We denote the corresponding quantities in the choice $(\xi_2,\xi_3)$ with [*tilde*]{} $\ \tilde{}\ $. For example, we denote the twist one-forms with respect to $(\xi_2,\xi_3)$ by $$\begin{aligned}
  \tilde V_{\tilde a}=*(\xi_2\wedge\xi_3\wedge d\xi_{\tilde a}) \,,\end{aligned}$$ where $\tilde a=2,3$. Then, using $
i_{\xi_I}d\psi_a=i_{\xi_I}d\mu=0, \ i_{\xi_I}d\tilde\psi_a=i_{\xi_I}d\tilde\mu=0
$, and eq. (\[eq:twistpotential\]), we show $ 
i_{\xi_I}dV_a=i_{\xi_I}d\tilde V_{\tilde a}=0
$, and hence have $$\begin{aligned}
*(\xi_1\wedge \xi_2\wedge \xi_3\wedge R(\xi_a))
&=& -i_{\xi_3}*(\xi_1\wedge\xi_2\wedge R(\xi_a)) 
  = -\frac{1}{2}i_{\xi_3}dV_a =0 \,, \end{aligned}$$ and $$\begin{aligned}
*(\xi_1\wedge \xi_2\wedge \xi_3\wedge R(\xi_3)) 
 &=& -i_{\xi_1}*(\xi_2\wedge\xi_3\wedge R(\xi_{3}))
  = -\frac{1}{2}i_{\xi_1}d\tilde V_{3}
 =0 \,. \end{aligned}$$ This implies that the condition $2$ holds in our present system (\[eq:action\]) with three commuting Killing vector fields. Furthermore, the axial symmetry of at least one of $\xi_I$ $(I=1,2,3)$ implies that the condition $1$ also holds on the axis of rotation. Therefore, the two-dimensional surface orthogonal to three $\xi_I$ is integrable.

Now, without loss of generality, we choose our three coordinates $(x^1,x^2,x^3)$ as the three Killing parameters, so that $\xi_3 = \partial/\partial t$ denotes the stationary (asymptotic time-translation) Killing vector field in our spacetimes and $\xi_1= \partial/ \partial \phi$ and $\xi_2 = \partial/\partial \psi$ are two independent axial Killing symmetries. Then, from the above observation, we can express the three dimensional metric $h_{ij}$ by $h=h_{pq}dx^pdx^q- \rho^2dt^2\ 