The $3$-dimensional Riemannian Schwarzschild manifold $(M,\bar{g})$ is given by $M=[s_0,\infty)\times\mathbb{S}^2$ with
\begin{align}
\bar{g}=\frac{1}{1-\frac{2m}{s}}ds\otimes ds+s^2g_{\mathbb{S}^2}
\end{align}
where $m>0$ and $g_{\mathbb{S}^2}$ is the standard round metric on the unit $2$-sphere. By a change of variable, it can be written as a warped product $M=[0,\infty)\times\mathbb{S}^2$ with
\begin{align}
\bar{g}=dr\otimes dr+u(r)^2g_{\mathbb{S}^2}
\end{align}
where $u:[0,\infty)\to\mathbb{R}$ is the smooth warping function satisfying
\begin{align}
u'(r):=\frac{du}{dr}=\sqrt{1-\frac{2m}{u(r)}}
\end{align}
In this question I am interested in the Willmore energy
\begin{align}
\int_{\Sigma_r}H^2d\mu
\end{align}
(my mean curvature here is $H=\lambda_1+\lambda_2$ w/o dividing by 4) of the slices $\{r\}\times\mathbb{S}^2$ in $(M,\bar{g})$.
Since $(M,\bar{g})$ is a warped product, I can compute that
\begin{align}
\int_{\Sigma_r}H^2d\mu&=16\pi u'(r)^2 \\
&=16\pi\left(1-\frac{2m}{u(r)}\right)
\end{align}
which is dependent on $r$.
On the other hand, it is also known that $(M,\bar{g})$ can be written as $(\mathbb{R}^3\setminus B_{2m}(0),\bar{g})$ with
\begin{align}
\bar{g}=\left(1+\frac{m}{2|x|}\right)^4\delta
\end{align}
where $\delta$ is the standard Euclidean flat metric, and $|\cdot|$ is the Euclidean norm. This shows that $\bar{g}$ is conformal to $\delta$. Now, since Willmore energy is a conformal invariant, and since $\Sigma_r$ is a round sphere, it follows from the standard result by Willmore himself that
\begin{align}
\int_{\Sigma_r}H^2d\mu=16\pi
\end{align}
which clearly is an absolute constant.
In summary, I obtain two different answers from different reasonings. At least one of them must contain some fallacy which I fail to notice. Hence, I would like to ask for an explanation on this.
Any comment and answer are welcomed and greatly appreciated.
Best Answer
Your mistake is in the interpretation of conformal invariance. The Willmore energy of a surface in $\mathbb{R}^3$ is conformally invariant in the sense that it is invariant under Möbius transformations.
A better way to describe the conformal invariance of the Willmore energy is as follows. Let $\Sigma$ be a hypersurface in a Riemannian manifold $(M^3,g)$. Define the Willmore energy of $\Sigma$ by $$ W(\Sigma,g) := \int_\Sigma \lvert A_0\rvert^2\,d\mu, $$ where $A_0$ is the trace-free part of the second fundamental form. This is conformally invariant, in the sense that $W(\Sigma,u^2g)=W(\Sigma,g)$ for any positive function $u$ on $M$, for the simple reason that $\lvert A_0\rvert_{u^2g}^2\,d\mu_{u^2g}=\lvert A_0\rvert_g^2\,d\mu_g$ (this uses the assumption $\dim M=3$).
Now, by the Gauss equation, $$ \label{e} \tag{$\ast$} W(\Sigma,g) + 4\pi\chi(\Sigma) = \frac{1}{2}\int_\Sigma H^2\,d\mu + \int_\Sigma \left( R - 2\mathrm{Ric}_{nn} \right)\,d\mu , $$ where $R$ is the scalar curvature of $g$ and $\mathrm{Ric}_{nn}$ is the normal component of the Ricci tensor of $g$. In your situation, $R=0$ but $\mathrm{Ric}_{nn}\not=0$, which explains the discrepancy.
(Note that \eqref{e} also gives the familiar formula for the Willmore energy of a surface $\Sigma\subset\mathbb{R}^3$ after pulling it back to $S^3$ by stereographic projection.)