The Kerr metric expressed in terms of polar coordinates $r,\theta,\phi$, such that $x = r\sin(\theta)\cos(\phi)$, $y = r\sin(\theta)\sin(\phi)$, $z = r\cos(\theta)$. Then the Kerr metric is given as \begin{align*} ds^2 = &-\left(1 - \frac{2GMr}{r^2+a^2\cos^2(\theta)}\right) dt^2 + \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 + \left(r^2+a^2\cos(\theta)\right) d\theta^2\\ &+ \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 - \left(\frac{4GMra\sin^2(\theta)}{r^2+a^2\cos^2(\theta)}\right) d\phi\, dt \end{align*} where $a \equiv S/M$ is the object's angular momentum per unit mass, and $G$ is the gravitational constant. This is an exact solution for the empty-space Einstein equation.
Say, If we are to consider the metric for a constant time, $t_0$. Is it then possible to define the Kerr metric on a submanifold of spacetime, say only in space? If so how can I accomlish this? Is it as simple as dropping the time dependent terms, i.e \begin{align*} ds^2 = & \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 + \left(r^2+a^2\cos(\theta)\right) d\theta^2\\ &+ \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 \end{align*} or do I need to use the induced metric to describe the metric on the submanifold?
Edit :
I solved the geodesic differential equations using a "time independent" Kerr metric, with a = 0
(i.e this reduces Kerr metric to the Schwarzschild metric), and the Schwarzschild radius to define the other parameters :
Most plots I got spiraled around a singularity at the origo.
Here is a plot where I set $\phi$ to a constant, the z-axis becomes the "time" :
Update :
I have found the following figure which seem to verify my first figure.
Strategies for Direct Visualization of Second-Rank Tensor Fields by Werner Benger and Hans-Christian Hege