Assume that the inner event horizon r_ = m - Sqrt[m^2-a^2] is greater than the Ring singularity radius a. That is, a^2 < (m - Sqrt[m^2-a^2])^2 which for positive mass leads to imaginary a. We therefore must have the event horizon inside the Ring, and a < m or J < m^2.
Note that if we further consider the inner ergosphere radius with respect to the Ring and upon demanding cos^2[theta] positive definite we get at the boundary a restriction on the angular momentum: J = m Sqrt[2m-cos^2]
This defines a band for J and M : m Sqrt[2m-1] < J < m Sqrt[2m].
See attached plot
http://arxiv.org/pdf/0706.0622.pdf
I guess the kerr solution reveals a naked singularity in case of a>m. In this case the radius of the inner and outer event horizon r=m-sqrt(m^2-a^2) and r=m+sqrt(m^2-a^2) are imaginary. Thus there will be no event horizons for a>m and one gets the problem with a naked singularity.
@Thomas Günther, you can check my thread on how to censor a black hole singularity -- https://www.researchgate.net/post/How_to_Censor_a_Black_Hole_Singularity -- where i argue that the diffraction pattern of the gravitational waves inside the black hole wouldn't let this happen.
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