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}