 Solution

For nonzero $a$, $\Sigma =0$ requires $r=0$, $\theta =\pi /2$. Oblate Cartesian coordinates satisfy $x^2+y^2=(r^2+a^2){\sin }^2\theta$, $z=r\cos \theta$, so this is the Kerr ring singularity of radius $|a|$. Curvature components and generic invariants diverge there. The whole $r=0$ surface is not singular: away from the ring it is a disk through which the extension continues to a sheet with $r<0$. This oblate radial parameter is not a nonnegative Euclidean distance.

In $G=c=1$ units, $\boxed{J=Ma}$. Horizon candidates solve $\Delta =0$, giving $r_{\pm }=M\pm \sqrt{M^2-a^2}$. For $J^2>M^4$, equivalently $|a|>M$, there are no real roots. The over-rotating Kerr solution has no event horizon hiding its naked ring singularity. The term black hole in this regime names the Kerr family, not a hidden-singularity black hole.