# Where is the black hole horizon in ingoing Kerr coordinates?

In ingoing Kerr coordinates, the metric of the Kerr black hole spacetime is:

$$$$ds^2 = -\left(1-\frac{2Mr}{\Sigma}\right)dv^2 +\frac{2Mr}{\Sigma}\left(dr-a \mathrm{sin}^2\theta d\phi\right)dv \\ +\left(1+\frac{2Mr}{\Sigma}\right)\left(dr^2 - 2a\mathrm{sin}^2\theta drd\phi\right)\\ +\Sigma d\theta^2 +\left(r^2+2Ma^2\frac{r \mathrm{sin}^2\theta}{\Sigma} + a^2\right)d\phi^2$$$$ where $$\Sigma = r^2 + a^2\mathrm{cos}^2\theta$$.

In Boyer-Lindquist coordinates, the two black hole event horizons are located at $$r_{\pm}=M\pm\sqrt{M^2-a^2}$$. In ingoing Kerr coordinates, it looks like the event horizons are located at $$r_{\pm} = M \pm \sqrt{M^2-a^2\mathrm{cos}^2\theta}$$. Is that correct? What I find odd about this relation is that then in the extremal limit ($$M=a$$), then $$r_{\pm}=M\left(1\pm\mathrm{cos}\theta\right)$$, so the inner and outer horizons would 'trade places' in terms of which is 'farther away' from the 'origin' $$r=0$$.

Turns out I wasn't properly computing the radial ingoing/outgoing charactersitics. I need to solve $$$$g^{vv}c^2 - 2 g^{vr}c + g^{rr}=0$$$$ which then gives me $$$$c_{\pm} = \left(1+\frac{2Mr}{\Sigma}\right)^{-1} \left( - \frac{2Mr}{\Sigma} \pm \left[ \frac{4M^2r^2}{\Sigma^2} + \left(1+\frac{2Mr}{\Sigma}\right)\frac{\Delta}{\Sigma} \right]^{1/2} \right) ,$$$$ where $$\Delta=r^2-2mr+a^2$$. We see that when $$\Delta=0$$ all the characteristics are ingoing/marginally trapped. So the black hole horizon is at the same radial radial coordinate point in both ingoing Kerr and Boyer-Lindquust coordinates.