Lemma: The adapted chart

1. ${\Phi }_{\pm }$ in (eq:chart) are linear automorphisms of ${\mathbb{R}}^{16}$.

2. On $\{d\in {\mathbb{R}}^4:d^0>0\}$ the map $\kappa (d)=(d^2,\mathbf{d})=(q,\zeta )$ is a diffeomorphism onto ${\Omega }_0=\{(q,\zeta )\in \mathbb{R}\times {\mathbb{R}}^3:q+|\zeta {|}^2>0\}$, with inverse $d^0=\sqrt{q+|\zeta {|}^2}$, $\mathbf{d}=\zeta$. The set $\partial J^{+}\setminus \{0\}$ is contained in $\{d^0>0\}$ and is $\{q=0\}$ there.

3. Hence, with ${\zeta }_{\pm }={\mathbf{d}}_{\pm }$, $w_{\pm }=(u_{\pm },{\zeta }_{\pm },q_{\pm },r_{\pm },y)$ is a global chart of $V_{\pm }:=\{{d_{\pm }}^0>0\}\subset {\mathbb{R}}^{16}$ onto ${\mathbb{R}}^4\times {\Omega }_0\times {\mathbb{R}}^4\times {\mathbb{R}}^4$, of dimension $4+3+1+4+4=16$, consistent with Ichart and with $\dim W=16$ Grelease.

4. On $V_{\pm }$,

   $$R(d_{\pm })=\frac{1}{2\pi }\delta (q_{\pm }),({\partial }_{\alpha }R)(d_{\pm })=\frac{1}{\pi }{d}_{\pm \alpha }{\delta }^{\prime }(q_{\pm }),({\partial }_{\mu }{\partial }_{\alpha }R)(d_{\pm })=\frac{1}{\pi }[{\eta }_{\mu \alpha }{\delta }^{\prime }(q_{\pm })+2d_{\pm \mu }{d}_{\pm \alpha }{\delta }^{\prime \prime }(q_{\pm })],$$

   where $d_{\pm \alpha }={\eta }_{\alpha \nu }{d_{\pm }}^{\nu }$ is smooth on the chart, and $\mathcal{B}(d_{\pm })=\frac{1}{8\pi }\Theta (q_{\pm })$.

Proof. (i) is immediate. (ii) $q=(d^0{)}^2-|\mathbf{d}{|}^2$ and $d^0>0$ give $d^0=\sqrt{q+|\zeta {|}^2}$, smooth on ${\Omega }_0$; bijectivity is clear. On $\partial J^{+}\setminus \{0\}$, $d^0=|\mathbf{d}|>0$ and $q=0$. (iii) follows from (i), (ii). (iv) On $V_{\pm }$, $\Theta (d^0)=1$, and $\delta (d^2)$ is the pullback of $\delta (q)$ under the diffeomorphism $\kappa$; the derivatives follow from the chain rule ${\partial }_{d^{\alpha }}=2d_{\alpha }{\partial }_q+{{\delta }_{\alpha }}^i{\partial }_{{\zeta }^i}$, since $\delta (q)$ is independent of $\zeta$. The statement for $\mathcal{B}$ is Lemma lem:RR.