Altogether, we therefore impose $∇_a A^a ≡0$ throughout $M$, which directly leads to the condition $Â_1 ≈0$. Using \cref{prop:residual_gauge_initial_surface}, we obtain a $χ^0_res.$ in a neighbourhood $𝒪^0$ of $Σ$ which sets $𝔞|_Σ= 0 = ∇ ·𝐀|_Σ$, and, using \cref{prop:residual_gauge_scri}, we obtain a $χ̂_res.^1$ in a neighbourhood $𝒪^1$ of $^+$ which sets $Â_1^[1] ≈0$. In general there is no reason for $χ^0_res.$ to be equal to $Ωχ̂_res.^1$, so we interpolate smoothly between the two in the region between $𝒪^0$ and $𝒪^1$. This procedure will break $χ_res. = 0$ in the interpolation region, and hence we will no longer satisfy the Lorenz gauge there. Therefore in this region we will work with the Maxwell field $F_ab$. This will present no difficulties as we will simply need to solve a regular Cauchy problem a finite time into the future (or past) here.

\begin{figure}[H]
\begin{tikzpicture}
\centering
\node[inner sep=0pt] (gauge_curved) at (7,0)

\node[label={[shift={(7.1,2.5)}]$i^+$}] {};

\node[label={[shift={(5.5, 1.2)}]$\mathscr{I}^+$}] {};
\node[label={[shift={(8.6, 1.2)}]$\mathscr{I}^+$}] {};

\node[label={[shift={(4.1,-0.3)}]$i^0$}]{};
\node[label={[shift={(9.9,-0.3)}]$i^0$}]{};

\node[label={[shift={(13.1,-1.3)}]$\mathcal{O}^0\colon$ Lorenz and $\mathfrak{a}|_\Sigma = 0 = \boldsymbol{\nabla}\cdot \mathbf{A}|_\Sigma$}]{};
\draw[->] (10.3,-0.85) .. controls (9.6,-0.7) and (8.6, -0.3) .. (8.2, 0.1);

\node[label={[shift={(11.8,2)}]$\mathcal{O}^1 \colon$ Lorenz and $\hat{A}_1^{[1]} \approx 0$}] {};
\draw[->] (9.7,2.45) .. controls (8.3,2.7) and (8, 2.5) .. (7.3,2.03);

\node[label={[shift={(7.05,-0.7)}]$\Sigma$}]{};

\end{tikzpicture} \vspace{-0.25cm}
\caption{Construction of the gauge on a generic Corvino--Schoen--Chru\'sciel--Delay spacetime.} \label{fig:gauge_curved}
\end{figure}

\subsection{Energy Estimates and Scattering Data}

\begin{theorem}  \label{thm:curvedestimates} For smooth compactly supported Maxwell data on $\Sigma$ one has the energy estimate
\begin{equation} \label{basicestimateMaxwellcurved} \mathcal{E}_{\scri^+} \simeq \int_{\scri^+} | \hat{F}_2 |^2 \, \widehat{\dvol}_{\scri^+} \simeq \int_{\Sigma} \left( |\mathbf{E}|^2 + |\mathbf{B}|^2 \right) \dvol_\Sigma \simeq \mathcal{E}_{\Sigma}.
\end{equation}
\end{theorem}

\begin{proof}
The proof is given in \cref{sec:Maxwell_energy_estimates_proof}.
\end{proof}

\subsubsection{Space of initial data} \label{sec:space_of_initial_data_curved}

We construct the space of initial data on $Σ$ by working with the physical potential $A_a$. We compute, in general, the expressions for the electric and magnetic fields
\begin{align}
\label{initial_fields_from_potential}
\begin{split}
\mathbf{E}_a &= -h^b_a \nabla_T \mathbf{A}_b - \mathfrak{a} h^b_a \nabla_T  T_b + \boldsymbol{\nabla}_a \mathfrak{a} - \mathbf{A}_b \kappa_a^{\phantom{a}b}, \\
\mathbf{B}_a &= \epsilon_a^{\phantom{a}bc} \boldsymbol{\nabla}_b \mathbf{A}_c,
\end{split}
\end{align}
where $ϵ_abc$ is the volume form on $Σ$. Noting that on $Σ$ we have $𝔞|_Σ= 0$, this gives
\begin{equation}
\label{electric_field_in_terms_of_potential_initial_surface}
\mathbf{E}_a|_\Sigma = \Big( -h^b_a \nabla_T \mathbf{A}_b -\mathbf{A}_b \kappa_a^{\phantom{a}b} \Big)\Big|_\Sigma.
\end{equation}
In turn, we find
\[ \int_\Sigma |\mathbf{E}|^2 \dvol_\Sigma = \int_\Sigma | \nabla_T \mathbf{A} - \mathbf{A} \cdot \kappa |^2  \dvol_\Sigma, \]
where $(𝐀 ·κ)_a = 𝐀_b κ_a^ab$, and the squares are with respect to the positive-definite metric $h_ab$ on $Σ$. Next, for $𝐀 ∈𝒞^∞_c(Σ)$ we have, using the Coulomb gauge on $Σ$,
\[ \int_\Sigma |\mathbf{B}|^2 \dvol_\Sigma = \int_{\Sigma} |\boldsymbol{\nabla} \mathbf{A}|^2 - \mathbf{R}_{ij} \mathbf{A}^i \mathbf{A}^j \dvol_\Sigma, \]
where $𝐑_ij$ is the Ricci curvature of $(Σ, h_ab)$. If the Ricci and extrinsic curvatures of $Σ$ are bounded, then it is easy to see that for $(𝐀, ∇_T 𝐀) ∈H^1(Σ) ⊕L^2(Σ)$ one has the estimate $𝐄^2_L^2(Σ) + 𝐁^2_L^2(Σ) 𝐀 ^2_H^1(Σ) + ∇_T 𝐀 ^2_L^2(Σ)$.

We claim that there is a one-to-one correspondence between $(𝐄, 𝐁) ∈L^2(Σ)^2$ and $(𝐀, ∇_T 𝐀)$ living in a suitable Hilbert space. Suppose we have $𝐁 ∈L^2(Σ)$ with $∇ ·𝐁 = 0$ in the sense of distributions. By the Poincar\'e lemma, we know that there exists $𝐀 ∈𝒟'(Σ)$ such that $𝐁 = ∇ ×𝐀$, where $(∇ ×𝐀 )_i = ϵ_i^jk∇_j 𝐀_k$. We impose that $∇ ·𝐀 = 0$ in the sense of distributions. By the above energy identity, we have
\[ \| \mathbf{A} \|^2_{\dot{H}^1} \leq \| \mathbf{B} \|^2_{L^2} + \int_\Sigma | \mathbf{R}_{ij} \mathbf{A}^i \mathbf{A}^j |\dvol_\Sigma. \]
This may be written, using Hardy's inequality on $Σ$ (\cite{DambrosioDipierro2013}, eq. (1.1)), as
\begin{equation} \label{estimate_potential_H1} \|\mathbf{A}\|^2_{\dot{H}^1} \leq \| \mathbf{B} \|^2_{L^2} +C \delta \|\mathbf{A}\|^2_{\dot{H}^1}, \end{equation}
\[ \delta \defeq \left\| r^2 \mathbf{R} \right\|_{L^\infty(\Sigma)}. \]
For $δ$ small enough this implies $𝐀 ∈Ḣ^1(Σ)$. That is, we assume that the Ricci curvature of $Σ$ is sufficiently small on all of $Σ$,
\begin{equation} \label{curvature_assumption} C \delta < 1.
\end{equation}
Since in the Schwarzschild sector the Ricci curvature is given by
\[ \mathbf{R}^i{}_j = \frac{m}{r^3} \left( \begin{array}{ccc} -2 & & \\ & 1 & \\ & & 1 \end{array} \right), \]
this is automatically satisfied for $r ≫1$, and amounts to a smallness assumption on $m$ if the Schwarzschild sector happens to contain a region of small $r$. Now differentiating the expression for $𝐁$ and using the Coulomb gauge, we find
\begin{equation}
\label{elliptic_equation_potential_sigma}
\boldsymbol{\Delta} \mathbf{A}_k + \mathbf{R}_{kj} \mathbf{A}^j = - (\boldsymbol{\nabla} \times \mathbf{B})_k .
\end{equation}
It is clear that $∇ ×𝐁  ∈H^-1(Σ)$, but in fact also $∇ ×𝐁 ∈Ḣ^-1(Σ)$, where $Ḣ^-1(Σ)$ is the dual space of $Ḣ^1(Σ)$. This follows from the fact that $𝒞^∞_c(Σ)$ is dense in $Ḣ^1(Σ)$ and integration by parts, as in analysis of equation \eqref{curlB}. Similarly, if $𝐀 ∈Ḣ^1(Σ)$, then $Δ 𝐀 ∈Ḣ^-1(Σ)$. Further, since $𝐑^k_j$ on the Schwarzschild sector decays like $∼m r^-3$, then by Hardy's inequality as before, we have $𝐑_kj 𝐀^j ∈Ḣ^-1(Σ)$. The operator $(𝐏 𝐀)_k - Δ 𝐀_k - 𝐑_kj 𝐀^j$ therefore maps $Ḣ^1(Σ) →Ḣ^-1(Σ)$, and is continuous, elliptic, formally self-adjoint and coercive (as a consequence of \eqref{curvature_assumption}),
\[ D(\mathbf{A}, \mathbf{A}) \geq (1- C \delta ) \|\mathbf{A}\|^2_{\dot{H}^1(\Sigma)} , \]
\[ D(\mathbf{U}, \mathbf{V}) \defeq \int_\Sigma \mathbf{U}^k (\mathbf{P} \mathbf{V})_k \dvol_\Sigma. \]
To get uniqueness of $𝐀$, it remains to investigate the kernel of the curl operator. We claim that on $Ḣ^1(Σ)$ this is equal to the kernel of $𝐏$, which in turn is trivial on $Ḣ^1(Σ)$ under the assumption \eqref{curvature_assumption}. Indeed, $ker 𝐏$ consists of those potentials $𝐀 ∈Ḣ^1(Σ)$ for which $∇ ×𝐁 = 0$, i.e. by the Poincar\'e lemma $𝐁 = ∇ ϕ$ for some $ϕ∈Ḣ^1(Σ)$ (precisely because $𝐁 ∈L^2(Σ)$). But one also has $∇ ·𝐁 = 0$, which together imply $Δ ϕ= 0$. Since $ϕ∈Ḣ^1(Σ)$, it can be approached by $ϕ_n ∈𝒞^∞_c(Σ)$ in $Ḣ^1(Σ)$, so, as $Δ ϕ∈Ḣ^-1(Σ)$,
\[ 0 = - \int_\Sigma \phi_n \boldsymbol{\Delta} \phi \dvol_\Sigma = \int_\Sigma \boldsymbol{\nabla} \phi_n \cdot \boldsymbol{\nabla} \phi \dvol_\Sigma \to \| \phi \|^2_{\dot{H}^1(\Sigma)}.  \]
Hence $∇ ϕ= 𝐁 = 0$. Then, for $𝐁 = 0$, $𝐀 = 0$ follows from the coercivity of $𝐏$.
We therefore define
\begin{equation}
\label{initial_surface_potential_space}
\dot{H}^1_C(\Sigma)^\text{curl} \defeq \big\{ \mathbf{A}^{(0)} \in \dot{H}^1(\Sigma) \, : \, \boldsymbol{\nabla} \cdot \mathbf{A}^{(0)} = 0, ~ \epsilon_i^{\phantom{i}jk} \boldsymbol{\nabla}_j \mathbf{A}^{(0)}_k \in L^2(\Sigma) \big\} .
\end{equation}
The equation \eqref{elliptic_equation_potential_sigma} therefore has a unique solution in $Ḣ^1_C(Σ)^curl$, which we write as $𝐀^(0) = 𝐏^-1(-∇ ×𝐁)$. Note that, by construction, $𝒞^∞_c(Σ)$ is dense in $Ḣ^1_C(Σ)^curl$ because $𝒞^∞_c(Σ)$ is dense in $L^2(Σ)$, and the choice of norm on $Ḣ^1_C(Σ)^curl$ is \emph{precisely} the norm on $𝐁$.

\begin{remark} For unrestricted $\delta$, due to the lack of positivity of the spacelike Ricci curvature and the fact that $\Sigma$ is unbounded, we expect the precise control of the kernel of $\mathbf{P}$ to be a very delicate question.
\end{remark}

Given $𝐀^(0) ∈Ḣ^1_C(Σ)^curl$ and $𝐄 ∈L^2(Σ)$, we then reconstruct the time derivative component of the initial data from \eqref{electric_field_in_terms_of_potential_initial_surface},
\[ \mathbf{A}^{(1)}_a \defeq -h^b_a \nabla_T \mathbf{A}_b = \mathbf{E}_a + (\mathbf{A}^{(0)} \cdot \kappa)_a.  \]
Since $𝐄 ∈L^2(Σ)$, $𝐀^(0) ∈L^2_loc(Σ)$ and $κ∈𝒞^k(Σ)$, we have $𝐀^(1) ∈L^2_loc(Σ)$. Moreover, since $κ$ vanishes in the Schwarzschild sector, in fact the $L^2$ norm of $𝐀^(1)$ is controlled by the $L^2$ norm of $𝐄$ plus a constant, i.e. $𝐀^(1) ∈L^2(Σ)$. We therefore have a bijection
\begin{align*}
\operatorname{Int} : L^2(\Sigma) \oplus L^2(\Sigma) &\longrightarrow \dot{H}^1_C(\Sigma)^\text{curl} \oplus L^2(\Sigma) \\
(\mathbf{B}, \mathbf{E}) &\longmapsto (\mathbf{A}^{(0)}, \mathbf{A}^{(1)}) = (\mathbf{P}^{-1} ( - \boldsymbol{\nabla} \times \mathbf{B})), ~ \mathbf{E} + \mathbf{A}^{(0)} \cdot \kappa ).
\end{align*}
Our space of initial data for the components $(𝐀, 𝐀̇)$ is therefore $Ḣ^1_C(Σ)^curl ⊕L^2(Σ)$, the elements of which, by construction, are in 1-to-1 correspondence with pairs of fields $(𝐄, 𝐁) ∈L^2(Σ)^2$.

For completeness, we also observe how to prescribe data for the component $𝔞$. Our gauge implies that
\begin{equation}
\label{little_a_initial_data}
(\mathfrak{a}, \nabla_T \mathfrak{a})|_\Sigma = (0, \, \mathbf{A}^{(0)} \cdot \boldsymbol{\nabla} \log N ).
\end{equation}
Certainly $(𝔞, ∇_T 𝔞) ∈𝒞^∞_c(Σ) ⊕L^2_loc(Σ)$. In the Schwarzschild sector in fact $logN = 1/2 log(1-2mr^-1)$, so $𝐀^(0) ·∇ logN ∼1/r 𝐀_r^(0)$, so $∇_T 𝔞$ decays one order faster than $𝐀^(0)$.

\subsubsection{Space of scattering data} \label{sec:scattering_data_potential}

As in \cref{sec:Maxwellpotentialsflatscattering}, the condition \eqref{Lorenz_gauge_first_order_reduction} also implies $ƪÂ_1 ≈0$, and by \cref{prop:spin_coefficients_near_scri}, the expression \eqref{F2components} for $F̂_2$ on $^+$ reduces to
\[ \hat{F}_2 \approx - \partial_u \bar{\hat{A}}_2. \]
\begin{equation}
\label{energy_on_scri}
\int_{\scri^+} |\hat{F}_2|^2 \, \widehat{\dvol}_{\scri^+} = \int_{\scri^+} |\partial_u \hat{A}_2|^2 \, \d u \wedge \dvol_{\mathbb{S}^2}.
\end{equation}
Suppose we have $F̂_2 ∈𝒞^∞(^+)$ supported away from $i^0$. We then put
\[ \hat{A}^+_2 \defeq - \int_{-\infty}^u \bar{\hat{F}}_2 \, \d u \in \mathcal{C}^\infty(\scri^+), \]
which remains supported away from $i^0$. Using the construction in \cref{sec:gauge_on_scri}, we then define
\[ \hat{A}_1^+ \equiv 0 \]
\[ \hat{A}_0^+ \defeq \int_{-\infty}^u 2 \operatorname{Re} (\hat{\eth} \bar{\hat{A}}_2^+ ) \, \d u \in \mathcal{C}^\infty(\scri^+). \]
We then define the space $ℋ̇^1(^+) ≃Ḣ^1(ℝ_u; L^2(𝕊^2))$ of scattering data by completing $(Â^+_0, Â_1^+, Â_2^+) ∈𝒞^∞(^+) ×𝒞^∞_c(^+) ×𝒞_c^∞(^+)$ in the norm \eqref{energy_on_scri}, as in \cref{def:characteristicdatapotential_Minkowski}.

\subsection{Trace and Scattering Operators} \label{sec:scattering_operator_curved_spacetimes}

\subsubsection{The forward Cauchy problem}

Suppose we are given $(𝐀^(0), 𝐀^(1)) ∈𝒞^∞_c(Σ)^2 ∩(Ḣ^1_C(Σ)^curl ⊕L^2(Σ))$ such that $∇ ·𝐀^(0) = 0$. We obtain $(𝔞, ∇_T 𝔞) ∈𝒞^∞_c(Σ)^2$ using \eqref{little_a_initial_data}, and then reconstruct the initial fields on $Σ$ using \eqref{initial_fields_from_potential}. We therefore obtain a smooth compactly supported $F̂_ab|_Σ$, which we propagate in $M̂$ as we did in \cref{thm:MaxwellpotentialflatCauchyproblem} (see Lemma 2.4 in \cite{MasonNicolas2004}). We thus obtain a smooth $F̂_ab$ on $^+$ which, by finite speed of propagation, is supported away from $i^0$, and which satisfies the estimate \eqref{basicestimateMaxwellcurved}. On $^+$, we reconstruct the potential as described in \cref{sec:scattering_data_potential}. We therefore obtain a linear operator
\begin{align*} \mathfrak{T}^+ : \mathcal{C}^\infty_c(\Sigma)^2 \cap (\dot{H}^1_C(\Sigma)^{\text{curl}} \oplus L^2(\Sigma)) &\longrightarrow \dot{\mathcal{H}}^1(\scri^+) \\
(\mathbf{A}^{0}, \mathbf{A}^{(1)}) &\longmapsto (\hat{A}_0^+, \hat{A}_1^+, \hat{A}_2^+) = \left( \int_{-\infty}^u 2 \operatorname{Re}(\hat{\eth}\bar{\hat{A}}_2^+) \, \d u, ~ 0, ~ \hat{A}_2^+ \right),
\end{align*}
which extends by density to a bounded linear operator
\begin{equation}
\label{forward_trace_operator_curved_spacetimes}
\mathfrak{T}^+ : \dot{H}^1_C(\Sigma)^{\text{curl}} \oplus L^2(\Sigma) \longrightarrow \dot{\mathcal{H}}^1(\scri^+).
\end{equation}

\subsubsection{The Goursat problem}

To show that the operator \eqref{forward_trace_operator_curved_spacetimes} is invertible, it now remains to show that we can solve the Goursat problem from data in a dense subspace of $ℋ̇^1(^+)$, and that the solution gives rise to a unique $(𝐀^(0), 𝐀^(1)) ∈Ḣ^1_C(Σ)^curl ⊕L^2(Σ)$. We proceed as in \cref{sec:Minkowski_Goursat_problem}. Suppose we are given data
\[ (\hat{A}_0^+, \hat{A}_1^+, \hat{A}_2^+) = \left( \int_{-\infty}^u 2 \operatorname{Re}(\hat{\eth}\bar{\hat{A}}_2^+) \, \d u, ~ 0, ~ \hat{A}_2^+ \right) \in \mathcal{C}^\infty(\scri^+) \times \mathcal{C}^\infty_c(\scri^+) \times \mathcal{C}_c^\infty(\scri^+) \]
supported away from $i^0$. Working first in the neighbourhood of $^+$ in which our gauge holds, and subsequently working with the field, we solve the Goursat problem in two steps. Introduce a short outgoing null hypersurface $ℋ$ which intersects $^+$ in the future of the support of $Â_2^+$, as shown in \cref{fig:outgoing_null_hypersurface_Minkowski}. In the future of $ℋ$, the data for $Â_1$ and $Â_2$ is identically zero, whereas the data for $Â_0$ is a constant, say $Â_0 ≈C_0$. Therefore the field $F̂_ab$ there has identically zero data on $^+$, $F̂_0^+ = F̂_1^+ = F̂_2^+ = 0$. Briefly changing conformal scale in $𝒪^+$ to bring $i^+$ to a finite distance, we may solve a wave equation for $F̂_ab$ in $𝒪^+$, the solution to which, by the uniqueness part of \cref{thm:Goursatproblem}, must be $F̂_ab ≡0$. Change conformal scale back to that of \cref{prop:spin_coefficients_near_scri}. We therefore have that in $𝒪^+$ the potential is pure gauge, $F̂_ab ≡0 Â_a = ∇̂_a χ$, where we use the Poincar\'e Lemma to obtain the existence of such a $χ∈H^2_loc(𝒪^+)$. A priori $χ$ is not unique, but we recall that our solution satisfies the gauge constructed in \eqref{sec:gauge_on_scri}. This imposes the conditions $χ= 0$, $'(Ω^-1 χ) ≈0$, and $Δ_𝕊^2 χ≈0$. We have already seen that the equation $χ= 0$ is equivalent to
\[ \widehat{\Box} \hat{\chi} + 2\hat{\mu}^{[1]} f \hat{\chi} = 0, \]
where $χ̂ = Ω^-1 χ$, so to obtain uniqueness we need only fix the data for $χ̂$ on $^+ ∩𝒪^+$. We compute
\[ \hat{\thorn}(\Omega^{-1} \chi ) = -f\Omega^{-1} \hat{\chi} + \Omega^{-1} \hat{\thorn} \chi, \]
which, noting that $ χ≈C_0$, implies that we should have
\[ \hat{\chi} \approx \frac{C_0}{f}, \]
and so this fixes $χ$ in $𝒪^+$.

By restriction to $ℋ$ of $ χ$, $' χ$ and $ƪ̂χ$, we therefore obtain $H^1_c$ data for $Â_0$, $Â_1$ and $Â_2$ on $ℋ$. Since the full data is now $H^1_c(ℋ∪(^+∩𝒪^-))$, we may apply \cref{thm:Goursatproblem} from $ℋ ∪(^+ ∩𝒪^-)$ to solve \eqref{Maxwells_equations_rescaled_physical_Lorenz} in the region $𝒪^-$. We thus obtain a solution $Â_a$ to \eqref{Maxwells_equations_rescaled_physical_Lorenz} in the neighbourhood $𝒪^+ ∪𝒪^- ⊂𝒪^1$ of $^+$ which has the regularity
\[ \hat{A}_a \in \mathcal{C}^0_{\text{sc}}(t(\mathcalboondox{M}); H^1(\Sigma_t)) \cap \mathcal{C}^1_{\text{sc}}(t(\mathcalboondox{M}); L^2(\Sigma_t)). \]
On a slice $Σ_t'$, $t' < ∞$, contained in $𝒪^+ ∪𝒪^-$, we now reconstruct the physical field $F_ab|_Σ_t' ∈L^2_c(Σ_t')$ from $Â_a = A_a$ and propagate $F_ab$ backwards in time to $Σ$. We thus obtain $(𝐄, 𝐁) ∈L^2_c(Σ)^2$ on $Σ$, and, using the operator $Int$ constructed in \cref{sec:space_of_initial_data_curved}, we obtain a unique
\[  (\mathbf{A}^{(0)}, \mathbf{A}^{(1)}) =  \operatorname{Int}(\mathbf{E}, \mathbf{B}) \in \dot{H}^1_C(\Sigma)^{\text{curl}} \oplus L^2(\Sigma). \]
We therefore obtain the inverse trace operator
\[ (\mathfrak{T}^+)^{-1} : \dot{\mathcal{H}}^1(\scri^+) \longrightarrow \dot{H}^1_C(\Sigma)^{\text{curl}} \oplus L^2(\Sigma). \]
As in the case of Minkowski space, we then define the \emph{scattering operator for Maxwell potentials} on $M$ to be the isomorphism of Hilbert spaces
\begin{align*}
\mathscr{S} \defeq \mathfrak{T}^+ \circ (\mathfrak{T}^-)^{-1} : \dot{\mathcal{H}}^1(\scri^-) & \longrightarrow \dot{\mathcal{H}}^1(\scri^+) \\
(\hat{A}^-_0, \hat{A}^-_1, \hat{A}_2^- ) &\longmapsto (\hat{A}^+_0, \hat{A}^+_1, \hat{A}_2^+ ),
\end{align*}
where $Â_1^+ ≡0 ≡Â_0^-$, and
\[ \hat{A}_0^+ = \int_{-\infty}^u 2 \operatorname{Re}(\hat{\eth} \bar{\hat{A}}_2^+) \, \d u \quad \text{and} \quad \hat{A}_1^- = \int_{-\infty}^v 2 \operatorname{Re}(\hat{\eth} \bar{\hat{A}}_2^-) \, \d v. \]

\section*{Acknowledgements}

For the purpose of open access, the authors have applied a Creative Commons Attribution (CC BY) licence to any Author Accepted Manuscript version arising from this submission. The authors thank Lionel Mason for useful discussions during the preparation of this article, and the Mathematical Institute at Oxford University for hospitality in Spring 2022.

\appendix
\section{Proof of Energy Estimates for Maxwell Fields}
\label{sec:Maxwell_energy_estimates_proof}

In this section we prove the energy estimate \eqref{basicestimateMaxwellcurved}. For convenience, here we shall use spinor notation. A real Maxwell field $F_ab$ can be decomposed into its self-dual and anti self-dual parts that are complex conjugates of one another,
\[ F_{ab} = \phi_{AB} \bar{\varepsilon}_{A'B'} + \bar{\phi}_{A'B'} \varepsilon_{AB} \, ,\]
with $ϕ_AB = ϕ_(AB)$; $ε_AB$ and its complex conjugate $ε̅_A'B'$ are the Levi-Civita symbols, the symplectic forms on the left- and right-handed spin bundles $𝕊^A$ and $𝕊^A'$, such that $g_ab = ε_AB ε_A'B'$. Under a conformal rescaling $ĝ_ab = Ω^2 g_ab$, $ϕ_AB$ transforms as $ϕ̂_AB = Ω^-1ϕ_AB$ and the Levi-Civita symbol as $ε̂_AB = Ωε_AB$. The Maxwell field is then invariant under conformal rescalings and we have $F̂_ab = F_ab = ϕ̂_AB ε̅̂̅_A'B' + ε̂_AB ϕ̅̂̅_A'B'$. In a normalized spin frame ${ ô^A, ι̂^A }$ the components of $ϕ̂_AB$ correspond exactly to the components of $F̂_ab$, $ϕ̂_i = F̂_i$, $i=0,   1,   2$, where
\[ \hat{\phi}_0 = \hat{\phi}_{AB} \hat{o}^A \hat{o}^B, \qquad \hat{\phi}_1 = \hat{\phi}_{AB} \hat{o}^A \hat\iota^B, \quad \text{and} \quad \hat{\phi}_2 = \hat{\phi}_{AB} \hat\iota^A \hat\iota^B. \]
Maxwell's equations reduce to equations on the self-dual part $ϕ_AB$ that are conformally invariant. On the compactified spacetime and for the rescaled self-dual Maxwell spinor, they have the following form
\begin{equation}\label{ASDMaxwell}
\hat{\nabla}^{AA'} \hat{\phi}_{AB} = 0 .
\end{equation}
The natural stress-energy tensor \eqref{stresstensor} has a very simple expression in terms of the spinors $ϕ̂_AB$ and $ϕ̂_A'B'$, given by
\begin{equation} \label{MaxwellSET}
\hat{\mathbf{T}}_{ab} = \hat{\phi}_{AB} \overline{\hat{{\phi}}}_{A'B'} \, .
\end{equation}
The tensor $𝐓̂_ab$ is symmetric, conserved on-shell, and conformally covariant with weight $-2$. Let $Σ= Σ_0$ be the ${ t=0 }$ slice in our asymptotically null foliation ${ Σ_t }$ of $M$. We denote by $M^+$ the future of $Σ$ in $M$ and by $M̂^+$ its closure in $M̂$. Let $τ^a$ be an observer (a timelike future-oriented vector field) on $M^+$ that aligns on $^+$ with its null generator. Defining the energy current
\begin{equation} \label{EnCurrentMaxwell}
\hat{J}_a \defeq \tau^b \hat{\mathbf{T}}_{ab} ,
\end{equation}
we have, for $ϕ̂_AB$ a solution to \eqref{ASDMaxwell}, the approximate conservation law
\begin{equation} \label{ACL}
\hat\nabla^a \hat{J}_a = \hat\nabla^{a} \big( \tau^{b} \hat\phi_{AB} \bar{\hat\phi}_{A'B'} \big) = \hat\nabla^{(a} \tau^{b)} \hat\phi_{AB} \bar{\hat\phi}_{A'B'} \, .
\end{equation}
The energy of the field on a given spacelike hypersurface $𝒮$ is simply the $L^2$ norm of $ϕ̂_AB$ on $𝒮$, with measure induced by $ĝ_ab$, and a weight associated to our choice of observer $τ^a$. More precisely,
\begin{equation} \label{MaxEnergyS}
\mathcal{E}_\mathcal{S} [\hat\phi] = \int_\mathcal{S} \hat{J}_a \nu^a ( v \intprod \widehat{\dvol}),
\end{equation}
where $$ is the $4$-volume measure associated to $ĝ_ab$, $ν^a$ is a normal vector field to $𝒮$ compatible with the orientation of $𝒮$ (i.e., future-pointing), and $v^a$ is a vector field transverse to $𝒮$ such that $ν^a v^b ĝ_ab =1$. Since the stress-energy tensor has conformal weight $-2$, $τ^a$ has weight $0$, $ν^a$ and $v^a$ can be chosen to have weight $-1$ and $$ has weight $4$, it follows that the energy flux \eqref{MaxEnergyS} is conformally invariant, i.e. conformally covariant with weight $0$ (see also\eqref{Maxwellconformalinvarianceenergies}).

For the purpose of proving the estimate \eqref{basicestimateMaxwellcurved}, we decompose $M^+$ into three distinct regions:
\begin{enumerate}[(i)]
\item a neighbourhood $U^0$ of $i^0$,
\[ U^0 = \{ u \leq u_0 \} \mbox{ for a given } u_0 \ll -1 \, ; \]
\item a neighbourhood $U^+$ of $i^+$,
\[ U^+ = \{ \tau \geq \tau_0 \} \mbox{ for a given } \tau_0 \gg 1, \]
where $\tau$ is the parameter of the foliation transverse to $\scri^+$ shown in Figure \ref{fig:transversefoliation};
\item an intermediate region $U \defeq \overline{\mathcalboondox{M}^+ \setminus (U^0 \cup U^+)}$.
\end{enumerate}
We obtain energy estimates in each region separately. Since the energy flux \eqref{MaxEnergyS} is conformally invariant, we may work with different conformal factors in the different regions. However, we must choose our observer $τ^a$ so that it is continuous and in fact smooth on $M̂^+$.

In the region $U^0$, we work with the conformal factor $Ω= 1/r$ which preserves the timelike Killing vector $K^a ∂_a = ∂_t = ∂_u$ on the Schwarzschildean neighbourhood of $i^0$. Here we choose $K^a$ for our observer,
\[ \tau^a \partial_a = \partial_u \, .\]
This gives immediate energy identities in the region $U^0$.

In the regions $U^+$ and $U$, we make use of special features of $^+$ and choose a conformal factor $Ω$ such that
\begin{enumerate}[(i)]
\item $i^+$ is a finite regular point of the compactified spacetime $(\hat{\mathcalboondox{M}}^+ , \hat{g}_{ab} )$;
\item $\hat{\mathrm{R}}_{ab}=0$ at $i^+$;
\item $\hat{\mathrm{R}}$ and $\hat{n}^a \hat{\mathrm{R}}_{ab}$ vanish on $\scri^+$, where $\hat{n}^a$ is the null generator of $\scri^+$;
\item $-\hat{\nabla}^a \Omega $ is timelike and future-oriented in $U^+ \setminus \scri^+$ and null and future-oriented on $U^+ \cap \scri^+$.
\end{enumerate}
The existence of such a conformal factor was established in \cite{MasonNicolas2004}, Lemma A.1. We work with a normalized spin frame ${ô^A, ι̂^A }$ and a Newman--Penrose tetrad $( ^a = ô^A ô̅^A',   ^a = ô^A ι̅̂̅^A',   ^a = ô̅^A' ι̂^A,   ^a = ι̂^A ι̅̂̅^A' )$ on $M^+$ such that $^a$ and $^a$ are real and future-oriented, smooth on $M^+ ∖i^+ $, bounded and non-vanishing at $i^+$, and $n̂^a$ is the null generator of $^+$. We assume in addition that the vector field $^a + ^a$ is hypersurface orthogonal. This is not a critical assumption, we may be easily remove it if desired, but it turns out to simplify the following estimates.

\subsection{Energy estimates in $U^+$}

In this region we put $τ^a -∇̂^a Ω$. This has the following decomposition along our null tetrad,
\[ -\hat{\nabla}^a \Omega = c_1 \hn^a + \Omega \big( c_0 \hl^a + c_2 \hm^a + \overline{c_2} \bhm^a \big)\, , \]
where $c_0$ $c_1$ and $c_2$ are smooth on $M^+$, $c_0$ and $c_1$ are real and positive on $M̂^+ ∖i ^+$ and $c_0$ vanishes at $i^+$. Here one might be tempted to work with the foliation by the level hypersurfaces of $Ω$. The energy density on these slices is given by the quadratic form
\begin{align*}
\hat{\mathbf{T}}_{ab} \hat{\nabla}^a \Omega \hat{\nabla}^b \Omega = c_1^2 \vert \hat\phi_2 \vert^2 + \mathcal{O} ( \Omega ) \, .
\end{align*}
This is a natural foliation to choose as in the limit $Ω→0$ it accumulates on $^+$; indeed, this is the foliation that was used in \cite{MasonNicolas2004}. However, this also means that the energy on the slices degenerates as $Ω→0$, and therefore in order to estimate the error term by the energy one needs to split the bulk integral of the error term and extract additional decay near $^+$ from the $3+1$ splitting of the $4$-volume form. It is much simpler to choose a foliation transverse to $^+$ whose normal vector field is given by
\[ \nu^a = \frac{1}{\sqrt{2}} ( \hl^a + \hn^a ) \, .\]
For this choice, the energy density becomes
\begin{equation} \label{EnergyDensityTransverseFoliationU+}
\hat{\mathbf{T}}_{ab} \tau^a \nu^b =  \frac{1}{\sqrt{2}} \big( c_1 \vert \hat\phi_2 \vert^2 + c_1 \vert \hat\phi_1 \vert^2 \big) + \mathcal{O}(\Omega ) \, .
\end{equation}
The advantage of such a foliation is that it does not degenerate near $^+$, and the associated $3+1$ splitting of the $4$-volume form does not induce any additional decay. It is therefore enough to show directly that the bulk error term is controlled by \eqref{EnergyDensityTransverseFoliationU+}.

The Killing form of $τ^a$ is
\[ \hat\nabla_{(a} \tau_{b)} = -\hat\nabla_{(a} \hat\nabla_{b)} \Omega = -\hat\nabla_a \hat\nabla_b \Omega, \]
and its behaviour can be understood using the conformal transformation law of the trace-free part of the Ricci tensor $Φ_ab$, and the asymptotic Einstein condition \eqref{asymptoticEinstein}. We first note that it splits into two parts,
\begin{align*}
\hat\nabla_a \tau_b = \hat\nabla_{(a} \tau_{b)} &= -\hat\nabla_{[A'\vert [A} \hat\nabla_{B]\vert B']} \Omega - \hat\nabla_{(A'\vert (A} \hat\nabla_{B)\vert B')} \Omega \\
&= - \frac14 (\widehat{\Box} \Omega ) \hat{g}_{ab} - \hat\nabla_{A'(A} \hat\nabla_{B)B'} \Omega \, .
\end{align*}
The first part will not appear in the divergence of the energy current. Indeed, due to the symmetry of $ϕ̂_AB$, \eqref{ACL}, we have
\[ \hat{g}^{ab} \hat\phi_{AB} \overline{\hat\phi}_{A'B'} = \varepsilon^{AB} \hat\phi_{AB} \varepsilon^{A'B'} \overline{\hat\phi}_{A'B'} = 0 \, .\]
Therefore \eqref{ACL} becomes
\[ \hat\nabla^a \hat{J}_a = -(\hat\nabla^{A'(A} \hat\nabla^{B)B'} \Omega ) \hat\phi_{AB} \bar{\hat\phi}_{A'B'} \, .\]
The conformal transformation law for $Φ_ab$ is given by
\[
\Phi_{ab} = \hat\Phi_{ab} + \Omega^{-1} \hat\nabla_{A'(A} \hat\nabla_{B)B'} \Omega,
\]
so---recalling the assumptions that $Φ_ab = 𝒪(Ω^2)$ and that $Φ̂_ab$ is smooth at $^+ ∪i^+$---we infer that the asymptotic Einstein condition \eqref{asymptoticEinstein} may be rewritten as
\begin{equation} \label{KFAsymptotics1U+}
-\hat\nabla_{A'(A} \hat\nabla_{B)B'} \Omega = \Omega \hat\Phi_{ab} - \Omega \Phi_{ab} = \Omega \hat\Phi_{ab} + \mathcal{O} (\Omega^3 ) \, .
\end{equation}
Decomposing $Φ̂_ab$ and $ϕ̂_AB$ along the spin frame ${ ô^A, ι̂^A }$, we can express the leading order part of the error term as follows (omitting the factor of $Ω$ for the moment),
\begin{align*}
\hat\Phi^{ab} \hat{\mathbf{T}}_{ab} &= \hat\Phi^{ab} \hat\phi_{AB} \bar{\hat\phi}_{A'B'} \\
&= \hat\Phi_{00} \vert \hat\phi_0 \vert^2 + 4 \hat\Phi_{11} \vert \hat\phi_1 \vert^2 + \hat\Phi_{22} \vert \hat\phi_{2} \vert^2 - 4 \Re ( \hat\Phi_{01} \hat\phi_0 \overline{\hat\phi_1} ) - 4 \Re ( \hat\Phi_{12} \hat\phi_1 \overline{\hat\phi_2} ) +2 \Re ( \hat\Phi_{02} \hat\phi_0 \overline{\hat\phi_2} ) \, .
\end{align*}
We now recall that our conformal factor $Ω$ was such that (ii) $Φ̂_ab$ and $R̂$ vanish at $i^+$, and (iii) $^a Φ̂_ab$ and $R̂$ vanish on $^+$. We have
\[ \hn^a \hat{\Phi}_{ab} = \hat{\Phi}_{11} \hn_b - \hat{\Phi}_{21} \hm_b - \bar{\hat{\Phi}}_{21} \bhm_b + \hat{\Phi}_{22} \hl_b \, ,\]
so $Φ̂_11$, $Φ̂_21$ and $Φ̂_22$ vanish on $^+$ and are therefore such that
\[ \hat{\Phi}_{11}^{[1]} \defeq \Omega^{-1} \hat{\Phi}_{11}, \qquad \hat{\Phi}_{21}^{[1]} \defeq \Omega^{-1} \hat{\Phi}_{21} \quad \text{and} \quad \hat{\Phi}_{22}^{[1]} \defeq \Omega^{-1} \hat{\Phi}_{22} \]
are smooth at $^+$. Hence, we can decompose the divergence of the energy current into terms that are of order $𝒪(Ω)$, $𝒪(Ω^2)$, and higher:
\begin{align*}
\hat\nabla^a \hat{J}_a &= \Omega \big( \hat\Phi_{00} \vert \hat\phi_0 \vert^2 - 4 \Re ( \hat\Phi_{01} \hat\phi_0 \overline{\hat\phi_1} ) +2 \Re ( \hat\Phi_{02} \hat\phi_0 \overline{\hat\phi_2} ) \big) \\
&\phantom{=} + \Omega^2 \big( 4 \hat{\Phi}^{[1]}_{11} \vert \hat\phi_1 \vert^2 + \hat{\Phi}^{[1]}_{22} \vert \hat\phi_2 \vert^2 - 4 \Re (\hat{\Phi}^{[1]}_{12} \hat\phi_1 \overline{\hat\phi_2} ) \big) \\
&\phantom{=} + \mathcal{O}(\Omega^3).
\end{align*} This error term is easily controlled by the energy density \eqref{EnergyDensityTransverseFoliationU+}. We can therefore obtain energy estimates in both directions on $U^+$ using Gr\"onwall's Lemma. This is done as follows. We start from the energy identity
\begin{equation} \label{EnId}
\mathcal{E}_{\scri^+_{\tau_{_0}}} - \mathcal{E}_{\Sigma_{\tau_{_0}}} = \int_{U^+} \hat\nabla^a \hat{J}_a \,\widehat{\dvol} ,
\end{equation}
where $^+_τ__0$ is the part of $^+$ in the future of $Σ_τ__0$.
\begin{enumerate}
\item {\bf Forward-in-time estimate.} Assuming that the parameter $\tau$ of the foliation $\{ \Sigma_\tau \}_\tau$ of $\mathcalboondox{M}^+$ ranges from $0$ to $T$, we introduce the hypersurfaces $S_\tau$, $\tau_0 \leq \tau \leq T$, in $\overline{\mathcalboondox{M}^+}$ that are the union of $\Sigma_\tau$ and the part of $\scri^+$ in the past of $\Sigma_\tau$ and in the future of $\Sigma_{\tau_{_0}}$. Then \eqref{EnId} can be rewritten as follows,
\[ \mathcal{E}_{S_{T}} = \mathcal{E}_{S_{\tau_{_0}}} + \int_{U^+} \hat\nabla^a \hat{J}_a \, \widehat{\dvol} \]
from which we infer an estimate using the control of the error terms by the energy density obtained above:
\[ \mathcal{E}_{S_{T}}
\leq  \mathcal{E}_{S_{\tau_{_0}}} + \int_{\tau_{_0}}^T \mathcal{E}_{S_{\tau}} \d \tau \]
We also have the intermediate inequalities
\[ \mathcal{E}_{S_{\tau}} - \mathcal{E}_{S_{\tau_{_0}}} \leq  \int_{\tau_{_0}}^\tau \mathcal{E}_{S_{\sigma}} \d \sigma \, .\]
Hence by Grönwall's inequality, we obtain,
\[ \mathcal{E}_{\scri^+_{\tau_{_0}}} \lesssim \mathcal{E}_{\Sigma_{\tau_{_0}}} \, .\]
\item {\bf Converse estimate.} In this direction, we simply use the foliation $\{ \Sigma_\tau \}_\tau$. From \eqref{EnId} and the control of the error terms, we have
\[ \Big\vert \mathcal{E}_{\scri^+_{\tau_{_0}}} - \mathcal{E}_{\Sigma_{\tau_{_0}}} \Big\vert \leq \int_{\tau_{_0}}^T \mathcal{E}_{\Sigma_{\tau}} \d \tau \, ,\]
\[  \mathcal{E}_{\Sigma_{\tau_{_0}}} \leq \mathcal{E}_{\scri^+_{\tau_{_0}}} + \int_{\tau_{_0}}^T \mathcal{E}_{\Sigma_{\tau}} \d \tau \, .\]
We also have the intermediate estimates for $\tau_0 \leq \tau \leq T$
\[  \mathcal{E}_{\Sigma_{\tau}} \leq \mathcal{E}_{\scri^+_{\tau_{_0}}} + \int_{\tau}^T \mathcal{E}_{\Sigma_{\sigma}} \d \sigma \, .\]
Grönwall's estimate therefore gives
\[ \mathcal{E}_{\Sigma_{\tau_{_0}}} \lesssim \mathcal{E}_{\scri^+_{\tau_{_0}}} \, .\]

\end{enumerate}

\subsection{Energy estimates in $U$}

On $U$, we choose our observer $τ^a$ to be
\[ \tau^a = d_1 \hn^a + \Omega \big( d_0 \hl^a + d_2 \hm^a + \overline{d_2} \bhm^a \big) \]
where $d_0$, $d_1$ and $d_2$ are smooth functions on $U$ that agree with $c_0$, $c_1$ and $c_2$ at the intersection with $U^+$, are such that $τ^a ∂_a = ∂_u$ on $U ∩U^0$ and $d_0$ and $d_1$ are positive on $U$. As we did on $U^+$, we choose a foliation transverse to $^+$ whose normal vector field is given by
\[ \nu^a = \frac{1}{\sqrt{2}} (\hl^a + \hn^a ) \, .\]
The associated energy density on the slices is then
\begin{equation} \label{EDTFU}
\hat{J}_a \nu^a = \hat{\mathbf{T}}_{ab} \tau^a \nu^b = \frac{1}{\sqrt{2}} \big( d_1 \vert \hat\phi_2 \vert^2 + d_1 \vert \hat\phi_1 \vert^2 \big) + \mathcal{O}(\Omega ) .
\end{equation}
The error term has the form
\begin{align*}
\hat\nabla^{(a} \tau^{b)} \hat{\mathbf{T}}_{ab} &= \hat\nabla^{(a} (  \Omega ( d_0 \hl^{b)} + d_2 \hm^{b)} + \overline{d_2} \bhm^{b)} ) + d_1 \hn^{b)} ) \hat{\mathbf{T}}_{ab} \\
&= \underset{\mathrm{I}}{\underbrace{\Omega \hat\nabla^a ( d_0 \hl^{b} + d_2 \hm^{b} + \overline{d_2} \bhm^{b} ) \hat{\mathbf{T}}_{ab}}} + \underset{\mathrm{II}}{\underbrace{(\hat\nabla^{a} \Omega ) ( d_0 \hl^{b} + d_2 \hm^{b} + \overline{d_2} \bhm^{b} ) \hat{\mathbf{T}}_{ab} }} \\
&\phantom{=} + \underset{\mathrm{III}}{\underbrace{(\hat\nabla^{(a} d_1 ) \hn^{b)} \hat{\mathbf{T}}_{ab} }} + \underset{\mathrm{IV}}{\underbrace{d_1 \hat\nabla^{(a} \hn^{b)} ) \hat{\mathbf{T}}_{ab}}} \, .
\end{align*}
In order to establish that the error terms are controlled by the energy density on the slices, we simply must show that no error term without a factor $Ω$ involves either $ϕ̂_0$ or $ϕ̂_0$. The term I is a quadratic form with bounded coefficients and an overall factor of $Ω$, so it is controlled by the energy density \eqref{EDTFU}. To estimate II, we decompose the gradient of $Ω$ along our Newman--Penrose tetrad,
\[ \hat\nabla^a \Omega = e_1 \hn^a + \Omega \big( e_1 \hl^a + e_2 \hm^a + \overline{e_2} \bhm^a \big) \, .\]
Since the only term in $∇̂^a Ω$ that does not have a factor of $Ω$ involves the tetrad vector $^a$, the terms from II without a factor $Ω$ also do not contain $ϕ̂_0$. Therefore, II is controlled by the energy density. The term III involves only $|ϕ̂_1 |^2$ and $|ϕ̂_2 |^2$ with bounded coefficients and is therefore also controlled by \eqref{EDTFU}. The fourth term requires the most care. We decompose $∇̂^a ^b$ along our NP tetrad,
\begin{equation} \label{gradhn}
\hat\nabla^a \hn^b = \hl^a \hat{\Delta} \hn^b + \hn^a \hat{D} \hn^b - \hm^a \bar{\hat{\delta}} \hn^b - \bhm^a \hat{\delta} \hn^b,
\end{equation}
and note the following transport equations along the NP tetrad for $n̂^a$ ((4.5.28), \cite{spinorsandspacetime1}):
\begin{eqnarray}
\hat{D} \hn^b &=& -(\hat{\epsilon} + \bar{\hat{\epsilon}} ) \hn^b + \hat{\pi} \hm^b + \bar{\hat{\pi}} \bhm^b , \label{Line1} \\
\hat{\delta} \hn^b &=& -(\hat{\beta} + \bar{\hat{\alpha}} ) \hn^b + \hat{\mu} \hm^b + \bar{\hat{\lambda}} \bhm^b , \label{Line2} \\
\bar{\hat{\delta}} \hn^b &=& - ( \hat{\alpha} + \bar{\hat{\beta}} ) \hn^b + \hat{\lambda} \hm^b + \bar{\hat{\mu}} \bhm^b , \label{Line3} \\
\hat{\Delta} \hat{n}^b &=& - (\hat{\gamma} + \bar{\hat{\gamma}}) \hat{n}^b + \hat{\nu} \hat{m}^b + \bar{\hat{\nu}} \bar{\hat{m}}^b . \label{Line4}
\end{eqnarray}
The right-hand side of \eqref{gradhn} therefore involves no $^a ^b$ term, so $|ϕ̂_0 |^2$ does not appear in term IV. However, any term involving $^a ^b$ or $^a ^b$ will produce a product involving either $ϕ̂_0$ or $ϕ̂_0$, and these terms must contain a factor of $Ω$ in order to be controlled by the energy density. We therefore need to take a closer look at the contributions of the first, third and fourth terms in \eqref{gradhn}.
The potentially dangerous terms are those involving $λ̅̂̅$ in \eqref{Line2}, $λ̂$ in \eqref{Line3}, and $ν̂$ and $ν̅̂̅$ in \eqref{Line4}. But both $λ̂$ and $ν̂$ vanish on $^+$, $λ̂ ≈0 ≈ν̂$, as a consequence of the asymptotic Einstein condition \eqref{asymptoticEinstein} (see \eqref{geodetictwistfreescri}, \eqref{shearfreescri}); indeed, the vanishing of the spin coefficients $ν̂$ and $λ̂$ on $^+$ simply restates the fact that $^+$ is geodetic and shear-free. Therefore these spin coefficients decay like $𝒪(Ω)$ towards $^+$, and it follows that the error term is controlled by the energy density on the slices. Then energy estimates on $U$ can be obtained by means of Grönwall's estimates in much the same way as on $U^+$.

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