${\bm{v}}$. Let $\text{exp}_{i}$ denote the exponential map at i,
$\text{exp}_{i}:\mathcal{T}\to\mathcal{S}^{\prime}$, defined such that
$\text{exp}_{i}({\bm{v}})=\gamma_{{\bm{v}}}(1)$. Then, the neighbourhood
$\mathcal{U}^{\prime}$ is said to be a normal neighbourhood of $i$ if
$\mathcal{U}^{\prime}=\text{exp}_{i}(\mathcal{Q})$, where $\mathcal{Q}\subset
T|_{i}(\mathcal{S}^{\prime})$ is a neighbourhood of the zero vector
${\bm{0}}$, and if for all $t\in[0,1]$,

${\bm{v}}\in\mathcal{Q}\rightarrow t{\bm{v}}\in\mathcal{Q}.$

For any point $p=\text{exp}_{i}({\bm{v}})\in\mathcal{U}^{\prime}$, the normal
coordinates $(z^{\alpha})$ is given by
$z^{\alpha}={\bm{e}}^{\alpha}{}_{{\bm{i}}}v^{{\bm{i}}}$, where $v^{{\bm{i}}}$
are the components of ${\bm{v}}$ with respect to an orthonormal basis
$\\{{\bm{e}}_{{\bm{i}}}\\}$ and ${\bm{e}}^{\alpha}{}_{{\bm{i}}}$ are defined
by

${\bm{e}}^{\alpha}{}_{{\bm{i}}}\equiv\langle{\bm{d}}{y}^{\alpha},{\bm{e}}_{{\bm{i}}}\rangle.$

Let $v^{\alpha}\equiv\langle{\bm{d}}{z}^{\alpha},{\bm{v}}\rangle$ denote the
components of ${\bm{v}}$ with respect to $(z^{\alpha})$, then the normal
coordinates for any point $p$ on $\gamma_{{\bm{v}}}(t)$ is given by
$x^{\alpha}(t)=tz^{\alpha}$. If
$\gamma^{\prime}_{\alpha}{}^{\beta}{}_{\gamma}$ denote the components of the
${\bm{h}}^{\prime}$-Levi-Civita connection coefficients with respect to
$(z^{\alpha})$, the geodesic equation can be written as

$\frac{d^{2}x^{\beta}}{dt^{2}}+\gamma^{\prime}_{\alpha}{}^{\beta}{}_{\gamma}\frac{dx^{\alpha}}{dt}\frac{dx^{\gamma}}{dt}=0,$
(99)

then one can show that $\gamma^{\prime}_{\alpha}{}^{\beta}{}_{\gamma}$ vanish
at $i$, i.e.,

$\gamma^{\prime}_{\alpha}{}^{\beta}{}_{\gamma}(i)=0.$

This implies that the components of the metric ${\bm{h}}^{\prime}$ in normal
coordinates satisfy

$h^{\prime}_{\alpha\beta,\gamma}=0\qquad\text{ at }i,$ (100)

where $h^{\prime}_{\alpha\beta,\gamma}$ is the derivative of
$h^{\prime}_{\alpha\beta}$ with respect to $z^{\gamma}$. Taylor expanding the
metric ${\bm{h}}^{\prime}$ around $i$ gives

$h^{\prime}_{\alpha\beta}(z)=h^{\prime(0)}_{\alpha\beta}+\frac{1}{2}h^{\prime}_{\alpha\beta,\gamma\delta}z^{\gamma}z^{\delta}+O(|z|^{3}),$
(101)

where $|z|^{2}\equiv\delta_{\alpha\beta}z^{\alpha}z^{\beta}$ and
$h^{\prime(0)}_{\alpha\beta}=-\delta_{\alpha\beta}$ is the metric at $i$ (i.e.
at $|z|=0$). The non-vanishing $O(\varrho)$ terms in the initial data for
${\bm{h}}^{\prime}$ given by eq. (98) implies that $(y^{\alpha})$ are not
normal coordinates. However, the discussion in [48] shows that the
transformation between a generic set of coordinates $(y^{\alpha})$ and normal
coordinates $(z^{\alpha})$ is given by

$y^{\alpha}=z^{\alpha}-\frac{1}{2}\gamma^{\prime(0)}_{\beta}{}^{\alpha}{}_{\gamma}z^{\beta}z^{\gamma}+O(|z|^{3}),$
(102)

where $\gamma^{\prime(0)}_{\beta}{}^{\alpha}{}_{\gamma}$ are the components of
the ${\bm{h}}^{\prime}$-Levi-Civita connection coefficients with respect to
$(y^{\alpha})$ evaluated at $\varrho=0$. In a slight abuse of notation, let
$h^{\prime}_{\beta\delta,\gamma}$ denote the derivative of
$h^{\prime}_{\beta\delta}$ with respect to $y^{\gamma}$, then
$\gamma^{\prime}_{\beta}{}^{\alpha}{}_{\gamma}$ is given by

$\gamma^{\prime}_{\beta}{}^{\alpha}{}_{\gamma}=\frac{1}{2}h^{\prime\alpha\delta}\left(h^{\prime}_{\beta\delta,\gamma}+h^{\prime}_{\delta\gamma,\beta}-h^{\prime}_{\beta\gamma,\delta}\right).$

From eq. (98), one has

$\displaystyle\gamma^{\prime(0)}_{\beta}{}^{\alpha}{}_{\gamma}=\xi_{;(\gamma}\left(\vartheta_{\beta)}\vartheta^{\alpha}-\frac{1}{2}\delta_{\beta)}{}^{\alpha}\right)-\frac{1}{2}\delta^{\alpha\delta}\xi_{;\delta}\left(\vartheta_{\beta}\vartheta_{\gamma}-\frac{1}{2}\delta_{\beta\gamma}\right)-\frac{1}{2}\vartheta_{\beta}\vartheta^{\alpha}\vartheta_{\gamma}\left(\vartheta^{\delta}\xi_{;\delta}+\xi\right)$
$\displaystyle\phantom{\gamma^{\prime(0)}_{\beta}{}^{\alpha}{}_{\gamma}}+\frac{1}{2}\vartheta^{\delta}\xi_{;\delta}\left(\vartheta_{(\gamma}\delta_{\beta)}{}^{\alpha}-\frac{1}{2}\vartheta^{\alpha}\delta_{\beta\gamma}\right)-\frac{\xi}{2}\left(\vartheta_{(\gamma}\delta_{\beta)}{}^{\alpha}+\frac{3}{2}\vartheta^{\alpha}\delta_{\beta\gamma}\right),$

where $\vartheta^{\alpha}\equiv y^{\alpha}/\varrho$. In the above,
$\xi_{;\gamma}$ is the derivative of $\xi$ with respect to the angular
coordinates $\vartheta^{\gamma}$ and (γβ) indicates a symmetrisation over
$\gamma$ and $\beta$. The transformation given by eq. (102) implies that the
components of ${\bm{h}}^{\prime}$ with respect to $(z^{\alpha})$ admit the
following expansion near $|z|=0$:

$h^{\prime}_{\alpha\beta}=-\delta_{\alpha\beta}+O(|z|^{2}).$

Moreover, the components of the extrinsic curvature $K^{\prime}_{\alpha\beta}$
with respect to $(z^{\alpha})$ can be written as

$K^{\prime}_{\alpha\beta}=-\frac{\zeta}{2}\delta_{\alpha\beta}-\frac{1}{2}\left(2B_{(\alpha}\vartheta_{\beta)}+\frac{1}{2}(B^{\gamma}\vartheta_{\gamma})\delta_{\alpha\beta}\right)+\zeta\vartheta_{\alpha}\vartheta_{\beta}-4(B^{\gamma}\vartheta_{\gamma})\vartheta_{\alpha}\vartheta_{\beta}+O(|z|).$

Taylor expanding the conformal factor $\Omega^{\prime}$ around $|z|=0$ gives

$\Omega^{\prime}=|z|^{2}-\frac{A}{2}|z|^{3}-\gamma^{\prime(0)}_{\beta}{}^{\alpha}{}_{\gamma}z^{\beta}z_{\alpha}z^{\gamma}+O(|z|^{4}).$

The next step is to exploit the conformal freedom in Definition 1 with the aim
of simplifying upcoming calculations. In particular, introduce the conformal
normal initial data $(\bar{{\bm{h}}},\bar{{\bm{K}}})$, related to
$({\bm{h}}^{\prime},{\bm{K}}^{\prime})$ by

$\bar{{\bm{h}}}=\varpi^{2}{\bm{h}}^{\prime},\qquad\bar{{\bm{K}}}=\varpi{\bm{K}}^{\prime}.$
(104)

In the following, let $l^{\prime}_{\alpha\beta}(i)$ denote the components of
the ${\bm{h}}^{\prime}$-Schouten tensor with respect to $(z^{\alpha})$
evaluated at $i$. If the conformal factor $\varpi$ is given by

$\varpi\equiv e^{f},\qquad\text{ with
}f=\frac{1}{2}l^{\prime}_{\alpha\beta}(i)z^{\alpha}z^{\beta},$ (105)

then one can show that the Riemann curvature tensor associated with
$\bar{{\bm{h}}}$ is vanishing at $i$. In particular, if ${\bm{D}}^{\prime}$
denote the ${\bm{h}}^{\prime}$-covariant derivative, the conformal factor
$\varpi$ can be shown to satisfy

$\varpi(i)=1,\qquad D^{\prime}_{\alpha}\varpi(i)=0,\qquad
D^{\prime}_{\alpha}D^{\prime}_{\beta}\varpi(i)=l^{\prime}_{\alpha\beta}(i).$

Accordingly, the conformal rescaling given by eq. (104) implies that

$\bar{l}_{\alpha\beta}(i)=0.$

However, the three-dimensional Riemann curvature tensor
$\bar{r}_{\alpha\beta\gamma\delta}$ is fully determined by
$\bar{l}_{\alpha\beta}$. Hence, one has

$\bar{r}_{\alpha\beta\gamma\delta}(i)=0.$

Note that the conformal rescaling in eq. (104) indicates that
$(\bar{{\bm{h}}},\bar{{\bm{K}}})$ are related to
$(\tilde{{\bm{h}}},\tilde{{\bm{K}}})$ by

$\bar{{\bm{h}}}=\Omega^{2}\tilde{{\bm{h}}},\qquad\bar{{\bm{K}}}=\Omega\tilde{{\bm{K}}},$

where

$\Omega\equiv\varpi\Omega^{\prime}.$

In a slight abuse of notation, let

$\vartheta^{\alpha}\equiv\frac{z^{\alpha}}{\rho},$

where $\rho\equiv|z|$. By Taylor expanding $\Omega$ around $\rho=0$, one gets

$\displaystyle\Omega=\rho^{2}-\left(\frac{A}{2}+\gamma^{\prime(0)}_{\beta}{}^{\alpha}{}_{\gamma}\vartheta^{\beta}\vartheta_{\alpha}\vartheta^{\gamma}\right)\rho^{3}+O(\rho^{4}),$
(106a)
$\displaystyle\phantom{\Omega}=\rho^{2}+\frac{1}{6}\Pi_{3}[\Omega]\rho^{3}+O(\rho^{4}).$

In the above, $\Pi_{3}[\Omega]$ is defined by

$\Pi_{3}[\Omega]\equiv-6\left(\frac{A}{2}+\gamma^{\prime(0)}_{\beta}{}^{\alpha}{}_{\gamma}\vartheta^{\beta}\vartheta_{\alpha}\vartheta^{\gamma}\right).$
(107)

A direct computation readily shows that

$\gamma^{\prime(0)}_{\beta}{}^{\alpha}{}_{\gamma}\vartheta^{\beta}\vartheta_{\alpha}\vartheta^{\gamma}=-\frac{7}{4}\xi,$

where it is recalled that $\xi\in C^{2}(\mathbb{S}^{2})$ is a freely
specifiable function on the 2-sphere. Thus, one has that

$\Pi_{3}[\Omega]=\frac{21}{2}\xi-3A.$ (108)

Accordingly, one sees that the coefficient $\Pi_{3}[\Omega]$ is completely
determined by the freely specifiable data $A$ and $\xi$.

######  Remark 8.

The $\Pi_{n}$ notation will be used frequently in later calculations. For any
smooth function $\chi$, we use $\Pi_{n}[\chi]$ to denote the coefficient of
$(1/n!)\rho^{n}$ in its Taylor series around $\rho=0$.

From the previous discussion, the components $\bar{{\bm{h}}}$ and
$\bar{{\bm{K}}}$ with respect to $(z^{\alpha})$ are given by

$\displaystyle\bar{h}_{\alpha\beta}=-\delta_{\alpha\beta}+O(\rho^{3}),$
$\displaystyle\bar{K}_{\alpha\beta}=-\frac{\zeta}{2}\delta_{\alpha\beta}-\frac{1}{2}\left(2B_{(\alpha}\vartheta_{\beta)}+\frac{1}{2}(B^{\gamma}\vartheta_{\gamma})\delta_{\alpha\beta}\right)+\zeta\vartheta_{\alpha}\vartheta_{\beta}-4(B^{\gamma}\vartheta_{\gamma})\vartheta_{\alpha}\vartheta_{\beta}+O(\rho).$

In other words, $(\bar{{\bm{h}}},\bar{{\bm{K}}})$ can be written as

$\displaystyle\bar{{\bm{h}}}=\bar{h}_{\alpha\beta}{\bm{d}}{z}^{\alpha}\otimes{\bm{d}}{z}^{\beta},$
$\displaystyle\bar{{\bm{K}}}=\bar{K}_{\alpha\beta}{\bm{d}}{z^{\alpha}}\otimes{\bm{d}}{z^{\beta}}.$

Observe that the leading order term in $\bar{K}_{\alpha\beta}$ is equivalent
to the leading order in $K^{\prime}_{\alpha\beta}$ since the contribution from
$\varpi$ is at higher orders. In upcoming calculations, the initial data
$(\bar{{\bm{h}}},\bar{{\bm{K}}})$ will be referred to as the conformal normal
initial data.

####  6.2.2 The regular initial data at spatial infinity

To analyse the conformal constraint eqs. (92), introduce the orthonormal frame
$\\{\bar{{\bm{e}}}_{{\bm{i}}}\\}$ and its dual
$\\{\bar{{\bm{\omega}}}^{{\bm{i}}}\\}$ related to
$\\{{\bm{\partial}}/{\bm{\partial}}z^{\alpha}\\}$ and
$\\{{\bm{d}}{z^{\alpha}}\\}$ by

$\bar{{\bm{e}}}_{{\bm{i}}}=\bar{{\bm{e}}}^{\alpha}{}_{{\bm{i}}}\frac{{\bm{\partial}}}{{\bm{\partial}}z^{\alpha}},\qquad\bar{{\bm{\omega}}}^{{\bm{i}}}=\bar{{\bm{\omega}}}_{\alpha}{}^{{\bm{i}}}{\bm{d}}{z^{\alpha}}.$

Given that
$\bar{{\bm{h}}}(\bar{{\bm{e}}}_{{\bm{i}}},\bar{{\bm{e}}}_{{\bm{j}}})=\delta_{{\bm{i}}{\bm{j}}}$,
one has

$\bar{{\bm{e}}}^{\alpha}{}_{{\bm{i}}}(i)=1,\qquad\bar{{\bm{\omega}}}_{\alpha}{}^{{\bm{i}}}(i)=1.$

In terms of $\\{\bar{{\bm{e}}}_{{\bm{i}}}\\}$, the conformal normal initial
data can be written as

$\displaystyle\bar{{\bm{h}}}=\bar{h}_{{\bm{i}}{\bm{j}}}\bar{{\bm{\omega}}}^{{\bm{i}}}\otimes\bar{{\bm{\omega}}}^{{\bm{j}}},$
$\displaystyle\bar{{\bm{K}}}=\bar{K}_{{\bm{i}}{\bm{j}}}\bar{{\bm{\omega}}}^{{\bm{i}}}\otimes\bar{{\bm{\omega}}}^{{\bm{j}}},$

with

$\displaystyle\bar{h}_{{\bm{i}}{\bm{j}}}=-\delta_{{\bm{i}}{\bm{j}}}+O(\rho^{3}),$
$\displaystyle\bar{K}_{{\bm{i}}{\bm{j}}}=-\frac{\zeta}{2}\delta_{{\bm{i}}{\bm{j}}}-\frac{1}{2}\left(2B_{({\bm{i}}}\vartheta_{{\bm{j}})}+\frac{1}{2}(B^{{\bm{k}}}\vartheta_{{\bm{k}}})\delta_{{\bm{i}}{\bm{j}}}\right)+\zeta\vartheta_{{\bm{i}}}\vartheta_{{\bm{j}}}-4(B^{{\bm{k}}}\vartheta_{{\bm{k}}})\vartheta_{{\bm{i}}}\vartheta_{{\bm{j}}}+O(\rho).$

Substituting in eq. (92), one can show that

$\bar{L}_{{\bm{i}}{\bm{j}}}=O(\rho^{-1}),\qquad\bar{d}_{{\bm{i}}{\bm{j}}{\bm{k}}}=O(\rho^{-3}),\qquad\bar{d}_{{\bm{i}}{\bm{j}}}=O(\rho^{-3}).$

Hence, the initial data for the conformal fields
$\bar{L}_{{\bm{i}}{\bm{j}}},\bar{d}_{{\bm{i}}{\bm{j}}},\ldots$ implied by the
conformal normal initial data is singular at $\rho=0$. Following Friedrich’s
formulation [39], the regular initial data for the ECFEs can be introduced by
considering the conformal rescaling

$\Omega\to\kappa^{-1}\Omega,$ (113)

where $\kappa=O(\rho)$ is the arbitrary function introduced in Section 3.
Then, introduce the rescaled frame fields $\\{{\bm{e}}_{{\bm{i}}}\\}$ and
their dual $\\{{\bm{\omega}}^{{\bm{i}}}\\}$ as

${\bm{e}}_{{\bm{i}}}=\kappa\bar{{\bm{e}}}_{{\bm{i}}},\qquad{\bm{\omega}}^{{\bm{i}}}=\kappa^{-1}\bar{{\bm{\omega}}}^{{\bm{i}}}.$

Then, the components of the rescaled metric
${\bm{h}}=\kappa^{-2}\bar{{\bm{h}}}$ with respect to
$\\{{\bm{e}}_{{\bm{i}}}\\}$ are related to $\bar{h}_{{\bm{i}}{\bm{j}}}$ by

$\displaystyle
h_{{\bm{i}}{\bm{j}}}={\bm{h}}({\bm{e}}_{{\bm{i}}},{\bm{e}}_{{\bm{j}}})$
$\displaystyle\phantom{h_{{\bm{i}}{\bm{j}}}}=\kappa^{-2}\bar{{\bm{h}}}(\kappa\bar{{\bm{e}}}_{{\bm{i}}},\kappa\bar{{\bm{e}}}_{{\bm{j}}})$
$\displaystyle\phantom{h_{{\bm{i}}{\bm{j}}}}=\bar{{\bm{h}}}(\bar{{\bm{e}}}_{{\bm{i}}},\bar{{\bm{e}}}_{{\bm{j}}})=\bar{h}_{{\bm{i}}{\bm{j}}}.$

Moreover, the components of the rescaled extrinsic curvature
${\bm{K}}=\kappa^{-1}\bar{{\bm{K}}}$ with respect to
$\\{{\bm{e}}_{{\bm{i}}}\\}$ are related to $\bar{K}_{{\bm{i}}{\bm{j}}}$ by

$K_{{\bm{i}}{\bm{j}}}={\bm{K}}({\bm{e}}_{{\bm{i}}},{\bm{e}}_{{\bm{j}}})=\kappa^{-1}\bar{{\bm{K}}}(\kappa\bar{{\bm{e}}}_{{\bm{i}}},\kappa\bar{{\bm{e}}}_{{\bm{j}}})=\kappa\bar{K}_{{\bm{i}}{\bm{j}}}.$

Finally, one can show that the components of the rescaled conformal fields
$(L_{{\bm{i}}{\bm{j}}},d_{{\bm{i}}{\bm{j}}{\bm{k}}},d_{{\bm{i}}{\bm{j}}})$ are
related to
$(\bar{L}_{{\bm{i}}{\bm{j}}},\bar{d}_{{\bm{i}}{\bm{j}}{\bm{k}}},\bar{d}_{{\bm{i}}{\bm{j}}})$
by

$L_{{\bm{i}}{\bm{j}}}=\kappa^{2}\bar{L}_{{\bm{i}}{\bm{j}}},\qquad
d_{{\bm{i}}{\bm{j}}{\bm{k}}}=\kappa^{3}\bar{d}_{{\bm{i}}{\bm{j}}{\bm{k}}},\qquad
d_{{\bm{i}}{\bm{j}}}=\kappa^{3}\bar{d}_{{\bm{i}}{\bm{j}}}.$

Hence, the conformal rescaling given by eq. (113) introduces regular initial
data for the ECFEs since

$K_{{\bm{i}}{\bm{j}}}=O(\rho),\qquad L_{{\bm{i}}{\bm{j}}}=O(\rho),\qquad
d_{{\bm{i}}{\bm{j}}{\bm{k}}}=O(1),\qquad d_{{\bm{i}}{\bm{j}}}=O(1),$

where the explicit form of these fields is omitted as they are not required
for upcoming calculations.

Note that the conformal freedom in Friedrich’s formulation is reflected by the
different choices of the conformal factor $\kappa$. In subsequent
calculations, assume

$\kappa=\omega,$ (115)

where $\omega$ is given by eq. (47). This particular choice of $\kappa$
introduces Friedrich’s horizontal representation of spatial infinity where
$\mathscr{I}^{\pm}_{a}$ are identified by $0<\rho<a$ and $\tau=\pm 1$ —see eq
(38a).

Using eq. (47) and Taylor expanding $\omega$ around $\rho=0$ gives

$\kappa=\rho+\tfrac{1}{2}\Pi_{2}[\omega]\rho^{2}+O(\rho^{3}),$ (116)

with

$\Pi_{2}[\omega]=-\frac{1}{6}\Pi_{3}[\Omega].$

Within the framework of the ECFEs, the conformal factor $\Theta$ relating the
spacetime metrics ${\bm{g}}$ and $\tilde{{\bm{g}}}$ is fixed by the initial
data. Given the above choice of $\kappa$ and using eq. (46), the conformal
factor $\Theta$ can be expanded near $\rho=0$ as

$\Theta=\rho(1-\tau^{2})+O(\rho^{2}).$ (117)

####  6.2.3 Conformally flat initial data for the space spinor fields

Given that the final form of the ECFEs is written in terms of spinors, one
needs to obtain initial data for the spinor fields appearing in eqs. (58) and
(59b). This discussion can be carried out for the general initial data
discussed in the previous section. However, the calculations in this work only
require expressions for the initial data at zero order. Accordingly, the
formulae for the conformally flat initial data given in [49] will be
sufficient for this analysis. In the following, assume the conformally flat
initial data for eqs. (58) and (59b), given by

$\displaystyle{\bm{e}}^{0}{}_{{\bm{A}}{\bm{B}}}=0,$
$\displaystyle{\bm{e}}^{1}{}_{{\bm{A}}{\bm{B}}}=\omega
x_{{\bm{A}}{\bm{B}}},\qquad{\bm{e}}^{2}{}_{{\bm{A}}{\bm{B}}}=\frac{\omega}{\rho}z_{{\bm{A}}{\bm{B}}},\qquad{\bm{e}}^{3}{}_{{\bm{A}}{\bm{B}}}=\frac{\omega}{\rho}y_{{\bm{A}}{\bm{B}}},$
$\displaystyle\xi_{{\bm{A}}{\bm{B}}{\bm{C}}{\bm{D}}}=\sqrt{2}\left(\frac{\omega}{\rho}(x_{{\bm{B}}{\bm{D}}}\epsilon_{{\bm{A}}{\bm{C}}}+x_{{\bm{A}}{\bm{C}}}\epsilon_{{\bm{B}}{\bm{D}}})\right)-\sqrt{2}(\epsilon_{{\bm{B}}{\bm{D}}}D_{{\bm{A}}{\bm{C}}}\omega+\epsilon_{{\bm{A}}{\bm{C}}}D_{{\bm{B}}{\bm{D}}}\omega),$
$\displaystyle f_{{\bm{A}}{\bm{B}}}=D_{{\bm{A}}{\bm{B}}}\omega,$
$\displaystyle\chi_{(({\bm{A}}{\bm{B}}){\bm{C}}{\bm{D}})}=0,$
$\displaystyle\Theta_{{\bm{A}}{\bm{B}}({\bm{C}}{\bm{D}})}=-\frac{\omega^{2}}{6\Omega}\left(D_{{\bm{A}}{\bm{B}}}D_{{\bm{C}}{\bm{D}}}\Omega+D_{{\bm{C}}{\bm{D}}}D_{{\bm{A}}{\bm{B}}}\Omega+2D_{{\bm{A}}({\bm{C}}}D_{{\bm{D}}){\bm{B}}}\Omega+2D_{{\bm{B}}({\bm{C}}}D_{{\bm{D}}){\bm{A}}}\Omega\right),$
$\displaystyle\Theta_{{\bm{A}}{\bm{B}}}=0,$
$\displaystyle\phi_{{\bm{A}}{\bm{B}}{\bm{C}}{\bm{D}}}=\frac{\omega^{3}}{6\Omega^{2}}\left(D_{{\bm{A}}{\bm{B}}}D_{{\bm{C}}{\bm{D}}}\Omega+D_{{\bm{C}}{\bm{D}}}D_{{\bm{A}}{\bm{B}}}\Omega+2D_{{\bm{A}}({\bm{C}}}D_{{\bm{D}}){\bm{B}}}\Omega+2D_{{\bm{B}}({\bm{C}}}D_{{\bm{D}}){\bm{A}}}\Omega\right).$

It is important to note that the initial data obtained from these equations
will not be consistent with those obtained from eqs. (92). However, the
crucial observation is that the zero-order expressions for the initial data
are indeed consistent, which is sufficient for the rest of the calculations in
this article. Using the xAct package, the non-vanishing conformally flat
initial data for the irreducible components of the fields appearing in eqs.
(58) and (59b) can be expressed as

$\displaystyle{\bm{e}}^{1}_{x\star}=\omega,\qquad{\bm{e}}^{2}_{z\star}=\frac{\omega}{\rho},\qquad{\bm{e}}^{3}_{y\star}=\frac{\omega}{\rho},$
$\displaystyle\xi_{x\star}=-\frac{\sqrt{2}\omega+\sqrt{2}\rho\partial_{\rho}(\omega)}{\rho},\qquad\xi_{y\star}=-\frac{\sqrt{2}X_{-}(\omega)}{\rho},\qquad\xi_{z\star}=-\frac{\sqrt{2}X_{+}(\omega)}{\rho},$
$\displaystyle f_{x\star}=\partial_{\rho}(\omega),\qquad
f_{y\star}=\frac{X_{-}(\omega)}{\rho},\qquad
f_{z\star}=\frac{X_{+}(\omega)}{\rho},$
$\displaystyle\Theta_{0\star}=-\frac{\omega^{2}X^{2}_{+}(\Omega)}{2\rho^{2}\Omega},\qquad\Theta_{1\star}=-\frac{2\rho\omega^{2}\partial_{\rho}(X_{+}(\Omega))-2\omega^{2}X_{+}(\Omega)}{\rho^{2}\Omega},$
$\displaystyle\Theta_{2\star}=\frac{-4\rho\omega^{2}\partial_{\rho}(\Omega)+4\rho^{2}\omega^{2}\partial^{2}_{\rho}(\Omega)-\omega^{2}X_{-}(X_{+}(\Omega))-\omega^{2}X_{+}(X_{-}(\Omega))}{2\rho^{2}\Omega},$
$\displaystyle\Theta_{3\star}=-\frac{-2\rho\omega^{2}\partial_{\rho}(X_{-}(\Omega))+2\omega^{2}X_{-}(\Omega)}{\rho^{2}\Omega},\qquad\Theta_{4\star}=-\frac{\omega^{2}X^{2}_{-}(\Omega)}{2\rho^{2}\Omega},$
$\displaystyle\phi_{0\star}=\frac{\omega^{3}X^{2}_{+}(\Omega)}{2\rho^{2}\Omega^{2}},\qquad\phi_{1\star}=\frac{-2\omega^{3}X_{+}(\Omega)}{\rho^{2}\Omega^{2}},$
$\displaystyle\phi_{2\star}=\frac{-4\rho\omega^{3}\partial_{\rho}(\Omega)+4\rho^{2}\omega^{3}\partial^{2}_{\rho}(\Omega)-\omega^{3}X_{-}(X_{+}(\Omega))-\omega^{3}X_{+}(X_{-}(\Omega))}{2\rho^{2}\Omega^{2}},$
$\displaystyle\phi_{3\star}=\frac{-2\rho\omega^{3}\partial_{\rho}(X_{-}(\Omega))+2\omega^{3}X_{-}(\Omega)}{\rho^{2}\Omega^{2}},\qquad\phi_{4\star}=\frac{\omega^{3}X^{2}_{-}(\Omega)}{2\rho^{2}\Omega^{2}},$

where $X_{\pm}^{2}(\Omega)\equiv X_{\pm}(X_{\pm}(\Omega))$ and
$\partial^{2}_{\rho}(\Omega)\equiv\partial_{\rho}(\partial_{\rho}(\Omega))$.
Using eqs. (106) and (116), the non-vanishing initial data for the zeroth-
order equations obtained from eqs. (58) and (59b) can be written as

$\displaystyle{\bm{e}}^{2(0)}_{z\star}=1,\qquad{\bm{e}}^{3(0)}_{y\star}=1,\qquad
f^{(0)}_{x\star}=1,$ (120a)
$\displaystyle\phi^{(0)}_{0\star}=\frac{1}{12}X^{2}_{+}(\Pi_{3}[\Omega]),\qquad\phi^{(0)}_{1\star}=\frac{2}{3}X_{+}(\Pi_{3}[\Omega]),$
(120b)
$\displaystyle\phi^{(0)}_{2\star}=\frac{1}{12}\left(12\Pi_{3}[\Omega]-2X_{-}(X_{+}(\Pi_{3}[\Omega]))\right),$
(120c)
$\displaystyle\phi^{(0)}_{3\star}=-\frac{2}{3}X_{-}(\Pi_{3}[\Omega]),\qquad\phi^{(0)}_{4\star}=\frac{1}{12}X^{2}_{-}(\Pi_{3}[\Omega]),$
(120d)

where the superscript (0) indicates that the initial data is evaluated at
$\rho=0$. Given that

${\bm{d}}=\Theta{\bm{f}}+{\bm{d}}{\Theta},$ (121)

one can show that the zero-order initial data for the irreducible components
of $d_{{\bm{A}}{\bm{B}}}$ is given by

$d_{x\star}^{(0)}=0,\qquad d_{y\star}^{(0)}=0,\qquad d_{z\star}^{(0)}=0.$

######  Remark 9.

The explicit calculation of the conformal factor $\theta$ and the
transformation matrices $\Lambda^{{\bm{A}}}{}_{{\bm{B}}}$ will require the
first-order solutions of ${\bm{e}}^{1}_{x},{\bm{e}}^{1}_{y}$ and
${\bm{e}}^{1}_{z}$. For this purpose, we list the first-order initial data for
these components —namely

${\bm{e}}^{1(1)}_{x\star}=1,\qquad{\bm{e}}^{1(1)}_{y\star}=0,\qquad{\bm{e}}^{1(1)}_{z\star}=0.$
(122)

Given the initial data for the spinor fields appearing in the ECFEs, it is now
possible to obtain solutions for the background evolution eqs. (58) and the
boundary-adapted evolution and constraint eqs. (59b)-(60).

##  7 Evaluating the BMS-supertranslation charges at the critical sets

This section aims to bring together this analysis’s various elements and
obtain an expression of BMS-supertranslation charges $\mathcal{Q}$ at the
critical sets $\mathcal{I}^{\pm}$. Recall that $\mathcal{Q}$ depends on the
solutions of eqs. (58) and (59b)-(60), the conformal factor $\theta$ and the
transformation matrices $\Lambda^{{\bm{A}}}{}_{{\bm{B}}}$ —see eq. (91). In
the following, the solutions of eqs. (58) and (59b)-(60) will be used to
obtain an explicit transformation from the NP-gauge to the F-gauge, allowing
us to assess the contribution from the background term $\sigma^{\bullet
ab}N^{\bullet}_{ab}$ and the components of $\phi_{ABCD}$ to $\mathcal{Q}$ at
zero-order. As will become evident, this analysis will reveal that the generic
initial data provided in [35] does not give rise to well-defined BMS-
asymptotic charges $\mathcal{Q}$ at $\mathcal{I}^{\pm}$.

###  7.1 Asymptotic solution of the extended conformal field equations

In subsequent calculations, the $\Pi_{n}$ notation introduced in Section 6.2.1
will be used to express the solutions of eqs. (58) and (59b)-(60). Using the
$\Pi_{n}$ notation, the expansion of the conformal factor $\Theta$ near
$\rho=0$ can be written as

$\Theta=\Pi_{0}[\Theta]+\Pi_{1}[\Theta]\rho+\frac{1}{2}\Pi_{2}[\Theta]\rho^{2}+O(\rho^{3}).$

Comparing with eq. (117), one has

$\Pi_{0}[\Theta]=0,\qquad\Pi_{1}[\Theta]=(1-\tau^{2}).$ (123)

From eq. (52), the components $d_{x},d_{y}$ and $d_{z}$ of the 1-form $d_{AB}$
can be shown to satisfy

$\displaystyle\Pi_{0}[d_{x}]=0,\qquad\Pi_{0}[d_{y}]=0,\qquad\Pi_{0}[d_{z}]=0,$
(124a)
$\displaystyle\Pi_{1}[d_{x}]=2,\qquad\Pi_{1}[d_{y}]=0,\qquad\Pi_{1}[d_{z}]=0.$
(124b)

Taylor expanding the spinor fields appearing in eqs. (58) and using the xAct
package, it is possible to show that eqs. (58) decouple from the boundary-
adapted evolution and constraint eqs. (59b)-(60) at zero-order. In other
words, the zero-order system for the background fields does not depend on
$\phi_{0},\phi_{1},\phi_{2},\phi_{3}$ and $\phi_{4}$. Integrating, one can
write zero-order solutions of eqs. (58) as

$\displaystyle\Pi_{0}[{\bm{e}}^{0}_{x}]=-\tau,\qquad\Pi_{0}[{\bm{e}}^{0}_{y}]=\Pi_{0}[{\bm{e}}^{0}_{z}]=\Pi_{0}[{\bm{e}}^{1}_{x}]=\Pi_{0}[{\bm{e}}^{1}_{y}]=\Pi_{0}[{\bm{e}}^{1}_{z}]=0,$
$\displaystyle\Pi_{0}[{\bm{e}}^{2}_{x}]=\Pi_{0}[{\bm{e}}^{2}_{y}]=0,\qquad\Pi_{0}[{\bm{e}}^{2}_{z}]=1,$
$\displaystyle\Pi_{0}[{\bm{e}}^{3}_{x}]=0,\qquad\Pi_{0}[{\bm{e}}^{3}_{y}]=1,\qquad\Pi_{0}[{\bm{e}}^{3}_{z}]=0,$
$\displaystyle\Pi_{0}[\xi_{0}]=\Pi_{0}[\xi_{1}]=\Pi_{0}[\xi_{2}]=\Pi_{0}[\xi_{3}]=\Pi_{0}[\xi_{4}]=\Pi_{0}[\xi_{h}]=0,$
$\displaystyle\Pi_{0}[\xi_{x}]=\Pi_{0}[\xi_{y}]=\Pi_{0}[\xi_{z}]=0,$
$\displaystyle\Pi_{0}[f_{x}]=1,\qquad\Pi_{0}[f_{y}]=\Pi_{0}[f_{z}]=0,$ (125)
$\displaystyle\Pi_{0}[\chi_{0}]=\Pi_{0}[\chi_{1}]=\Pi_{0}[\chi_{2}]=\Pi_{0}[\chi_{3}]=\Pi_{0}[\chi_{4}]=\Pi_{0}[\chi_{h}]=0,$
$\displaystyle\Pi_{0}[\chi_{x}]=\Pi_{0}[\chi_{y}]=\Pi_{0}[\chi_{z}]=0,$
$\displaystyle\Pi_{0}[\Theta_{0}]=\Pi_{0}[\Theta_{1}]=\Pi_{0}[\Theta_{2}]=\Pi_{0}[\Theta_{3}]=\Pi_{0}[\Theta_{4}]=\Pi_{0}[\Theta_{h}]=0,$
$\displaystyle\Pi_{0}[\Theta_{x}]=\Pi_{0}[\Theta_{y}]=\Pi_{0}[\Theta_{z}]=0,$
$\displaystyle\Pi_{0}[\theta_{x}]=\Pi_{0}[\theta_{y}]=\Pi_{0}[\theta_{z}]=0.$

For the first-order system, only the equations for
${\bm{e}}^{1}_{x},{\bm{e}}^{1}_{y}$ and ${\bm{e}}^{1}_{z}$ will decouple from
the boundary-adapted evolution and constraint equations. Integrating, it can
be shown that $\Pi_{1}[{\bm{e}}^{1}_{x}],\Pi_{1}[{\bm{e}}^{1}_{y}]$ and
$\Pi_{1}[{\bm{e}}^{1}_{z}]$ are constants and fixed by the initial data given
in eq. (122). Then, the first-order solution for
${\bm{e}}^{1}_{x},{\bm{e}}^{1}_{y}$ and ${\bm{e}}^{1}_{z}$ is given by

$\Pi_{1}[{\bm{e}}^{1}_{x}]=1,\qquad\Pi_{1}[{\bm{e}}^{1}_{y}]=0,\qquad\Pi_{1}[{\bm{e}}^{1}_{z}]=0.$

From the above discussion, the solutions of eqs. (58) can be written as

$\displaystyle{\bm{e}}^{0}_{x}=-\tau+O(\rho),\qquad{\bm{e}}^{1}_{x}=\rho+O(\rho^{2}),$
(126a)
$\displaystyle{\bm{e}}^{2}_{z}=1+O(\rho),\qquad{\bm{e}}^{3}_{y}=1+O(\rho),$
(126b) $\displaystyle f_{x}=1+O(\rho),$ (126c)

where all other components are $O(\rho)$ or higher order.

The next step in this analysis is to examine the zero-order boundary-adapted
evolution and constraint eqs. (59b)-(60). By Taylor-expanding all the
components of the spinor fields in eqs. (59b)-(60) and substituting the zero-
order solution of the background fields eqs. (125) and using eqs. (123)-(124),
the zero-order boundary-adapted evolution system can be written as

$\displaystyle\sqrt{2}(1+\tau)\partial_{\tau}(\Pi_{0}[\phi_{0}])+\frac{1}{2\sqrt{2}}X_{+}(\Pi_{0}[\phi_{1}])=-2\sqrt{2}\Pi_{0}[\phi_{0}],$
(127a)
$\displaystyle\frac{1}{2\sqrt{2}}\partial_{\tau}(\Pi_{0}[\phi_{1}])+\frac{1}{\sqrt{2}}X_{-}(\Pi_{0}[\phi_{0}])+\frac{1}{6\sqrt{2}}X_{+}(\Pi_{0}[\phi_{2}])=-\frac{1}{2\sqrt{2}}\Pi_{0}[\phi_{1}],\quad$
(127b)
$\displaystyle\frac{1}{3\sqrt{2}}\partial_{\tau}(\Pi_{0}[\phi_{2}])+\frac{1}{4\sqrt{2}}X_{-}(\Pi_{0}[\phi_{1}])+\frac{1}{4\sqrt{2}}X_{+}(\Pi_{0}[\phi_{3}])=0,$
(127c)
$\displaystyle\frac{1}{2\sqrt{2}}\partial_{\tau}(\Pi_{0}[\phi_{3}])+\frac{1}{6\sqrt{2}}X_{-}(\Pi_{0}[\phi_{2}])+\frac{1}{\sqrt{2}}X_{+}(\Pi_{0}[\phi_{4}])=\frac{1}{2\sqrt{2}}\Pi_{0}[\phi_{3}],$
(127d)
$\displaystyle\sqrt{2}(1-\tau)\partial_{\tau}(\Pi_{0}[\phi_{4}])+\frac{1}{2\sqrt{2}}X_{-}(\Pi_{0}[\phi_{3}])=2\sqrt{2}\Pi_{0}[\phi_{4}],$
(127e)

while the zero-order boundary-adapted constraint system is given by

$\displaystyle\frac{1}{2\sqrt{2}}\tau\partial_{\tau}(\Pi_{0}[\phi_{1}])-\frac{1}{\sqrt{2}}X_{-}(\Pi_{0}[\phi_{0}])+\frac{1}{6\sqrt{2}}X_{+}(\Pi_{0}[\phi_{2}])=0,$
(128a)
$\displaystyle\frac{1}{3\sqrt{2}}\tau\partial_{\tau}(\Pi_{0}[\phi_{2}])-\frac{1}{4\sqrt{2}}X_{-}(\Pi_{0}[\phi_{1}])+\frac{1}{4\sqrt{2}}X_{+}(\Pi_{0}[\phi_{3}])=0,$
(128b)
$\displaystyle\frac{1}{2\sqrt{2}}\tau\partial_{\tau}(\Pi_{0}[\phi_{3}])-\frac{1}{6\sqrt{2}}X_{-}(\Pi_{0}[\phi_{2}])+\frac{1}{\sqrt{2}}X_{+}(\Pi_{0}[\phi_{4}])=0.$
(128c)

To simplify subsequent calculations, define the complex-valued functions
$T_{m}{}^{j}{}_{k}$ as follows

$\displaystyle T_{m}{}^{j}{}_{k}:SU(2,\mathbb{C})$
$\displaystyle\to\mathbb{C},$
$\displaystyle\phantom{T_{m}{}^{j}{}_{k}:}{\bm{t}}$ $\displaystyle\mapsto
T_{m}{}^{j}{}_{k}({\bm{t}}),$

where

$T_{m}{}^{j}{}_{k}({\bm{t}})=\begin{pmatrix}m\\\
j\end{pmatrix}^{1/2}\begin{pmatrix}m\\\
k\end{pmatrix}^{1/2}{\bm{t}}^{(b_{1}}{}_{(a_{1}}\ldots{\bm{t}}^{b_{m})}{}_{a_{m})_{k}},\qquad
T_{0}{}^{0}{}_{0}({\bm{t}})=1.$

Here, $m=0,1,2,\ldots$, and $j,k=0,\ldots,m$ and

$\begin{pmatrix}m\\\ j\end{pmatrix}=\frac{m!}{j!(m-j)!}.$

Note that the complex conjugate of $T_{m}{}^{j}{}_{k}({\bm{t}})$ is defined by

$\overline{T_{m}{}^{j}{}_{k}({\bm{t}})}=(-1)^{j+k}T_{m}{}^{m-j}{}_{m-k}({\bm{t}}),$

while the action of $X,X_{+}$ and $X_{-}$ on $T_{m}{}^{j}{}_{k}$ is given by

$\displaystyle X(T_{m}{}^{j}{}_{k})=(m-2k)T_{m}{}^{j}{}_{k},\qquad
X_{+}(T_{m}{}^{j}{}_{k})=\sqrt{k(1-k+m)}T_{m}{}^{j}{}_{k-1},$ $\displaystyle
X_{-}(T_{m}{}^{j}{}_{k})=-\sqrt{(1+k)(m-k)}T_{m}{}^{j}{}_{k+1}.$

######  Definition 4 (spin-weight).

A function $f$ is said to be of spin-weight $s$ if it satisfies $X(f)=2sf$,
where $s$ is an integer or half-integer.

######  Remark 10.

The functions $T_{m}{}^{j}{}_{k}({\bm{t}})$ on $SU(2,\mathbb{C})$ are closely
related to the standard spin-weighted harmonics on $\mathbb{S}^{2}$ —see e.g.
[40]

######  Remark 11.

By construction, functions on $\mathcal{M}_{a,\kappa}$ will have a well-
defined spin-weight.

The components $\Pi_{0}[\phi_{n}]$ are functions on $\mathbb{R}\times
SU(2,\mathbb{C})$. Hence, they admit a decomposition in terms of
$T_{m}{}^{j}{}_{k}$. In the following, assume that:

######  Assumption 1.

The components $\Pi_{0}[\phi_{n}]$ admit an expansion of the form

$\Pi_{0}[\phi_{n}]=\sum_{l=|2-n|}^{\infty}\sum_{m=0}^{2l}a_{n;2l,m}(\tau)T_{2l}{}^{m}{}_{l-2+n},\qquad\tau\in[-1,1],$
(129)

where $a_{n;2l,m}(\tau):\mathbb{R}\to\mathbb{C}$.

Furthermore, note that $\Pi_{3}[\Omega]$ at $\rho=0$ can be decomposed as

$\Pi_{3}[\Omega]|_{\rho=0}=\sum_{l=0}^{\infty}\sum_{m=0}^{2l}\Pi_{3}[\Omega]_{2l,m}T_{2l}{}^{m}{}_{l}.$
(130)

######  Remark 12.

The argument of $a_{n;2l,m}(\tau)$ will be omitted in upcoming calculations.
i.e., write $a_{n;2l,m}$ instead of $a_{n;2l,m}(\tau)$. Moreover, we use
$a_{n;2l,m\star}$ to refer to $a_{n;2l,m}(0)$.

For $l=0,m=0$, one can use eq. (129) in eqs. (127)-(128) to show that
$a_{2;0,0}$ is constant and is given by

$a_{2;0,0}=\Pi_{3}[\Omega]_{0,0}.$ (131)

For $l=1,m=0,1,2$, the evolution eqs. (127) imply

$\displaystyle\frac{1}{2\sqrt{2}}\dot{a}_{1;2,m}+\frac{1}{6}a_{2;2,m}=-\frac{1}{2\sqrt{2}}a_{1;2,m},$
(132a)
$\displaystyle\frac{1}{3\sqrt{2}}\dot{a}_{2;2,m}-\frac{1}{4}a_{1;2,m}+\frac{1}{4}a_{3;2,m}=0,$
(132b)
$\displaystyle\frac{1}{2\sqrt{2}}\dot{a}_{3;2,m}-\frac{1}{6}a_{2;2,m}=\frac{1}{2\sqrt{2}}a_{3;2,m},$
(132c)

while the constraint eqs. (128) give

$\displaystyle\frac{\tau}{2\sqrt{2}}\dot{a}_{1;2,m}+\frac{1}{6}a_{2;2,m}=0,$
(133a)
$\displaystyle\frac{\tau}{3\sqrt{2}}\dot{a}_{2;2,m}+\frac{1}{4}a_{1;2,m}+\frac{1}{4}a_{3;2,m}=0,$
(133b)
$\displaystyle\frac{\tau}{2\sqrt{2}}\dot{a}_{3;2,m}+\frac{1}{6}a_{2;2,m}=0,$
(133c)

where $\dot{a}_{n;2l,m}\equiv\partial_{\tau}(a_{n;2l,m})$. By multiplying eqs.
(132) with $\tau$ and substracting from eqs. (133), one obtain linear
equations to be solved for $a_{1;2,m},a_{2;2,m}$ and $a_{3;2,m}$. Substituting
back into the evolution eqs. (132) and simplifying, one gets

$\displaystyle(-1+\tau)\dot{a}_{1;2,m}-a_{1;2,m}=0.$ (134a)
$\displaystyle\tau\dot{a}_{2;2,m}-a_{2;2,m}=0,$ (134b)
$\displaystyle(1+\tau)\dot{a}_{3;2,m}-a_{3;2,m}=0.$ (134c)

Then, the solution for $a_{1;2,m},a_{2;2,m}$ and $a_{3;2,m}$ is given by

$a_{1;2,m}=-\mathcal{C}_{1,m}(1-\tau),\qquad
a_{2;2,m}=\mathcal{C}_{2,m}\tau,\qquad a_{3;2,m}=\mathcal{C}_{3,m}(1+\tau),$

where $\mathcal{C}_{1,m},\mathcal{C}_{2,m}$ and $\mathcal{C}_{3,m}$ are some
constants that depend on $m$ with $m\in\\{0,1,2\\}$. The initial data for
$\Pi_{0}[\phi_{1}]$, given by eq. (120), indicates that

$\mathcal{C}_{1,m}=-\frac{2\sqrt{2}}{3}\Pi_{3}[\Omega]_{2,m},\qquad m=0,1,2.$

Using eq. (133b), we have

$\mathcal{C}_{3,m}=\mathcal{C}_{1,m},\qquad m=0,1,2.$

Finally, eq. (132b) gives

$\mathcal{C}_{2,m}=-\frac{3}{\sqrt{2}}\mathcal{C}_{1,m},\qquad m=0,1,2.$

Therefore, $a_{1;2,m},a_{2;2,m}$ and $a_{3;2,m}$ can be written as

$\displaystyle
a_{1;2,m}=\frac{2\sqrt{2}}{3}\Pi_{3}[\Omega]_{2,m}(1-\tau),\qquad
a_{2;2,m}=2\Pi_{3}[\Omega]_{2,m}\tau,$ (135a) $\displaystyle
a_{3;2,m}=-\frac{2\sqrt{2}}{3}\Pi_{3}[\Omega]_{2,m}(1+\tau).$ (135b)

where $m\in\\{0,1,2\\}$.

For $l\geq 2,0\leq m\leq 2l$, the evolution eqs. (127) imply the ODEs

$\displaystyle(1+\tau)\dot{a}_{0}+2a_{0}+\frac{1}{4}\sqrt{(2+l)(l-1)}a_{1}=0,$
(136a)
$\displaystyle\dot{a}_{1}+a_{1}+\frac{1}{3}\sqrt{l(l+1)}a_{2}-2\sqrt{(2+l)(l-1)}a_{0}=0,$
(136b) $\displaystyle\dot{a}_{2}+\frac{3}{4}(a_{3}-a_{1})=0,$ (136c)
$\displaystyle\dot{a}_{3}-a_{3}-\frac{1}{3}\sqrt{l(l+1)}a_{2}+2\sqrt{(2+l)(l-1)}a_{4}=0,$
(136d)
$\displaystyle(1-\tau)\dot{a}_{4}-2a_{4}-\frac{1}{4}\sqrt{(2+l)(l-1)}a_{3}=0,$
(136e)

where $a_{n}$ and $\dot{a}_{n}$ refer to $a_{n;2l,m}$ and $\dot{a}_{n;2l,m}$,
respectively. The constraint eqs. (128) give

$\displaystyle\tau\dot{a}_{1}+\frac{1}{3}\sqrt{l(l+1)}a_{2}+2\sqrt{(2+l)(l-1)}a_{0}=0,$
(137a) $\displaystyle\tau\dot{a}_{2}+\frac{3}{4}\sqrt{l(l+1)}(a_{3}+a_{1})=0,$
(137b)
$\displaystyle\tau\dot{a}_{3}+\frac{1}{3}\sqrt{l(l+1)}a_{2}+2\sqrt{(2+l)(l-1)}a_{4}=0.$
(137c)

After some manipulations, the equations for $a_{1},a_{2}$ and $a_{3}$ can be
decoupled. In particular, one obtains the following second-order ODEs for
$a_{1},a_{2}$ and $a_{3}$:

$\displaystyle(1-\tau^{2})\ddot{a}_{1}+2(1-\tau)\dot{a}_{1}+l(l+1)a_{1}=0,$
(138a) $\displaystyle(1-\tau^{2})\ddot{a}_{2}-2\tau\dot{a}_{2}+l(l+1)a_{2}=0,$
(138b)
$\displaystyle(1-\tau^{2})\ddot{a}_{3}-2(1+\tau)\dot{a}_{3}+l(l+1)a_{3}=0.$
(138c)

Note that the equations for $a_{0}$ and $a_{4}$ will not decouple. This can be
confirmed by solving eqs. (136b), (136c) and (136d) for
$\dot{a}_{1},\dot{a}_{2}$ and $\dot{a}_{3}$, respectively, and substituting in
eq. (137) to obtain an algebraic system that can be written as

$A\begin{pmatrix}a_{1}\\\ a_{2}\\\
a_{3}\end{pmatrix}=6\sqrt{(2+l)(l-1)}\begin{pmatrix}(1+\tau)a_{0}\\\ 0\\\
-(1-\tau)a_{4}\end{pmatrix},$

where $A$ is a $3\times 3$ matrix given by

$A=\begin{pmatrix}3\tau&(-1+\tau)\sqrt{l(l+1)}&0\\\
(1+\tau)\sqrt{l(l+1)}&0&-(-1+\tau)\sqrt{l(l+1)}\\\
0&(1+\tau)\sqrt{l(l+1)}&3\tau\end{pmatrix}.$

One can confirm that $\text{det}(A)=0$. Hence, the equations for $a_{0}$ and
$a_{4}$ will not decouple. A solution for $a_{0}$ and $a_{4}$ can be
determined by treating the terms involving $a_{1}$ and $a_{3}$ as source
terms. Remarkably, as it will be seen in the next subsection, the value of the
coefficients $a_{0}$ and $a_{4}$ are not required in the computation of the
BMS asymptotic charges at the critical sets.

###  7.2 The explicit transformation from the NP to the F-gauge

In this section, the solution of the background fields given by eqs. (125)
will be used to obtain asymptotic expansions of the conformal factor $\theta$
and the transformation matrices $\Lambda^{{\bm{A}}}{}_{{\bm{B}}}$ relating the
NP-gauge and the F-gauge. The analysis in this section will be concerned with
the relation between the F-gauge frame
$\\{{\bm{e}}_{{\bm{A}}{\bm{A}}^{\prime}}\\}$ and the NP-gauge frame
$\\{{\bm{e}}^{\bullet}_{{\bm{A}}{\bm{A}}^{\prime}}\\}$ at future null infinity
$\mathscr{I}^{+}$. An analogous calculation can be carried out for the
transformation at $\mathscr{I}^{-}$. The discussion in this section follows
that of the transformation from the NP-gauge to the F-gauge originally
presented in [40]. However, our initial data differ from those considered in
[40], which results in differences in the asymptotic expansions of some of the
fields in this section.

Consider the adapted frame
$\\{{\bm{e}}^{\circ}_{{\bm{A}}{\bm{A}}^{\prime}}\\}$ at $\mathscr{I}^{+}$
satisfying the conditions introduced in Section 4, then the vector field
${\bm{e}}^{\circ}_{{\bm{1}}{\bm{1}}^{\prime}}$ must satisfy

${\bm{e}}^{\circ\mu}{}_{{\bm{1}}{\bm{1}}^{\prime}}=f\nabla^{\mu}\Theta,$ (139)

where $\nabla$ is the Levi-Civita connection associated with ${\bm{g}}$ and
$\Theta$ is the conformal factor defining $\mathscr{I}^{+}$ —see eqs. (117)
and (38a). From eq. (115), the set $\mathscr{I}^{+}$ can be identified by
$\tau=1$. To determine the function $f$ in eq. (139), consider the parallel
propagation condition
${\bm{e}}^{\circ\mu}{}_{{\bm{1}}{\bm{1}}^{\prime}}\nabla_{\mu}{\bm{e}}^{\circ\nu}{}_{{\bm{1}}{\bm{1}}^{\prime}}=0$
satisfied by ${\bm{e}}^{\circ}_{{\bm{1}}{\bm{1}}^{\prime}}$ which can be
written as

$f\nabla^{\mu}\Theta\nabla_{\mu}f\nabla^{\nu}\Theta+f^{2}\nabla^{\nu}(\frac{1}{2}\nabla_{\mu}\Theta\nabla^{\mu}\Theta)=0.$

A contraction with a vector field ${\bm{Z}}$ transverse to $\mathscr{I}^{+}$
yields

$\nabla^{\mu}\Theta\nabla_{\mu}(\log{f})=-\frac{{\bm{Z}}\left(\frac{1}{2}\nabla_{\mu}\Theta\nabla^{\mu}\Theta\right)}{{{\bm{Z}}}(\Theta)}.$
(140)

By setting ${\bm{Z}}=\partial_{\tau}$, one can show that

$\frac{{\bm{Z}}\left(\frac{1}{2}\nabla_{\mu}\Theta\nabla^{\mu}\Theta\right)}{{\bm{Z}}(\Theta)}=2\Pi_{1}[f_{x}]\rho^{2}+O(\rho^{3}).$
(141)

Moreover, the LHS of eq. (140) can be written as

$\displaystyle\nabla^{\mu}\Theta\nabla_{\mu}(\log{f})=\left(-2\rho^{2}+O(\rho^{3})\right)\partial_{\rho}(\log{f})+$
(142)
$\displaystyle\phantom{\nabla^{\mu}\Theta\nabla_{\mu}(\log{f})=}\left(-\frac{1}{2}\left(4\Pi_{1}[{\bm{e}}^{2}_{x}]+\frac{1}{6}X_{-}(\Pi_{3}[\Omega])\right)\rho^{2}+O(\rho^{3})\right)X_{-}(\log{f})+$
$\displaystyle\phantom{\nabla^{\mu}\Theta\nabla_{\mu}(\log{f})=}\left(-\frac{1}{2}\left(4\Pi_{1}[{\bm{e}}^{3}_{x}]+\frac{1}{6}X_{+}(\Pi_{3}[\Omega])\right)\rho^{2}+O(\rho^{3})\right)X_{+}(\log{f}).$

To solve for $f$, assume $\log{f}$ can be expanded as

$\log{f}=F_{0}+F_{1}\rho+O(\rho^{2}),$

with $X_{-}(F_{0})=X_{+}(F_{0})=0$, $X_{-}(F_{1})\neq 0$ and $X_{+}(F_{1})\neq
0$. Then, $\partial_{\rho}(\log{f}),X_{-}(\log{f})$ and $X_{+}(\log{f})$ can
be written as

$\displaystyle\partial_{\rho}(\log{f})=F_{1}+O(\rho),$ $\displaystyle
X_{-}(\log{f})=X_{-}(F_{1})\rho+O(\rho^{2}),$ $\displaystyle
X_{+}(\log{f})=X_{+}(F_{1})\rho+O(\rho^{2}).$

Using the above and eqs. (140), (141) and (142), one has

$F_{1}=-\Pi_{1}[f_{x}].$

Next, consider the expansion

$f=f_{0}+f_{1}\rho+O(\rho^{2}).$

But $f=e^{\log{f}}$, so

$f_{0}=e^{F_{0}},\qquad f_{1}=e^{F_{0}}F_{1}.$

In the following, the coefficient $F_{0}$ is chosen such that

$f_{0}=-\frac{1}{2\sqrt{2}},\qquad
f_{1}=-\frac{F_{1}}{2\sqrt{2}}=\frac{\Pi_{1}[f_{x}]}{2\sqrt{2}}.$

Thus, the function $f$ can be written as

$f=-\frac{1}{2\sqrt{2}}-\frac{\Pi_{1}[f_{x}]}{2\sqrt{2}}\rho+O(\rho^{2}).$

######  Remark 13.

While the function $f$ depends on the first-order solution of $f_{x}$, it will
become evident that the explicit solution will not be required for the
following calculations. In the rest of this analysis, first-order solutions
for the background fields will be treated as unknowns, and it will be shown
that the final expression of $\mathcal{Q}$ will only depend on the zero-order
solution.

In the following, assume that the adapted frame
$\\{{\bm{e}}^{\circ}_{{\bm{A}}{\bm{A}}^{\prime}}\\}$ and the F-gauge frame
$\\{{\bm{e}}_{{\bm{A}}{\bm{A}}^{\prime}}\\}$ are related by

${\bm{e}}^{\circ}_{{\bm{A}}{\bm{A}}^{\prime}}=\lambda^{{\bm{B}}}{}_{{\bm{A}}}\bar{\lambda}^{{\bm{B}}^{\prime}}{}_{{\bm{A}}^{\prime}}{\bm{e}}_{{\bm{B}}{\bm{B}}^{\prime}},$
(143)

where $\lambda^{{\bm{A}}}{}_{{\bm{B}}}$ denote an $SL(2,\mathbb{C})$
transformation matrix. To determine $\lambda^{{\bm{B}}}{}_{{\bm{A}}}$, note
that eq. (139) can be written as

$\displaystyle{\bm{e}}^{\circ\mu}{}_{{\bm{1}}{\bm{1}}^{\prime}}=f{\bm{e}}^{\circ\mu}{}_{{\bm{A}}{\bm{A}}^{\prime}}{\bm{e}}^{\circ{\bm{A}}{\bm{A}}^{\prime}}(\Theta)$
$\displaystyle\phantom{{\bm{e}}^{\circ\mu}{}_{{\bm{1}}{\bm{1}}^{\prime}}}=f{\bm{e}}^{\circ\mu}{}_{{\bm{A}}{\bm{A}}^{\prime}}\epsilon^{{\bm{A}}{\bm{B}}}\epsilon^{{\bm{A}}^{\prime}{\bm{B}}^{\prime}}{\bm{e}}^{\circ}_{{\bm{B}}{\bm{B}}^{\prime}}(\Theta).$

Using eq. (143) and substituting in the above, one obtains

$\displaystyle\lambda^{{\bm{0}}}{}_{{\bm{1}}}\bar{\lambda}^{{\bm{0}}^{\prime}}{}_{{\bm{1}}^{\prime}}=f{\bm{e}}_{{\bm{1}}{\bm{1}}^{\prime}}(\Theta),$
(144a)
$\displaystyle\lambda^{{\bm{0}}}{}_{{\bm{1}}}\bar{\lambda}^{{\bm{1}}^{\prime}}{}_{{\bm{1}}^{\prime}}=-f{\bm{e}}_{{\bm{1}}{\bm{0}}^{\prime}}(\Theta),$
(144b)
$\displaystyle\lambda^{{\bm{1}}}{}_{{\bm{1}}}\bar{\lambda}^{{\bm{1}}}{}_{{\bm{1}}^{\prime}}=f{\bm{e}}_{{\bm{0}}{\bm{0}}^{\prime}}(\Theta).$
(144c)

But,

$\displaystyle{\bm{e}}_{{\bm{0}}{\bm{0}}^{\prime}}(\Theta)=\frac{\Pi_{4}[\Omega]}{6\sqrt{2}}\rho^{3}+O(\rho^{4}),$
$\displaystyle{\bm{e}}_{{\bm{0}}{\bm{1}}^{\prime}}(\Theta)=-\frac{X_{+}(\Pi_{3}[\Omega])}{6\sqrt{2}}\rho^{2}+O(\rho^{3}),$
$\displaystyle{\bm{e}}_{{\bm{1}}{\bm{0}}^{\prime}}(\Theta)=-\frac{X_{-}(\Pi_{3}[\Omega])}{6\sqrt{2}}\rho^{2}+O(\rho^{3}),$
$\displaystyle{\bm{e}}_{{\bm{1}}{\bm{1}}^{\prime}}(\Theta)=-2\sqrt{2}\rho-\frac{\Pi_{3}[\Omega]}{\sqrt{2}}\rho^{2}+O(\rho^{3}),$

Hence, eq. (144a) yields

$|\lambda^{{\bm{0}}}{}_{{\bm{1}}}|^{2}=\rho+\frac{1}{2}\left(\frac{\Pi_{3}[\Omega]}{2}-2\Pi_{1}[f_{x}]\right)\rho^{2}+O(\rho^{3}).$

By writing $\lambda^{{\bm{0}}}{}_{{\bm{1}}}$ as
$\lambda^{{\bm{0}}}{}_{{\bm{1}}}=|\lambda^{{\bm{0}}}{}_{{\bm{1}}}|e^{i\omega_{1}}$,
where $\omega_{1}$ is the phase of $\lambda^{{\bm{0}}}{}_{{\bm{1}}}$, and
choosing $\omega_{1}$ such that $e^{i\omega_{1}}=1$, one shows that

$\lambda^{{\bm{0}}}{}_{{\bm{1}}}=\sqrt{\rho}+\frac{1}{8}\left(\Pi_{3}[\Omega]-4\Pi_{1}[f_{x}]\right)\rho^{3/2}+O(\rho^{5/2}),$

From eq. (144c), one has

$|\lambda^{{\bm{1}}}{}_{{\bm{1}}}|^{2}=-\frac{1}{24}\Pi_{4}[\Omega]\rho^{3}+O(\rho^{4}).$

Assume
$\lambda^{{\bm{1}}}{}_{{\bm{1}}}=|\lambda^{{\bm{1}}}{}_{{\bm{1}}}|e^{i\omega_{2}}$,
and make use of eq. (144b) to fix $e^{i\omega_{2}}$. Then,
$\lambda^{{\bm{1}}}{}_{{\bm{1}}}$ can be written as

$\lambda^{{\bm{1}}}{}_{{\bm{1}}}=\frac{\Pi_{4}[\Omega]}{X_{-}(\Pi_{3}[\Omega])}\rho^{3/2}+O(\rho^{5/2}).$

Note that eq. (143) implies

$\lambda^{{\bm{0}}}{}_{{\bm{1}}}{\bm{e}}^{\circ}_{{\bm{0}}{\bm{1}}^{\prime}}-\lambda^{{\bm{0}}}{}_{{\bm{0}}}{\bm{e}}^{\circ}_{{\bm{1}}{\bm{1}}^{\prime}}=-\bar{\lambda}^{{\bm{0}}^{\prime}}{}_{{\bm{1}}^{\prime}}{\bm{e}}_{{\bm{1}}{\bm{0}}^{\prime}}-\bar{\lambda}^{{\bm{1}}^{\prime}}{}_{{\bm{1}}^{\prime}}{\bm{e}}_{{\bm{1}}^{\prime}{\bm{1}}^{\prime}}.$

By applying the above to the affine parameter $u^{\circ}$ and by making use of
the conditions ${\bm{e}}^{\circ}_{{\bm{0}}{\bm{1}}^{\prime}}(u^{\circ})=0$ and
${\bm{e}}^{\circ}_{{\bm{1}}{\bm{1}}^{\prime}}(u^{\circ})=1$ on
$\mathscr{I}^{+}$, one can show that

$\lambda^{{\bm{0}}}{}_{{\bm{0}}}=\bar{\lambda}^{{\bm{0}}^{\prime}}{}_{{\bm{1}}^{\prime}}{\bm{e}}_{{\bm{1}}{\bm{0}}^{\prime}}(u^{\circ})+\bar{\lambda}^{{\bm{1}}^{\prime}}{}_{{\bm{1}}^{\prime}}{\bm{e}}_{{\bm{1}}{\bm{1}}^{\prime}}(u^{\circ}).$
(146)

The above equation allow us to determine $\lambda^{{\bm{0}}}{}_{{\bm{0}}}$
given
$\bar{\lambda}^{{\bm{0}}^{\prime}}{}_{{\bm{1}}^{\prime}},\bar{\lambda}^{{\bm{1}}^{\prime}}{}_{{\bm{1}}^{\prime}},{\bm{e}}_{{\bm{1}}{\bm{0}}^{\prime}}(u^{\circ})$
and ${\bm{e}}_{{\bm{1}}{\bm{1}}^{\prime}}(u^{\circ})$. Then, the condition
$\text{det}(\lambda^{{\bm{A}}}{}_{{\bm{B}}})=1$ can be used to determine
$\lambda^{{\bm{1}}}{}_{{\bm{0}}}$.

To obtain an asymptotic expression for the affine parameter $u^{\circ}$, note
that eq. (143) implies that

$\displaystyle{\bm{e}}^{\circ}_{{\bm{1}}{\bm{1}}^{\prime}}=\left(\frac{\Pi_{1}[{\bm{e}}^{0}_{x}]}{\sqrt{2}}\rho^{2}+O(\rho^{3})\right)\partial_{\tau}+\left(\frac{1}{\sqrt{2}}\rho^{2}+O(\rho^{3})\right)\partial_{\rho}+$
$\displaystyle\phantom{{\bm{e}}^{\circ}_{{\bm{1}}{\bm{1}}^{\prime}}=}\left(\frac{1}{24\sqrt{2}}\left(24\Pi_{1}[{\bm{e}}^{2}_{x}]+X_{-}(\Pi_{3}[\Omega])\right)\rho^{2}+O(\rho^{3})\right)X_{+}+$
$\displaystyle\phantom{{\bm{e}}^{\circ}_{{\bm{1}}{\bm{1}}^{\prime}}=}\left(\frac{1}{24\sqrt{2}}\left(24\Pi_{1}[{\bm{e}}^{3}_{x}]+X_{+}(\Pi_{3}[\Omega])\right)\rho^{2}+O(\rho^{3})\right)X_{-}.$

Consider the ansatz

$u^{\circ}=\frac{A^{\circ}}{\rho}+\mathcal{W}^{\circ}+B^{\circ}\log{\rho},$

where $\partial_{\rho}(A^{\circ})=\partial_{\rho}(B^{\circ})=0$ and
$\mathcal{W}^{\circ}$ can be written as

$\mathcal{W}^{\circ}=u^{\circ}_{\star}+O(\rho).$

Then, one has

$\displaystyle\partial_{\tau}(u^{\circ})=\frac{\partial_{\tau}(A^{\circ})}{\rho}+\partial_{\tau}(\mathcal{W}^{\circ})+\partial_{\tau}(B^{\circ})\log{\rho},$
$\displaystyle\partial_{\rho}(u^{\circ})=-\frac{A^{\circ}}{\rho^{2}}+\frac{B^{\circ}}{\rho}+\partial_{\rho}(\mathcal{W}^{\circ}),$
$\displaystyle
X_{+}(u^{\circ})=\frac{X_{+}(A^{\circ})}{\rho}+X_{+}(\mathcal{W}^{\circ})+X_{+}(B^{\circ})\log{\rho},$
$\displaystyle
X_{-}(u^{\circ})=\frac{X_{-}(A^{\circ})}{\rho}+X_{-}(\mathcal{W}^{\circ})+X_{-}(B^{\circ})\log{\rho}.$

Moreover, the condition
${\bm{e}}^{\circ}_{{\bm{1}}{\bm{1}}^{\prime}}(u^{\circ})=1$ on
$\mathscr{I}^{+}$ yields

$A^{\circ}=-\frac{1}{\sqrt{2}},\qquad B^{\circ}=0.$

From the above, one has

$u^{\circ}=-\frac{1}{\sqrt{2}\rho}+\mathcal{W}^{\circ}=-\frac{1}{\sqrt{2}\rho}+u^{\circ}_{\star}+O(\rho).$
(148)

######  Remark 14.

In the following, we use $\Pi_{0}[\mathcal{W}^{\circ}]$ to refer to
$u^{\circ}_{\star}$.

Using eq. (148), we have

$\displaystyle{\bm{e}}_{{\bm{1}}{\bm{0}}^{\prime}}(u^{\circ})=\left(-\frac{1}{4}\Pi_{2}[{\bm{e}}^{1}_{y}]-\frac{1}{\sqrt{2}}X_{-}(\Pi_{0}[\mathcal{W}^{\circ}])\right)+O(\rho),$
$\displaystyle{\bm{e}}_{{\bm{1}}{\bm{1}}^{\prime}}(u^{\circ})=-\frac{1}{2\rho}+\left(\frac{2\partial_{\tau}(\Pi_{0}[\mathcal{W}^{\circ}])}{\sqrt{2}}-\frac{1}{4}\Pi_{2}[{\bm{e}}^{2}_{x}]\right)+O(\rho).$

Substituting in eq. (146) and using
$\text{det}(\lambda^{{\bm{A}}}{}_{{\bm{B}}})=1$ yields

$\lambda^{{\bm{0}}}{}_{{\bm{0}}}=O(\rho^{1/2}),\qquad\lambda^{{\bm{1}}}{}_{{\bm{0}}}=\frac{1}{\sqrt{\rho}}+\left(\frac{\Pi_{3}[\Omega]}{8}-\frac{\Pi_{1}[f_{x}]}{2}\right)+O(\rho^{3/2}).$

From eq. (75), the relation between the NP frame
$\\{{\bm{e}}^{\bullet}_{{\bm{A}}{\bm{A}}^{\prime}}\\}$ and the F-gauge frame
$\\{{\bm{e}}^{\bullet}_{{\bm{A}}{\bm{A}}^{\prime}}\\}$ is given by

${\bm{e}}^{\bullet}_{{\bm{A}}{\bm{A}}^{\prime}}=\Lambda^{{\bm{B}}}{}_{{\bm{A}}}\bar{\Lambda}^{{\bm{B}}^{\prime}}{}_{{\bm{A}}^{\prime}}{\bm{e}}_{{\bm{B}}{\bm{B}}^{\prime}}.$
(149)

Comparing eq. (149) with eq. (143), one can show that

$\Lambda^{{\bm{0}}}{}_{{\bm{1}}}=\theta^{-1/2}\lambda^{{\bm{0}}}{}_{{\bm{1}}}e^{ic},\qquad\Lambda^{{\bm{1}}}{}_{{\bm{1}}}=\theta^{-1/2}\lambda^{{\bm{1}}}{}_{{\bm{1}}}e^{ic},$
(150)

where $c$ is a function encoding the phase freedom. From eq. (69), one can
obtain an asymptotic expansion for $\Phi^{\circ}_{22}$. Then, eq. (70) with
initial data

$\lim_{\rho\to 0}\theta=1,$

can be used to obtain an asymptotic expansion for $\theta$. However, the
strategy in this analysis is to use eq. (70) to confirm the expansion of
$\theta$ obtained in [40] with the assumptions that $\Phi^{\circ}_{22}$ can be
written as

$\Phi^{\circ}_{22}=\Pi_{0}[\Phi^{\circ}_{22}]+\Pi_{1}[\Phi^{\circ}_{22}]\rho+\frac{1}{2}\Pi_{2}[\Phi^{\circ}_{22}]\rho^{2}+\frac{1}{6}\Pi_{3}[\Phi^{\circ}_{22}]\rho^{3}+O(\rho^{4}).$

From eq. (70), one can confirm that

$\Pi_{0}[\Phi^{\circ}_{22}]=\Pi_{1}[\Phi^{\circ}_{22}]=\Pi_{2}[\Phi^{\circ}_{22}]=0,\qquad\Pi_{1}[\theta]=\frac{1}{6}\Pi_{3}[\Phi^{\circ}_{22}].$

Then, the conformal factor $\theta$ can be written as

$\theta=1+\frac{1}{6}\Pi_{3}[\Phi^{\circ}_{22}]\rho+O(\rho^{2}).$

Using the above and substituting in eqs. (150), one can show that
$\Lambda^{{\bm{0}}}{}_{{\bm{1}}}$ and $\Lambda^{{\bm{1}}}{}_{{\bm{1}}}$ are
given by

$\displaystyle\Lambda^{{\bm{0}}}{}_{{\bm{1}}}=\left(\sqrt{\rho}-\frac{1}{24}\left(-3\Pi_{3}[\Omega]+2\Pi_{3}[\Phi^{\circ}_{22}]+12\Pi_{1}[f_{x}]\right)\rho^{3/2}+O(\rho^{5/2})\right)e^{ic},\qquad$
(151a)
$\displaystyle\Lambda^{{\bm{1}}}{}_{{\bm{1}}}=\left(\frac{\Pi_{4}[\Omega]}{X_{-}(\Pi_{3}[\Omega])}\rho^{3/2}+O(\rho^{5/2})\right)e^{ic}.$
(151b)

Then, eq. (149) show that the frame
${\bm{e}}^{\bullet}_{{\bm{1}}{\bm{1}}^{\prime}}$ can be written as

$\displaystyle{\bm{e}}^{\bullet}_{{\bm{1}}{\bm{1}}^{\prime}}=\left(\frac{\Pi_{1}[{\bm{e}}^{0}_{x}]}{\sqrt{2}}\rho^{2}+O(\rho^{3})\right)\partial_{\tau}+\left(\frac{1}{\sqrt{2}}\rho^{2}+O(\rho^{3})\right)\partial_{\rho}+$
$\displaystyle\phantom{{\bm{e}}^{\bullet}_{{\bm{1}}{\bm{1}}^{\prime}}=}\left(\left(\frac{\Pi_{1}[{\bm{e}}^{2}_{x}]}{\sqrt{2}}-\frac{\overline{\Pi_{4}[\Omega]}}{\sqrt{2}X_{+}(\overline{\Pi_{3}[\Omega]})}\right)\rho^{2}+O(\rho^{3})\right)X_{+}+$
$\displaystyle\phantom{{\bm{e}}^{\bullet}_{{\bm{1}}{\bm{1}}^{\prime}}=}\left(\left(\frac{\Pi_{1}[{\bm{e}}^{3}_{x}]}{\sqrt{2}}-\frac{\Pi_{4}[\Omega]}{\sqrt{2}X_{-}(\Pi_{3}[\Omega])}\right)\rho^{2}+O(\rho^{3})\right)X_{-}.$

Applying the above to the affine parameter $u^{\bullet}$ and using the
conditions ${\bm{e}}^{\bullet}_{{\bm{0}}{\bm{1}}^{\prime}}(u^{\bullet})=0$ and
${\bm{e}}^{\bullet}_{{\bm{1}}{\bm{1}}^{\prime}}(u^{\bullet})=1$ on
$\mathscr{I}^{+}$, one can show that

$\Lambda^{{\bm{0}}}{}_{{\bm{0}}}=\bar{\Lambda}^{{\bm{0}}^{\prime}}{}_{{\bm{1}}^{\prime}}{\bm{e}}_{{\bm{1}}{\bm{0}}^{\prime}}(u^{\bullet})+\bar{\Lambda}^{{\bm{1}}^{\prime}}{}_{{\bm{1}}^{\prime}}{\bm{e}}_{{\bm{1}}{\bm{1}}^{\prime}}(u^{\bullet}).$
(152)

Then, the condition $\text{det}(\Lambda^{{\bm{A}}}{}_{{\bm{B}}})=1$ can be
used to solve for $\Lambda^{{\bm{1}}}{}_{{\bm{0}}}$. To determine
$u^{\bullet}$, consider the ansatz

$u^{\bullet}=\frac{A^{\bullet}}{\rho}+\mathcal{W}^{\bullet}+B^{\bullet}\log{\rho},$

with $\partial_{\rho}(A^{\bullet})=\partial_{\rho}(B^{\bullet})=0$. Then, the
condition ${\bm{e}}^{\bullet}_{{\bm{1}}{\bm{1}}^{\prime}}(u^{\bullet})=1$ on
$\mathscr{I}^{+}$ can be used to show that

$A^{\bullet}=-\sqrt{2},\qquad B^{\bullet}=0.$

Given the above, the affine parameter $u^{\bullet}$ can be written as

$u^{\bullet}=-\frac{\sqrt{2}}{\rho}+\mathcal{W}^{\bullet}.$

Assume $\mathcal{W}^{\bullet}$ can be written as
$\mathcal{W}^{\bullet}=u^{\bullet}_{\star}+O(\rho)$, then

$u^{\bullet}=-\frac{\sqrt{2}}{\rho}+u^{\bullet}_{\star}+O(\rho).$

Similar to earlier discussion, one has
$\Pi_{0}[\mathcal{W}^{\bullet}]=u^{\bullet}_{\star}$. Using eq. (152) and the
condition $\text{det}(\Lambda^{{\bm{A}}}{}_{{\bm{B}}})=1$ yields

$\displaystyle\Lambda^{{\bm{0}}}{}_{{\bm{0}}}=\left(-\frac{1}{2}\left(\Pi_{2}[{\bm{e}}^{1}_{y}]+\sqrt{2}X_{-}(\Pi_{0}[\mathcal{W}^{\bullet}])\right)\sqrt{\rho}+O(\rho^{3/2})\right)e^{-ic},$
(153a)
$\displaystyle\Lambda^{{\bm{1}}}{}_{{\bm{0}}}=\left(-\frac{1}{\sqrt{\rho}}+O(\rho^{1/2})\right)e^{-ic}.$
(153b)

######  Remark 15.

The phase parameter $c$ can be determined using eq. (71) and the initial data
$c=0$ on $\mathcal{C}$. For our purpose, it will be sufficient to note that

$e^{ic}=1+O(\rho).$

Given the expansions of the components of the matrix
$\Lambda^{{\bm{A}}}{}_{{\bm{B}}}$, one can show that the components of the
inverse matrix $\Lambda_{{\bm{A}}}{}^{{\bm{B}}}$ can be written as

$\displaystyle\Lambda_{{\bm{0}}}{}^{{\bm{0}}}=\left(-\frac{\Pi_{4}[\Omega]}{X_{-}(\Pi_{3}[\Omega])}\rho^{3/2}+O(\rho^{5/2})\right)e^{ic},$
$\displaystyle\Lambda_{{\bm{0}}}{}^{{\bm{1}}}=\left(-\frac{1}{\sqrt{\rho}}-\left(\frac{\Pi_{3}[\Omega]}{8}+\frac{\Pi_{3}[\Phi^{\circ}_{22}]}{12}+\frac{\Pi_{1}[f_{x}]}{2}\right)\sqrt{2}+O(\rho^{3/2})\right)e^{-ic},$
$\displaystyle\Lambda_{{\bm{1}}}{}^{{\bm{0}}}=\left(\sqrt{\rho}-\frac{1}{24}\left(-3\Pi_{3}[\Omega]+2\Pi_{3}[\Phi^{\circ}_{22}]+12\Pi_{1}[f_{x}]\right)\rho^{3/2}+O(\rho^{5/2})\right)e^{ic},$
$\displaystyle\Lambda_{{\bm{1}}}{}^{{\bm{1}}}=\left(\frac{1}{2}\left(\Pi_{2}[{\bm{e}}^{1}_{y}]+\sqrt{2}X_{-}(\Pi_{0}[\mathcal{W}^{\bullet}])\right)+O(\rho^{3/2})\right)e^{-ic}.$

Given eqs. (7.2), (151)-(153) and eqs. (89), it is possible to obtain
asymptotic expansions for the NP-connection coefficients
$\sigma^{\bullet},\mu^{\bullet}$ and $\gamma^{\bullet}$. Then, eq. (86a) can
be used to confirm whether $\sigma^{\bullet ab}N^{\bullet}_{ab}$ will
contribute to $\mathcal{Q}$ at $\mathcal{I}^{+}$. Using eq. (90), one can also
determine which components of $\phi_{ABCD}$ will appear in the expression of
$\mathcal{Q}$ at $\mathcal{I}^{+}$. From eqs. (89), one can show that

$\displaystyle\mu^{\bullet}=O(\rho^{2}),$
$\displaystyle\gamma^{\bullet}=-\frac{\rho}{2\sqrt{2}}+O(\rho^{2}),$
$\displaystyle\sigma^{\bullet}=-\frac{1}{2\sqrt{2}}\left(X_{-}(\Pi_{2}[{\bm{e}}^{1}_{y}])+\sqrt{2}X_{-}(X_{-}(\Pi_{0}[\mathcal{W}^{\bullet}]))\right)+O(\rho).$

Then, the term $\Delta|\sigma^{\bullet}|^{2}$ satisfies

$\Delta|\sigma^{\bullet}|^{2}=O(\rho^{2}).$

This readily implies that

$\lim_{\rho\to 0}\sigma^{\bullet ab}N^{\bullet}_{ab}\simeq 0.$

From eq. (90), one can show that

$\bar{\phi}^{\bullet}_{2}=-\Pi_{0}[\bar{\phi}_{2}]+O(\rho).$

Hence, the charges $\mathcal{Q}$ at $\mathcal{I}^{+}$ can be written as

$\mathcal{Q}(f,\mathcal{C})|_{\mathcal{I}^{+}}=\oint_{\mathcal{C}}-2\bm{\varepsilon}_{2}f\Pi_{0}[\bar{\phi}_{2}],$
(156)

where $\Pi_{0}[\bar{\phi}_{2}]$ is evaluated at $\tau=1$. A similar analysis
for the transformation between the NP-gauge frame
$\\{{\bm{e}}^{\bullet}_{{\bm{A}}{\bm{A}}^{\prime}}\\}$ and the F-gauge frame
$\\{{\bm{e}}_{{\bm{A}}{\bm{A}}^{\prime}}\\}$ at $\mathscr{I}^{-}$ reveals that
the background term $\sigma^{\bullet ab}N^{\bullet}_{ab}$ given by eq. (86b)
will not contribute to $\mathcal{Q}$ at $\mathcal{I}^{-}$ and that
$\bar{\phi}^{\bullet}_{2}=-\Pi_{0}[\bar{\phi}_{2}]$ at $\rho=0$. Thus,

$\mathcal{Q}(f,\mathcal{C})|_{\mathcal{I}^{-}}=\oint_{\mathcal{C}}-2\bm{\varepsilon}_{2}f\Pi_{0}[\bar{\phi}_{2}],$
(157)

where $\Pi_{0}[\bar{\phi}_{2}]$ is evaluated at $\tau=-1$.

######  Remark 16.

It should be highlighted the limited amount of explicit information about the
asymptotic expansions, which is required for the evaluation of the BMS
asymptotic charges at $\mathcal{I}^{\pm}$. In particular, expressions in eqs.
(156) and (157) are formally identical to their spin-2 equivalents given in
[27].

###  7.3 BMS-supertranslation charges at the critical sets

The discussion in the previous section indicates that in order to evaluate
$\mathcal{Q}$ at $\mathcal{I}^{\pm}$, we require a solution for
$\Pi_{0}[\bar{\phi}_{2}]$ at $\tau=\pm 1$. Given the second-order ODE for
$a_{2}$ in eq. (138), one can show that

######  Proposition 3.

For $l\geq 2$ and $0\leq m\leq 2l$, the solution to eq. (138b) is given by:

$a_{2}(\tau)=\mathcal{A}_{l,m}P_{l}(\tau)+\mathcal{B}_{l,m}Q_{l}(\tau)$ (158)

where $P_{l}(\tau)$ is the Legendre polynomial of order $l$ and $Q_{l}(\tau)$
is the Legendre function of the second kind of order $l$, $\mathcal{A}_{l,m}$
and $\mathcal{B}_{l,m}$ are constants that can be expressed in terms of the
coefficients $\Pi_{3}[\Omega]_{2l,m}$ appearing in eq. (130). In particular,
one has

$\displaystyle\mathcal{A}_{l,m}=-\frac{(6+l+l^{2})Q_{l+1}(0)\Pi_{3}[\Omega]_{2l,m}}{6P_{l+1}(0)Q_{l}(0)-6P_{l}(0)Q_{l+1}(0)},$
(159a)
$\displaystyle\mathcal{B}_{l,m}=\frac{(6+l+l^{2})P_{l+1}(0)\Pi_{3}[\Omega]_{2l,m}}{6P_{l+1}(0)Q_{l}(0)-6P_{l}(0)Q_{l+1}(0)}.$
(159b)

Note that the recurrence relation of $Q_{l}(\tau)$ implies that $Q_{l}(\tau)$
diverges logarithmically near $\tau=\pm 1$, i.e.,

$Q_{l}(\tau)=\mathcal{C}_{l}\ln(1\pm\tau)+O(1),\mbox{for some constant
$\mathcal{C}_{l}$.}$

Thus, it is straightforward to see that the solution for $a_{2}$ given in eq.
(158) will diverge at the critical sets unless $\mathcal{B}_{l,m}=0$. From eq.
(159b), we have that $\mathcal{B}_{l,m}=0$ for even $l$. For odd $l$, one must
restrict the initial data set to ensure that $\mathcal{B}_{l,m}=0$.

######  Lemma 1.

The solution in Proposition 159 is regular at $\tau=\pm 1$ if and only if the
coefficients $\Pi_{3}[\Omega]_{2l,m}$ satisfy

$\Pi_{3}[\Omega]_{2l,m}=0,\qquad\text{for odd }l\geq 2,\quad 0\leq m\leq 2l.$

To simplify the integrand in eqs. (156)-(157), we rewrite the expansion of
$\Pi_{0}[\phi_{2}]$ as

$\Pi_{0}[\phi_{2}]=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}a_{2;2l,m+l}(\tau)T_{2l}{}^{m+l}{}_{l},$
(160)

and the expansion for $\Pi_{3}[\Omega]$ at $\rho=0$ as

$\Pi_{3}[\Omega]|_{\rho=0}=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\Pi_{3}[\Omega]_{2l,m+l}T_{2l}{}^{m+l}{}_{l}.$
(161)

Then, the regularity condition in Lemma 1 can be written as:

######  Lemma 2.

The solution in Proposition 159 is regular at $\tau=\pm 1$ if and only if the
coefficients $\Pi_{3}[\Omega]_{2l,m+l}$ satisfy

$\Pi_{3}[\Omega]_{2l,m+l}=0,\qquad\text{for odd }l\geq 2\quad-l\leq m\leq l.$

######  Remark 17.

Recalling eq. (108), one readily sees that the regularity condition in Lemma 2
is, in fact, a statement about the multipolar structure of the freely
specifiable function $\xi$ on $\mathbb{S}^{2}$. More precisely, the condition
excludes from $\xi$ the modes with odd parity. This condition is, to the best
of our knowledge, new.

Given initial data that satisfy the above regularity condition, the solution
for $a_{2;2l,m+l}$ for $l\geq 2$ and $-l\leq m\leq l$ can be written as

$a_{2;2l,m+l}(\tau)=\mathcal{A}_{l,m+l}P_{l}(\tau).$ (162)

Note that $\mathcal{A}_{l,m+l}=0$ for odd $l$. Therefore, for $l\geq 2$, the
solution for $a_{2;2l,m+l}$ is only non-vanishing for even $l$. Substituting
in eqs. (156)-(157) using eqs. (160), we have

$\mathcal{Q}(f,\mathcal{C})|_{\mathcal{I}^{\pm}}=-2\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\bar{a}_{2;2l,m+l}(\pm
1)\oint_{\mathcal{C}}\bm{\varepsilon}_{2}f\bar{T}_{2l}{}^{m+l}{}_{l}.$ (163)

But, the functions $\bar{T}_{2l}{}^{m+l}{}_{l}$ are related to the complex
conjugate of the spherical harmonics $\bar{Y}_{l,m}$. In particular, we have
the correspondence

$\bar{T}_{2l}{}^{m+l}{}_{l}\mapsto C_{l,m}\bar{Y}_{l,m},$

where $C_{l,m}$ is a constant that depends on $l$ and $m$ and whose specific
form is not required for our discussion —see [49]. Given that $f$ is a
function on $\mathbb{S}^{2}$, one can always write $f$ as

$f=\frac{1}{C_{l^{\prime},m^{\prime}}}Y_{l^{\prime}m^{\prime}}.$ (164)

Substituting in eq. (163), we get

$\displaystyle\mathcal{Q}(f,\mathcal{C})|_{\mathcal{I}^{\pm}}=-2\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\frac{C_{l,m}}{C_{l^{\prime},m^{\prime}}}\bar{a}_{2;2l,m+l}(\pm
1)\oint_{\mathcal{C}}\bm{\varepsilon}_{2}Y_{l^{\prime}m^{\prime}}\bar{Y}_{l,m}$
$\displaystyle\phantom{\mathcal{Q}(f,\mathcal{C})|_{\mathcal{I}^{\pm}}}=-2\bar{a}_{2;2l^{\prime},m^{\prime}+l^{\prime}}(\pm
1)$

Then, using eqs. (131), (135a) and (162), one has

$\mathcal{Q}|_{\mathcal{I}^{\pm}}=\begin{cases}-2\Pi_{3}[\Omega]_{0,0},\qquad&\text{for
}l=0,m=0,\\\ \mp 4\Pi_{3}[\Omega]_{2,m+1},\qquad&\text{for }l=1,-1\leq m\leq
1,\\\ 0,\qquad&\text{for odd }l\geq 2,-l\leq m\leq l,\\\
-2\mathcal{A}_{l,m+l},\qquad&\text{for even }l\geq 2,-l\leq m\leq
l,\end{cases}$

where $\mathcal{A}_{l,m+l}$ is given by

$\mathcal{A}_{l,m+l}=-\frac{(6+l+l^{2})Q_{l+1}(0)\Pi_{3}[\Omega]_{2l,m+l}}{6P_{l+1}(0)Q_{l}(0)-6P_{l}(0)Q_{l+1}(0)}.$

Using eqs. (108) and (161), and expanding the freely specifiable data $\xi$ as

$\xi=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\xi_{2l,m+l}T_{2l}{}^{m+l}{}_{l},$

one can see that

$\displaystyle\Pi_{3}[\Omega]_{0,0}=\frac{21}{2}\xi_{0,0}-3A,$
$\displaystyle\Pi_{3}[\Omega]_{2l,m+l}=\frac{21}{2}\xi_{2l,m+l}.$

Then, the final result of our analysis can be summarised in the following:

######  Theorem 1.

Given the generic initial data in Proposition 95, the asymptotic BMS-
supertranslation charges $\mathcal{Q}$ are not well-defined at
$\mathcal{I}^{\pm}$ unless the freely specifiable data satisfy the regularity
condition:

$\xi_{2l,m+l}=0\qquad\text{for odd }l\geq 2\quad-l\leq m\leq l.$ (166)

Given initial data that satisfy the extra regularity condition, the charges
$\mathcal{Q}$ at $\mathcal{I}^{\pm}$ can be expressed in terms of the freely
specifiable data $A$ and $\xi$:

$\mathcal{Q}|_{\mathcal{I}^{\pm}}=\begin{cases}-21\xi_{0,0}+6A,\qquad&\text{for
}l=0,m=0,\\\ \mp 42\xi_{2,m+1},\qquad&\text{for }l=1,-1\leq m\leq 1,\\\
0,\qquad&\text{for odd }l\geq 2,-l\leq m\leq l,\\\
-2\mathcal{A}^{\prime}_{l,m+l},\qquad&\text{for even }l\geq 2,-l\leq m\leq
l,\end{cases}$

where $\mathcal{A}^{\prime}_{l,m+l}$ is given by

$\mathcal{A}^{\prime}_{l,m+l}=-\frac{21(6+l+l^{2})Q_{l+1}(0)\xi_{2l,m+l}}{12P_{l+1}(0)Q_{l}(0)-12P_{l}(0)Q_{l+1}(0)}.$

Moreover, there is a natural correspondence between the charges at
$\mathcal{I}^{+}$ and $\mathcal{I}^{-}$ expressed as follows:

$\mathcal{Q}|_{\mathcal{I}^{+}}=(-1)^{l}\mathcal{Q}|_{\mathcal{I}^{-}}.$

######  Remark 18.

It should be stressed that the above results express the BMS asymptotic
charges at $\mathcal{I}^{\pm}$ in terms of freely specifiable data coming from
the initial metric. Remarkably, there is no contribution to the charges coming
from the extrinsic curvature.

######  Remark 19.

As a consequence of the regularity condition in Lemma 1, one finds that for
$l\geq 2$, only the even parity charges have a non-trivial content. Observe
that the regularity condition given in Lemma 1 are necessary conditions for
the BMS charges to be well-defined at the critical sets $\mathcal{I}^{\pm}$.
The regularity condition eliminates the odd parity modes for $l\geq 2$ in the
function $\xi$ over $\mathbb{S}^{2}$. Thus, compared with the equivalent
conditions arising from analysing the spin-2 field on the Minkowski spacetime
in [27], one finds that the full non-linear GR situation is much more
restrictive. This should not be surprising as the Einstein constraints are
known to be more rigid than their linearised counterpart.

##  8 Conclusions

In this article, we presented an analysis of BMS asymptotic charges using
Friedrich’s formulation of spatial infinity. Our analysis indicates that the
initial data given in [35] does not give rise to well-defined asymptotic
charges at $\mathcal{I}^{\pm}$ unless the initial data satisfies an extra
regularity condition. The regularity condition eliminates the odd $l\geq 2$
modes on the freely specifiable leading term of the initial metric. If the
initial data are chosen to satisfy the regularity condition in eq. (166), BMS
asymptotic charges are well-defined at $\mathcal{I}^{\pm}$. Then, if the
function $f\in C^{\infty}(\mathbb{S}^{2})$ is written as eq. (164), one can
show that BMS asymptotic charges at $\mathcal{I}^{\pm}$ can be fully expressed
in terms of the freely specifiable data and that there exists a natural
correspondence between the charges at $\mathcal{I}^{+}$ and $\mathcal{I}^{-}$.
As a consequence of the regularity condition on the initial data, only even
parity BMS asymptotic charges for $l\geq 2$ have a non-trivial content at the
critical sets $\mathcal{I}^{\pm}$. Observe that the vanishing of the odd
parity BMS asymptotic charges for $l\geq 2$ does not necessarily mean that the
corresponding BMS asymptotic charges vanish everywhere on $\mathscr{I}^{\pm}$
and generically they will not be conserved.

Our result expands the discussion in [50] that showed that the component of
the Weyl tensor appearing in the Bondi mass at $\mathscr{I}^{+}$ match
antipodally with that appearing at $\mathscr{I}^{-}$. In [50], the antipodal
matching of these components also follows from the regularity conditions on
$\mathscr{I}^{\pm}$ and at $i^{0}$. This provides further evidence that the
antipodal matching is a regularity statement, i.e., it follows directly from
the regularity condition on the initial data. It would be very interesting to
prove a converse of the latter —namely, that the antipodal matching implies
regularity conditions on initial data similar to the ones obtained in the
present analysis. This question goes beyond the scope of this article and will
be considered elsewhere.

##  9 Acknowledgements

M.M.A.Mohamed would like to thank Mahdi Godazgar, Grigalius Taujanskas and
Marica Minucci for the useful discussions while working on this project. This
work was supported in part by the NSF grant PHY-2107939 to the University of
California, Santa Barbara. Calculations in this project used the computer
algebra system Mathematica with the package xAct —see [47].

##  Appendix A The boundary adapted evolution and constraint equations

In the main text of this article, we present the analysis of the zero-order
boundary adapted evolution and constraint eqs. (59b)-(59b). The zero-order
system in given in eqs. (127)-(128) while the full system is presented in this
section.

The boundary-adapted evolution system eq. (59b) can be written as

$\displaystyle\frac{1}{8}\left(4\sqrt{2}\xi_{2}\phi_{0}-24\sqrt{2}\xi_{h}\phi_{0}-16\xi_{x}\phi_{0}-3\sqrt{2}\xi_{1}\phi_{1}+\right.$
$\displaystyle
4\xi_{z}\phi_{1}+4\sqrt{2}\xi_{0}\phi_{2}+16\sqrt{2}\phi_{0}f_{x}-4\sqrt{2}\phi_{1}f_{z}-4\sqrt{2}\phi_{2}\chi_{0}+$
$\displaystyle
3\sqrt{2}\phi_{1}\chi_{1}-4\sqrt{2}\phi_{0}\chi_{2}+24\sqrt{2}\phi_{0}\chi_{h}+16\phi_{0}\chi_{x}-4\phi_{1}\chi_{z}+$
$\displaystyle
2\sqrt{2}\left(-4\partial_{\tau}\left[\phi_{0}\right]\left(-1+{\bm{e}}_{x}^{0}\right)-4\partial_{\rho}\left[\phi_{0}\right]{\bm{e}}_{x}^{1}+\partial_{\tau}\left[\phi_{1}\right]{\bm{e}}_{z}^{0}+\partial_{\rho}\left[\phi_{1}\right]{\bm{e}}_{z}^{1}-\right.$
$\displaystyle\left.\left.4{\bm{e}}_{x}^{3}X_{-}\left[\phi_{0}\right]+{\bm{e}}_{z}^{3}X_{-}\left[\phi_{1}\right]-4{\bm{e}}_{x}^{2}X_{+}\left[\phi_{0}\right]+{\bm{e}}_{z}^{2}X_{+}\left[\phi_{1}\right]\right)\right)=0$
(167a)
$\displaystyle\frac{1}{6\sqrt{2}}\left(3\partial_{\tau}\left[\phi_{1}\right]+6\partial_{\tau}\left[\phi_{0}\right]{\bm{e}}_{y}^{0}+6\partial_{\rho}\left[\phi_{0}\right]{\bm{e}}_{y}^{1}+\partial_{\tau}\left[\phi_{2}\right]\bm{{\bm{e}}}_{z}^{0}+\right.$
$\displaystyle\left.\partial_{\rho}\left[\phi_{2}\right]{\bm{e}}_{z}^{1}+6{\bm{e}}_{y}^{3}X_{-}\left[\phi_{0}\right]+{\bm{e}}_{z}^{3}X_{-}\left[\phi_{2}\right]+6{\bm{e}}_{y}^{2}X_{+}\left[\phi_{0}\right]+{\bm{e}}_{z}^{2}X_{+}\left[\phi_{2}\right]\right)=$
$\displaystyle\frac{1}{8}\left(-4\sqrt{2}\xi_{3}\phi_{0}-16\xi_{y}\phi_{0}+\sqrt{2}\xi_{2}\phi_{1}+6\sqrt{2}\xi_{h}\phi_{1}-4\xi_{x}\phi_{1}-\right.$
$\displaystyle
2\sqrt{2}\xi_{0}\phi_{3}-2\sqrt{2}\phi_{1}f_{x}+4\sqrt{2}\phi_{0}f_{y}+2\sqrt{2}\phi_{2}f_{z}+2\sqrt{2}\phi_{3}\chi_{0}-$
$\displaystyle\left.\sqrt{2}\phi_{1}\chi_{2}+4\sqrt{2}\phi_{0}\chi_{3}-6\sqrt{2}\phi_{1}\chi_{\mathrm{h}}+4\phi_{1}\chi_{x}+16\phi_{0}\chi_{y}\right)$
(167b)
$\displaystyle\frac{1}{24}\left(\frac{1}{2}\left(24\sqrt{2}\phi_{4}\left(\xi_{0}-\chi_{0}\right)+24\sqrt{2}\phi_{0}\left(\xi_{4}-\chi_{4}\right)-\right.\right.$
$\displaystyle
4\sqrt{2}\phi_{2}\left(\xi_{2}+6\xi_{h}-\chi_{2}-6\chi_{h}\right)+3\phi_{1}\left(\sqrt{2}\xi_{3}+4\xi_{y}-4\sqrt{2}f_{y}-\sqrt{2}\chi_{3}-4\chi_{y}\right)+$
$\displaystyle\left.3\phi_{3}\left(\sqrt{2}\xi_{1}+4\xi_{z}-4\sqrt{2}f_{z}-\sqrt{2}\chi_{1}-4\chi_{z}\right)\right)+$
$\displaystyle\sqrt{2}\left(4\partial_{\tau}\left[\phi_{2}\right]+3\left(\partial_{\tau}\left[\phi_{1}\right]{\bm{e}}_{y}^{0}+\partial_{\rho}\left[\phi_{1}\right]{\bm{e}}_{y}^{1}+\partial_{\tau}\left[\phi_{3}\right]{\bm{e}}_{z}^{0}+\partial_{\rho}\left[\phi_{3}\right]{\bm{e}}_{z}^{1}+\right.\right.$
$\displaystyle\left.\left.\left.{\bm{e}}_{y}^{3}X_{-}\left[\phi_{1}\right]+{\bm{e}}_{z}^{3}X_{-}\left[\phi_{3}\right]+{\bm{e}}_{y^{2}}X_{+}\left[\phi_{1}\right]+{\bm{e}}_{z}^{2}X_{+}\left[\phi_{3}\right]\right)\right)\right)=0$
(167c)
$\displaystyle\frac{1}{6\sqrt{2}}\left(3\partial_{\tau}\left[\phi_{3}\right]+\partial_{\tau}\left[\phi_{2}\right]{\bm{e}}_{y}^{0}+\partial_{\rho}\left[\phi_{2}\right]\quad{\bm{e}}_{y}^{1}+6\partial_{\tau}\left[\phi_{4}\right]\bm{{\bm{e}}}_{z}^{0}+\right.$
$\displaystyle\left.6\partial_{\rho}\left[\phi_{4}\right]\bm{{\bm{e}}}_{z}^{1}+{\bm{e}}_{y}^{3}X_{-}\left[\phi_{2}\right]+6{\bm{e}}_{z}^{3}X_{-}\left[\phi_{4}\right]+{\bm{e}}_{y}^{2}X_{+}\left[\phi_{2}\right]+6{\bm{e}}_{z}^{2}X_{+}\left[\phi_{4}\right]\right)=$
$\displaystyle\frac{1}{8}\left(-2\sqrt{2}\xi_{4}\phi_{1}+\sqrt{2}\xi_{2}\phi_{3}+6\sqrt{2}\xi_{h}\phi_{3}+4\xi_{x}\phi_{3}-4\sqrt{2}\xi_{1}\phi_{4}-\right.$
$\displaystyle
16\xi_{z}\phi_{4}+2\sqrt{2}\phi_{3}f_{x}+2\sqrt{2}\phi_{2}f_{y}+4\sqrt{2}\phi_{4}f_{z}+4\sqrt{2}\phi_{4}\chi_{1}-$
$\displaystyle\left.\sqrt{2}\phi_{3}\chi_{2}+2\sqrt{2}\phi_{1}\chi_{4}-6\sqrt{2}\phi_{3}\chi_{\mathrm{h}}-4\phi_{3}\chi_{x}+16\phi_{4}\chi_{z}\right)$
(167d)
$\displaystyle\frac{1}{2\sqrt{2}}\left(4\partial_{\tau}\left[\phi_{4}\right]\left(1+{\bm{e}}_{x}^{0}\right)+4\partial_{\rho}\left[\phi_{4}\right]{\bm{e}}_{x}^{1}+\partial_{\tau}\left[\phi_{3}\right]{\bm{e}}_{y}^{0}+\right.$
$\displaystyle\left.\partial_{\rho}\left[\phi_{3}\right]{\bm{e}}_{y}^{1}+{\bm{e}}_{y}^{3}X_{-}\left[\phi_{3}\right]+4{\bm{e}}_{x}^{3}X_{-}\left[\phi_{4}\right]+{\bm{e}}_{y}^{2}X_{+}\left[\phi_{3}\right]+4{\bm{e}}_{x}^{2}X_{+}\left[\phi_{4}\right]\right)=$
$\displaystyle\frac{1}{8}\left(-4\sqrt{2}\xi_{4}\phi_{2}+3\sqrt{2}\xi_{3}\phi_{3}-4\xi_{y}\phi_{3}-4\sqrt{2}\xi_{2}\phi_{4}+24\sqrt{2}\xi_{h}\phi_{4}-\right.$
$\displaystyle
16\xi_{x}\phi_{4}+16\sqrt{2}\phi_{4}f_{x}+4\sqrt{2}\phi_{3}f_{y}+4\sqrt{2}\phi_{4}\chi_{2}-$
$\displaystyle\left.3\sqrt{2}\phi_{3}\chi_{3}+4\sqrt{2}\phi_{2}\chi_{4}-24\sqrt{2}\phi_{4}\chi_{h}+16\phi_{4}\chi_{x}+4\phi_{3}\chi_{y}\right)$
(167e)

The constraint equations eq. (60) can be written as

$\displaystyle\frac{1}{6\sqrt{2}}\left(-3\partial_{\tau}\left[\phi_{1}\right]{\bm{e}}_{x}^{0}-3\partial_{\rho}\left[\phi_{1}\right]{\bm{e}}_{x}^{1}-6\partial_{\tau}\left[\phi_{0}\right]{\bm{e}}_{y}^{0}-\right.$
$\displaystyle
6\partial_{\rho}\left[\phi_{0}\right]{\bm{e}}_{y}^{1}+\partial_{\tau}\left[\phi_{2}\right]{\bm{e}}_{z}^{0}+\partial_{\rho}\left[\phi_{2}\right]{\bm{e}}_{z}^{1}-6{\bm{e}}_{y}^{3}X_{-}\left[\phi_{0}\right]-3{\bm{e}}_{x}^{3}X_{-}\left[\phi_{1}\right]+$
$\displaystyle\left.{\bm{e}}_{z}^{3}X_{-}\left[\phi_{2}\right]-6{\bm{e}}_{y}^{2}X_{+}\left[\phi_{0}\right]-3{\bm{e}}_{x}^{2}X_{+}\left[\phi_{1}\right]+{\bm{e}}_{z}^{2}X_{+}\left[\phi_{2}\right]\right)=$
$\displaystyle\frac{1}{8}\left(2\sqrt{2}\xi_{3}\phi_{0}+24\xi_{y}\phi_{0}-\sqrt{2}\xi_{2}\phi_{1}+12\xi_{x}\phi_{1}+\sqrt{2}\xi_{1}\phi_{2}-4\xi_{z}\phi_{2}-2\sqrt{2}\xi_{0}\phi_{3}+\right.$
$\displaystyle\left.2\sqrt{2}\phi_{3}\chi_{0}-\sqrt{2}\phi_{2}\chi_{1}+\sqrt{2}\phi_{1}\chi_{2}-2\sqrt{2}\phi_{0}\chi_{3}-12\phi_{1}\chi_{x}-24\phi_{0}\chi_{y}+4\phi_{2}\chi_{z}\right)$
(168a)
$\displaystyle-\frac{1}{12\sqrt{2}}\left(4\partial_{\tau}\left[\phi_{2}\right]{\bm{e}}_{x}^{0}+4\partial_{\rho}\left[\phi_{2}\right]{\bm{e}}_{x}^{1}+3\partial_{\tau}\left[\phi_{1}\right]{\bm{e}}_{y}^{0}+3\partial_{\rho}\left[\phi_{1}\right]{\bm{e}}_{y}^{1}-\right.$
$\displaystyle
3\partial_{\tau}\left[\phi_{3}\right]{\bm{e}}_{z}^{0}-3\partial_{\rho}\left[\phi_{3}\right]{\bm{e}}_{z}^{1}+3{\bm{e}}_{y}^{3}X_{-}\left[\phi_{1}\right]+4{\bm{e}}_{x}^{3}X_{-}\left[\phi_{2}\right]-$
$\displaystyle\left.3{\bm{e}}_{z}^{3}X_{-}\left[\phi_{3}\right]+3{\bm{e}}_{y}^{2}X_{+}\left[\phi_{1}\right]+4{\bm{e}}_{x}^{2}X_{+}\left[\phi_{2}\right]-3{\bm{e}}_{z}^{2}X_{+}\left[\phi_{3}\right]\right)=$
$\displaystyle\frac{1}{16}\left(8\sqrt{2}\xi_{4}\phi_{0}-\sqrt{2}\xi_{3}\phi_{1}+12\xi_{y}\phi_{1}+16\xi_{x}\phi_{2}+\sqrt{2}\xi_{1}\phi_{3}-12\xi_{z}\phi_{3}-8\sqrt{2}\xi_{0}\phi_{4}+\right.$
$\displaystyle\left.8\sqrt{2}\phi_{4}\chi_{0}-\sqrt{2}\phi_{3}\chi_{1}+\sqrt{2}\phi_{1}\chi_{3}-8\sqrt{2}\phi_{0}\chi_{4}-16\phi_{2}\chi_{x}-12\phi_{1}\chi_{y}+12\phi_{3}\chi_{z}\right)$
(168b)
$\displaystyle-\frac{1}{6\sqrt{2}}\left(3\partial_{\tau}\left[\phi_{3}\right]{\bm{e}}_{x}^{0}+3\partial_{\rho}\left[\phi_{3}\right]{\bm{e}}_{x}^{1}+\partial_{\tau}\left[\phi_{2}\right]{\bm{e}}_{y}^{0}+\partial_{\rho}\left[\phi_{2}\right]{\bm{e}}_{y}^{1}\right.$
$\displaystyle
6\partial_{\tau}\left[\phi_{4}\right]{\bm{e}}_{z}^{0}-6\partial_{\rho}\left[\phi_{4}\right]{\bm{e}}_{z}^{1}+{\bm{e}}_{y}^{3}X_{-}\left[\phi_{2}\right]+3{\bm{e}}_{x}^{3}X_{-}\left[\phi_{3}\right]-$
$\displaystyle\left.6{\bm{e}}_{z}^{3}X_{-}\left[\phi_{4}\right]+{\bm{e}}_{y}^{2}X_{+}\left[\phi_{2}\right]+3{\bm{e}}_{x}^{2}X_{+}\left[\phi_{3}\right]-6{\bm{e}}_{z}^{2}X_{+}\left[\phi_{4}\right]\right)=$
$\displaystyle\frac{1}{8}\left(2\sqrt{2}\xi_{4}\phi_{1}-\sqrt{2}\xi_{3}\phi_{2}+4\xi_{y}\phi_{2}+\sqrt{2}\xi_{2}\phi_{3}+12\xi_{x}\phi_{3}-2\sqrt{2}\xi_{1}\phi_{4}-24\xi_{z}\phi_{4}+\right.$
$\displaystyle\left.2\sqrt{2}\phi_{4}\chi_{1}-\sqrt{2}\phi_{3}\chi_{2}+\sqrt{2}\phi_{2}\chi_{3}-2\sqrt{2}\phi_{1}\chi_{4}-12\phi_{3}\chi_{x}-4\phi_{2}\chi_{y}+24\phi_{4}\chi_{z}\right)$
(168c)

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