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For the superextreme Kerr metric,for all real $r$, so there are no coordinate singularities at horizon roots. The only stipulated curvature singularity is $\Sigma =0$, which forces $r=0$, $\theta =\pi /2$. In the supplied Cartesian coordinates this is the Kerr ring singularity $x^2+y^2=a^2$, $z=0$.

Write ${\rho }^2=x^2+y^2$. The coordinate transformation impliesand henceThe positive-radius Cartesian sheet usesIts zero set is the disk $z=0$, $\rho \le a$. At an interior disk point, $\rho <a$, one has $\cos \theta \ne 0$, so $\Sigma =a^2{\cos }^2\theta >0$. The disk interior is therefore not a singularity: in signed $(r,\theta ,\varphi )$ coordinates, the metric is analytic through $r=0$. In particular, $g_{rr}=\Sigma /\Delta$ is finite and the metric determinant $-{\Sigma }^2{\sin }^2\theta$ is nonzero away from the ordinary polar coordinate axis. The polar axes are regular using Cartesian angular charts and $2\pi$ periodicity.

The required two-sheeted extension of superextremal Kerr spacetime is constructed by taking a second Cartesian sheet with the negative root $r\le 0$, cutting both sheets along the disk, and gluing the upper bank of one to the lower bank of the other, and conversely. Signed $r$ then passes continuously through zero as the disk is crossed. The ring is omitted. A useful check near an interior disk point isso signed $r$ is a smooth transverse coordinate there. The apparent factors $z/r$ in the Cartesian metric represent $\cos \theta$ and have regular signed-coordinate limits; they do not justify deleting the whole disk. Identifying the two signs of $r$ on a single Cartesian copy would also identify two physically different metric values, since the Kerr-Schild coefficient changes sign.

Continue the two sheets out to $r\to \pm \infty$. Both ends are asymptotically flat; at the negative-radius end the mass term has the opposite sign, as $g_{tt}=-1+2M/r+O(r^{-2})$. There are no further finite-radius horizon boundaries requiring additional blocks. Removing only the curvature-singular ring, and removing coordinate artifacts with the charts described above, gives the standard maximal analytic extension. The ring cannot be filled by an analytic nondegenerate metric because its curvature diverges. This extension is not globally hyperbolic; analyticity of an extension should not be confused with unique evolution from Cauchy data.

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.

For $0<M<a$, $\Delta =(r-M{)}^2+a^2-M^2>0$, so no horizon blocks need adjoining. The oblate Cartesian coordinates of the Kerr metric satisfy the displayed relation, where ${\rho }^2=x^2+y^2$. Keeping only the positive root $r$ leaves a cut along the disk $z=0$, $\rho <a$. Continue $r$ through zero away from its boundary, the Kerr ring singularity, onto a negative-radius sheet. Equivalently cut two Cartesian sheets along the disk and join opposite banks, so $r$ changes sign continuously. The metric is analytic at the disk interior in signed oblate coordinates since $\Sigma =a^2{\cos }^2\theta \ne 0$ there. The extension has asymptotically flat ends at $r\to +\infty$ and $r\to -\infty$; only the singular ring remains inextendible by an analytic Lorentz metric.