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Choose spin orientation so that $0<a<M$; the printed $M/a$ threshold assumes nonzero positive $a$. Put $H=2Mr/(r^2+a^2)$ and $F=1-H$. The axial metric has components $(-F,H,1+H)$ and determinant $-1$. Conserved energy and timelike normalization giveEliminating $\dot{\tilde{t}}$ gives the Kerr axial timelike effective potentialIt extends through the zeros of $F$ in the regular original coordinates. Since $V_{\mathrm{eff}}(0)=1$, reaching zero requires ${\epsilon }^2\ge 1$. At equality the infalling orbit reaches zero with zero radial velocity and turns back, because the nearby negative sheet is forbidden. Along the axis zero is regular, unlike the off-axis ring.

For $r<0$, $V_{\mathrm{eff}}^{\prime }=2M(r^2-a^2)/(r^2+a^2{)}^2$ and the global maximum is $V_{\mathrm{eff}}(-a)=1+M/a$. If ${\epsilon }^2>1+M/a$, an inward particle has no negative-sheet turning point. It passes through zero and continues toward $\boxed{r\to -\infty }$, another asymptotically flat end, with $\dot{r}\to -\sqrt{{\epsilon }^2-1}$ and infinite proper time to infinity. It first crosses the outer and inner horizons. Perturbative Cauchy-horizon instability can invalidate this ideal Kerr axial analytic extension physically.

This is the two-dimensional axis restriction. The dashed $r=0$ curve is regular; the four-dimensional ring lies off-axis and is not an axial singular boundary.