How compute the mass of AdS-Schwarzschild by ADM mass formula? I want to compute the mass of AdS schwarzschild by ADM mass formula but I could not find where I am wrong.
AdS schwarzschild line element is :
$$
ds^2 =-f dt^2 +\frac{dr^2}{f} +r^2 d\sigma^2_{d-1}
$$
where:
$$
f=k+\frac{r^2}{L^2}-\frac{\omega^{d-2}}{r^{d-2}}
$$
ADM mass formula is
$$
M=\int (k-k_0)\sqrt{\sigma}d^{d-1}x
$$
$k$ is extrinsic curvature of $S_{t=cte,r=cte}$ and $k_0$ the extrinsic curvature $S_{t=cte,r=cte}$ in pure AdS.
$$
k=\sigma^{\alpha\beta}k_{\alpha\beta}=\frac{1}{r^2}\Gamma^r_{\alpha\beta}n_r
$$
$n$ is normal vector to the surface $r=cte$, $n_\alpha=(\frac{1}{\sqrt{f}},0,...,0)$.
$$
k=\frac{1}{r^2}\frac{1}{2}g^{rr}\partial_rg_{\alpha\beta}\frac{1}{\sqrt{f}}=\frac{\sqrt{f}}{r}
$$
$k_0$ has the same relation of $k$ but $f=k+\frac{r^2}{L^2}$.
$$
M=\lim _{r->\infty}\int (k-k_0)\sqrt{\sigma}d^{d-1}x=V_{d-1} r^{d-1} (\frac{\sqrt{k+\frac{r^2}{L^2}-\frac{\omega^{d-2}}{r^{d-2}}}}{r}-\frac{\sqrt{k+\frac{r^2}{L^2}}}{r})=V_{d-1} r^{d-1} L((1+\frac{kL^2}{r^2}-\frac{\omega^{d-2}L^2}{r^d})^{\frac{1}{2}}-(1+\frac{kL^2}{r^2})^{\frac{1}{2}}=V_{d-1} r^{d-1}L (-\frac{\omega^{d-2}L^2}{r^d})=\lim_{r->\infty}V_{d-1} L (-\frac{\omega^{d-2}L^2}{r})=0
$$
I dont know where it is wrong.

I think the issue is that you’re using a formula for the ADM mass which assumes unit lapse. Consider a spherically symmetric spacetime of the form
\begin{align}
{\rm d}s^2 & = -f(r) \, {\rm d}t^2 + h(r)\, {\rm d}r^2 + r^2 ({\rm d}\theta^2 + \sin^2\theta\, {\rm d}\varphi^2)
\end{align}
From the outward pointing unit normal 3-vector $(s^r,s^\theta,s^\varphi)=(h(r)^{-1/2},0,0)$ we compute the trace of the extrinsic curvature,
\begin{align}
k & = D_i s^i \\
& = \partial_r s^r + (\Gamma^r_{rr}+\Gamma^\theta_{\theta r} + \Gamma^\varphi_{\varphi r})s^r \\
& = -\frac{1}{2}\frac{h’(r)}{h(r)^{3/2}} + \left[\frac{h’(r)}{2h(r)} + \frac{1}{r} + \frac{1}{r}\right]\frac{1}{\sqrt{h(r)}} \\
& = \frac{2}{r}\frac{1}{\sqrt{h(r)}}
\end{align}
Specializing to the case of Schwarzschild-AdS$_4$,
\begin{align}
f(r) & = 1 - \frac{2m}{r} + a^2 r^2 \\
h(r) & = \frac{1}{1 - \frac{2m}{r} + a^2 r^2}
\end{align}
we obtain
\begin{align}
k & = \frac{2}{r}\left(1-\frac{2m}{r}+a^2r^2\right)^{1/2} \\
k_0 & =  \frac{2}{r}\left(1 + a^2 r^2\right)^{1/2}
\end{align}
The ADM mass is
\begin{align}
M_{\rm ADM} & =
-\frac{1}{8\pi}\int {\rm d}^2 x \sqrt{\sigma} N(k-k_0) \\
&  = -\frac{1}{8\pi} \lim_{r\to\infty} (4\pi r^2) \left(1 - \frac{2m}{r} + a^2 r^2\right)^{1/2}
\frac{2}{r}\left[\left(1 - \frac{2m}{r} + a^2 r^2\right)^{1/2} - \left(1 + a^2 r^2\right)^{1/2}\right] \\
& = m
\end{align}
