\Psi\simeq O(\rho) \,. \end{aligned}$$

\(ii) Horizon: $\partial \Sigma_{\cal H}=\{(\rho,z)|\ \rho=0, \ -k^2<z<k^2 \}$. The regularity on the horizon requires that for $\rho \to 0$, $$\begin{aligned}
&&\lambda_{ab}\simeq {\cal O}(1),\quad \omega_{a}\simeq {\cal O}(1) \,,\\
&&\psi_{a}\simeq {\cal O}(1), \quad \mu\simeq {\cal O}(1) \,. \end{aligned}$$ Therefore, for $\rho \to 0$, $\rho\ {\rm tr}\Psi$ behaves as $$\begin{aligned}
\rho\ \partial_z{\rm tr}\ \Psi\simeq O(\rho) \,.  \end{aligned}$$

\(iii) $\psi$-invariant plane: $\partial\Sigma_\psi=\{(\rho,z)|\rho=0,\ -\infty<z<-k^2\}$. Similarly to the case (i), the regularity on the $\phi$-invariant plane requires $$\begin{aligned}
&&\lambda_{\psi\psi}\simeq {\cal O}(\rho^2) \, ,\label{eq:b1}\\
&&\lambda_{\phi\phi}\simeq {\cal O}(1) \,, \label{eq:b2}\\
&&\lambda_{\phi\psi}\simeq {\cal O}(\rho^2) \,, \label{eq:b3}\\
&&\omega_{\phi}\simeq \frac{2J_\phi}{\pi}+{\cal O}(\rho^2) \,, \label{eq:b4}\\
&&\omega_{\psi}\simeq \frac{2J_\psi}{\pi}+{\cal O}(\rho^2) \,, \label{eq:b5}\end{aligned}$$ and $$\begin{aligned}
&&\psi_{\phi}\simeq {\cal O}(1) \,, \label{eq:b6} \\
&&\psi_{\psi}\simeq {\cal O}(\rho^2) \,,\label{eq:b7} \\
&&\mu\simeq -\frac{2Q}{\sqrt{3}\pi}+{\cal O}(\rho^2) \,. \label{eq:b8}\end{aligned}$$ Therefore, for $\rho \to 0$, $\rho\ {\rm tr}\Psi$ behaves as $$\begin{aligned}
\rho\ \partial_z{\rm tr}\ \Psi \simeq O(\rho) \,. \end{aligned}$$

\(iv) Infinity: $\partial\Sigma_\infty=\{(\rho,z)|\ \sqrt{\rho^2+z^2}\to\infty$ with $z/\sqrt{\rho^2+z^2}$ finite $\}$. Recall that the three-dimensional metric $g=(g_{IJ})\ (I,J=t,\phi,\psi)$ is subject to the constraint $$\begin{aligned}
{\rm det}(g)=-\rho^2 \,. \end{aligned}$$ Therefore, using the constraint and the formula, $$\begin{aligned}
{\rm det}(g+\delta g) 
 &=&{\rm det}[g(1+g^{-1}\delta g)] 
\nonumber \\ 
 &=&-\rho^2\left(1+{\rm tr}(g^{-1}\delta g)+{\rm det}(g^{-1}\delta g)\right) 
\nonumber \\
 &\simeq&-\rho^2\left(1+{\rm tr}(g^{-1}\delta g)\right) \,, \end{aligned}$$ we can see in the next order that 