0,1]\times\R^3$ with the initial data of Proposition \ref{prop ID main}. Moreover, the tensors $F^{(1)}_\A$, $F^{(2,1)}_\A$, $F^{(2,2)}_\A$ and $\F_\A$ are supported in $J^+_0(B_R)$ and there exists $C=C(N,\de,R)>0$ such that
\begin{align}
\l F^{(1)}_\A \r_{H^N} + \l F^{(2,1)}_\A \r_{H^{N-2}} + \l F^{(2,2)}_\A \r_{H^{N-2}} & \leq C \e, \label{estim F1 F21 F22}
\\ \l \F_\A \r_{L^2} + \max_{r\in\llbracket 1,6\rrbracket}\lambda^{r-1} \l \nabla^r \F_\A \r_{L^2}+ \max_{r\in\llbracket 0,5\rrbracket}\lambda^{r} \l \dr_t \nabla^r \F_\A \r_{L^2} + \max_{r\in\llbracket 0,4\rrbracket}\lambda^{r+1} \l \dr_t^2 \nabla^r \F_\A \r_{L^2} & \leq C\e, \label{estim F}
\\ \max_{r\in\llbracket 0,4\rrbracket} \lambda^r \l \nabla^r \tBox_{\g_\la} \F_\A \r_{L^2} & \leq C \e,\label{estim box F}
\\ \max_{r\in\llbracket 0,4\rrbracket}\lambda^r\pth{ \l \dr_t \nabla^r \h_\la \r_{L^2_{\delta+1+r}} + \l \nabla \nabla^r \h_\la \r_{L^2_{\delta+1+r}} } + \max_{r\in\llbracket 0,3\rrbracket} \lambda^{r+1} \l \dr_t^2 \nabla^r \h_\la \r_{L^2_{\delta+2+r}} & \leq C \e.  \label{estim h}
\end{align}
\end{prop}

The proof of Proposition \ref{prop system} follows exactly the same lines as the corresponding statements in the singlephase case of [28], namely Theorems 6.1 and 7.1 there. Indeed, the reasons why well-posedness of the system \eqref{eq F1}-\eqref{eq h} does not follow from simple arguments are not linked with the directions of oscillations and thus don't differ from the singlephase to the the multiphase case. Therefore, we won't reproduce the argument and refer the reader to Sections 6 and 7 of [28]. We will however provide an outline of what is done there. As in the singlephase case, proving well-posedness is done in two steps: we first consider the equations \eqref{eq F1}-\eqref{eq F22} and then the equations \eqref{eq F}-\eqref{eq h}.

\paragraph{Solving for $F^{(1)}_\A$, $F^{(2,1)}_\A$ and $F^{(2,2)}_\A$.} The equations \eqref{eq F1}-\eqref{eq F22} form a triangular system for the unknowns $F^(1)_$, $F^(2,1)_$ and $F^(2,2)_$. Thanks to \eqref{Q 0} and \eqref{Rtilde 1 1}-\eqref{Rtilde 1 2} the equations \eqref{eq F1}-\eqref{eq F22} rewrite schematically
\begin{align*}
\Ll_\A F^{(1)}_\A & = 0,
\\ \Ll_\A F^{(2,1)}_\A & = \cro{  \dr^{\leq 2} F^{(1)} + (F^{(1)})^2 + (F^{(1)})^3 },
\\ \Ll_\A F^{(2,2)}_\A & = \cro{ F^{(1)}\dr F^{(1)} + F^{(2,1)}F^{(1)} +  (F^{(1)})^2  }.
\end{align*}
Since the transport operator $_$ only depends on the background spacetime, a combination of the characteristics method and the energy estimate for the vector field $L_$ easily implies the existence and uniqueness of solutions $F^(1)_$, $F^(2,1)_$ and $F^(2,2)_$ to these equations on $[0,1]×^3$. The estimate \eqref{estim F1 F21 F22} follows from the following consideration: it holds on $_0$ (see Proposition \ref{prop ID main}) and is easily propagated for $F^(1)_$, for $F^(2,1)_$ we lose two derivatives because of the $^≤2F^(1)$ on the RHS, and because of the $F^(2,1)$ on the RHS of the equation for $F^(2,2)_$ it lives at the same level of regularity. The only harmless difference with the singlephase case is that there is as many equations as number of directions of oscillation, and since we don't keep track on how they mix in the equations (meaning for instance that one could see a $F^(1)_$ in the equation for $F^(2,1)_$ with $≠$) the estimate \eqref{estim F1 F21 F22} is far from being uniform in $|𝒜|$.

\paragraph{Solving for $\F_\A$ and $\h_\la$.} Once $F^(1)_$, $F^(2,1)_$ and $F^(2,2)_$ are defined, the equations \eqref{eq F}-\eqref{eq h} form an independent coupled system for $_$ and $_$. Besides the fact that there is a transport equation for each $∈𝒜$, the only difference between the multiphase and singlephase case is the presence of $__≫^(3,e)_v,T$ on the RHS of \eqref{eq h}. Indeed this term is not present in the singlephase case since its purpose is precisely to deal with the mixed harmonics in $R^(1)_$. However, this new term dos not change the structure of the system. Indeed, thanks to \eqref{def g3e} and \eqref{R 1 e}, $≫^(3,e)_v,T$ contains terms depending on $F^(1)_$, $F^(2,1)_$ and $F^(2,2)_$ and a linear term in $_$, the latter being the only one of interest at this stage. Since the term $__≫^(3,e)_v,T$ comes with an extra $$ power in \eqref{eq h} this new term is at least better than the $___$ in \eqref{eq h} and already present in the singlephase case. In conclusion, one can say that the argument (which we will briefly outline below) of Section 7 of [28] applies also here without any change.

\saut
The major issue with the coupled system \eqref{eq F}-\eqref{eq h} is the \textit{a priori} loss of one derivative, which prevents a straightforward proof of well-posedness. Indeed, from an energy estimate on \eqref{eq F} one concludes that $_$ lives at the same level of regularity as $_$, and from an energy estimate on \eqref{eq h} one concludes that one derivative of $_$ lives at the same level of regularity as two derivatives of $_$ (because of $___$ on the RHS). The system \eqref{eq F}-\eqref{eq h} is thus \textit{a priori} ill-posed. The operators $Π_≤$ and $Π_≥$ have been introduced precisely to solve this problem. Indeed, applying the Bernstein estimate
\begin{align}\label{bernstein}
\l \nab \Pi_\leq (u) \r_{L^2} &\lesssim \frac{1}{\la} \l u \r_{L^2}
\end{align}
to \eqref{eq F} we can estimate two derivatives of $_$ by one derivative of $_$ at the cost of the loss of one $$ power. This would however prove well-posedness of \eqref{eq F}-\eqref{eq h} only on a time scale of order $$, which is far from satisfactory if one is interested in the limit $→0$. Therefore, one absolutely needs to use the structure of the term $___$ on the RHS of \eqref{eq h}. This is done by decomposing it as follows
\begin{align}\label{box F}
\tBox_{\g_\la}\F_\A & = \tBox_{\g_0}\F_\A + \pth{\g_\la^{\mu\nu}-\g_0^{\mu\nu}}\dr_\mu \dr_\nu \F_\A.
\end{align}
We explain how the two terms in \eqref{box F} can be estimated:
\begin{itemize}
\item  The first term in \eqref{box F} can be estimated by commuting \eqref{eq F} and using the estimate (proved in Appendix B of [28])
\begin{align*}
\l [L_\A,\Box_{\g_0}]f \r_{L^2} \leq C(C_0)\pth{ \l \dr L_\A f \r_{L^2} + \l \Box_{\g_0}f \r_{L^2} + \l \dr f \r_{L^2} },
\end{align*}
where it is crucial that the only second order derivatives appearing on the RHS include a $L_\A$ or are directly a wave operator. This implies that $\Box_{\g_0}\F_\A$ can be estimated roughly by $\dr \h_\la$ and $\Box_{\g_0}\Pi_\leq(\h)$. The former is itself controlled by $\Box_{\g_\la}\F_\A$ thanks to \eqref{eq h} and the latter requires to commute $\Box_{\g_0}$ and $\Pi_\leq$. In Appendix C of [28] we prove the following estimate
\begin{align*}
\l \left[ u , \Pi_\leq  \right]\nabla v \r_{L^2} & \lesssim  \left( \l \nabla  u \r_{L^\infty}  +   \lambda \l u \r_{H^{\frac{7}{2}}}  \right)\l v \r_{L^2}.
\end{align*}
This estimates shows in particular how one gain a derivative at top order.
\item For the second term in \eqref{box F}, instead of relying on the structure of the derivatives we benefit from the fact that $\g_\la^{\mu\nu}-\g_0^{\mu\nu}=\GO{\la}$. This extra factor $\la$ makes now possible the use of the Bernstein estimate \eqref{bernstein}.
\end{itemize}
Integrating this procedure in a bootstrap argument with \eqref{estim F}-\eqref{estim h} as bootstrap assumptions allows us to prove well-posedness for the system \eqref{eq F}-\eqref{eq h} and obtain a unique solution $_$ and $_$ satisfying \eqref{estim F}-\eqref{estim h} on the desired time scale. This concludes our outline of the proof of Proposition \ref{prop system}, and we again refer to [28] for all the technical details.

\section{Propagation of polarization and gauge conditions}\label{section propagation}

In this section, we conclude the proof of Theorem \ref{theo evol} by showing that the metric $_$ on $[0,1]×^3$ defined by \eqref{expansion metric} where $F^(1)_,F^(2,1)_,F^(2,2)_,_,_$ are constructed in Section \ref{section solving the system} and where $F^(2,±)_$, $≫^(3,e)_v,T$ and $≫^(3,h)_u,T$ are respectively defined in \eqref{def F2pm}, \eqref{def g3e} and Remark \ref{remark g3h}, is a solution of the Einstein vacuum equations \eqref{EVE}. As first stated in Section \ref{section reformulation}, this will be done by showing that the algebraic conditions \eqref{pola F1}-\eqref{GWC} holds on $[0,1]×^3$ (\eqref{pola g3h} already holds thanks to the definition of $≫^(3,h)_u,T$, see Remark \ref{remark g3h}), relying obviously on the fact that they all hold on $_0$ thanks to Proposition \ref{prop ID main}. These conditions are separated in two groups:
\begin{itemize}
\item First, the conditions on $F^{(1)}_\A$ and $\F_\A$, i.e \eqref{pola F1}-\eqref{pola F}, \eqref{energy F1}, which are all propagated by directly using the transport equations that $F^{(1)}_\A$ and $\F_\A$ satisfy on $[0,1]\times\R^3$, i.e \eqref{eq F1} and \eqref{eq F}. This is done in Section \ref{section propa 1}.
\item Second, the conditions on $F^{(2,1)}_\A$ and $F^{(2,2)}_\A$, i.e \eqref{pola F21}-\eqref{pola F22}, together with the generalised wave gauge condition \eqref{GWC}, which are propagated using the contracted Bianchi identities. This is done in Section \ref{section propa 2}.
\end{itemize}

\begin{remark}
As explained in Remark \ref{remark Q0}, the conditions \eqref{pola F21}-\eqref{pola F22} are not gauge conditions since they don't depend on coordinates. However we treat them as gauge conditions and propagate them with the contracted Bianchi identities.
\end{remark}

\subsection{Propagation via the system} \label{section propa 1}

The proof of the following lemma is identical to the corresponding ones in [28], we only sketch its proof and refer the reader to Section 8.1 in [28] for more details.

\begin{lemma}\label{lem pola F1 F}
The polarization conditions \eqref{pola F1}-\eqref{pola F} and the backreaction condition \eqref{energy F1} hold on $[0,1]\times \R^3$.
\end{lemma}

\begin{proof}
By contracting the transport equation \eqref{eq F1} with various vector fields we obtain the transport system
\begin{equation}\label{system pola F1}
\begin{aligned}
\Ll_\A \pth{ \Pol{F^{(1)}_\A}{u_\A}_{L_\A}} & = 0,
\\ \Ll_\A \pth{ \Pol{F^{(1)}_\A}{u_\A}_{e^{(\mathbf{k})}_\A}} & = - \g_0\pth{ \D_{L_\A}\Lb_\A,e^{(\mathbf{k})}_\A } \Pol{F^{(1)}_\A}{u_\A}_{L_\A}
\\&\quad - 2 \sum_{\i,\j=1,2} \de_{\i\j}\g_0\pth{ \D_{L_\A}e^{(\mathbf{k})}_\A,e^{(\i)}_\A} \Pol{F^{(1)}_\A}{u_\A}_{e^{(\j)}_\A},
\\ \Ll_\A \pth{ \Pol{F^{(1)}_\A}{u_\A}_{\Lb_\A}} & =  -2 \sum_{\i,\j=1,2} \de_{\i\j} \g_0\pth{\D_{L_\A}\Lb_\A,e^{(\i)}_\A} \Pol{F^{(1)}_\A}{u_\A}_{e^{(\j)}_\A},
\\ \Ll_\A \pth{ (F^{(1)}_\A)_{L_\A\Lb_\A} } & = 2 \sum_{\i,\j=1,2} \de_{\i\j} \g_0\pth{\D_{L_\A}\Lb_\A,e^{(\i)}_\A} \Pol{F^{(1)}_\A}{u_\A}_{e^{(\j)}_\A}.
\end{aligned}
\end{equation}
Since $\Pol{F^{(1)}_\A}{u_\A}_{|_{\Si_0}}=0$ and $(F^{(1)}_\A)_{L_\A\Lb_\A|_{\Si_0}}=0$, the system \eqref{system pola F1} implies that \eqref{pola F1}-\eqref{pola F1 bis} hold on $[0,1]\times \R^3$. Similarly, using now \eqref{pola F1}-\eqref{pola F1 bis} we deduce from \eqref{eq F1} the following transport equation
\begin{align*}
\pth{ - L_\A + \Box_{\g_0}u_\A} \left| F^{(1)}_\A \right|_{\g_0}^2 & = 0.
\end{align*}
Using now the background transport equation satisfied by $F_\A$ (recall \eqref{BG system}), we obtain
\begin{align}\label{transport energy}
\pth{ - L_\A + \Box_{\g_0}u_\A} \pth{ \left| F^{(1)}_\A \right|_{\g_0}^2 - 8 F_\A^2} & = 0.
\end{align}
Since we have
\begin{align*}
\pth{ \left| F^{(1)}_\A \right|_{\g_0}^2 - 8 F_\A^2}_{|_{\Si_0}} =0,
\end{align*}
the equation \eqref{transport energy} implies that \eqref{energy F1} holds on $[0,1]\times \R^3$. Finally, using again \eqref{pola F1} we deduce from \eqref{eq F} that $\Pol{\F_\A}{u_\A}$ satisfies the same system as $\Pol{F^{(1)}_\A}{u_\A}$, i.e \eqref{system pola F1} with $F^{(1)}_\A$ replaced by $\F_\A$. Since $\Pol{\F_\A}{u_\A}_{|_{\Si_0}}=0$, we again obtain that \eqref{pola F} holds on $[0,1]\times \R^3$, which concludes the proof of the lemma.
\end{proof}

\subsection{Propagation via the contracted Bianchi identities} \label{section propa 2}

We now propagate from $_0$ the conditions \eqref{pola F21}-\eqref{pola F22} and \eqref{GWC}. This is done by deriving extra equations for the quantities $V^(2,1)_$, $V^(2,2)_$ and $Υ^ρ$ from the contracted Bianchi identities, which state that the Einstein tensor of any Lorentzian metric is divergence free. We first compute the Einstein tensor of $_$.

\begin{lemma}\label{lem G}
The Einstein tensor of $\g_\la$ admits the decomposition
\begin{align*}
G_{\a\b}(\g_\la) & = G(V)_{\a\b} + G(\Upsilon)_{\a\b}.
\end{align*}
The term $G(V)_{\a\b}$ is given by
\begin{align}\label{expansion G(V)}
G(V)_{\a\b} & = G^{(0)}_{\a\b} + \la G^{(1)}_{\a\b} + \GO{\la^2},
\end{align}
\begin{align}
G^{(0)}_{\a\b} & = -\half \sum_\A \sin\pth{\frac{u_\A}{\la}} \pth{ \dr_{(\a} u_\A (V_\A^{(2,1)})_{\b)} + (V^{(2,1)}_\A)_{L_\A} (\g_0)_{\a\b} } \label{G 0}
\\&\quad -2 \sum_\A \cos\pth{\frac{2u_\A}{\la}} \pth{ \dr_{(\a} u_\A (V_\A^{(2,2)})_{\b)} + (V^{(2,2)}_\A)_{L_\A} (\g_0)_{\a\b} } ,\non
\\ G^{(1)}_{\a\b} & =  \frac{1}{2}\sum_\A \cos\pth{\frac{u_\A}{\lambda}} \pth{ \D_{(\a} (V^{(2,1)}_\A)_{\b)} - \div_{g_0} V^{(2,1)}_\A (\g_0)_{\a\b} } \label{G 1}
\\&\quad  - \sum_\A\sin\pth{\frac{2u_\A}{\lambda}} \pth{ \D_{(\a} (V^{(2,2)}_\A)_{\b)}  - \div_{g_0}V^{(2,2)}_\A (\g_0)_{\a\b}}\non
\\&\quad + \sum_{\substack{\A , k=1,2,3\\ T\in\{\cos,\sin\}}} T\pth{\frac{ku_\A}{\la}} \cro{V^{(2,1)} + V^{(2,2)}} (F^{(1)}_\A)_{\a\b} \non
\\&\quad + \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cro{  (V^{(2,1)} + V^{(2,2)}) F^{(1)} }_{\a\b}.\non
\end{align}
The term $G(\Upsilon)_{\a\b}$ is given by
\begin{align}\label{def G upsilon}
G(\Upsilon)_{\a\b} & = \frac{\la^2}{2} \bigg( (\g_\la)_{\rho(\a}\dr_{\b)}\Upsilon^\rho - (\g_\la)_{\a\b}\dr_\rho\Upsilon^\rho + \Upsilon^\rho \dr_\rho (\g_\la)_{\a\b} - \half \g_\la^{\mu\nu}\Upsilon^\rho \dr_\rho (\g_\la)_{\mu\nu}(\g_\la)_{\a\b} \bigg).
\end{align}
\end{lemma}

\begin{proof}
Thanks to Lemma \ref{lem pola F1 F}, the assumptions \eqref{assumption} are satisfied and we can use the results of Proposition \ref{prop expression of the ricci}. Thanks to Lemma \ref{lem pola F1 F} again, the non-oscillating term in \eqref{R 0} vanish and thanks to \eqref{R 0 mixed}, \eqref{def F2pm} and Lemma \ref{lem F2pm} we have $(R^{(0)}_{\mathrm{mixed}})_{\a\b}=0$. Therefore we have $R^{(0)}_{\a\b} = (R^{(0)}_{\mathrm{null}})_{\a\b}$. Moreover, the transport equation \eqref{eq F1}, the polarization condition \eqref{pola F} (which holds thanks to Lemma \ref{lem pola F1 F}) and the definitions \eqref{def 1 V 2 1}-\eqref{def 1 V 2 2} imply that
\begin{align}\label{R0 final}
R^{(0)}_{\a\b} & = - \half \sum_\A \sin\pth{\frac{u_\A}{\la}}  \dr_{(\a}u_\A (V^{(2,1)}_\A )_{\b)} - 2  \sum_\A \cos\pth{\frac{2u_\A}{\la}}  \dr_{(\a}u_\A (V^{(2,2)}_\A )_{\b)}.
\end{align}
We now compute $R^{(1)}_{\a\b}$. The definition \eqref{def g3e} of $\gg^{(3,e)}_{v,T}$ precisely ensures that $(R^{(1)}_{\mathrm{mixed}})_{\a\b}=0$ so that
\begin{align}\label{R1 final}
R^{(1)}_{\a\b} = (R^{(1)}_{\mathrm{null}})_{\a\b} + \half (\g_0)_{\rho(\a}(\dr_{\b)}\Upsilon^\rho)^{(-1)}.
\end{align}
Moreover, the transport equations \eqref{eq F21}-\eqref{eq F} and the conditions \eqref{pola F} and \eqref{pola g3h} imply
\begin{align}\label{R1 final bis}
(R^{(1)}_{\mathrm{null}})_{\a\b} & = \half\sum_\A \cos\pth{\frac{u_\A}{\lambda}} \bigg(  -  \Pi_\geq \pth{ (\h_\lambda)_{L_\A L_\A} }(F^{(1)}_\A)_{\alpha\beta}   + \D_{(\a} (V^{(2,1)}_\A)_{\b)}  \bigg) \non
\\&\quad - \sum_\A\sin\pth{\frac{2u_\A}{\lambda}}  \D_{(\a} (V^{(2,2)}_\A)_{\b)}
\\&\quad - \half\sum_\A\bigg( 2  \sin\pth{\frac{u_\A}{\la}}  \sin\pth{\frac{2u_\A}{\la}}  - \cos\pth{\frac{u_\A}{\lambda}} \cos\pth{ \frac{2u_\A}{\lambda} } \bigg) (V^{(2,2)}_\A)_{L_\A} (F^{(1)}_\A)_{\alpha\beta}  ,\non
\end{align}
where we used \eqref{R 1 null bis}. Finally, thanks to the wave equation \eqref{eq h} we have from \eqref{R 2}:
\begin{align}\label{R2 final}
R^{(\geq 2)}_{\a\b} & =  \frac{1}{2\la}\sum_\A \cos\pth{\frac{u_\A}{\la}}\Pi_\geq\pth{(\h_\lambda)_{L_\A L_\A}}(F^{(1)}_\A)_{\a\b} +\half\pth{ \Upsilon^\rho \dr_\rho (\g_\la)_{\a\b} + (\g_\la)_{\rho(\a} \dr_{\b)}\Upsilon^\rho }^{(\geq 0)} .
\end{align}
Combining \eqref{R0 final}-\eqref{R2 final} we obtain the decomposition
\begin{align*}
R_{\a\b}(\g_\la) & =  R(V)_{\a\b} + R(\Upsilon)_{\a\b}
\end{align*}
\begin{align}
R(V)_{\a\b} & \vcentcolon = - \half \sum_\A \sin\pth{\frac{u_\A}{\la}}  \dr_{(\a}u_\A (V^{(2,1)}_\A )_{\b)} - 2  \sum_\A \cos\pth{\frac{2u_\A}{\la}}  \dr_{(\a}u_\A (V^{(2,2)}_\A )_{\b)} \non
\\&\quad +  \frac{\la}{2}\sum_\A \cos\pth{\frac{u_\A}{\lambda}} \D_{(\a} (V^{(2,1)}_\A)_{\b)}  - \la\sum_\A\sin\pth{\frac{2u_\A}{\lambda}}  \D_{(\a} (V^{(2,2)}_\A)_{\b)}   \label{R V}
\\&\quad - \frac{\la}{2}\sum_\A\bigg( 2  \sin\pth{\frac{u_\A}{\la}}  \sin\pth{\frac{2u_\A}{\la}}  - \cos\pth{\frac{u_\A}{\lambda}} \cos\pth{ \frac{2u_\A}{\lambda} } \bigg) (V^{(2,2)}_\A)_{L_\A} (F^{(1)}_\A)_{\alpha\beta},  \non
\\ R(\Upsilon)_{\a\b} & \vcentcolon = \frac{\la^2}{2}\pth{ \Upsilon^\rho \dr_\rho (\g_\la)_{\a\b} + (\g_\la)_{\rho(\a} \dr_{\b)}\Upsilon^\rho }. \label{R Upsilon}
\end{align}
We define
\begin{align*}
G(V)_{\a\b} & \vcentcolon = R(V)_{\a\b} - \half \g_\la^{\mu\nu}R(V)_{\mu\nu}(\g_\la)_{\a\b},
\\ G(\Upsilon)_{\a\b} & \vcentcolon = R(\Upsilon)_{\a\b} - \half \g_\la^{\mu\nu}R(\Upsilon)_{\mu\nu}(\g_\la)_{\a\b},
\end{align*}
so that the Einstein tensor of $\g_\la$ satisfies $G_{\a\b}(\g_\la) = G(V)_{\a\b} + G(\Upsilon)_{\a\b}$. Moreover, $G(\Upsilon)_{\a\b}$ is indeed given by \eqref{def G upsilon}. It remains to compute $G(V)_{\a\b}$, for that we compute the trace $\g_\la^{\mu\nu}R(V)_{\mu\nu}$:
\begin{align*}
\g_\la^{\mu\nu}R(V)_{\mu\nu} & =  \sum_\A \sin\pth{\frac{u_\A}{\la}}  (V^{(2,1)}_\A )_{L_\A} + 4  \sum_\A \cos\pth{\frac{2u_\A}{\la}}  (V^{(2,2)}_\A )_{L_\A}
\\&\quad + \la\sum_\A \cos\pth{\frac{u_\A}{\lambda}} \div_{g_0} V^{(2,1)}_\A  - 2\la\sum_\A\sin\pth{\frac{2u_\A}{\lambda}} \div_{g_0}V^{(2,2)}_\A
\\&\quad + \la \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cro{ F^{(1)} (V^{(2,1)} + V^{(2,2)}) } + \GO{\la^2}.
\end{align*}
Plugging this into the definition of $G(V)_{\a\b}$, we obtain the decomposition \eqref{expansion G(V)}.
\end{proof}

The contracted Bianchi identities applied to the metric $_$ now read
\begin{align}\label{cBI}
\div_{\g_\la} G(V) + \div_{\g_\la} G(\Upsilon) & = 0.
\end{align}
To deduce extra equations from \eqref{cBI} we compute the divergence of $G(Υ)$ and $G(V)$ in the two following lemmas.

\begin{lemma}\label{lem divG(Ups)}
Let $K$ be any compact set of $\R^3$.  We have
\begin{align*}
\dive_{\g_\la} G(\Upsilon) & = \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}} \cro{\gg^{(3,e)}} + \mathcal{R}_1
\end{align*}
\begin{align}\label{reste 1}
\l \mathcal{R}_1 \r_{L^2(K)} \lesssim  \la.
\end{align}
\end{lemma}

\begin{proof}
From \eqref{def G upsilon} we obtain
\begin{align*}
\dive_{\g_\la} G(\Upsilon)_\a & = \frac{\la^2}{2} \pth{ (\g_\la)_{\rho\a}\tBox_{\g_\la}\Upsilon^\rho + \mathcal{B}_\a }
\end{align*}
where $\mathcal{B}_\a$ is schematically of the form
\begin{align*}
\mathcal{B} & =\pth{ 1 + \g_\la^{-1}\g_\la } \pth{ \g_\la^{-1} \dr \Upsilon \dr \g_\la + \g_\la^{-1}  \Upsilon \dr^2\g_\la + \g_\la^{-1}  \g_\la^{-1}\Upsilon \dr\g_\la\dr \g_\la  }.
\end{align*}
Since $\Upsilon=\GO{1}$ and $\g_\la=\GO{1}$ both with oscillating coefficients, the worst terms in $\mathcal{B}$ are $\dr\Upsilon$ and $\dr^2\g_\la$, and note that they are not multiplied together. Using the regularity and the estimates stated in Proposition \ref{prop system}, we conclude easily that
\begin{align*}
\l \mathcal{B} \r_{L^2(K)} \lesssim  \frac{1}{\la}.
\end{align*}
We now look at the term $\tBox_{\g_\la}\Upsilon^\rho$. From \eqref{Upsilon} we obtain the following schematic expression of $\Upsilon^\rho$:
\begin{align*}
\Upsilon & \vcentcolon =    \cro{\dr\h_\la} +  \sum_\A  \sin\left(\frac{u_\A}{\lambda} \right) \cro{\dr\F_\A + F^{(1)}} +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T'\pth{\frac{v}{\la}}\cro{\gg^{(3,e)}} + \text{better terms},
\end{align*}
where for clarity we choose not to write the $\g_\la^{-1}$ factors (which are lower order terms from the point of view of the $\la$ behaviour) and where the better terms includes terms with the same amounts of derivatives but with better $\la$ behaviour (for instance, terms of the form $\la\dr\gg^{(3,e)}$ or $\la\dr\gg^{(3,h)}$). Therefore, using \eqref{Box T f} we obtain
\begin{equation}\label{box upsilon}
\begin{aligned}
\tBox_{\g_\la}\Upsilon & \vcentcolon =    \cro{\dr\tBox_{\g_\la}\h_\la}  +  \cro{ \dr\tBox_{\g_\la}\F + \dr^2F^{(1)} }^\osc + \frac{1}{\la} \cro{ \dr L_\A  \F_\A + \dr^{\leq 1}\F_\A + \dr^{\leq 1}F^{(1)}}^\osc
\\&\quad + \frac{1}{\la^2}  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cro{\gg^{(3,e)}} + \frac{1}{\la} \cro{\dr^{\leq 1}\gg^{(3,e)}}^\osc + \cro{ \tBox_{\g_\la}\gg^{(3,e)}}^\osc + \text{better terms},
\end{aligned}
\end{equation}
where we used the fact that the commutators $\left[ \tBox_{\g_\la}, \dr \right]$ and $[L_\A,\dr]$ are better terms. Compared to the singlephase construction of [28], the only new term in \eqref{box upsilon} is \[\frac{1}{\la^2}  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cro{\gg^{(3,e)}},\] which we keep as it is. The other terms can be estimated as in Lemma 8.2 of [28] (using in particular the fact that the worst term in $\gg^{(3,e)}$ is a linear term in $\F$), we don't write the details and obtain that there are all bounded by $\frac{1}{\la}$ in $L^2(K)$. This proves the lemma.
\end{proof}

\begin{lemma}\label{lem divG(V)}
We have
\begin{align*}
\dive_{\g_\la}G(V)_\a & = \half \sum_\A \sin\pth{\frac{u_\A}{\la}}\Big( 2\D_{L_\A} (V^{(2,1)}_\A)_{\a} +  \div_{\g_0}L_\A (V^{(2,1)}_\A)_{\a}    + (V^{(2,1)}_\A)^\b\D_{[\b}(L_\A)_{\a]}   \Big)
\\&\quad + 2 \sum_\A \cos\pth{\frac{2u_\A}{\la}} \Big(  2 \D_{L_\A} (V^{(2,2)}_\A)_{\a} + \div_{\g_0}L_\A (V^{(2,2)}_\A)_{\a} + (V^{(2,2)}_\A)^\b\D_{[\b}(L_\A)_{\a]} \Big)
\\&\quad +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cro{  (V^{(2,1)} + V^{(2,2)}) F^{(1)} }_{\a\b} + (\mathcal{R}_2)_\a
\end{align*}
\begin{align}\label{reste 2}
\l \mathcal{R}_2 \r_{L^2(K)} \lesssim  \la.
\end{align}
\end{lemma}

\begin{proof}
Since $G(V)$ contains oscillating terms at the order $\la^0$ we have
\begin{align*}
\dive_{\g_\la}G(V)_\a & = \frac{1}{\la} \pth{ \dive_{\g_\la}G(V)_\a }^{(-1)} + \pth{ \dive_{\g_\la}G(V)_\a }^{(0)} + \GO{\la}.
\end{align*}
Recall that if $T$ is a symmetric 2-tensor we have in coordinates
\begin{align*}
\dive_{\g_\la}T_\a & = \g_\la^{\rho\b}  \dr_\rho T_{\a\b} - \g_\la^{\rho\b}\Ga(\g_\la)_{\rho(\a}^{\mu}T_{\mu\b)} .
\end{align*}
Therefore, we have thanks to Lemma \ref{lem G}:
\begin{align*}
\pth{ \dive_{\g_\la}G(V)_\a }^{(-1)} & = -\half \sum_\A \cos\pth{\frac{u_\A}{\la}} \dr^\b u_\A\pth{ \dr_{(\a} u_\A (V_\A^{(2,1)})_{\b)} + (V^{(2,1)}_\A)_{L_\A} (\g_0)_{\a\b} }
\\&\quad + 4 \sum_\A \sin\pth{\frac{2u_\A}{\la}} \dr^\b u_\A\pth{ \dr_{(\a} u_\A (V_\A^{(2,2)})_{\b)} + (V^{(2,2)}_\A)_{L_\A} (\g_0)_{\a\b} }
\\& = 0.
\end{align*}
Moreover, we have
\begin{align*}
\pth{ \dive_{\g_\la}G(V)_\a }^{(0)} & = \pth{\div_{g_0} G^{(0)}_{\a}}^{(0)}   +  \g_0^{\rho\b} \pth{ \dr_\rho G^{(1)}_{\a\b} }^{(-1)}   + \pth{ \g_\la^{\rho\b}}^{(1)} \pth{ \dr_\rho G(V)_{\a\b} }^{(-1)}  - \g_0^{\rho\b} (\tilde{\Ga}^{(0)})_{\rho(\a}^{\mu} G^{(0)}_{\mu\b)}
\\& \vcentcolon= I + II + III + IV,
\end{align*}
\begin{align*}
(\tilde{\Ga}^{(0)})_{\rho\a}^{\mu} & = -\half\sum_\B  \sin\pth{\frac{u_\B}{\la}} \pth{\dr_{(\rho} u_\B (F^{(1)}_\B)_{\a)}^\mu - \dr^\mu  u_\B (F^{(1)}_\B)_{\rho\a} }.
\end{align*}
For $I$, \eqref{G 0} implies
\begin{align*}
I & = \half \sum_\A \sin\pth{\frac{u_\A}{\la}} \Big( (L_\A)_{\a} \div_{\g_0} (V^{(2,1)}_\A) + (V^{(2,1)}_\A)_{\a}\div_{\g_0}L_\A
\\&\hspace{4cm} +  \D_{L_\A} (V^{(2,1)}_\A)_{\a} + (V^{(2,1)}_\A)^\rho\D_\rho(L_\A)_{\a} - \dr_\a(V^{(2,1)}_\A)_{L_\A}  \Big)
\\&\quad + 2 \sum_\A \cos\pth{\frac{2u_\A}{\la}} \Big( (L_\A)_{\a} \div_{\g_0} (V^{(2,2)}_\A) + (V^{(2,2)}_\A)_{\a}\div_{\g_0}L_\A
\\&\hspace{5cm} +  \D_{L_\A} (V^{(2,2)}_\A)_{\a} + (V^{(2,2)}_\A)^\rho\D_\rho(L_\A)_{\a} - \dr_\a(V^{(2,2)}_\A)_{L_\A}  \Big).
\end{align*}
For $II$, \eqref{G 1} implies
\begin{align*}
II & = - \frac{1}{2}\sum_\A \sin\pth{\frac{u_\A}{\lambda}}  \pth{ - L_\A^\b \D_{\a} (V^{(2,1)}_\A)_{\b} - \D_{L_\A} (V^{(2,1)}_\A)_{\a}  - \div_{\g_0} V^{(2,1)}_\A \dr_\a u_\A }
\\&\quad  - 2\sum_\A\cos\pth{\frac{2u_\A}{\lambda}}  \pth{- L_\A^\b \D_{\a} (V^{(2,2)}_\A)_{\b} - \D_{L_\A} (V^{(2,2)}_\A)_{\a}  - \div_{\g_0}V^{(2,2)}_\A \dr_\a u_\A }
\\&\quad + \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cro{  (V^{(2,1)} + V^{(2,2)}) F^{(1)} }_{\a\b}
\end{align*}
For $III$ and $IV$, \eqref{G 0} simply implies
\begin{align*}
III + IV & = \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cro{  (V^{(2,1)} + V^{(2,2)}) F^{(1)} }_{\a\b}.
\end{align*}
Putting everything together, we obtain the lemma, using in particular the fact that all the $\la$ terms in $G(V)$ are oscillating with amplitudes depending only on $V^{(2,1)}_\A$, $V^{(2,2)}_\A$ and $F^{(1)}$ which can be bounded independently of $\la$, thus producing the remainder term $\mathcal{R}_2$.
\end{proof}

\begin{lemma}
The polarization conditions \eqref{pola F21}-\eqref{pola F22} holds on $[0,1]\times \R^3$.
\end{lemma}

\begin{proof}
Thanks to the contracted Bianchi identities \eqref{cBI} and Lemmas \ref{lem divG(Ups)} and \ref{lem divG(V)} we have
\begin{align*}
0 & = \half \sum_\A \sin\pth{\frac{u_\A}{\la}}\Big( 2\D_{L_\A} (V^{(2,1)}_\A)_{\a} +  \div_{\g_0}L_\A (V^{(2,1)}_\A)_{\a}    + (V^{(2,1)}_\A)^\b\D_{[\b}(L_\A)_{\a]}   \Big)
\\&\quad + 2 \sum_\A \cos\pth{\frac{2u_\A}{\la}} \Big(  2 \D_{L_\A} (V^{(2,2)}_\A)_{\a}  + \div_{\g_0}L_\A (V^{(2,2)}_\A)_{\a} + (V^{(2,2)}_\A)^\b\D_{[\b}(L_\A)_{\a]} \Big)
\\&\quad +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cro{  (V^{(2,1)} + V^{(2,2)}) F^{(1)} + \gg^{(3,e)}}_{\a} + \mathcal{R}_1 + \mathcal{R}_2.
\end{align*}
We multiply this equality by $\sin\pth{\frac{u_\B}{\la}}$ and obtain
\begin{equation}\label{div G a}
\begin{aligned}
0 & = \half \sin^2\pth{\frac{u_\B}{\la}} \pth{   (V^{(2,1)}_\B)_{\a}\div_{\g_0}L_\B +  2\D_{L_\B} (V^{(2,1)}_\B)_{\a} + (V^{(2,1)}_\B)^\b\D_{[\b}(L_\B)_{\a]}   }
\\&\quad +  \frac{1}{4} \sum_{\A\in\mathcal{A}\setminus\{\B\},\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}} \cro{\dr^{\leq 1}V^{(2,1)}}_\a +  \sum_{\A,\pm} \sin\pth{\frac{u_\B\pm 2u_\A}{\la}}  \cro{\dr^{\leq 1}V^{(2,2)}}_\a
\\&\quad +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\sin\pth{\frac{u_\B}{\la}}\cro{  (V^{(2,1)} + V^{(2,2)}) F^{(1)} + \gg^{(3,e)} }_{\a}
\\&\quad +\sin\pth{\frac{u_\B}{\la}}\pth{\mathcal{R}_1 + \mathcal{R}_2}.
\end{aligned}
\end{equation}
We will use the following claim: for all $T:\R\longrightarrow\R$ smooth, $2\pi$-periodic with $\int_0^{2\pi}T=0$ and for all $z\in\mathcal{Z}$ the sequence of functions $\pth{T\pth{\frac{z}{\la}}}_{\la\in(0,1]}$ converges weakly to 0 in $L^2(K)$ when $\la\to 0$ and where $K$ is any compact subset of $\R^3$. This follows from \eqref{spatial} and was already mentioned in Section \ref{section BG}. Here, we use this claim on each $\Si_t$ for $t\in[0,1]$ and $K=\Si_t\cap J^+_0(B_R)$, i.e the support of $V^{(2,1)}_\A$ and $V^{(2,2)}_\A$. Thus, we have the following weak limits in $L^2(K)$ when $\la\to 0$:
\begin{align*}
\sin^2\pth{\frac{u_\B}{\la}} & \rightharpoonup \half,
\\ \cos\pth{\frac{u_\A\pm u_\B}{\la}} & \rightharpoonup 0,
\\ \sin\pth{\frac{u_\B\pm 2u_\A}{\la}}  & \rightharpoonup 0,
\\ T\pth{\frac{v}{\la}}\sin\pth{\frac{u_\B}{\la}} & \rightharpoonup 0,
\end{align*}
where we used that $ T\pth{\frac{v}{\la}}\sin\pth{\frac{u_\B}{\la}}$ with $v\in\mathcal{I}$ and $T\in\{\cos,\sin\}$ is a linear combination of functions of the form $S\pth{\frac{z}{\la}}$ for $z\in\mathcal{Z}$ and $S$ periodic with zero mean. Since $\F_\A$ is uniformly bounded in $H^1_{loc}$ with respect to $\la$ (see \eqref{estim F}), we still get the weak convergence
\begin{align*}
T\pth{\frac{v}{\la}}\sin\pth{\frac{u_\B}{\la}} \cro{\gg^{(3,e)}} \rightharpoonup 0,
\end{align*}
where we again used the fact that the worst term in $\gg^{(3,e)}$ is a linear term in $\F_\A$. Using now \eqref{reste 1} and \eqref{reste 2}, we can take the weak limit in $L^2(K)$ when $\la\to 0$ of the equality \eqref{div G a} and get for all $\B\in\mathcal{A}$
\begin{align*}
2\D_{L_\B} (V^{(2,1)}_\B)_{\a}  & =  (V^{(2,1)}_\B)^\b\D_{[\a}(L_\B)_{\b]} - (V^{(2,1)}_\B)_{\a}\div_{\g_0}L_\B.
\end{align*}
Since $V^{(2,1)}_{\B\;\;|_{\Si_0}}=0$ this implies $V^{(2,1)}_\B=0$ everywhere, i.e that \eqref{pola F21} holds on $[0,1]\times \R^3$. We come back to the contracted Bianchi identities (plugging in particular $V^{(2,1)}_\C=0$):
\begin{align*}
0 & =  2 \sum_\A \cos\pth{\frac{2u_\A}{\la}} \Big(  2 \D_{L_\A} (V^{(2,2)}_\A)_{\a}  + \div_{\g_0}L_\A (V^{(2,2)}_\A)_{\a} + (V^{(2,2)}_\A)^\b\D_{[\b}(L_\A)_{\a]} \Big)
\\&\quad +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cro{ V^{(2,2)} F^{(1)} + \gg^{(3,e)}}_{\a} + \mathcal{R}_1 + \mathcal{R}_2.
\end{align*}
We multiply this equality by $\cos\pth{\frac{2u_\B}{\la}}$:
\begin{align*}
0 & =  2 \cos^2\pth{\frac{2u_\B}{\la}} \Big(  2 \D_{L_\B} (V^{(2,2)}_\B)_{\a}  + \div_{\g_0}L_\B (V^{(2,2)}_\B)_{\a} + (V^{(2,2)}_\B)^\b\D_{[\b}(L_\B)_{\a]} \Big)
\\&\quad +  \sum_{\A\in\mathcal{A}\setminus\{\B\},\pm} \cos\pth{\frac{2(u_\A\pm u_\B)}{\la}}\Big(  2 \D_{L_\A} (V^{(2,2)}_\A)_{\a}  + \div_{\g_0}L_\A (V^{(2,2)}_\A)_{\a} + (V^{(2,2)}_\A)^\b\D_{[\b}(L_\A)_{\a]} \Big)
\\&\quad +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\cos\pth{\frac{2u_\B}{\la}}\cro{V^{(2,2)} F^{(1)}+ \gg^{(3,e)} }_{\a\b} +\cos\pth{\frac{2u_\B}{\la}}\pth{ \mathcal{R}_1 + \mathcal{R}_2}.
\end{align*}
Using the claim as above, taking the weak limit in $L^2(K)$ when $\la\to 0$ of this expression implies
\begin{align*}
2\D_{L_\C} (V^{(2,2)}_\C)_{\a}  & =  (V^{(2,2)}_\C)^\b\D_{[\a}(L_\C)_{\b]} - (V^{(2,2)}_\C)_{\a}\div_{\g_0}L_\C.
\end{align*}
Since $V^{(2,2)}_{\C\;\;|_{\Si_0}}=0$ this implies $V^{(2,2)}_\C=0$ everywhere. This concludes the proof of the lemma.
\end{proof}

\begin{lemma}
The generalised wave gauge condition \eqref{GWC} holds on $[0,1]\times \R^3$.
\end{lemma}

\begin{proof}
Thanks to the previous proposition we have $V^{(2,1)}_\A=V^{(2,2)}_\A=0$ so that \eqref{R V} implies that $R(V)=0$ and that the Einstein tensor of $\g_\la$ is given by
\begin{align}\label{final expression G}
G_{\a\b}(\g_\la) & = \frac{\la^2}{2} \bigg( (\g_\la)_{\rho(\a}\dr_{\b)}\Upsilon^\rho - (\g_\la)_{\a\b}\dr_\rho\Upsilon^\rho  + \Upsilon^\rho \dr_\rho (\g_\la)_{\a\b} - \half \g_\la^{\mu\nu}\Upsilon^\rho \dr_\rho (\g_\la)_{\mu\nu}(\g_\la)_{\a\b} \bigg).
\end{align}
From there, it is standard to deduce from the contracted Bianchi identities that $\Upsilon^\rho$ satisfies a linear system of the form
\begin{align}\label{system Ups}
\tBox_{\g_\la} \Upsilon^\rho & = A^{\rho\mu}_\si \dr_\mu \Upsilon^\si + B^\rho_\si \Upsilon^\si,
\end{align}
with $A^{\rho\mu}_\si$ and $B^{\rho}_\si$ regular coefficients. From Proposition \ref{prop ID main} recall that $\Upsilon^\rho_{|_{\Si_0}}=0$. Moreover again from Proposition \ref{prop ID main} we know that the constraint equations are solved by the induced metric and the second fundamental form so that $G(\g_\la)_{\T_\la \T_\la|_{\Si_0}} = 0$ and $G(\g_\la)_{\T_\la i|_{\Si_0}} = 0$, where $\T_\la$ is the future-directed unit normal for $\g_\la$ to $\Si_0$. From \eqref{final expression G} we have
\begin{align*}
G(\g_\la)_{\T_\la i|_{\Si_0}} & = \frac{\la^2}{2}(\g_\la)_{\rho i} \pth{ \T_\la\Upsilon^\rho - \T_\la^\rho \dr_t\Upsilon^0  },
\\ G(\g_\la)_{\T_\la \T_\la|_{\Si_0}} & = \frac{\la^2}{2} \pth{ 2(\T_\la)_\rho \T_\la\Upsilon^\rho + \dr_t\Upsilon^0  },
\end{align*}
where we used $\Upsilon^\rho_{|_{\Si_0}}=0$. The constraint equations being solved thus implies
\begin{align}
\T_\la\Upsilon^\rho & = \T_\la^\rho \dr_t\Upsilon^0 , \label{C1}
\\  \dr_t\Upsilon^0 & = - 2(\T_\la)_\rho \T_\la\Upsilon^\rho .\label{C2}
\end{align}
Thanks to \eqref{C1} and $\g_\la(\T_\la,\T_\la)=-1$, \eqref{C2} implies $\dr_t\Upsilon^0  = 0 $ so that \eqref{C1} becomes in turn $\T_\la\Upsilon^\rho=0$. Therefore, $\Upsilon^\rho$ satisfies the linear wave system \eqref{system Ups} and $\Upsilon^\rho_{|_{\Si_0}}=\T_\la\Upsilon^\rho_{|_{\Si_0}}=0$, which concludes the proof of the proposition.
\end{proof}

This concludes the proof of Theorem \ref{theo evol}, after the identifications
\begin{align*}
\sum_{\substack{w\in\mathcal{N}_1\cup\mathcal{N}_2\cup\mathcal{I}_2\\T\in\{\cos,\sin\}}} T\pth{\frac{w}{\la}}F^{(2,w,T)}_\la & =   \sum_\A \pth{ \sin\pth{\frac{u_\A}{\la}} \pth{ \F_\A + F^{(2,1)}_\A } + \cos\pth{\frac{2u_\A}{\la}} F^{(2,2)}_\A }
\\&\quad + \la^2 \sum_{\A\neq \B,\pm}   \cos\pth{\frac{u_\A\pm u_\B}{\la}} F^{(2,\pm)}_{\A\B},
\\ \widetilde{\g_\la} & =   \h_\la  + \la \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T\pth{\frac{u}{\la}} \gg^{(3,h)}_{u,T} + \la \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}} \gg^{(3,e)}_{v,T}.
\end{align*}
The estimates \eqref{estim theo 1}-\eqref{estim theo 3} can then be derived from \eqref{estim F1 F21 F22}-\eqref{estim F} and \eqref{estim h}.

\appendix

\section{Proofs of Section \ref{section expansion Ricci}}\label{appendix expansion Ricci}

In the following sections we will use the following trigonometric identities:
\begin{align}
\cos(a)\cos(b)  & = \half \sum_{\pm} \cos(a\pm b) ,\label{trigo 1}
\\ \sin(a) \sin(b)  & = \half \sum_{\pm}\mp \cos(a\pm b) , \label{trigo 2}
\\ \sin(a)\cos(b)  & = \half \sum_{\pm}\sin(a\pm b).\label{trigo 3}
\end{align}
We will also use the following identities on double sums:
\begin{align}
\sum_{\A\neq \B,\pm}  \cos\pth{\frac{u_\A\pm u_\B}{\la}} F^{(\pm)}_{\A\B} & = \half \sum_{\A\neq \B,\pm}  \cos\pth{\frac{u_\A\pm u_\B}{\la}}\pth{ F^{(\pm)}_{\A\B} + F^{(\pm)}_{\B\A}}, \label{double sum 1}
\\ \sum_{\A\neq \B,\pm} \sin\pth{\frac{u_\A\pm u_\B}{\la}} F^{(\pm)}_{\A\B} & = \half \sum_{\A\neq \B,\pm} \sin\pth{\frac{u_\A\pm u_\B}{\la}}\pth{ F^{(\pm)}_{\A\B} \pm F^{(\pm)}_{\B\A}}, \label{double sum 2}
\end{align}
which simply follow from
\begin{align*}
\sum_{\A\neq \B} T_{\A\B} = \half \sum_{\A\neq \B}(T_{\A\B} + T_{\B\A}).
\end{align*}
Moreover, from \eqref{expansion metric} we can obtain the inverse of $_λ$:
\begin{align*}
\g_\la^{\mu\nu} & =  \g_0^{\mu\nu} + \la (\g_\la^{\mu\nu})^{(1)}  + \la^2 (\g_\la^{\mu\nu})^{(2)}  + \la^3 (\g_\la^{\mu\nu})^{(\geq 3)} ,
\end{align*}
\begin{align}
(\g_\lambda^{\mu\nu})^{(1)}  & = - \sum_\A \cos\pth{\frac{u_\A}{\lambda}}(F^{(1)}_\A)^{\mu\nu},\label{inverse 1}
\\ (\g_\lambda^{\mu\nu})^{(2)} & = - \h_\lambda^{\mu\nu} -  \sum_\A \pth{\sin\left(\frac{u_\A}{\lambda} \right) \left( (\F_\A)^{\mu\nu} + (F^{(2,1)}_\A)^{\mu\nu} \right) + \cos\left( \frac{2u_\A}{\lambda}\right) (F^{(2,2)}_\A)^{\mu\nu}}\label{inverse 2}
\\&\quad +  \sum_{\A\neq\B,\pm} \cos\left(\frac{u_\A\pm u_\B}{\lambda}\right)\pth{  \half (F^{(1)}_\A)^{\nu\sigma} (F^{(1)}_\B)^\mu_{\sigma}  - (F^{(2,\pm)}_{\A\B} )^{\mu\nu}}\non
\\&\quad +  \sum_{\A} \cos^2\pth{\frac{u_\A}{\lambda}}  (F^{(1)}_\A)^{\nu\sigma} (F^{(1)}_\A)^\mu_{\sigma},\non
\end{align}
and where on the RHS of \eqref{inverse 1} and \eqref{inverse 2} the indexes are moved with respect to the background metric $_0$.

\subsection{Proof of Lemma \ref{lemma W}}\label{appendix proof lemma W}

The expansion of the quasi-linear wave operator $__λ(_λ)_αβ$ follows from a systematic use of the exact formula
\begin{equation} \label{Box T f}
\begin{aligned}
\tBox_g \left( T\left( \frac{v}{\lambda} \right) f \right) & = \frac{1}{\la^2}T''\pth{ \frac{v}{\la}} g^{-1}(\d v, \d v) f  +  \frac{1}{\la}T'\pth{ \frac{v}{\la}} \pth{ 2g^{\mu\nu}\dr_\mu v \dr_\nu f + (\tBox_g v)f  }   + T\pth{\frac{v}{\la}}\tBox_g f ,
\end{aligned}
\end{equation}
which holds for any Lorentzian metric $g$, real function $T$ and scalar functions on the manifold $v$ and $f$. From \eqref{inverse 1} and \eqref{inverse 2} we can also obtain the expansion
\begin{equation}\label{expansion eikonal}
\begin{aligned}
\g_\la^{-1}(\d u_\A,\d u_\A) & = - \la \sum_\B \cos\pth{\frac{u_\B}{\lambda}}(F^{(1)}_\B)_{L_\A L_\A}
\\&\quad - \la^2  (\h_\la)_{L_\A L_\A}  - \la^2  \sum_\B \sin\pth{\frac{u_\B}{\la}} \pth{ (\F_\B)_{L_\A L_\A}  + (F^{(2,1)}_\B)_{L_\A L_\A}  }
\\&\quad  - \la^2  \sum_\B \cos\pth{ \frac{2u_\B}{\la}} (F^{(2,2)}_\B)_{L_\A L_\A}
\\&\quad + \la^2  \sum_{\B\neq\C,\pm} \cos\pth{\frac{u_\B\pm u_\C}{\la}} \pth{  \half (F^{(1)}_\B)^{\si}_{L_\A} (F^{(1)}_\C)_{L_\A\si}   - (F^{(2,\pm)}_{\B\C} )_{L_\A L_\A} }
\\&\quad + \la^2  \sum_{\B} \cos^2\pth{\frac{u_\B}{\la}}  (F^{(1)}_\B)^{\si}_{L_\A} (F^{(1)}_\B)_{L_\A\si}  + \la^3 (\g_\la^{\mu\nu})^{(\geq 3)} \dr_\mu u_\A\dr_\nu u_\A.
\end{aligned}
\end{equation}
Thanks to \eqref{expansion metric} we have
\begin{equation}\label{exp box g}
\begin{aligned}
\tBox_{\g_\la}(\g_\la)_{\a\b} & = \tBox_{\g_\la}(\g_0)_{\a\b} + \la \sum_\A  \tBox_{\g_\la}\pth{\cos\pth{ \frac{u_\A}{\la} } (F^{(1)}_\A)_{\a\b}  }
\\&\quad + \la^2 \sum_\A \tBox_{\g_\la}\pth{\sin\pth{\frac{u_\A}{\la}} \pth{ (\F_\A)_{\a\b}  + (F^{(2,1)}_\A)_{\a\b} }}
\\&\quad + \la^2 \sum_\A \tBox_{\g_\la}\pth{\cos\pth{ \frac{2u_\A}{\la}} (F^{(2,2)}_\A)_{\a\b} }
\\&\quad + \la^2 \sum_{\A\neq \B,\pm}   \tBox_{\g_\la}\pth{\cos\pth{\frac{u_\A\pm u_\B}{\la}} (F^{(2,\pm)}_{\A\B})_{\a\b} }
\\&\quad + \la^2 \tBox_{\g_\la}(\h_\la)_{\a\b} + \la^3 \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}\tBox_{\g_\la}\pth{T\pth{\frac{u}{\la}} (\gg^{(3,h)}_{u,T})_{\a\b}}
\\&\quad + \la^3 \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} \tBox_{\g_\la}\pth{T\pth{\frac{v}{\la}} (\gg^{(3,e)}_{v,T})_{\a\b}}.
\end{aligned}
\end{equation}
We expand each term in this expression, using without mention \eqref{Box T f}, \eqref{expansion eikonal} and more generally \eqref{inverse 1} and \eqref{inverse 2}. We will also use without mention the assumptions \eqref{assumption}. For the first term, we simply have
\begin{align}
\tBox_{\g_\la}(\g_0)_{\a\b} & = \tBox_{\g_0}(\g_0)_{\a\b} + \la\cro{ (\g_\la^{-1})^{(\geq 1)} \dr^2\g_0}_{\a\b}. \label{wave proof 1}
\end{align}
For the second term, we first apply \eqref{Box T f}
\begin{align*}
\la \sum_\A  \tBox_{\g_\la}\pth{\cos\pth{ \frac{u_\A}{\la} } (F^{(1)}_\A)_{\a\b}  } & =  - \frac{1}{\la}\sum_\A \cos\pth{\frac{u_\A}{\la}} \g_\la^{-1}(\d u_\A, \d u_\A) (F^{(1)}_\A)_{\a\b}
\\&\quad - \sum_\A \sin\pth{\frac{u_\A}{\la}} \pth{ 2\g_\la^{\mu\nu}\dr_\mu u_\A \dr_\nu (F^{(1)}_\A)_{\a\b} + (\tBox_{\g_\la} u_\A)(F^{(1)}_\A)_{\a\b}  }
\\&\quad + \la \sum_\A  \cos\pth{\frac{u_\A}{\la}}\tBox_{\g_\la} (F^{(1)}_\A)_{\a\b}.
\end{align*}
Using now \eqref{expansion eikonal}, \eqref{trigo 1}, \eqref{trigo 3} and \eqref{double sum 1} we obtain
\begin{equation}\label{wave proof 2}
\begin{aligned}
&\la \sum_\A  \tBox_{\g_\la}\pth{\cos\pth{ \frac{u_\A}{\la} } (F^{(1)}_\A)_{\a\b}  }
\\& = - \sum_\A \sin\pth{\frac{u_\A}{\la}} \pth{ -2 L_\A (F^{(1)}_\A)_{\a\b} + (\tBox_{\g_0} u_\A)(F^{(1)}_\A)_{\a\b}  }
\\&\quad + \frac{1}{4} \sum_{\A\neq\B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}} \pth{ (F^{(1)}_\B)_{L_\A L_\A}(F^{(1)}_\A)_{\a\b} + (F^{(1)}_\A)_{L_\B L_\B}(F^{(1)}_\B)_{\a\b}  }
\\&\quad +\la\sum_\A \cos\pth{\frac{u_\A}{\la}} \pth{  (\h_\la)_{L_\A L_\A}  (F^{(1)}_\A)_{\a\b}  + \cro{\pth{F^{(2,\pm)} + (F^{(1)})^2 }F^{(1)} + \dr^2 F^{(1)}}_{\a\b}}
\\&\quad + \frac{\la}{2} \sum_{\A} \sin\pth{\frac{2u_\A}{\la}} (\F_\A)_{L_\A L_\A} (F^{(1)}_\A)_{\a\b}
\\&\quad + \la \sum_\A \sin\pth{\frac{2u_\A}{\la}} \cro{ \pth{ F^{(2,1)} + \dr F^{(1)} + F^{(1)} }F^{(1)} }_{\a\b}
\\&\quad + \la\sum_{\A} \cos\pth{\frac{u_\A}{\la}} \cos\pth{ \frac{2u_\A}{\la}} (F^{(2,2)}_\A)_{L_\A L_\A} (F^{(1)}_\A)_{\a\b}
\\&\quad + \la \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{\frac{v}{\la}} \Big\{ \pth{ \F + F^{(2,1)} + F^{(2,2)} + F^{(2,\pm)} }F^{(1)}
\\&\hspace{6cm} + \pth{ (F^{(1)})^2 + F^{(1)} + \dr F^{(1)} }F^{(1)} \Big\}_{\a\b}
\\&\quad + \la^2 \cro{ (\g_\la^{-1})^{(\geq 2)} \dr^{\leq 2} F^{(1)} }_{\a\b}^\osc.
\end{aligned}
\end{equation}
For the third, fourth and fifth term in \eqref{exp box g} we simply apply \eqref{Box T f}. We obtain
\begin{equation}\label{wave proof 3}
\begin{aligned}
&\la^2 \sum_\A \tBox_{\g_\la}\pth{\sin\pth{\frac{u_\A}{\la}} \pth{ (\F_\A)_{\a\b}  + (F^{(2,1)}_\A)_{\a\b} }}
\\& =   \la \sum_\A \cos\pth{ \frac{u_\A}{\la}} \pth{ -2L_\A  + \tBox_{\g_0} u_\A  } \pth{ (\F_\A)_{\a\b}  + (F^{(2,1)}_\A)_{\a\b} }
\\&\quad +\la \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{\frac{v}{\la}} \cro{ \pth{ \F  + F^{(2,1)} } F^{(1)} }_{\a\b} + \la^2 \sum_\A \sin\pth{\frac{u_\A}{\la}}\tBox_{\g_\la} (\F_\A)_{\a\b}
\\&\quad + \la^2  \cro{ (\g_\la^{-1})^{(\geq 1)} \dr^{\leq 1} \pth{ \F  + F^{(2,1)}  }+ \g_\la^{-1} \dr^2F^{(2,1)}}_{\a\b}^\osc,
\end{aligned}
\end{equation}
\begin{equation}\label{wave proof 4}
\begin{aligned}
& \la^2 \sum_\A \tBox_{\g_\la}\pth{\cos\pth{ \frac{2u_\A}{\la}} (F^{(2,2)}_\A)_{\a\b} }
\\& =  -2 \la \sum_\A \sin\pth{ \frac{2u_\A}{\la}} \pth{ -2L_\A  + \tBox_{\g_\la} u_\A  } (F^{(2,2)}_\A)_{\a\b}
\\&\quad + \la \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{\frac{v}{\la}} \cro{ F^{(2,2)} F^{(1)} }_{\a\b} + \la^2  \cro{ \g_\la^{-1} \dr^{\leq 2}F^{(2,2)}}_{\a\b}^\osc,
\end{aligned}
\end{equation}
\begin{equation}\label{wave proof 5}
\begin{aligned}
&\la^2 \sum_{\A\neq \B,\pm}   \tBox_{\g_\la}\pth{\cos\pth{\frac{u_\A\pm u_\B}{\la}} (F^{(2,\pm)}_{\A\B})_{\a\b} }
\\& = -  \sum_{\A\neq \B,\pm} \cos\pth{ \frac{u_\A\pm u_\B}{\la}} \g_0^{-1}(\d (u_\A\pm u_\B), \d (u_\A\pm u_\B)) (F^{(2,\pm)}_{\A\B})_{\a\b}
\\&\quad  + \la \sum_{\A\neq \B,\pm_1} \cos\pth{ \frac{u_\A}{\la}}  (F^{(1)}_\B)_{L_\A L_\A} (F^{(2,\pm_1)}_{\A\B})_{\a\b}
\\&\quad +  \la \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{\frac{v}{\la}} \cro{\dr^{\leq 1}F^{(2,\pm)} + F^{(1)} F^{(2,\pm)}}_{\a\b}  + \la^2 \cro{ \g_\la^{-1}\dr^{\leq 2} F^{(2,\pm)}}_{\a\b}^\osc.
\end{aligned}
\end{equation}
The sixth term is left as it is. For the seventh and eighth we apply \eqref{Box T f} and obtain
\begin{equation}\label{wave proof 7}
\begin{aligned}
\la^3 \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}\tBox_{\g_\la}\pth{T\pth{\frac{u}{\la}} (\gg^{(3,h)}_{u,T})_{\a\b}} & = \la^2  \cro{ \g_\la^{-1}\dr^{\leq 1}\gg^{(3,h)} }_{\a\b}^\osc
\\&\quad  + \la^3 \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T\pth{\frac{u}{\la}}\tBox_{\g_\la} (\gg^{(3,h)}_{u,T})_{\a\b},
\end{aligned}
\end{equation}
\begin{equation}\label{wave proof 8}
\begin{aligned}
\la^3 \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} \tBox_{\g_\la}\pth{T\pth{\frac{v}{\la}} (\gg^{(3,e)}_{v,T})_{\a\b}}& = - \la \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{ \frac{v}{\la}} \g_0^{-1}(\d v, \d v) (\gg^{(3,e)}_{v,T})_{\a\b}
\\&\quad +  \la^2\cro{ \g_\la^{-1}\dr^{\leq 1}\gg^{(3,e)} }_{\a\b}^\osc
\\&\quad + \la^3 \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{\frac{v}{\la}}\tBox_{\g_\la} (\gg^{(3,e)}_{v,T})_{\a\b},
\end{aligned}
\end{equation}
where we used that $T”=-T$ if $T∈{cos,sin}$. Collecting \eqref{wave proof 1}-\eqref{wave proof 8} concludes the proof of Lemma \ref{lemma W}.

\subsection{Proof of Lemma \ref{lemma P}}\label{appendix proof lemma P}

We recall \eqref{P dg dg}, which gives the expression of $P_αβ(_)(_λ,_λ)$. It is of the form $_^-1_^-1__$ so that schematically we have
\begin{align*}
\pth{P_{\alpha\beta}(\g_\la)(\dr \g_\lambda ,\dr \g_\lambda)}^{(0)} & = \g_0^{-1}\g_0^{-1} \pth{\dr\g_\la}^{(0)} \pth{\dr\g_\la}^{(0)}.
\end{align*}
Thanks to \eqref{expansion metric} we have
\begin{align}\label{dg0}
\pth{ \dr_\a (\g_\la)_{\mu\nu}}^{(0)} & = \dr_\a (\g_0)_{\mu\nu} - \sum_\A \sin\pth{\frac{u_\A}{\la}}\dr_\a u_\A (F^{(1)}_\A)_{\mu\nu}.
\end{align}
With \eqref{P dg dg}, \eqref{trigo 2} and \eqref{double sum 1} this implies
\begin{equation}\label{P0 proof}
\begin{aligned}
&\pth{P_{\alpha\beta}(\g_\la)(\dr \g_\lambda ,\dr \g_\lambda)}^{(0)}
\\& = P_{\alpha\beta}(\g_0)(\dr \g_0 ,\dr \g_0) - \frac{1}{4}  \sum_{\A} \dr_\alpha u_\A \dr_\beta u_\A  \left| F^{(1)}_\A \right|^2_{\g_0}
\\&\quad  + \sum_\A \sin\pth{\frac{u_\A}{\la}} \bigg(- 2\Ga(\g_0)^\mu_{(\a \rho} \dr^\rho u_\A (F^{(1)}_\A)_{\beta)\mu} - (F^{(1)}_\A)^{\mu\nu} \dr_{(\a}u_\A \pth{ \dr_\mu (\g_0)_{\b)\nu}  - \frac{1}{2} \dr_{\b)} (\g_0)_{\mu\nu} }  \bigg)
\\&\quad + \frac{1}{4}  \sum_{\A} \cos\pth{\frac{2u_\A}{\la}} \dr_\alpha u_\A \dr_\beta u_\A  \left| F^{(1)}_\A \right|^2_{\g_0}
\\&\quad + \frac{1}{4} \sum_{\A\neq\B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}  \bigg( \pm\dr_{(\alpha}u_\A (F^{(1)}_\A)_{L_\B\si} (F^{(1)}_\B)_{\beta)}^\si  \pm\dr_{(\alpha}u_\B (F^{(1)}_\B)_{L_\A\si} (F^{(1)}_\A)_{\beta)}^\si
\\&\hspace{5cm} \pm\frac{1}{2} \dr_{(\alpha} u_\A\dr_{\beta)} u_\B   \left| F^{(1)}_\A\cdot F^{(1)}_\B \right|_{\g_0}  \pm (F^{(1)}_\A)_{(\alpha L_\B} (F^{(1)}_\B)_{\beta) L_\A}
\\&\hspace{8.5cm}  \mp \g_0^{-1}(\d u_\A,\d u_\B)  (F^{(1)}_\A)_{(\alpha}^\nu (F^{(1)}_\B)_{\nu\beta)}   \bigg),
\end{aligned}
\end{equation}
where we also used \eqref{assumption}. Moreover we also have schematically
\begin{align*}
\pth{P_{\alpha\beta}(\g_\la)(\dr \g_\lambda ,\dr \g_\lambda)}^{(1)} & =\underbrace{ \g_0^{-1}\g_0^{-1} \pth{\dr\g_\la}^{(0)} \pth{\dr\g_\la}^{(1)}}_{\vcentcolon = A} + \underbrace{\pth{\g_\la^{-1}}^{(1)}\g_0^{-1} \pth{\dr\g_\la}^{(0)} \pth{\dr\g_\la}^{(0)}}_{\vcentcolon=B} .
\end{align*}
We start with $A_$. Thanks to \eqref{expansion metric} we have
\begin{equation}\label{dg1}
\begin{aligned}
\pth{ \dr_\a (\g_\la)_{\mu\nu}}^{(1)} & = \sum_\A \cos\pth{\frac{u_\A}{\la}} \dr_\a (F^{(1)}_\A)_{\mu\nu}  +  \sum_\A  \cos\left(\frac{u_\A}{\lambda} \right) \dr_\a u_\A \left( (\F_\A)_{\mu\nu} + (F^{(2,1)}_\A)_{\mu\nu} \right)
\\&\quad -2 \sum_\A  \sin\left( \frac{2u_\A}{\lambda}\right) \dr_\a u_\A (F^{(2,2)}_\A)_{\mu\nu}
\\&\quad - \sum_{\A\neq \B,\pm}   \sin\left(\frac{u_\A\pm u_\B}{\lambda}\right) \dr_\a(u_\A \pm u_\B)  (F^{(2,\pm)}_{\A\B})_{\mu\nu} .
\end{aligned}
\end{equation}
Now, \eqref{P dg dg}, \eqref{dg0} and \eqref{dg1} imply
\begin{equation}\label{P1 A proof}
\begin{aligned}
A_{\a\b} & =  2 \sum_\A  \cos\pth{\frac{u_\A}{\la}}\Big( \Ga(\g_0)^{\mu}_{(\a\rho}  \dr^\rho u_\A\pth{(\F_\A)_{\beta)\mu} +  (F^{(2,1)}_\A)_{\beta)\mu}}   +  \cro{ \dr F^{(1)} + F^{(1)} F^{(2,\pm)} }_{\a\b}  \Big)
\\&\quad - 4 \sum_\A  \sin\pth{ \frac{2u_\A}{\la}} \pth{ \Ga(\g_0)^{\mu}_{(\a\rho}  \dr^\rho u_\A (F^{(2,2)}_\A)_{\beta)\mu} + \cro{F^{(1)} \dr F^{(1)}}_{\a\b}}
\\&\quad  + \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T\pth{\frac{u}{\la}}\dr_{(\a}u\cro{\pth{1+F^{(1)}}\pth{\F + F^{(2,1)} +  F^{(2,2)}}}_{\b)}
\\&\quad + \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{ \frac{v}{\la}} \cro{F^{(2,\pm)} + F^{(1)} \pth{\dr F^{(1)} + \F + F^{(2,1)} + F^{(2,2)} + F^{(2,\pm)} } }_{\a\b}.
\end{aligned}
\end{equation}
It remains to compute $B_$:
\begin{equation}\label{P1 B proof}
\begin{aligned}
B_{\a\b} & =\sum_\A  \cos\pth{\frac{u_\A}{\la}}\cro{F^{(1)}+(F^{(1)})^3}_{\a\b} + \sum_\A \sin\pth{\frac{2u_\A}{\la}}\cro{ (F^{(1)})^2 }_{\a\b}
\\&\quad +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{ \frac{v}{\la}} \cro{ (F^{(1)})^2+(F^{(1)})^3 }_{\a\b} + \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T\pth{\frac{u}{\la}}\dr_{(\a}u\cro{(F^{(1)})^3}_{\b)}.
\end{aligned}
\end{equation}
Collecting \eqref{P0 proof}, \eqref{P1 A proof} and \eqref{P1 B proof} concludes the proof of Lemma \ref{lemma P}.

\subsection{Proof of Lemma \ref{lemma H}}\label{appendix proof lemma H}

We recall that
\begin{align*}
H^\rho & = \g_\la^{\rho\sigma} \g_\la^{\mu\nu} \pth{ \dr_\mu (\g_\la)_{\sigma\nu} - \half \dr_\sigma (\g_\la)_{\mu\nu} }.
\end{align*}
We will first expand the derivatives and then use the expansion of the inverse (see \eqref{inverse 1}-\eqref{inverse 2}) to obtain $H^ρ$. Thanks to \eqref{expansion metric} we obtain
\begin{align*}
\dr_\mu (\g_\la)_{\sigma\nu} - \half \dr_\sigma (\g_\la)_{\mu\nu} & = D_{\mu\nu\si}^{(0)} + \la D_{\mu\nu\si}^{(1)} + \la^2 D_{\mu\nu\si}^{(2)} + \la^3 D_{\mu\nu\si}^{(3)},
\end{align*}
\begin{align*}
D_{\mu\nu\si}^{(0)} & = \dr_\mu (\g_0)_{\sigma\nu} - \half \dr_\sigma (\g_0)_{\mu\nu} -  \sum_\A  \sin\left( \frac{u_\A}{\lambda} \right) \pth{ \dr_\mu u_\A (F^{(1)}_\A)_{\sigma\nu} - \half \dr_\sigma u_\A (F^{(1)}_\A)_{\mu\nu}  }  ,
\end{align*}
\begin{align*}
D_{\mu\nu\si}^{(1)} & =   \sum_\A  \cos\left( \frac{u_\A}{\lambda} \right)\pth{ \dr_\mu (F^{(1)}_\A)_{\sigma\nu} - \half \dr_\sigma (F^{(1)}_\A)_{\mu\nu} }
\\&\quad +   \sum_\A \cos\left(\frac{u_\A}{\lambda} \right)\pth{\dr_\mu u_\A(\F_\A + F^{(2,1)}_\A)_{\sigma\nu} - \half \dr_\sigma u_\A(\F_\A + F^{(2,1)}_\A)_{\mu\nu}   }
\\&\quad -2  \sum_\A  \sin\left( \frac{2u_\A}{\lambda}\right) \pth{\dr_\mu u_\A(F^{(2,2)}_\A)_{\sigma\nu} - \half \dr_\sigma u_\A(F^{(2,2)}_\A)_{\mu\nu}}
\\&\quad - \sum_{\A\neq \B,\pm}   \sin\left(\frac{u_\A\pm u_\B}{\lambda}\right)\pth{ \dr_\mu (u_\A\pm u_\B) (F^{(2,\pm)}_{\A\B})_{\sigma\nu} - \half \dr_\sigma (u_\A\pm u_\B) (F^{(2,\pm)}_{\A\B})_{\mu\nu} },
\end{align*}
\begin{align*}
D_{\mu\nu\si}^{(2)} & =   \sum_\A  \sin\left(\frac{u_\A}{\lambda} \right) \pth{\dr_\mu (\F_\A)_{\sigma\nu} - \half \dr_\sigma (\F_\A)_{\mu\nu} }
\\&\quad +  \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}} T\pth{\frac{u}{\la}} \cro{ \dr F^{(2,1)} + \dr F^{(2,2)} }_{\mu\nu\si} +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{\frac{v}{\la}} \cro{ \dr F^{(2,\pm)} }_{\mu\nu\si}
\\&\quad +   \dr_\mu (\h_\lambda)_{\sigma\nu} - \half \dr_\sigma (\h_\lambda)_{\mu\nu}
+  \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T'\pth{\frac{u}{\la}}\pth{ \dr_\mu u(\gg^{(3,h)}_{u,T})_{\sigma\nu} - \half \dr_\sigma u(\gg^{(3,h)}_{u,T})_{\mu\nu} }
\\&\quad +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T'\pth{\frac{v}{\la}}\pth{ \dr_\mu v(\gg^{(3,e)}_{v,T})_{\sigma\nu} - \half \dr_\sigma v(\gg^{(3,e)}_{v,T})_{\mu\nu} } ,
\end{align*}
\begin{align*}
D_{\mu\nu\si}^{(3)} & =   \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T\pth{\frac{u}{\la}} \pth{\dr_\mu (\gg^{(3,h)}_{u,T})_{\sigma\nu} - \half \dr_\sigma (\gg^{(3,h)}_{u,T})_{\mu\nu}}
\\&\quad +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{\frac{v}{\la}} \pth{\dr_\mu (\gg^{(3,e)}_{v,T})_{\sigma\nu} - \half \dr_\sigma (\gg^{(3,e)}_{v,T})_{\mu\nu}}.
\end{align*}
We have
\begin{align}\label{H rho somme}
H^\rho & = \g_\la^{\rho\sigma} \g_\la^{\mu\nu} \pth{ D_{\mu\nu\si}^{(0)} + \la D_{\mu\nu\si}^{(1)} + \la^2 D_{\mu\nu\si}^{(2)} + \la^3 D_{\mu\nu\si}^{(3)} } .
\end{align}
Using the background wave coordinate condition \eqref{wave condition BG}, \eqref{assumption}, \eqref{trigo 3} and \eqref{double sum 2} we first obtain
\begin{equation}\label{ggD0}
\begin{aligned}
&\g_\la^{\rho\sigma} \g_\la^{\mu\nu} D_{\mu\nu\si}^{(0)}
\\& = - \la \sum_\A\cos\pth{\frac{u_\A}{\la}} (F^{(1)}_\A)^{\mu\nu} \g_0^{\rho\sigma} \pth{\dr_\mu (\g_0)_{\sigma\nu} - \half \dr_\sigma (\g_0)_{\mu\nu} }
\\&\quad - \frac{\la}{4}  \sum_{\A}\sin\pth{\frac{2u_\A}{\la}}  \dr^\rho u_\A  \left| F^{(1)}_\A\right|^2_{\g_0}
\\&\quad - \frac{\la}{4}  \sum_{\A\neq\B,\pm} \sin\pth{ \frac{u_\A\pm u_\B}{\la}} \bigg( (F^{(1)}_\B)^{\nu}_{L_\A} (F^{(1)}_\A)_{\nu}^\rho \pm (F^{(1)}_\A)^{\nu}_{L_\B} (F^{(1)}_\B)_{\nu}^\rho
\\&\hspace{7cm} + \half \dr^\rho (u_\A\pm u_\B) \left| F^{(1)}_\A\cdot F^{(1)}_\B\right|_{\g_0}    \bigg)
\\&\quad - \la^2 \g_0^{\rho\sigma} \h_\la^{\mu\nu} \Bigg(\dr_\mu (\g_0)_{\sigma\nu} - \half \dr_\sigma (\g_0)_{\mu\nu}  -  \sum_\A  \sin\left( \frac{u_\A}{\lambda} \right) \pth{ \dr_\mu u_\A (F^{(1)}_\A)_{\sigma\nu} - \half \dr_\sigma u_\A (F^{(1)}_\A)_{\mu\nu}  }\Bigg)
\\&\quad + \la^2 \sum_{\substack{w\in \mathcal{N}\cup\mathcal{I}\\T\in\{1,\cos,\sin\}}}T\pth{ \frac{w}{\la}} \cro{  \pth{ \F + F^{(2,1)} + F^{(2,2)} + F^{(2,\pm)} + (F^{(1)})^2 }\pth{1+F^{(1)}} }^\rho
\\&\quad + \la^3  \cro{ \pth{\g_\la^{-1} (\g_\la)^{(\geq 3)} + (\g_\la)^{(\geq 1)} (\g_\la)^{(\geq 2)} } \pth{ 1 + F^{(1)}} }^{\rho,\osc}.
\end{aligned}
\end{equation}
For the remaining terms we obtain
\begin{equation}\label{ggD1}
\begin{aligned}
&\g_\la^{\rho\sigma} \g_\la^{\mu\nu} D_{\mu\nu\si}^{(1)}
\\& =   \sum_\A  \cos\left( \frac{u_\A}{\lambda} \right)\g_0^{\rho\sigma} \bigg(\Pol{\F_\A + F^{(2,1)}_\A}{u_\A}_\si
+  \g_0^{\mu\nu} \pth{ \dr_\mu (F^{(1)}_\A)_{\sigma\nu} - \half \dr_\sigma (F^{(1)}_\A)_{\mu\nu} }   \bigg)
\\&\quad -2  \sum_\A  \sin\left( \frac{2u_\A}{\lambda}\right)\g_0^{\rho\sigma} \Pol{F^{(2,2)}_\A}{u_\A}_\si - \sum_{\A\neq \B,\pm}   \sin\left(\frac{u_\A\pm u_\B}{\lambda}\right)\g_0^{\rho\sigma} \Pol{F^{(2,\pm)}_{\A\B}}{u_\A\pm u_\B}_\si
\\&\quad +\la \sum_{\substack{w\in \mathcal{N}\cup\mathcal{I}\\T\in\{1,\cos,\sin\}}}T\pth{ \frac{w}{\la}} \cro{  F^{(1)} \pth{ \dr F^{(1)}  + \F + F^{(2,1)} + F^{(2,2)} + F^{(2,\pm)} } }^\rho
\\&\quad +\la^2 \Big\{ \pth{ (\g_\la^{-1})^{(\geq 2)} + (\g_\la^{-1})^{(\geq 1)}(\g_\la^{-1})^{(\geq 1)}} \pth{ \dr F^{(1)}  + \F + F^{(2,1)} + F^{(2,2)} + F^{(2,\pm)} } \Big\}^{\rho,\osc}
\end{aligned}
\end{equation}
\begin{equation}\label{ggD2}
\begin{aligned}
&\g_\la^{\rho\sigma} \g_\la^{\mu\nu} D_{\mu\nu\si}^{(2)}
\\& =   \g_\la^{\rho\sigma} \g_\la^{\mu\nu} \pth{ \dr_\mu (\h_\lambda)_{\sigma\nu} - \half \dr_\sigma (\h_\lambda)_{\mu\nu}  +  \sum_\A  \sin\left(\frac{u_\A}{\lambda} \right) \pth{\dr_\mu (\F_\A)_{\sigma\nu} - \half \dr_\sigma (\F_\A)_{\mu\nu} }  }
\\&\quad +    \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T'\pth{\frac{u}{\la}}\g_0^{\rho\sigma}  \Pol{\gg^{(3,h)}_{u,T}}{u}_\si
\\&\quad +  \g_\la^{\rho\sigma} \g_\la^{\mu\nu}  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T'\pth{\frac{v}{\la}}\pth{ \dr_\mu v(\gg^{(3,e)}_{v,T})_{\sigma\nu} - \half \dr_\sigma v(\gg^{(3,e)}_{v,T})_{\mu\nu} }
\\&\quad +  \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}} T\pth{\frac{u}{\la}} \cro{ \dr F^{(2,1)} + \dr F^{(2,2)} }^\rho +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{\frac{v}{\la}} \cro{ \dr F^{(2,\pm)} }^\rho
\\&\quad + \la   \cro{ \g_\la^{-1}(\g_\la)^{(\geq 1)} \pth{\dr F^{(2,1)} + \dr F^{(2,2)} + \dr F^{(2,\pm)} + \gg^{(3,h)}} }^{\rho,\osc} .
\end{aligned}
\end{equation}
The term $_^ρσ _^μνD_μν^(3) $ is left as it is. Collecting \eqref{ggD0}-\eqref{ggD2} and plugging them into \eqref{H rho somme} concludes the proof of Lemma \ref{lemma H}.

\subsection{Proof of Proposition \ref{prop expression of the ricci}}\label{appendix proof prop ricci}

We start with $R^(0)_$. From \eqref{Ricci tensor GWC}, Lemmas \ref{lemma W} and \ref{lemma P} we have
\begin{align*}
2R^{(0)}_{\a\b} & = - \tBox_{\g_0}(\g_0)_{\a\b} - \sum_\A \sin\pth{\frac{u_\A}{\la}} (W^{(0,1)}_\A)_{\a\b}  -  \sum_{\A\neq\B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}  (W^{(0,\pm)}_{\A\B})_{\a\b}
\\&\quad + P_{\alpha\beta}(\g_0)(\dr \g_0, \dr \g_0)  - \frac{1}{4}  \sum_{\A} \dr_\alpha u_\A \dr_\beta u_\A  \left| F^{(1)}_\A \right|^2_{\g_0}
\\&\quad + \sum_\A \sin\pth{\frac{u_\A}{\la}} (P^{(0,1)}_\A)_{\a\b} + \sum_\A \cos\pth{\frac{2u_\A}{\la}} (P^{(0,2)}_\A)_{\a\b}
\\&\quad + \sum_{\A\neq \B,\pm}\cos\pth{\frac{u_\A\pm u_\B}{\la}} (P^{(0,\pm)}_{\A\B})_{\a\b} + \pth{ H^\rho \dr_\rho (\g_\la)_{\a\b} + (\g_\la)_{\rho(\a} \dr_{\b)}H^\rho }^{(0)}.
\end{align*}
We use Lemma \ref{lemma H} to compute the gauge terms:
\begin{align*}
&\pth{ H^\rho \dr_\rho (\g_\la)_{\a\b} + (\g_\la)_{\rho(\a} \dr_{\b)}H^\rho }^{(0)}
\\& =   \sum_\A\pth{-\sin\pth{\frac{u_\A}{\la}} (\g_0)_{\rho(\a}  \dr_{\b)}u_\A (H^{(1,1)}_\A)^\rho +2 \cos\pth{\frac{2u_\A}{\la}} (\g_0)_{\rho(\a} \dr_{\b)} u_\A (H^{(1,2)}_\A)^\rho}
\\&\quad + \sum_{\A\neq\B,\pm}\cos\pth{\frac{u_\A\pm u_\B}{\la}}(\g_0)_{\rho(\a} \dr_{\b)}(u_\A\pm u_\B)(H^{(1,\pm)}_{\A\B})^\rho .
\end{align*}
Therefore we obtain
\begin{align*}
2R^{(0)}_{\a\b} & = \sum_\A \pth{2 F_\A^2 -  \frac{1}{4} \left| F^{(1)}_\A \right|^2_{\g_0} } \dr_\mu u_\A \dr_\nu u_\A
\\&\quad + \sum_\A \sin\pth{\frac{u_\A}{\la}}\pth{ -  (W^{(0,1)}_\A)_{\a\b} + (P^{(0,1)}_\A)_{\a\b} - (\g_0)_{\rho(\a}  \dr_{\b)}u_\A (H^{(1,1)}_\A)^\rho}
\\&\quad + \sum_\A \cos\pth{\frac{2u_\A}{\la}} \pth{ (P^{(0,2)}_\A)_{\a\b} + 2(\g_0)_{\rho(\a} \dr_{\b)} u_\A (H^{(1,2)}_\A)^\rho}
\\&\quad + \sum_{\A\neq\B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}\Big( -  (W^{(0,\pm)}_{\A\B})_{\a\b} + (P^{(0,\pm)}_{\A\B})_{\a\b}
+(\g_0)_{\rho(\a} \dr_{\b)}(u_\A\pm u_\B)(H^{(1,\pm)}_{\A\B})^\rho \Big),
\end{align*}
where we also used \eqref{eq BG coordinates}. Using now \eqref{W 0 1}, \eqref{P 0 1}, \eqref{H 1 1}, \eqref{P 0 2}, \eqref{H 1 2}, \eqref{W 0 pm} and \eqref{H 1 pm} we obtain \eqref{R 0}-\eqref{I 0 pm}.

\saut
We now compute $R^(1)_$. From \eqref{Ricci tensor GWC}, Lemmas \ref{lemma W} and \ref{lemma P} we have
\begin{equation}\label{R1 proof}
\begin{aligned}
2R^{(1)}_{\a\b} & =  \sum_\A\cos\pth{\frac{u_\A}{\la}}\pth{ - (W^{(1,1)}_\A)_{\a\b} + (P^{(1,1)}_\A)_{\a\b} }
\\&\quad + \sum_\A \sin\pth{\frac{2u_\A}{\la}}\pth{ - (W^{(1,2)}_\A)_{\a\b} + (P^{(1,2)}_\A)_{\a\b} }
\\&\quad - \sum_\A \cos\pth{\frac{u_\A}{\lambda}} \cos\pth{ \frac{2u_\A}{\lambda} } (F^{(2,2)}_\A)_{L_\A L_\A} (F^{(1)}_\A)_{\alpha\beta}
\\&\quad - \half \sum_\A \sin\pth{\frac{2u_\A}{\la}} (\F_\A)_{L_\A L_\A} (F^{(1)}_\A)_{\a\b}
\\&\quad + \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{v}{\la}}\pth{ - (W^{(1)}_{v,T})_{\a\b} + (P^{(1)}_{v,T})_{\a\b} }
\\&\quad + \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T\pth{\frac{u}{\la}} \dr_{(\a}u (\hat{P}^{(1)}_{u,T})_{\b)}  + \pth{ H^\rho \dr_\rho (\g_\la)_{\a\b} + (\g_\la)_{\rho(\a} \dr_{\b)}H^\rho }^{(1)}.
\end{aligned}
\end{equation}
We use Lemma \ref{lemma H} (and in particular \eqref{H 1 1}-\eqref{H 2}) to compute the gauge terms. We have
\begin{align*}
&\pth{ (\g_\la)_{\rho(\a} \dr_{\b)}H^\rho }^{(1)}
\\& =  \sum_\A\cos\pth{\frac{u_\A}{\la}}\pth{ (\g_0)_{\rho(\a} \dr_{\b)}\pth{ \g_0^{\rho\sigma} \Pol{\F_\A + F^{(2,1)}_\A}{u_\A}_\si} + \cro{\dr^{\leq 2}F^{(1)} + (F^{(1)})^2}^\rho }
\\&\quad -2 \sum_\A\sin\pth{\frac{2u_\A}{\la}}\pth{ (\g_0)_{\rho(\a}\dr_{\b)}\pth{ \g_0^{\rho\sigma} \Pol{F^{(2,2)}_\A}{u_\A}_\si } + \cro{F^{(1)}\dr F^{(1)}}^\rho }
\\&\quad -  \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T\pth{\frac{u}{\la}} \dr_{(\a}u
\\&\hspace{2.5cm} \times \bigg( \Pol{\gg^{(3,h)}_{u,T}}{u}_{\b)}
\\&\hspace{3.5cm} + \Big\{\dr F^{(2,1)} + \dr F^{(2,2)}
\\&\hspace{4.5cm} +  \pth{ \F + F^{(2,1)} + F^{(2,2)} + F^{(2,\pm)} + (F^{(1)})^2 + \dr F^{(1)} }\pth{1+F^{(1)}} \Big\}_{\b)} \bigg)
\\&\quad +  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{ \frac{v}{\la}} \Big\{ \dr F^{(2,\pm)}
+  \pth{ \F + F^{(2,1)} + F^{(2,2)} + F^{(2,\pm)} + (F^{(1)})^2 + \dr F^{(1)} }\pth{1+F^{(1)}} \Big\}_{\a\b}
\\&\quad +(\g_0)_{\rho(\a} \pth{\dr_{\b)}\Upsilon^\rho }^{(-1)}
\end{align*}
\begin{align*}
&\pth{ H^\rho \dr_\rho (\g_\la)_{\a\b} }^{(1)}
\\& =   \sum_\A \cos\pth{\frac{u_\A}{\la}} \pth{ \dr_\rho (\g_0)_{\a\b} \g_0^{\rho\sigma} \Pol{\F_\A + F^{(2,1)}_\A}{u_\A}_\si  + \cro{\dr^{\leq 1}F^{(1)}+(F^{(1)})^3}_{\a\b}}
\\&\quad -2 \sum_\A\sin\pth{\frac{2u_\A}{\la}} \pth{ \dr_\rho (\g_0)_{\a\b}  \g_0^{\rho\sigma} \Pol{F^{(2,2)}_\A}{u_\A}_\si  + \cro{F^{(1)}\pth{F^{(2,1)} + \dr^{\leq 1}F^{(1)}}}_{\a\b}}
\\&\quad + \half \sum_\A \sin\pth{\frac{2u_\A}{\la}} \Pol{\F_\A}{u_\A}_{L_\A} (F^{(1)}_\A)_{\a\b}
+ 2 \sum_\A \sin\pth{\frac{u_\A}{\la}}\sin\pth{\frac{2u_\A}{\la}} (F^{(2,2)}_\A)_{L_\A L_\A} (F^{(1)}_\A)_{\a\b}
\\&\quad + \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}}T\pth{ \frac{v}{\la}} \cro{ F^{(1)} \pth{F^{(1)} + \F + F^{(2,1)} + F^{(2,2)} + \dr^{\leq 1}F^{(1)} + (F^{(1)})^2 } }_{\a\b}.
\end{align*}
We plug these two computations into \eqref{R1 proof}, use \eqref{W 1 1}, \eqref{P 1 1}, \eqref{W 1 2}, \eqref{P 1 2}, \eqref{W 1 v T}, \eqref{P 1 v T} and \eqref{Phat 1 u T} and obtain \eqref{R 1}-\eqref{R 1 e}.

\saut
We now compute $R^(≥2)_$. From Lemmas \ref{lemma W} and \ref{lemma P} we have
\begin{align*}
2 R^{(\geq 2)}_{\a\b} & = - W^{(\geq 2)}_{\a\b} + P^{(\geq 2)}_{\a\b}  +\pth{ H^\rho \dr_\rho (\g_\la)_{\a\b} + (\g_\la)_{\rho(\a} \dr_{\b)}H^\rho }^{(\geq 2)}.
\end{align*}
We use \eqref{W 2} and \eqref{P 2} to obtain
\begin{align*}
2 R^{(\geq 2)}_{\a\b} & = -\tBox_{\g_\la}(\h_\la)_{\a\b}  - \sum_\A \sin\pth{\frac{u_\A}{\la}}\tBox_{\g_\la}(\F_\A)_{\a\b}
\\&\quad -  \la \sum_{\substack{u\in\mathcal{N}\\T\in\{\cos,\sin\}}}T\pth{\frac{u}{\la}}\tBox_{\g_\la} (\gg^{(3,h)}_{u,T})_{\a\b}  - \la  \sum_{\substack{v\in\mathcal{I}\\T\in\{\cos,\sin\}}} T\left( \frac{v}{\lambda} \right)\tBox_{\g_\la} (\gg^{(3,e)}_{v,T})_{\a\b}
\\&\quad +  \Big\{\pth{ \g_\la^{-1}\g_\la^{-1}\dr\g_\la\dr\g_\la }^{(\geq 2)} + \g_\la^{-1}\big(\dr^{\leq 1}\gg^{(3,h)} + \dr^{\leq 1}\gg^{(3,e)} + \dr^{\leq 2} F^{(1)} + \dr^{\leq 1} \F
\\&\hspace{7cm}+ \dr^{\leq 2}F^{(2,1)} + \dr^{\leq 2}F^{(2,2)} + \dr^{\leq 2} F^{(2,\pm)} \big)  \Big\}_{\a\b}^\osc
\\&\quad  +\pth{ H^\rho \dr_\rho (\g_\la)_{\a\b} + (\g_\la)_{\rho(\a} \dr_{\b)}H^\rho }^{(\geq 2)},
\end{align*}
which gives \eqref{R 2}. This concludes the proof of Proposition \ref{prop expression of the ricci}.

\section{Proof of Proposition \ref{prop constraint main}}\label{appendix proof prop constraint main}

\subsection{Conformal formulation of the constraint equations}\label{section conformal}

We introduce the notations
\begin{align*}
\HH(g,k) & \vcentcolon= R(g) - |k|^2_g + (\tr_g k)^2,
\\ \MM(g,k) & \vcentcolon= \div_g k - \d \tr_g k,
\end{align*}
In order to solve the constraint equations, we take inspiration from [8] and look for $(g_,k_)$ under the form
\begin{equation}\label{conformal ansatz}
\begin{aligned}
g_\la & = \ffi^4_\la \ga_\la,
\\ k_\la & = \ffi^2_\la \pth{  \ka_\la  +  \L_{\ga_\la} X_\la},
\end{aligned}
\end{equation}
\begin{align}\label{def LX}
\L_g X \vcentcolon =  \LL_{X}g -  \half \pth{\div_{g}X} g ,
\end{align}
with $_Xg$ is the Lie derivative. A straightforward computation implies that the constraint equations for $(g_,k_)$ rewrite
\begin{align}
8 \De_{\ga_\la}\ffi_\la & = \HH\pth{\ga_\la,\ka_\la} + \RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la},\label{hamiltonian}
\\ (\vDe_{\ga_\la} X_\la)_i& = -  \MM(\ga_\la,\ka_\la)_i  +  \RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}_i ,\label{momentum}
\end{align}
\begin{align*}
\vDe_{g} X_i \vcentcolon= \De_g X_i + R_{ij}(g)X^{j}
\end{align*}
with $_g$ now acting on vector fields and $R_ij(g)$ the Ricci tensor of $g$, and where the remainders are given by
\begin{align}
\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la} & \vcentcolon= ( \ffi_\la-1)\HH\pth{\ga_\la,\ka_\la} \non
\\&\quad +\ffi_\la \bigg( - \left|  \LL_{X_\la} \ga_\la \right|^2_{\ga_\la} -2 \left| \ka_\la \cdot \LL_{X_\la} \ga_\la \right|_{\ga_\la} \label{reste H}
\\&\hspace{3cm}  + \frac{3}{2} \pth{\div_{\ga_\la} X_\la}^2 +2 \pth{\tr_{\ga_\la}\ka_\la}\pth{ \div_{\ga_\la} X_\la}\bigg),\non
\\ \RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}_i & \vcentcolon= - 4 \ffi_\la^{-1} \ga_\la^{ab} \pth{ \ka_\la+ \LL_{X_\la}\ga_\la }_{ai}\dr_b \ffi_\la + 2 \ffi_\la^{-1} \pth{\div_{\ga_\la}X_\la}\dr_i \ffi_\la. \label{reste M}
\end{align}
The parameters $_$ and $_$ of the conformal formulation are defined by
\begin{align}
\ga_\la & \vcentcolon= g_0 + \la \sum_\A \cos\pth{\frac{u_\A}{\la}} \bar{F}^{(1)}_\A + \la^2 \sum_{\A\neq \B,\pm}\cos\pth{\frac{u_\A\pm u_\B}{\la}}\ga^{(2,\pm)}_{\A\B},\label{ga la}
\\ \ka_\la & \vcentcolon= k_0 + \sum_\A \sin\pth{ \frac{u_\A}{\la}}  \half |\nab u_\A| \bar{F}^{(1)}_\A  + \la\sum_\A\pth{\cos\pth{\frac{u_\A}{\la}} \ka^{(1,1)}_\A + \sin\pth{\frac{2u_\A}{\la}} \ka^{(1,2)}_\A} \label{ka la}
\\&\quad + \la\sum_{\A\neq\B,\pm}\sin\pth{\frac{u_\A\pm u_\B}{\la}}\ka^{(1,\pm)}_{\A\B}.\nonumber
\end{align}
The unknowns $_$ and $X_$ of the conformal formulation are of the form
\begin{align}
\ffi_\la & = 1 + \la^2 \ffi^{(2)} + \la^2 \tffi_\la + \la^3 \ffi^{(3)}\label{def ffi la},
\\ X_\la & = \la^2 X^{(2)} + \la^2 \tX_\la + \la^3 X^{(3)}.\label{def X la}
\end{align}
\begin{align}
\ffi^{(2)} & =  \sum_\A \pth{ \sin\pth{\frac{u_\A}{\la}} \ffi^{(2,1)}_\A + \cos\pth{\frac{2u_\A}{\la}}\ffi^{(2,2)}_\A } +  \sum_{\A\neq \B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}\ffi^{(2,\pm)}_{\A\B},\label{def ffi 2}
\\ X^{(2)} & =  \sum_\A \pth{ \sin\pth{\frac{u_\A}{\la}} X^{(2,1)}_\A + \cos\pth{\frac{2u_\A}{\la}}X^{(2,2)}_\A }  + \sum_{\A\neq \B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}X^{(2,\pm)}_{\A\B}.\label{def X 2}
\end{align}
In \eqref{def ffi la}-\eqref{def X la}, $_$ and $_$ are non-oscillating remainders, while $^(3)$ and $X^(3)$ are oscillating but we don't need to be precise on their oscillating behaviour.

\subsection{The constraint hierarchy}\label{section constraint hierarchy}

We first compute useful expansions related to $_$. The inverse of $_$ satisfies
\begin{align}\label{ga-1}
\ga_\la^{ij} = g_0^{ij} + \la(\ga_\la^{ij})^{(1)} + \la^2(\ga_\la^{ij})^{(2)} + \GO{\la^3}
\end{align}
\begin{align}
(\ga_\la^{ij})^{(1)} & = - \sum_\A \cos\pth{\frac{u_\A}{\la}} (\bar{F}^{(1)}_\A)^{ij},\label{ga-1 1}
\\ (\ga_\la^{ij})^{(2)}  & = -  \sum_{\A\neq \B,\pm}\cos\pth{\frac{u_\A\pm u_\B}{\la}} (\ga^{(2,\pm)}_{\A\B})^{ij} -  \sum_{\A,\B} \cos\pth{\frac{u_\A}{\la}} \cos\pth{\frac{u_\B}{\la}} (\bar{F}^{(1)}_\B)^{ik} (\bar{F}^{(1)}_\A)_{k}^j. \label{ga-1 2}
\end{align}
The Christofel symbols of $_$ satisfy
\begin{align}
\Ga(\ga_\la)^k_{ij} & =  \Ga(g_0)^k_{ij} + (\tilde{\Ga}^{(0)})^k_{ij} + \la (\Ga^{(1)})^k_{ij}  + \GO{\la^2}\label{christofel symbols}
\end{align}
\begin{align}
(\tilde{\Ga}^{(0)})^k_{ij} & = -\half\sum_\A  \sin\pth{\frac{u_\A}{\la}} \pth{\dr_{(i} u_\A (\bar{F}^{(1)}_\A)_{j)}^k - \dr^k  u_\A (\bar{F}^{(1)}_\A)_{ij} }, \label{Gatilde 0}
\\ (\Ga^{(1)})^k_{ij} & = \sum_\A \cos\pth{\frac{u_\A}{\la}} (\bar{Q}_\A)^k_{ij}  + \half \sum_\A \cos\pth{\frac{u_\A}{\la}} g_0^{k\ell} \pth{ \dr_{(i} u_\A (\ga^{(2,1)}_\A)_{j)\ell} - \dr_\ell u_\A (\ga^{(2,1)}_\A)_{ij} }\non
\\&\quad - \half  \sum_{\A\neq \B,\pm}\sin\pth{\frac{u_\A\pm u_\B}{\la}}g_0^{k\ell}\Big( \dr_{(i} (u_\A\pm u_\B)(\ga^{(2,\pm)}_{\A\B})_{j)\ell} \label{Ga 1}
- \dr_\ell (u_\A\pm u_\B)(\ga^{(2,\pm)}_{\A\B})_{ij}\Big)
\\&\quad + \half \sum_{\A,\B} \cos\pth{\frac{u_\A}{\la}}\sin\pth{\frac{u_\B}{\la}} (\bar{F}^{(1)}_\A)^{k\ell} \pth{ \dr_{(i} u_\B (\bar{F}^{(1)}_\B)_{j)\ell} - \dr_\ell u_\B (\bar{F}^{(1)}_\B)_{ij}},\non
\end{align}
\begin{align*}
(\bar{Q}_\A)^k_{ij} & =  \half  \pth{ g_0^{k\ell}\pth{ \dr_{(i} (\bar{F}^{(1)}_\A)_{j)\ell} - \dr_\ell (\bar{F}^{(1)}_\A)_{ij}} - (\bar{F}^{(1)}_\A)^{k\ell}\pth{ \dr_{(i} (g_0)_{j)\ell} - \dr_\ell (g_0)_{ij}} }.
\end{align*}
Note that \eqref{assum F1} implies
\begin{equation}\label{barQ}
\begin{aligned}
(\bar{Q}_\A)^k_{kj} & = 0,
\\ g_0^{ij}\dr_k u_\A (\bar{Q}_\A)^k_{ij} & =(\bar{F}^{(1)}_\A)_{ij}  \pth{ -   \dr^i \dr^j u_\A + \half\dr^\ell u_\A\dr_\ell g_0^{ij}}.
\end{aligned}
\end{equation}
We are now ready to compute the main terms in the equations \eqref{hamiltonian}-\eqref{momentum}.

\begin{lemma}\label{lem constraint expansion H}
We have
\begin{align*}
\HH(\ga_\la,\ka_\la) & = \pth{ \HH(\ga_\la,\ka_\la)}^{(0)} + \la \pth{ \HH(\ga_\la,\ka_\la)}^{(1)} + \la^2 \pth{ \HH(\ga_\la,\ka_\la)}^{(\geq 2)},
\end{align*}
\begin{equation}\label{H 0}
\begin{aligned}
&\pth{ \HH(\ga_\la,\ka_\la)}^{(0)}
\\& =  \sum_\A \sin\pth{\frac{u_\A}{\la}}  (\bar{F}^{(1)}_\A)_{ij}  \pth{ \dr^i \dr^j u_\A - \half\dr^\ell u_\A\dr_\ell g_0^{ij} - |\nab_{g_0} u_\A|_{g_0}k_0^{ij}}
\\&\quad - 6 \sum_{\A}\cos\pth{\frac{2u_\A}{\la}}  |\nab_{g_0} u_\A|_{g_0}^2 F_\A^2
\\&\quad + \sum_{\A\neq \B,\pm}\cos\pth{\frac{u_\A\pm u_\B}{\la}}\Big( |\nab_{g_0}(u_\A\pm u_\B)|^2_{g_0}\tr_{g_0}\ga^{(2,\pm)}_{\A\B} -(\ga^{(2,\pm)}_{\A\B})_{\nab_{g_0}(u_\A\pm u_\B)\nab_{g_0}(u_\A\pm u_\B)} \Big)
\\&\quad + \frac{1}{8}\sum_{\A\neq\B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}} \bigg( \pm 2 (\bar{F}^{(1)}_\A)_{i \nab_{g_0} u_\B} (\bar{F}^{(1)}_\B)_{\nab_{g_0} u_\A}^i
\\&\hspace{5.5cm} + \left| \bar{F}^{(1)}_\A \cdot \bar{F}^{(1)}_\B \right|_{g_0} \Big( -2 |\nab_{g_0} u_\A|_{g_0}^2 -2 |\nab_{g_0} u_\B|_{g_0}^2
\\&\hspace{9cm} \mp 3 |\nab_{g_0} u_\A\cdot \nab_{g_0} u_\B|_{g_0}
\\&\hspace{9cm}\pm |\nab_{g_0} u_\A|_{g_0} |\nab_{g_0} u_\B|_{g_0} \Big)  \bigg).
\end{aligned}
\end{equation}
Moreover, the higher order terms satisfy
\begin{align}
\pth{ \HH(\ga_\la,\ka_\la)}^{(1)} & = \sum_{\substack{w\in\mathcal{N}\cup\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{w}{\la}} \cro{ \pth{ \ga_\la^{-1}\dr^2\ga_\la+ \ga_\la^{-1}\ga_\la^{-1}\pth{(\dr\ga_\la)^2 + (\ka_\la)^2} }^{(1)} }      ,\label{H 1}
\\\pth{ \HH(\ga_\la,\ka_\la)}^{(\geq 2)} & = \cro{ \pth{ \ga_\la^{-1}\dr^2\ga_\la+ \ga_\la^{-1}\ga_\la^{-1}\pth{(\dr\ga_\la)^2 + (\ka_\la)^2} }^{(\geq 2)} }^\osc.\label{H 2}
\end{align}
\end{lemma}

\begin{proof}
We start with the scalar curvature, using its expression in coordinates:
\begin{align*}
R(\ga) & = \ga^{ij} \pth{ \dr_k \Ga(\ga)^k_{ij} - \dr_i  \Ga(\ga)^k_{jk} +  \Ga(\ga)^k_{k\ell} \Ga(\ga)^\ell_{ij} -  \Ga(\ga)^k_{i\ell} \Ga(\ga)^\ell_{jk} }.
\end{align*}
Using \eqref{assum F1} we first obtain
\begin{align*}
\pth{ R(\ga_\la) }^{(-1)} & = g_0^{ij} \pth{ \dr_k (\tilde{\Ga}^{(0)})^k_{ij} - \dr_i  (\tilde{\Ga}^{(0)})^k_{jk} }^{(-1)}.
\\& =- \sum_\A  \cos\pth{\frac{u_\A}{\la}}\pth{ (\bar{F}^{(1)}_\A)_{\nab_{g_0} u_\A \nab_{g_0} u_\A} -|\nab_{g_0} u_\A|_{g_0}^2 \tr_{g_0}\bar{F}^{(1)}_\A  }
\\& = 0.
\end{align*}
We have $\pth{ R(\ga_\la) }^{(0)} = I + II + III$ with
\begin{align*}
I & \vcentcolon = g_0^{ij} \pth{  \dr_k \Ga(\ga_\la)^k_{ij} - \dr_i  \Ga(\ga_\la)^k_{jk}  }^{(0)},
\\ II & \vcentcolon = \pth{\ga_\la^{ij}}^{(1)} \pth{  \dr_k (\tilde{\Ga}^{(0)})^k_{ij} - \dr_i (\tilde{\Ga}^{(0)})^k_{jk}  }^{(-1)},
\\ III & \vcentcolon = g_0^{ij} \pth{\Ga(\ga_\la)^k_{k\ell} \Ga(\ga_\la)^\ell_{ij} -  \Ga(\ga_\la)^k_{i\ell} \Ga(\ga_\la)^\ell_{jk}}^{(0)}.
\end{align*}
We start with $I$. Using $(\tilde{\Ga}^{(0)})^k_{jk} =0$ and $g_0^{ij}(\tilde{\Ga}^{(0)})^k_{ij}=0$ (which follow from \eqref{Gatilde 0} and \eqref{assum F1}) we have
\begin{equation}\label{R0 I}
\begin{aligned}
I & = g_0^{ij} \pth{  \dr_k \Ga(g_0)^k_{ij} - \dr_i  \Ga(g_0)^k_{jk}  }  -  (\tilde{\Ga}^{(0)})^k_{ij} \dr_k g_0^{ij}  + g_0^{ij} \pth{  \dr_k (\Ga^{(1)})^k_{ij} - \dr_i (\Ga^{(1)})^k_{jk}  }^{(-1)}
\\& =  g_0^{ij} \pth{  \dr_k \Ga(g_0)^k_{ij} - \dr_i  \Ga(g_0)^k_{jk}  }  -  (\tilde{\Ga}^{(0)})^k_{ij} \dr_k g_0^{ij}
\\&\quad + \sum_\A \sin\pth{\frac{u_\A}{\la}}  (\bar{F}^{(1)}_\A)_{ij}  \pth{ \dr^i \dr^j u_\A - \half\dr^\ell u_\A\dr_\ell g_0^{ij}}
\\&\quad + \sum_{\A}\sin\pth{\frac{u_\A}{\la}}\pth{ |\nab u_\A|^2_{g_0}\tr_{g_0}\ga^{(2,1)}_{\A} -(\ga^{(2,1)}_{\A})_{\nab_{g_0} u_\A \nab_{g_0} u_\A} }
\\&\quad + \sum_{\A\neq \B,\pm}\cos\pth{\frac{u_\A\pm u_\B}{\la}}\Big( |\nab_{g_0}(u_\A\pm u_\B)|^2_{g_0}\tr_{g_0}\ga^{(2,\pm)}_{\A\B} -(\ga^{(2,\pm)}_{\A\B})_{\nab_{g_0}(u_\A\pm u_\B)\nab_{g_0}(u_\A\pm u_\B)} \Big)
\\&\quad - 4 \sum_{\A}\cos\pth{\frac{2u_\A}{\la}}  |\nab_{g_0} u_\A|_{g_0}^2 F_\A^2
\\&\quad -  \frac{1}{8} \sum_{\A\neq\B,\pm}\cos\pth{\frac{u_\A\pm u_\B}{\la}}  |\nab_{g_0}(u_\A\pm u_\B)|^2_{g_0}\left| \bar{F}^{(1)}_\A \cdot \bar{F}^{(1)}_\B \right|_{g_0} ,
\end{aligned}
\end{equation}
where we have also used \eqref{Ga 1} and \eqref{barQ}. We now compute $II$, using \eqref{ga-1 1} and \eqref{Gatilde 0}:
\begin{equation}\label{R0 II}
\begin{aligned}
II & = -2 \sum_{\A}  |\nab_{g_0} u_\A|_{g_0}^2F_\A^2 -2 \sum_{\A} \cos\pth{\frac{2u_\A}{\la}}  |\nab_{g_0} u_\A|_{g_0}^2F_\A^2
\\&\quad  -\frac{1}{8}  \sum_{\A\neq\B,\pm}  \cos\pth{\frac{u_\A\pm u_\B}{\la}} \pth{ |\nab_{g_0} u_\A|_{g_0}^2 + |\nab_{g_0} u_\B|_{g_0}^2 } \left| \bar{F}^{(1)}_\A \cdot \bar{F}^{(1)}_\B \right|_{g_0} .
\end{aligned}
\end{equation}
We now compute $III$ using \eqref{Gatilde 0}:
\begin{equation}\label{R0 III}
\begin{aligned}
III & = g_0^{ij} \pth{\Ga(g_0)^k_{k\ell} \Ga(g_0)^\ell_{ij} -  \Ga(g_0)^k_{i\ell} \Ga(g_0)^\ell_{jk}} - 2g_0^{ij} (\tilde{\Ga}^{(0)})^\ell_{ik} \Ga(g_0)^k_{j\ell}   - g_0^{ij}  (\tilde{\Ga}^{(0)})^k_{i\ell} (\tilde{\Ga}^{(0)})^\ell_{jk}
\\& = g_0^{ij} \pth{\Ga(g_0)^k_{k\ell} \Ga(g_0)^\ell_{ij} -  \Ga(g_0)^k_{i\ell} \Ga(g_0)^\ell_{jk}} - 2g_0^{ij} (\tilde{\Ga}^{(0)})^\ell_{ik} \Ga(g_0)^k_{j\ell}
\\&\quad + \sum_{\A}  |\nab_{g_0} u_\A |_{g_0}^2 F_\A^2 - \sum_{\A} \cos\pth{\frac{2u_\A}{\la}}  |\nab_{g_0} u_\A |_{g_0}^2 F_\A^2
\\&\quad + \frac{1}{8}\sum_{\A\neq\B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}} \bigg( \pm 2 (\bar{F}^{(1)}_\A)_{i \nab_{g_0} u_\B} (\bar{F}^{(1)}_\B)_{\nab_{g_0} u_\A}^i
\\&\hspace{7cm} \mp  |\nab_{g_0} u_\A \cdot \nab_{g_0} u_\B |_{g_0}  \left| \bar{F}^{(1)}_\A \cdot \bar{F}^{(1)}_\B \right|_{g_0} \bigg).
\end{aligned}
\end{equation}
We now compute the $\la^0$ contribution of the quadratic terms in $\ka_\la$. From \eqref{assum F1} we have $\tr_{\ga_\la} \ka_\la = \tr_{g_0}k_0 + \GO{\la}$ so that
\begin{align}\label{tr2 0}
\pth{ \pth{\tr_{\ga_\la} \ka_\la}^2}^{(0)} = (\tr_{g_0}k_0)^2
\end{align}
\begin{equation}\label{k2 0}
\begin{aligned}
\pth{ |\ka_\la|^2_{\ga_\la} }^{(0)} & = |k_0|^2_{g_0} + \sum_\A \sin\pth{ \frac{u_\A}{\la}}|\nab_{g_0} u_\A|_{g_0} \left| \bar{F}^{(1)}_\A \cdot k_0 \right|_{g_0}
\\&\quad +  \sum_{\A} |\nab_{g_0} u_\A|_{g_0}^2  F_\A^2  -  \sum_{\A} \cos\pth{\frac{2u_\A}{\la}} |\nab_{g_0} u_\A|_{g_0}^2  F_\A^2
\\&\quad +  \frac{1}{8} \sum_{\A\neq\B,\pm}\mp\cos\pth{ \frac{u_\A\pm u_\B}{\la}}  |\nab_{g_0} u_\A|_{g_0} |\nab_{g_0} u_\B|_{g_0} \left| \bar{F}^{(1)}_\A \cdot \bar{F}^{(1)}_\B  \right|_{g_0}.
\end{aligned}
\end{equation}
Since the quadratic terms in $\ka_\la$ don't contribute to $\la^{-1}$ order we have proved that \[\pth{ \HH(\ga_\la,\ka_\la)}^{(-1)} =0.\] Moreover we have also proved that $\pth{ \HH(\ga_\la,\ka_\la)}^{(0)}$ is indeed given by \eqref{H 0}, putting together \eqref{R0 I}-\eqref{k2 0} and using in particular \eqref{BG hamiltonian} and \eqref{eikonal init} to cancel the resonant term. It only remains to prove \eqref{H 1}, since \eqref{H 2} follows from the definition of the Hamiltonian constraint operator $\HH$. In order to prove \eqref{H 1}, we only need to check that $\pth{ \HH(\ga_\la,\ka_\la)}^{(1)}$ is purely oscillating, i.e that is does not contain any resonant term. We use a schematic expression of $\pth{ \HH(\ga_\la,\ka_\la)}^{(1)}$ which follows from the definition of $\HH$:
\begin{equation}\label{H 1 rough}
\begin{aligned}
\pth{ \HH(\ga_\la,\ka_\la)}^{(1)} & = \pth{\ga_\la^{-1}}^{(0)}\pth{\dr^2\ga_\la}^{(1)} + \pth{\ga_\la^{-1}}^{(1)}\pth{\dr^2\ga_\la}^{(0)} + \pth{\ga_\la^{-1}}^{(2)}\pth{\dr^2\ga_\la}^{(-1)}
\\&\quad + \pth{\Ga(\ga_\la)}^{(0)}\pth{ \Ga(\ga_\la)}^{(1)} + \pth{ \ga_\la^{-1}\ka_\la}^{(0)}  \pth{ \ga_\la^{-1}\ka_\la}^{(1)} .
\end{aligned}
\end{equation}
We list every oscillating functions appearing in the five terms of the schematic expression \eqref{H 1 rough}:
\begin{itemize}
\item Thanks to \eqref{ga-1}, $\pth{\ga_\la^{-1}}^{(0)}\pth{\dr^2\ga_\la}^{(1)}$ contains $\cos\pth{\frac{u_\A}{\la}}$ and $\sin\pth{ \frac{u_\A\pm u_\B}{\la}}$.
\item Thanks to \eqref{ga-1 1}-\eqref{ga-1 2}, $ \pth{\ga_\la^{-1}}^{(1)}\pth{\dr^2\ga_\la}^{(0)}$ and $ \pth{\ga_\la^{-1}}^{(2)}\pth{\dr^2\ga_\la}^{(-1)}$ contain $\cos\pth{\frac{u_\A}{\la}}$, $\sin\pth{\frac{2u_\A}{\la}}$, $\sin\pth{ \frac{u_\A\pm u_\B}{\la}}$, $\cos\pth{ \frac{u_\A\pm2 u_\B}{\la}}$ and $\cos\pth{ \frac{u_\A\pm_1 u_\B \pm_2 u_\C}{\la}}$.
\item Thanks to \eqref{Gatilde 0}-\eqref{Ga 1}, $\pth{\Ga(\ga_\la)}^{(0)}\pth{ \Ga(\ga_\la)}^{(1)}$ contains $\cos\pth{\frac{u_\A}{\la}}$, $\sin\pth{\frac{2u_\A}{\la}}$, $\sin\pth{\frac{u_\A\pm u_\B}{\la}}$, $\cos\pth{ \frac{u_\A\pm2 u_\B}{\la}}$ and $\cos\pth{ \frac{u_\A\pm_1 u_\B \pm_2 u_\C}{\la}}$.
\item Thanks to \eqref{ga la}-\eqref{ka la} we see that the oscillating behaviour of $\ga_\la^{-1}\ka_\la$ is the same as the oscillating behaviour of $\Ga(\ga_\la)$ so that $\pth{ \ga_\la^{-1}\ka_\la}^{(0)}  \pth{ \ga_\la^{-1}\ka_\la}^{(1)}$ behaves as $\pth{\Ga(\ga_\la)}^{(0)}\pth{ \Ga(\ga_\la)}^{(1)}$.
\end{itemize}
This concludes the proof of the lemma.
\end{proof}

\begin{lemma}\label{lem constraint expansion M}
We have
\begin{align*}
\MM(\ga_\la,\ka_\la) & =  \pth{ \MM(\ga_\la,\ka_\la)}^{(0)} + \la \pth{ \MM(\ga_\la,\ka_\la)}^{(1)} + \la^2 \pth{ \MM(\ga_\la,\ka_\la)}^{(\geq 2)},
\end{align*}
\begin{equation}\label{M 0}
\begin{aligned}
&\pth{ \MM(\ga_\la,\ka_\la)_i}^{(0)}
\\& =  \sum_\A\sin\pth{ \frac{u_\A}{\la}} \bigg(  \dr_i u_\A \tr_{g_0}\ka^{(1,1)}_\A - (\ka^{(1,1)}_\A)_{\nab_{g_0} u_\A i}
\\&\quad\hspace{2.5cm} + \half  \dr^b \pth{|\nab_{g_0} u_\A|_{g_0} \bar{F}^{(1)}_\A}_{bi} +\frac{1}{4} |\nab_{g_0} u_\A|_{g_0} (\bar{F}^{(1)}_\A)_{bc} \dr_i g_0^{bc}
\\&\quad\hspace{2.5cm} +  \half |\nab_{g_0} u_\A|_{g_0} \pth{ \dr_a g_0^{ca} + \half g_0^{ab}g_0^{cd}\dr_d (g_0)_{ab} } (\bar{F}^{(1)}_\A)_{ic}  - \half \dr_i u_\A (\bar{F}^{(1)}_\A)_{bc} (k_0)^{bc} \bigg)
\\&\quad + \sum_\A\cos\pth{\frac{2u_\A}{\la}} \pth{ 2 (\ka^{(1,2)}_\A)_{\nab_{g_0} u_\A i} - 2 \dr_i u_\A \tr_{g_0}\ka^{(1,2)}_\A + 3 |\nab_{g_0} u_\A|_{g_0}  \dr_i u_\A F_\A^2}
\\&\quad +\sum_{\A\neq\B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}\pth{ (\ka^{(1,\pm)}_{\A\B})_{(\nab_{g_0} u_\A \pm \nab_{g_0} u_\B)i} - \dr_i(u_\A\pm u_\B)\tr_{g_0}  \ka^{(1,\pm)}_{\A\B}}
\\&\quad - \half \sum_{\A\neq\B} \cos\pth{\frac{u_\A}{\la}}  \cos\pth{ \frac{u_\B}{\la}} |\nab_{g_0} u_\B|_{g_0} \bigg((\bar{F}^{(1)}_\A)^{b}_{\nab_{g_0} u_\B} (\bar{F}^{(1)}_\B)_{bi}  - \dr_i u_\B \left| \bar{F}^{(1)}_\A \cdot \bar{F}^{(1)}_\B\right|_{g_0}\bigg)
\\&\quad -\frac{1}{4} \sum_{\A\neq\B} \sin\pth{\frac{u_\A}{\la}} \sin\pth{ \frac{u_\B}{\la}}   |\nab_{g_0} u_\B|_{g_0}   \dr_i u_\A \left| \bar{F}^{(1)}_\A \cdot \bar{F}^{(1)}_\B \right|_{g_0}.
\end{aligned}
\end{equation}
Moreover, the higher order terms satisfy
\begin{align}
\pth{ \MM(\ga_\la,\ka_\la)}^{(1)} & = \sum_{\substack{w\in\mathcal{N}\cup\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{w}{\la}}   \cro{\pth{ \ga_\la^{-1}\dr\ka_\la + \ga_\la^{-1}\ga_\la^{-1} \ka_\la\dr\ga_\la }^{(1)}} \label{M 1}
\\ \pth{ \MM(\ga_\la,\ka_\la)}^{(\geq 2)} & = \cro{\pth{ \ga_\la^{-1}\dr\ka_\la + \ga_\la^{-1}\ga_\la^{-1} \ka_\la\dr\ga_\la }^{(\geq 2)}}^\osc.  \label{M 2}
\end{align}
\end{lemma}

\begin{proof}
In coordinates we have
\begin{align*}
\MM(g,k)_i & =  g^{ab}\pth{ \dr_a k_{bi} - \dr_i  k_{ab}} + \pth{ \dr_a g^{ca} + \half g^{ab}g^{cd}\dr_d g_{ab} } k_{ic} - \half \dr_i g^{bc} k_{bc}   .
\end{align*}
Using again \eqref{assum F1} we easily obtain $\pth{ \MM(\ga_\la,\ka_\la)_i  }^{(-1)} = 0$. We now have $\pth{ \MM(\ga_\la,\ka_\la)_i}^{(0)}=I+II+III$ where
\begin{align*}
I & \vcentcolon = g_0^{ab}\pth{ \dr_a (\ka_\la)_{bi} - \dr_i  (\ka_\la)_{ab}}^{(0)},
\\ II & \vcentcolon = (\ga_\la^{ab})^{(1)} \pth{ \dr_a (\ka_\la)_{bi} - \dr_i  (\ka_\la)_{ab}}^{(-1)},
\\ III & \vcentcolon =  \pth{ \dr_a \ga_\la^{ca} + \half g_0^{ab}g_0^{cd}\dr_d (\ga_\la)_{ab} }^{(0)} \pth{(\ka_\la)_{ic}}^{(0)} - \half \pth{\dr_i \ga_\la^{bc}}^{(0)} \pth{(\ka_\la)_{bc}}^{(0)} .
\end{align*}
We start with $I$:
\begin{equation}\label{M0 I}
\begin{aligned}
I & = g_0^{ab}\pth{ \dr_a (k_0)_{bi} - \dr_i  (k_0)_{ab}}
\\&\quad + \sum_\A\sin\pth{ \frac{u_\A}{\la}} \bigg( \dr_i u_\A \tr_{g_0}\ka^{(1,1)}_\A - (\ka^{(1,1)}_\A)_{\nab_{g_0} u_\A i}
\\&\hspace{4cm} + \half \pth{ \dr^b \pth{|\nab_{g_0} u_\A|_{g_0} \bar{F}^{(1)}_\A}_{bi} + |\nab_{g_0} u_\A|_{g_0} (\bar{F}^{(1)}_\A)_{ab} \dr_i g_0^{ab}  } \bigg)
\\&\quad + 2\sum_\A\cos\pth{\frac{2u_\A}{\la}} \pth{  (\ka^{(1,2)}_\A)_{\nab_{g_0} u_\A i} - \dr_i u_\A \tr_{g_0}\ka^{(1,2)}_\A}
\\&\quad +\sum_{\A\neq\B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}\pth{ (\ka^{(1,\pm)}_{\A\B})_{(\nab_{g_0} u_\A \pm \nab_{g_0} u_\B)i} - \dr_i(u_\A\pm u_\B)\tr_{g_0}  \ka^{(1,\pm)}_{\A\B}}.
\end{aligned}
\end{equation}
We now compute $II$ and $III$ using \eqref{ga-1 1}:
\begin{align}
II & =  2 \sum_{\A} |\nab_{g_0} u_\A|_{g_0}  \dr_i u_\A F_\A^2 +  2 \sum_{\A} \cos\pth{\frac{2u_\A}{\la}} |\nab_{g_0} u_\A|_{g_0}  \dr_i u_\A F_\A^2 \label{M0 III}
\\&\quad - \frac{1}{4} \sum_{\A\neq\B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}  |\nab_{g_0} u_\B|_{g_0} \pth{(\bar{F}^{(1)}_\A)^{b}_{\nab_{g_0} u_\B} (\bar{F}^{(1)}_\B)_{bi} - \dr_i u_\B \left| \bar{F}^{(1)}_\A \cdot \bar{F}^{(1)}_\B\right|_{g_0}}. \non
\\ III & =  \pth{ \dr_a g_0^{ca} + \half g_0^{ab}g_0^{cd}\dr_d (g_0)_{ab} } (k_0)_{ic} - \half \dr_i g_0^{bc} (k_0)_{bc} -  \sum_{\A}  |\nab_{g_0} u_\A|_{g_0}   \dr_i u_\A F_\A^2 \non
\\&\quad +  \half \sum_\A \sin\pth{ \frac{u_\A}{\la}} \bigg( |\nab_{g_0} u_\A|_{g_0} \pth{ \dr_a g_0^{ca} + \half g_0^{ab}g_0^{cd}\dr_d (g_0)_{ab} } (\bar{F}^{(1)}_\A)_{ic} \non
\\&\hspace{5cm} - \half  |\nab_{g_0} u_\A|_{g_0} (\bar{F}^{(1)}_\A)_{bc} \dr_i g_0^{bc}  - \dr_i u_\A (\bar{F}^{(1)}_\A)_{bc} (k_0)^{bc} \bigg) \label{M0 III}
\\&\quad  +   \sum_{\A}  \cos\pth{\frac{2u_\A}{\la}}   |\nab_{g_0} u_\A|_{g_0}   \dr_i u_\A F_\A^2 \non
\\&\quad -\frac{1}{4} \sum_{\A\neq\B} \sin\pth{\frac{u_\A}{\la}} \sin\pth{ \frac{u_\B}{\la}}   |\nab_{g_0} u_\B|_{g_0}   \dr_i u_\A \left| \bar{F}^{(1)}_\A \cdot \bar{F}^{(1)}_\B \right|_{g_0}. \non
\end{align}
Collecting \eqref{M0 I}-\eqref{M0 III} and using \eqref{BG momentum} and \eqref{eikonal init} to cancel the non-oscillating term we have proved that $\pth{ \MM(\ga_\la,\ka_\la)_i}^{(0)}$ is indeed given by \eqref{M 0}. It only remains to prove \eqref{M 1}, since \eqref{M 2} follows from the definition of the momentum constraint operator $\MM$. This actually follows from \eqref{H 1} since $\ka_\la$ has the same oscillating behaviour than a derivative of $\ga_\la$. In particular $\pth{ \MM(\ga_\la,\ka_\la)}^{(1)}$ and $\pth{ \HH(\ga_\la,\ka_\la)}^{(1)}$ share the same oscillating behaviour. This concludes the proof of the lemma.
\end{proof}

We now estimate the remainders in \eqref{hamiltonian}-\eqref{momentum}.

\begin{lemma}
We have
\begin{align*}
\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la} & = \la\pth{\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} + \la^2\pth{\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(\geq 2)},
\\ \RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la} & = \la\pth{\RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} + \la^2\pth{\RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(\geq 2)},
\end{align*}
\begin{align}
\pth{\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} & = \sum_{\substack{w\in\mathcal{N}\cup\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{w}{\la}} \cro{ \pth{1+\bar{F}^{(1)}} \pth{X^{(2,1)} + X^{(2,2)} + X^{(2,\pm)}} }  \label{reste H 1},
\\ \pth{\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(\geq 2)} & =  \pth{\ffi^{(2)} + \tffi_\la + \la\ffi^{(3)} }\HH\pth{\ga_\la,\ka_\la}  \label{reste H 2}
\\&\hspace{0.5cm} +\Big\{ \pth{ \ffi_\la (\ga_\la^{-1})^2 \pth{ \dr X_\la + \ga_\la^{-1} \dr\ga_\la X_\la} ( \ka_\la + \dr X_\la + \ga_\la^{-1} \dr\ga_\la X_\la)  }^{(\geq 2)}\Big\}^\osc \non,
\end{align}
\begin{align}
\pth{\RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} & = \sum_{\substack{w\in\mathcal{N}\cup\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{w}{\la}} \cro{ \pth{1+\bar{F}^{(1)}} \pth{\ffi^{(2,1)} + \ffi^{(2,2)} + \ffi^{(2,\pm)}} },  \label{reste M 1}
\\ \pth{\RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(\geq 2)} & = \cro{ \pth{\ffi_\la^{-1} \ga_\la^{-1} \dr \ffi_\la (\ka_\la + \dr X_\la + \ga_\la^{-1} \dr\ga_\la X_\la)}^{(\geq 2)} }^\osc. \label{reste M 2}
\end{align}
\end{lemma}

\begin{proof}
We start with $\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la} $ defined in \eqref{reste H}. Since $\ffi_\la-1=\GO{\la^2}$ (see \eqref{def ffi la}) and $\HH\pth{\ga_\la,\ka_\la}=\GO{1}$ (see Lemma \ref{lem constraint expansion H}), $( \ffi_\la-1)\HH\pth{\ga_\la,\ka_\la}$ does not contribute to $\pth{\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} $. We find that $\pth{\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} $ is of the schematic form
\begin{align*}
\pth{\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} & = \pth{ \ffi_\la (\ga_\la^{-1})^2 \pth{ \dr X_\la + \ga_\la^{-1} \dr\ga_\la X_\la} ( \ka_\la + \dr X_\la + \ga_\la^{-1} \dr\ga_\la X_\la)  }^{(1)}
\\& = (g_0^{-1})^2 \pth{ \dr X_\la }^{(1)} ( \ka_\la)^{(0)}
\end{align*}
where we also used $\ga_\la^{-1} \dr\ga_\la X_\la=\GO{\la^2}$ and $\dr X_\la=\GO{\la}$. Using now $\pth{ \dr X_\la }^{(1)}= \pth{ \dr X^{(2)} }^{(-1)}$, \eqref{def X 2} and \eqref{ka la} we finally obtain \eqref{reste H 1}. Moreover, \eqref{reste H 2} can be deduced from the schematic expression of $\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la} $. We now turn to $\RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la} $ defined in \eqref{reste M}.
\begin{align*}
\pth{\RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} & = \pth{\ffi_\la^{-1} \ga_\la^{-1} \dr \ffi_\la (\ka_\la + \dr X_\la + \ga_\la^{-1} \dr\ga_\la X_\la) }^{(1)}
\\& = g_0^{-1} \pth{ \dr \ffi_\la}^{(1)} (\ka_\la)^{(0)}
\end{align*}
where we used $\dr\ffi_\la = \GO{\la}$. Using now $\pth{ \dr \ffi_\la }^{(1)}= \pth{ \dr \ffi^{(2)} }^{(-1)}$, \eqref{def ffi 2} and \eqref{ka la} we finally obtain \eqref{reste M 1}. Moreover, \eqref{reste H 2} can be deduced from the schematic expression of $\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la} $. This concludes the proof of the lemma.
\end{proof}

Finally we expand the elliptic operators appearing on the RHS of \eqref{hamiltonian}-\eqref{momentum}.

\begin{lemma}\label{lem exp elliptic}
Let $f$ and $v$ be scalar functions and $Z$ a vector field on $\Si_0$ and let $T\in\{\cos,\sin\}$.
\begin{itemize}
\item[(i)] We have
\begin{align}\label{De ga la f}
\De_{\ga_\la} f & = \De_e f  + \cro{ (\ga_\la^{-1}-e^{-1})\dr^2 f + \ga_\la^{-1}\Ga(\ga_\la) \dr f }^\osc.
\end{align}
\begin{align}
\pth{\De_{\ga_\la}\pth{ T\pth{\frac{v}{\la}}f}}^{(-2)} & = T''\pth{\frac{v}{\la}} |\nab_{g_0}v|^2_{g_0} f \label{De ga la -2},
\\ \pth{\De_{\ga_\la}\pth{ T\pth{\frac{v}{\la}}f}}^{(-1)} & = T'\pth{\frac{v}{\la}}  \cro{ \dr^{\leq 1} f   } + \sum_\A T\pth{\frac{v}{\la}}\cos\pth{\frac{u_\A}{\la}} \cro{\bar{F}^{(1)} f} ,\label{De ga la -1}
\\ \pth{\De_{\ga_\la}\pth{ T\pth{\frac{v}{\la}}f}}^{(\geq 0)} & = \cro{\pth{\ga_\la^{-1}\dr^2 f + \ga_\la^{-1}\ga_\la^{-1}\dr\ga_\la\dr f}^{(\geq 0)}}^\osc.\label{De ga la 0}
\end{align}
\item[(ii)] We have
\begin{align}
\pth{\vDe_{\ga_\la} Z_i}^{(-1)} & = \sum_\A  \cos\pth{\frac{u_\A}{\la}}|\nab_{g_0} u_\A|^2_{g_0} (\bar{F}^{(1)}_\A)_{ij} Z^j , \label{vDe Z -1}
\\ \pth{\vDe_{\ga_\la} Z_i}^{(\geq 0)} & =  \De_eZ_i  + \Big\{(\ga_\la^{-1}-e^{-1})\dr^2 Z + \ga_\la^{-1}\Ga(\ga_\la) \dr Z \label{vDe Z 0}
\\&\hspace{2cm} + \pth{\pth{\ga_\la^{-1}\dr\Ga(\ga_\la)}^{(\geq 0)} + \ga_\la^{-1} \Ga(\ga_\la)\Ga(\ga_\la)} Z \Big\}^\osc_i. \non
\end{align}
\begin{align}
\pth{\vDe_{\ga_\la}\pth{T\pth{\frac{v}{\la}}Z}_i }^{(-2)} & = T''\pth{\frac{v}{\la}} |\nab_{g_0}v|^2_{g_0} Z_i \label{vDe osc -2}
\\ \pth{\vDe_{\ga_\la}\pth{T\pth{\frac{v}{\la}}Z}_i }^{(-1)} & =  T'\pth{\frac{v}{\la}}\cro{ \dr^{\leq 1} Z }_i + \sum_\A \cos\pth{\frac{u_\A}{\la}} T\pth{\frac{v}{\la}}\cro{ Z \bar{F}^{(1)} }_i \label{vDe osc -1}
\end{align}
\begin{equation}\label{vDe osc 0}
\begin{aligned}
&\pth{\vDe_{\ga_\la}\pth{T\pth{\frac{v}{\la}}Z}_i }^{(\geq 0)}
\\& =   \Big\{ \pth{ \pth{\ga_\la^{-1}}^{(\geq 2)} + \pth{ \ga_\la^{-1} \Ga(\ga_\la)}^{(\geq 1)} + \pth{\ga_\la^{-1} \dr \Ga(\ga_\la)}^{(\geq 0)} + \ga_\la^{-1}\Ga(\ga_\la)\Ga(\ga_\la)} Z
\\&\hspace{5cm} +\pth{ \pth{\ga_\la^{-1}}^{(\geq 1)} + \ga_\la^{-1} \Ga(\ga_\la)}\dr^{\leq 1} Z  + \ga_\la^{-1}\dr^2 Z \Big\}_i^{\osc} .
\end{aligned}
\end{equation}
\end{itemize}
\end{lemma}

\begin{proof}
The identity \eqref{De ga la f} follows directly from the expression of the Laplace-Beltrami operator in coordinates. The following computation is the elliptic equivalent of \eqref{Box T f}:
\begin{align*}
\De_{\ga_\la}\pth{ T\pth{\frac{v}{\la}}f} & =  \frac{1}{\la^2} T''\pth{\frac{v}{\la}} \ga_\la^{-1}(\d v,\d v) f +  \frac{1}{\la}T'\pth{\frac{v}{\la}}  \pth{ 2\ga_\la^{ij}\dr_j v \dr_i f + (\De_{\ga_\la} v) f   }  +T\pth{\frac{v}{\la}}\De_{\ga_\la}f.
\end{align*}
We use the expansion of the inverse of $\ga_\la$ given by \eqref{ga-1} to obtain \eqref{De ga la -2}-\eqref{De ga la 0}. We now turn to the elliptic operator $\vDe_{\ga_\la}$ acting on vector fields. Thanks to the expression of the Ricci tensor in coordinates we find
\begin{equation}\label{vDe coordinates}
\begin{aligned}
\vDe_{g}Z_i & = g^{k\ell}  \dr_k \dr_\ell Z_i  - 2g^{k\ell} \Ga(g)_{\ell i}^a \dr_k Z_a - g^{k\ell} Z_a \pth{ \dr_k  \Ga(g)_{\ell i}^a - \Ga(g)_{ki}^{b}\Ga(g)_{\ell b}^a  }
\\&\quad + Z^{j}\dr_a \Ga(g)^a_{ij} - Z^{j} \dr_j \Ga(g)^a_{ai} + Z^{j}\Ga(g)^a_{ab} \Ga(g)^b_{ij} - Z^{j}\Ga(g)^a_{ib} \Ga(g)^b_{aj}.
\end{aligned}
\end{equation}
This already gives
\begin{align*}
\pth{ \vDe_{\ga_\la}Z_i }^{(-1)} & =  Z_b\pth{ - g_0^{k\ell}  \dr_k (\tilde{\Ga}^{(0)})_{\ell i}^b + g_0^{jb}\pth{\dr_a (\tilde{\Ga}^{(0)})^a_{ij} - \dr_j (\tilde{\Ga}^{(0)})^a_{ai}}}^{(-1)}
\\& = \sum_\A  \cos\pth{\frac{u_\A}{\la}}|\nab u_\A|^2 (\bar{F}^{(1)}_\A)_{ij} Z^j.
\end{align*}
Moreover this also gives \eqref{vDe Z 0}. Thanks to \eqref{vDe coordinates} we finally obtain
\begin{align*}
\vDe_{\ga_\la}\pth{T\pth{\frac{v}{\la}}Z}_i & = \frac{1}{\la^2} \ga_\la^{k\ell}  T''\pth{\frac{v}{\la}}\dr_k v \dr_\ell v Z_i
\\&\quad +  \frac{1}{\la}T'\pth{\frac{v}{\la}}\pth{2 \ga_\la^{k\ell} \dr_\ell v \dr_k Z_i + \ga_\la^{k\ell}  \dr_k\dr_\ell v Z_i - 2 \ga_\la^{k\ell} \Ga(\ga_\la)_{\ell i}^a \dr_k v Z_a }
\\&\quad +  T\pth{\frac{v}{\la}}\pth{Z^{j}\dr_a \Ga(\ga_\la)^a_{ij} - Z^{j} \dr_j \Ga(\ga_\la)^a_{ai}  - \ga_\la^{k\ell} Z_a  \dr_k  \Ga(\ga_\la)_{\ell i}^a  }
\\&\quad + T\pth{\frac{v}{\la}}\pth{\ga_\la^{k\ell} Z_a \Ga(\ga_\la)_{ki}^{b}\Ga(\ga_\la)_{\ell b}^a  + Z^{j}\Ga(\ga_\la)^a_{ab} \Ga(\ga_\la)^b_{ij} -Z^{j}\Ga(\ga_\la)^a_{ib} \Ga(\ga_\la)^b_{aj}}
\\&\quad + T\pth{\frac{v}{\la}}\pth{ - 2\ga_\la^{k\ell} \Ga(\ga_\la)_{\ell i}^a \dr_kZ_a  + \ga_\la^{k\ell} \dr_k\dr_\ell Z_i }.
\end{align*}
We use the expansion of the inverse of $\ga_\la$ given by \eqref{ga-1} and of the Christofel symbols given by \eqref{christofel symbols} to deduce \eqref{vDe osc -2}-\eqref{vDe osc 0}. This concludes the proof of the lemma.
\end{proof}

We are now ready to define the constraint hierarchy, i.e to deduce from Lemmas \ref{lem constraint expansion H}-\ref{lem exp elliptic} a hierarchy of equations for the various terms in $_$ and $X_$ such that they solve \eqref{hamiltonian}-\eqref{momentum}. The hierarchy associated to the Hamiltonian constraint is
\begin{align}
8 \pth{ \De_{\ga_\la}\ffi^{(2)}}^{(-2)} & = \pth{ \HH(\ga_\la,\ka_\la) }^{(0)},\label{eq ffi2}
\\ 8 \pth{ \De_{\ga_\la}\ffi^{(3)}}^{(-2)} & =  \pth{ \HH(\ga_\la,\ka_\la) }^{(1)} + \pth{\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} - 8 \pth{ \De_{\ga_\la}\ffi^{(2)}}^{(-1)},\label{eq ffi3}
\\ 8 \De_{\ga_\la}\tffi_\la & = \pth{\HH\pth{\ga_\la,\ka_\la}}^{(\geq 2)} + \pth{\RR_\HH\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(\geq 2)} - 8 \pth{ \De_{\ga_\la}\ffi^{(2)}}^{(\geq 0)}  \label{eq tffi}
\\&\quad - 8 \pth{ \De_{\ga_\la}\ffi^{(3)}}^{(\geq -1)}.\non
\end{align}
The hierarchy associated to the momentum constraint is
\begin{align}
\pth{\vDe_{\ga_\la}X^{(2)} }^{(-2)} & = - \pth{ \MM(\ga_\la,\ka_\la) }^{(0)}, \label{eq X2}
\\ \pth{\vDe_{\ga_\la}X^{(3)} }^{(-2)} & = - \pth{ \MM(\ga_\la,\ka_\la) }^{(1)} + \pth{\RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la}}^{(1)} - \pth{\vDe_{\ga_\la}X^{(2)} }^{(-1)} \label{eq X3}
\\&\quad - \pth{\vDe_{\ga_\la}\tX_\la}^{(-1)}, \non
\\ \pth{\vDe_{\ga_\la}\tX_\la}^{(\geq 0)} & =  -\pth{\MM(\ga_\la,\ka_\la)}^{(\geq 2)}  + \pth{\RR_\MM\pth{ \ga_\la,\ka_\la,\ffi_\la,X_\la} }^{(\geq 2)} - \pth{\vDe_{\ga_\la}X^{(2)} }^{(\geq 0)} \label{eq tX}
\\&\quad  - \pth{\vDe_{\ga_\la}X^{(3)} }^{(\geq -1)} .\non
\end{align}

\begin{remark}\label{remark couplage}
The momentum hierarchy \eqref{eq X2}-\eqref{eq tX} differs from the Hamiltonian hierarchy \eqref{eq ffi2}-\eqref{eq tffi} because of the term $\pth{\vDe_{\ga_\la}\tX_\la}^{(-1)}$ in \eqref{eq X3}. The presence of this term is due to the fact that the operator $\vDe_{\ga_\la}$ acting on vector fields contains second derivatives of $\ga_\la$ and thus looses one power of $\la$ even when applied to a non-oscillating vector field such as $\tX_\la$. This is not the case for the operator $\De_{\ga_\la}$ acting on functions since it only contains up to first derivatives of $\ga_\la$.
\end{remark}

\subsection{Solving the hierarchy}\label{section solving constraint}

In this section we construct a solution $^(2), ^(3), , X^(2), X^(3), $ of the equations \eqref{eq ffi2}-\eqref{eq tX}.

\paragraph{The $\la^0$ equations.}

We start by defining $^(2)$ and $X^(2)$ such that they solve \eqref{eq ffi2} and \eqref{eq X2}. On the one hand, \eqref{def ffi 2}-\eqref{def X 2} together with \eqref{De ga la -2} and \eqref{vDe osc -2} imply
\begin{align*}
8 \pth{ \De_{\ga_\la}\ffi^{(2)}}^{(-2)} & = - 8 \sum_\A  \sin\pth{\frac{u_\A}{\la}} |\nab_{g_0}u_\A|^2_{g_0} \ffi^{(2,1)}_\A -32 \sum_\A   \cos\pth{\frac{2u_\A}{\la}} |\nab_{g_0}u_\A|^2_{g_0} \ffi^{(2,2)}_\A
\\&\quad - 8 \sum_{\A\neq \B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}} |\nab_{g_0}(u_\A\pm u_\B)|^2_{g_0} \ffi^{(2,\pm)}_{\A\B},
\\ \pth{\vDe_{\ga_\la}X^{(2)} }^{(-2)}_i & = -\sum_\A  \sin\pth{\frac{u_\A}{\la}} |\nab_{g_0}u_\A|^2_{g_0} (X^{(2,1)}_\A)_i  -4 \sum_\A \cos\pth{\frac{2u_\A}{\la}}|\nab_{g_0}u_\A|^2_{g_0}(X^{(2,2)}_\A)_i
\\&\quad - \sum_{\A\neq \B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}|\nab_{g_0}(u_\A\pm u_\B)|^2_{g_0}(X^{(2,\pm)}_{\A\B})_i.
\end{align*}
On the other hand, \eqref{H 0} and \eqref{M 0} give the expression of the RHS of \eqref{eq ffi2} and \eqref{eq X2}. Therefore, solving \eqref{eq ffi2} and \eqref{eq X2} is equivalent to defining $^(2,1)_$, $ ^(2,2)_$, $^(2,±)_$ by
\begin{align}
\ffi^{(2,1)}_\A & \vcentcolon = \wc{\ffi}^{(2,1)}_{\A} , \label{ffi 2 1 def}
\\  \ffi^{(2,2)}_\A & \vcentcolon =  \frac{3}{16}  F_\A^2 , \label{ffi 2 2 def}
\\   \ffi^{(2,\pm)}_{\A\B} & \vcentcolon = -\frac{1}{8} \pth{\tr_{g_0}\ga^{(2,\pm)}_{\A\B} -(\ga^{(2,\pm)}_{\A\B})_{N^{(\pm)}_{\A\B}N^{(\pm)}_{\A\B}}} +  \wc{\ffi}^{(2,\pm)}_{\A\B}, \label{ffi 2 pm def}
\end{align}
and $(X^(2,1)_)_i$, $(X^(2,2)_)_i  $,  $(X^(2,±)_)_i$ by
\begin{align}
(X^{(2,1)}_\A)_i & \vcentcolon = \frac{1}{ |\nab_{g_0}u_\A|^2_{g_0}} \pth{ \dr_i u_\A \tr_{g_0}\ka^{(1,1)}_\A - (\ka^{(1,1)}_\A)_{\nab_{g_0} u_\A i}  +  (\wc{X}^{(2,1)}_\A)_i }, \label{X 2 1 def}
\\ (X^{(2,2)}_\A)_i  & \vcentcolon = \frac{1}{4 |\nab_{g_0}u_\A|^2_{g_0}} \pth{ 2 (\ka^{(1,2)}_\A)_{\nab_{g_0} u_\A i} - 2 \dr_i u_\A \tr_{g_0}\ka^{(1,2)}_\A + 3 |\nab_{g_0} u_\A|_{g_0}  \dr_i u_\A F_\A^2},
\\ (X^{(2,\pm)}_{\A\B})_i & \vcentcolon = \frac{1}{|\nab_{g_0}(u_\A\pm u_\B)|^2_{g_0}}\pth{ (\ka^{(1,\pm)}_{\A\B})_{(\nab_{g_0} u_\A \pm \nab_{g_0} u_\B)i} - \dr_i(u_\A\pm u_\B)\tr_{g_0}  \ka^{(1,\pm)}_{\A\B} + (\wc{X}^{(2,\pm)}_{\A\B})_i } . \label{X 2 pm def}
\end{align}
where we recognized the expression of $^(2,1)_$, $ ^(2,±)_$, $X^(2,1)_$ and $X^(2,±)_$ given by \eqref{wcffi 21}-\eqref{wcX 2pm}.

\paragraph{The $\la^1$ equations.} We now define $^(3)$ and $X^(3)$ such that they solve \eqref{eq ffi3} and \eqref{eq X3}. Thanks to \eqref{H 1}, \eqref{reste H 1}, \eqref{def ffi 2} and \eqref{De ga la -1} the equation \eqref{eq ffi3} rewrites
\begin{align}
\pth{ \De_{\ga_\la}\ffi^{(3)}}^{(-2)} & = \sum_{\substack{w\in\mathcal{N}\cup\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{w}{\la}} S^\HH_{w,T} , \label{eq ffi3 bis}
\end{align}
where each $S^_w,T$ is of the form
\begin{align*}
S^\HH_{w,T} & = \bigg\{ \pth{ \ga_\la^{-1}\dr^2\ga_\la+ \ga_\la^{-1}\ga_\la^{-1}\pth{(\dr\ga_\la)^2 + (\ka_\la)^2} }^{(1)}  +  \dr\pth{  \ffi^{(2,1)}  +  \ffi^{(2,2)} + \ffi^{(2,\pm)} }
\\&\hspace{2cm} +  \pth{1+\bar{F}^{(1)}} \pth{X^{(2,1)} + X^{(2,2)} + X^{(2,\pm)} + \ffi^{(2,1)}  +  \ffi^{(2,2)} + \ffi^{(2,\pm)}}    \bigg\}.
\end{align*}
We now define $^(3)$ by
\begin{align}
\ffi^{(3)} & \vcentcolon = - \sum_{\substack{w\in\mathcal{N}\cup\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{w}{\la}}  \frac{S^\HH_{w,T}}{|\nab_{g_0}w|^2_{g_0} } . \label{def ffi 3}
\end{align}
Using \eqref{De ga la -2} and $T”=-T$ for $T∈{cos,sin}$ one can check that $^(3)$ solves \eqref{eq ffi3 bis} and thus \eqref{eq ffi3}. We now turn to \eqref{eq X3}, which thanks to \eqref{M 1}, \eqref{reste M 1}, \eqref{def X 2}, \eqref{vDe Z -1} and \eqref{vDe osc -1} rewrites
\begin{align}\label{eq X3 bis}
\pth{\vDe_{\ga_\la}X^{(3)} }^{(-2)}_i & = \sum_{\substack{w\in\mathcal{N}\cup\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{w}{\la}}  (S^\MM_{w,T})_i - \sum_\A  \cos\pth{\frac{u_\A}{\la}}|\nab_{g_0} u_\A|^2_{g_0} (\bar{F}^{(1)}_\A)_{ij} \tX_\la^j ,
\end{align}
where each $S^_w,T$ is of the form
\begin{align*}
S^\MM_{w,T} & =  \bigg\{ \pth{ \ga_\la^{-1}\dr\ka_\la + \ga_\la^{-1}\ga_\la^{-1} \ka_\la\dr\ga_\la }^{(1)}+  \dr\pth{  X^{(2,1)}  +  X^{(2,2)} + X^{(2,\pm)}}
\\&\hspace{1cm} +  \pth{1+\bar{F}^{(1)}} \pth{X^{(2,1)} + X^{(2,2)} + X^{(2,\pm)}+\ffi^{(2,1)} + \ffi^{(2,2)} + \ffi^{(2,\pm)}}  \bigg\}.
\end{align*}
In order to solve this equation, we further decompose $X^(3)$ into
\begin{align}\label{def X 3}
X^{(3)}_i & \vcentcolon = - \sum_{\substack{w\in\mathcal{N}\cup\mathcal{I}\\T\in\{\cos,\sin\}}} T\pth{\frac{w}{\la}} \frac{ (S^\MM_{w,T})_i }{ |\nab_{g_0}w|^2_{g_0}} + \sum_\A \cos\pth{\frac{u_\A}{\la}} (\bar{F}^{(1)}_\A)_{ij} \tX_\la^j .
\end{align}
Using \eqref{vDe osc -2} and $T”=-T$ for $T∈{cos,sin}$ one can check that $X^(3)$ solves \eqref{eq X3 bis} and thus \eqref{eq X3}. Following Remark \ref{remark couplage}, note that $X^(3)$ depends on the remainder $_$.

\paragraph{The $\la^2$ equations.} We conclude the resolution of \eqref{hamiltonian}-\eqref{momentum} by solving for $_$ and $_$. This step is identical to the singlephase case considered in [29] since the multiphase aspect of the current construction does not change the nature of the equations for the remainders. We point out two harmless difference:
\begin{itemize}
\item The conformal formulation here is slightly different from the one used in [29]. In particular the elliptic operator for the vector equation is different but as in [29] we will benefit from \eqref{vDe Z 0} and invert $\De_e$.
\item The free tensors $\ga^{(2,\pm)}_{\A\B}$, $\ka^{(1,1)}_\A$, $\ka^{(1,2)}_\A$ and $\ka^{(1,\pm)}_{\A\B}$ are not strictly speaking present in [29] but they can easily be estimated with the assumption \eqref{estim ga kappa param}.
\end{itemize}
Therefore, following the same steps as Section 6 of [29] we obtain the existence of a unique couple $_,_∈H^N-3_^2$ solving \eqref{hamiltonian}-\eqref{momentum} and satisfying moreover
\begin{align}\label{estim tffi tX}
\l \tffi_\la \r_{H^{k+2}_\de} + \l \tX_\la \r_{H^{k+2}_\de} \lesssim \frac{\e}{\la^k}
\end{align}
for $k∈0,N-5$.

\subsection{Final form of the solution}

In the previous section we have a constructed a solution $(_,X_)$ to the system \eqref{hamiltonian}-\eqref{momentum}. Therefore we have constructed a solution of the constraint equations $(g_,k_)$ of the form \eqref{conformal ansatz}. To conclude the proof of Proposition \ref{prop constraint main} we need to recover the expressions \eqref{g la}-\eqref{k la} as well as the estimates \eqref{estim reste cons 1}-\eqref{estim reste cons 2}.

\paragraph{The induced metric.} We have $g_=_^4_$, so that \eqref{ga la} and \eqref{def ffi la} implies
\begin{align*}
g_\la & = g_0 + \la \sum_\A \cos\pth{\frac{u_\A}{\la}} \bar{F}^{(1)}_\A + 4\la^2\pth{ \ffi^{(2)} + \tffi_\la}g_0 + \la^2 \sum_{\A\neq \B,\pm}\cos\pth{\frac{u_\A\pm u_\B}{\la}}\ga^{(2,\pm)}_{\A\B}
\\&\quad + \la^3 \cro{ \pth{\ffi^{(2)} + \tffi_\la + \pth{\ffi_\la^4}^{(\geq 3)}}\pth{\bar{F}^{(1)}+\ga^{(2,\pm)}}  }^\osc.
\end{align*}
Using now \eqref{def ffi 2} we obtain
\begin{align*}
g_\la & = g_0 + \la \sum_\A \cos\pth{\frac{u_\A}{\la}} \bar{F}^{(1)}_\A + \la^2 \sum_\A  \sin\pth{\frac{u_\A}{\la}} 4\ffi^{(2,1)}_\A g_0 + \la^2 \sum_\A \cos\pth{\frac{2u_\A}{\la}}4\ffi^{(2,2)}_\A g_0
\\&\quad + \la^2 \sum_{\A\neq \B,\pm} \cos\pth{\frac{u_\A\pm u_\B}{\la}}\pth{ \ga^{(2,\pm)}_{\A\B} + 4\ffi^{(2,\pm)}_{\A\B}g_0 }
\\&\quad + \la^2 \tffi_\la g_0 + \la^3 \cro{ \pth{\ffi^{(2)} + \tffi_\la + \pth{\ffi_\la^4}^{(\geq 3)}}\pth{\bar{F}^{(1)}+\ga^{(2,\pm)}}  }^\osc.
\end{align*}
Using now \eqref{ffi 2 pm def} we obtain
\begin{align*}
\ga^{(2,\pm)}_{\A\B} + 4\ffi^{(2,\pm)}_{\A\B}g_0 & = \bar{\PP}^{[1]}_{u_\A\pm u_\B}\pth{\ga^{(2,\pm)}_{\A\B}} + 4 \wc{\ffi}^{(2,\pm)}_{\A\B} g_0,
\end{align*}
where we recognized the operator $^[1]_u_±u_$ defined by \eqref{def P1(T)}. Using also \eqref{ffi 2 1 def} and \eqref{ffi 2 2 def}, and defining
\begin{align*}
h_\la & \vcentcolon = \tffi_\la g_0 + \la \cro{ \pth{\ffi^{(2)} + \tffi_\la + \pth{\ffi_\la^4}^{(\geq 3)}}\pth{\bar{F}^{(1)}+\ga^{(2,\pm)}}  }^\osc
\end{align*}
allows us to recover the expression \eqref{g la} for $g_$. The estimate \eqref{estim reste cons 1} follows from \eqref{estim seed}, \eqref{estim ga kappa param}, the definition of $^(2)$ and $^(3)$ and \eqref{estim tffi tX}.

\paragraph{The second fundamental form.} We have
\begin{align*}
k_\la & = \ffi^2_\la\pth{  \ka_\la  +  \L_{\ga_\la} X_\la} .
\end{align*}
We first expand $Ł__X_$. Thanks to \eqref{def LX}, \eqref{def X la} and \eqref{def X 2} we have
\begin{align*}
\L_{\ga_\la}X_\la & =   \la \sum_\A \cos\pth{\frac{u_\A}{\la}} \nab_{g_0}u_\A \tilde{\otimes} X^{(2,1)}_\A -2\la\sum_\A \sin\pth{\frac{2u_\A}{\la}}\nab_{g_0}u_\A \tilde{\otimes} X^{(2,2)}_\A
\\&\quad  - \la \sum_{\A\neq \B,\pm} \sin\pth{\frac{u_\A\pm u_\B}{\la}} \nab_{g_0}(u_\A\pm u_\B) \tilde{\otimes} X^{(2,1)}_\A
\\&\quad + \la^2 \cro{\pth{ \pth{1+\ga_\la^{-1}\ga_\la}\pth{\dr X_\la + \ga_\la^{-1} X_\la\dr\ga_\la} }^{(\geq 2)}}^\osc,
\end{align*}
where we also used the notation introduced in \eqref{def produit}. Using now also \eqref{ka la} we obtain
\begin{align*}
k_\la & = k_0 + \sum_\A \sin\pth{ \frac{u_\A}{\la}}  \half |\nab u_\A| \bar{F}^{(1)}_\A   + \la\sum_\A\cos\pth{\frac{u_\A}{\la}}\pth{ \ka^{(1,1)}_\A  + \nab_{g_0}u_\A \tilde{\otimes} X^{(2,1)}_\A}
\\&\quad + \la\sum_\A \sin\pth{\frac{2u_\A}{\la}} \pth{ \ka^{(1,2)}_\A -2 \nab_{g_0}u_\A \tilde{\otimes} X^{(2,2)}_\A}
\\&\quad + \la\sum_{\A\neq\B,\pm}\sin\pth{\frac{u_\A\pm u_\B}{\la}}\pth{ \ka^{(1,\pm)}_{\A\B}  - \nab_{g_0}(u_\A\pm u_\B) \tilde{\otimes} X^{(2,1)}_\A }
\\&\quad + \la^2 \cro{ \pth{ \ka_\la  +  \L_{\ga_\la} X_\la}^{(\geq 2)} +\pth{\ffi^2_\la}^{(\geq 2)}\pth{ \ka_\la  +  \L_{\ga_\la} X_\la}  }^\osc.
\end{align*}
We use now \eqref{X 2 1 def}-\eqref{X 2 pm def} to obtain
\begin{align*}
\ka^{(1,1)}_\A  + \nab_{g_0}u_\A \tilde{\otimes} X^{(2,1)}_\A & = \bar{\PP}^{[2]}_{u_\A}\pth{\ka^{(1,1)}_\A} + \frac{N_\A \tilde{\otimes} \wc{X}^{(2,1)}_\A}{ |\nab_{g_0}u_\A|_{g_0}} ,
\\ \ka^{(1,2)}_\A -2 \nab_{g_0}u_\A \tilde{\otimes} X^{(2,2)}_\A & = \bar{\PP}^{[2]}_{u_\A}\pth{\ka^{(1,2)}_\A}  -\frac{3}{2} |\nab_{g_0} u_\A|_{g_0} F_\A^2 N_\A\tilde{\otimes} N_\A,
\\ \ka^{(1,\pm)}_{\A\B}  - \nab_{g_0}(u_\A\pm u_\B) \tilde{\otimes} X^{(2,1)}_\A & =  \bar{\PP}^{[2]}_{u_\A\pm u_\B}\pth{\ka^{(1,\pm)}_{\A\B}} - \frac{N^{(\pm)}_{\A\B}\tilde{\otimes} \wc{X}^{(2,\pm)}_{\A\B}}{|\nab_{g_0}\pth{ u_\A\pm u_\B}|_{g_0}},
\end{align*}
where we recognized the operators $^[2]_u_$ and $^[2]_u_±u_$ defined by \eqref{def P2(T)}. Defining
\begin{align*}
\tilde{k}^{cons}_\la & \vcentcolon = \cro{ \pth{ \ka_\la  +  \L_{\ga_\la} X_\la}^{(\geq 2)} +\pth{\ffi^2_\la}^{(\geq 2)}\pth{ \ka_\la  +  \L_{\ga_\la} X_\la}  }^\osc
\end{align*}
allows us to recover the expression \eqref{k la} for $k_$. The estimate \eqref{estim reste cons 2} follows from \eqref{estim seed}, \eqref{estim ga kappa param}, the definition of $^(2)$, $^(3)$, $X^(2)$, $X^(3)$ and \eqref{estim tffi tX}.

\begin{thebibliography}{MTW73}

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\end{document}