horizon: $\partial \Sigma_{\cal H}=\{(\rho,z)|\ \rho=0,-k^2<z<k^2\}$ ,

3.  the $\psi$-invariant plane: $\partial \Sigma_\psi=\{(\rho,z)|\rho=0,-\infty<z<-k^2\}$ with the rod vector $v=(0,0,1)$ ,

4.  the infinity: $\partial \Sigma_\infty 
    = \{(\rho,z)|\sqrt{\rho^2+z^2}\to 
    \infty\ {\rm with}\ z/\sqrt{\rho^2+z^2}\ {\rm finite}  \}$ ,

where here and hereafter ${\cal H}$ denotes a spatial cross-section of the event horizon. Accordingly, the boundary integral in the left-hand side of the Mazur identity, eq. (\[eq:id\]), is decomposed into the integrals over the three rods (i)–(iii), and the integral at infinity (iv), as $$\begin{aligned}
\int_{\partial \Sigma}\rho\partial_p{\rm tr}\Psi dS^p 
&=& \int_{-\infty}^{-k^2}\rho\frac{\partial {\rm tr}\Psi }{\partial z}dz 
   +\int_{-k^2}^{k^2}\rho\frac{\partial {\rm tr}\Psi }{\partial z}dz 
\nonumber\\
 && +\int_{k^2}^{\infty}\rho\frac{\partial {\rm tr}\Psi }{\partial z}dz 
    +\int_{\partial\Sigma_\infty}\rho\partial_a{\rm tr}\Psi dS^a \,.   
\label{eq:integral} \end{aligned}$$

In order to evaluate this boundary integral, let us first consider the integrals of the twist one-forms $d\omega_a$ along the $z$-axis. By definition, the partial derivatives with respect to $z$ of the twist potentials $\omega_a$ vanish on both rotation axes. This means that the twist potentials $\omega_a$ are constant over the $\phi$-invariant plane and the $\psi$-invariant plane. Therefore, the integral can be written as $$\begin{aligned}
\int_{-\infty}^{\infty}\omega_{a,z}dz
 &=& \int_{-k^2}^{k^2}\omega_{a,z}dz \\ 
 &=& \biggl[\omega_a\biggr]_{z=-k^2}^{z=k^2} \,. 
\label{eq:int} \end{aligned}$$ On the other hand, by Stokes’s theorem, the integral of $d\omega_a$ on the horizon is evaluated as $$\begin{aligned}
\int_{\partial\Sigma_{\cal 