I am trying to construct the Penrose diagram of a black hole in AdS space. Now, I thought I was on a good track, my diagram looked like this:
The grey lines are the surfaces of constant $t$ and constant $r$ of the coordinate chart in which the metric reads
$$g = - f(r) dt^2 + f(r)^{-1} dr^2 \quad , \qquad f(r) = 1 - 2m/r + r^2.$$
Now this looks good so far. However, according to my calculations, a surface of constant r within the event horizon looks like the blue line.
This seems awfully wrong. I cannot pinpoint the error in my calculations, so please have a look at them. I proceed as follows:
1. Tortoise coordinate: Calculate the tortoise coordinate, which is defined via the equation $dr^{\ast} = f(r)^{-1} dr$.
$$r^{\ast}(r) = \frac{-1}{f'(r_+)} \bigg( \ln \sqrt{1 + \frac{3rr_+ + 1}{(r - r_+)^2}} - \frac{A^2-2}{A} \arctan \frac{2r + r_+}{A} + \mathrm{const} \bigg)$$
with $r_+$ being the event horizon and with $A = \sqrt{r_+ f'(r_+) + 3}$.
2. Eddington-Finkelstein coordinates: Define $u = t - r^{\ast}$ and $v = t + r^{\ast}$, such that
$$g = - f dudv$$
This metric is degenerate at the event horizon, so another coordinate change is in order:
3. Another coordinate change: Define $\hat{u} = - \operatorname{sgn}(f) e^{- \eta u}$ and $\hat{v} = e^{\eta v}$, where $\eta$ is chosen such that the degeneracy vanishes. Choose $\eta = f'(r_+)/2$, the metric then is
$$ g = - \operatorname{sgn}(f) \frac{f}{\eta^2} e^{-2\eta r^{\ast}} d\hat{u} d\hat{v} $$
4. Limits: The value of the constant in $r^{\ast}$ is still unspecified. At the event horizon, the $r^{\ast}$ diverges and so the constant may be chosen for both sides of the event horizon independently (denote $C_{1,2}$):
$$ \lim_{r \rightarrow \infty} \hat{u} \hat{v} \overset{!}{=} -1 \quad \Longrightarrow \quad C_1 = \frac{A^2 - 2}{A} \frac{\pi}{2}$$
$$\lim_{r \rightarrow 0} \hat{u} \hat{v} \overset{!}{=} 1 \quad \Longrightarrow \quad C_2 = -\ln \sqrt{ 1 + \frac{1}{r_+^2}} + \frac{A^2-2}{A} \arctan \frac{r_+}{A} $$
5. Compactify: Define $\hat{\hat{u}} = \arctan \hat{u}$ and $\hat{\hat{v}} = \arctan \hat{v}$, then
$$g = - \operatorname{sgn}(f) \frac{f}{\eta^2} \frac{e^{-2\eta r^{\ast}}}{\cos^2 \hat{\hat{u}} \cos^2 \hat{\hat{v}}} d\hat{\hat{u}} d\hat{\hat{v}}$$
6. Rotate: Define $\hat{t} = (\hat{\hat{u}} + \hat{\hat{v}})/2$ and $\hat{x} = -(\hat{\hat{u}} - \hat{\hat{v}})/2$. The metric is
$$g = \operatorname{sgn}(f) \frac{f}{\eta^2} \frac{e^{-2\eta r^{\ast}}}{\cos^2 (\hat{t}-\hat{x}) \cos^2 (\hat{t}+\hat{x})} \big( - d\hat{t} + d\hat{x} \big)$$
7. Finally: The surfaces of constant $r$ (or $t$, respectively), are the parametrised curves such that
$$\boxed{\hat{t} = \frac{1}{2} \Big( \arctan \big[ - \operatorname{sgn}(f) e^{-\eta (t - r^{\ast})} \big] + \arctan \big[ e^{\eta(t + r^{\ast})} \big] \Big) \Big|_{r=const.}}$$
$$\boxed{\hat{x} = \frac{-1}{2} \Big( \arctan \big[ - \operatorname{sgn}(f) e^{-\eta (t - r^{\ast})} \big] - \arctan \big[ e^{\eta(t + r^{\ast})} \big]\Big) \Big|_{r=const.}}$$
However, these equations produce the blue curves that seem so wrong. Any ideas where I have gone wrong?