is spacelike everywhere on $\Sigma$ in the second case. In the both cases (i) and (ii), the strings are rigidly rotating[@FSZH]. The solutions in the case (i) give the stationary rotating strings.

![Parameter regions. The shaded areas are the parameter regions for the rigidly rotating strings, the cases (i) and (ii). The dark shaded area is the parameter region for the stationary rotating strings, the case (i). []{data-label="lq"}](parameter.eps){width="6cm"}

There are three singularities of $h^{red}_{AB}$ given by , $\rho=0$, $\rho=\rho_q :=\sqrt{1-q^2}/|\Omega|$ and $\rho=1/|\Omega|$. For the stationary string, $0 \leq  \rho \leq \rho_{max} \leq \rho_q$. The point $ \rho=0$ is a fixed point of the rotational isometry generated by the Killing vector $\partial/\partial \varphi$. When $q\neq 0$ the point $ \rho=\rho_q$ is also a fixed point of the rotation, that is, there are two centers in this space. When $q=0$ the point $ \rho=\rho_q$ becomes a circle.

The scalar curvature of the metric is calculated as $$\begin{aligned}
    R=\frac{-6 q^4 + 2 \left(\Omega^2 \rho^2+4\right)\left(1-\Omega^2 \rho^2\right)
   q^2-2 \left(1-\Omega^2 \rho^2\right)^3}
        {\left(1-\Omega^2 \rho^2\right)^2 \left(1-q^2-\Omega^2 \rho^2\right)^3}.
\end{aligned}$$ One of the centers $ \rho=\rho_q$ is the curvature singularity, while the other center $ \rho=0$ is regular point. The scalar curvature vanishes at $ \rho=0$ when $q^2=1/3$. When $q^2>1/3$, 