H}}d\omega_a 
   &=& \int_{\partial\Sigma_\infty}d\omega_a\nonumber \\ 
   &=& \int_{\partial\Sigma_\infty}V_a 
    +\int_{\partial\Sigma_\infty}\psi_a(3d\mu+\epsilon^{bc}\psi_bd\psi_c) \,. 
\label{eq:int3}\end{aligned}$$ We find that the first integral in the right-hand side of eq. (\[eq:int3\]) is proportional to the angular momenta $J_a$, defined by eq. (\[def:Ja\]) above. As will be seen later, the second integral vanishes at infinity. Hence, using the degrees of freedom in adding a constant to $\omega_a$, we can always set the value of $\omega_a$ on the two rotation axes to be $$\begin{aligned}
\omega_a(z)=-\frac{2J_a}{\pi} \,, \label{eq:twist}\end{aligned}$$ for $z\in [k^2,\infty]$, and $$\begin{aligned}
\omega_a(z)=\frac{2J_a}{\pi} \,, \label{eq:twist2} \end{aligned}$$ for $z\in [-\infty,-k^2]$.

Next, consider the integral of $\mu_{,z}$ on the horizon $\partial \Sigma_{\cal H}$. The derivative of the potential, $d\mu$, vanishes on the two rotation axes by definition. Hence the integral along the $z$-axis becomes $$\begin{aligned}
\int^{\infty}_{-\infty}\mu_{,z}dz=\int^{k^2}_{-k^2}\mu_{,z}dz=\biggl[\ \mu\ \biggr]_{z=-k^2}^{z=k^2}.\end{aligned}$$ We find that this integral is proportional to the electric charge $Q$ defined by $$\begin{aligned}
Q&=&\frac{1}{16\pi}\int_{\cal H}\left(*F+\frac{1}{\sqrt{3}}A\wedge F\right) \,. \end{aligned}$$ In fact, straightforward calculation shows $$\begin{aligned}
&&\frac{1}{16\pi}\int_{\cal H}\left(*F+\frac{1}{\sqrt{3}}A\wedge F\right)\nonumber\\ 
 &=&\frac{\pi}{4}\int^{k^2}_{-k^2}
    \left[
          \frac{\tau}{\rho}(A_{t,\rho}-a^\phi{}_tA_{\phi,\rho}
          -a^\psi{}_tA_{\psi,\rho}) 
          -\frac{1}{\sqrt{3}}(A_\phi A_{\psi,z}-A_\psi A_{\phi,z}) 
    \right]dz 
\nonumber \\
 &=&\frac{\pi\sqrt{3}}{4}\int^{k^2}_{-k^2} \mu_{,z}dz \,.  \end{aligned}$$ Hence, without loss of generality, $\mu$ can be set to be $$\begin{aligned}
\mu=-\frac{2Q}{\sqrt{3}\pi} \,, 
\label{eq:charge1}\end{aligned}$$ for $\rho=0,\ z\in [-\infty,-k^2]$, and $$\begin{aligned}
\mu=\frac{2Q}{\sqrt{3}\pi} \,, 
\label{eq:charge2} \end{aligned}$$ for $\rho=0,\ z\in [k^2,\infty]$.

Now we would 