Convergence of ADM mass on asymptotically flat manifolds Hey dear mathematicians,
I just read about Riemannian geometry and didn't understand this:
Given a $n$-dimensional Riemannian manifold $(M,g)$ with a specific asymptotic such that the scalar curvature $R$ is given as
$R=\sum_{i,j} \left( \partial_{i}\partial_{j}g_{ij} - \partial_{j}\partial_{j}g_{ii} \right) + \mathcal{O}(|x|^{-2p-2})$,
where $2p+2>n$. Now the divergence theorem ensures the existence of 
$\lim\limits_{r \to \infty} \int_{\mathbb{S}_{r}} \sum_{i,j} \left( \partial_{i}g_{ij} - \partial_{j}g_{ii} \right)\nu_{j}\text{d}\xi(r)$,
where $\nu=x/r$ is the Euclidean unit normal to $\mathbb{S}_r$ and $\text{d}\xi(r)$ is the Euclidean area element of $\mathbb{S}_r$.
I don't understand this implication, can someone clarify this?
Thanks in advance! 
Sincerely, schoeni
 A: $\def\p{\partial}$Since we are taking the surface integral of a (Euclidean) dot product with the (Euclidean) unit normal, the divergence theorem tells us that the integral can be written as
$$ \int_{S_r} (\p_ig_{ij}-\partial_j g_{ii}) \nu_j\, d\xi= \int_{B_r} \p_j(\p_i g_{ij}-\p_jg_{ii})\,dV$$
where $B_r$ is the ball of radius $r$ and $dV$ is the Euclidean volume element. We now see that the integrand differs from the curvature $R$ by some error $E(x)$ which has asymptotic $O(|x|^{-k})$ with $k>n$. To put it precisely, once $|x|$ is larger than some cutoff radius $r_0$ we have a bound of the form $|E(x)| \le C|x|^{-k}$.
The integral of the error over the ball $B_{r_0}$ is finite by compactness (and continuity of $R$ and the ADM mass integrand). The integral of the error everywhere else can be shown to be finite using the asymptotic: we can estimate
$$ \left|\int_{M \setminus B_{r_0}} E\right| \le C \int_{r_0}^\infty \int_{S_r} r^{-k}\, d\xi\, d r = CA_n\int_{r_0}^\infty r^{n-1-k} dr,$$ where $A_n$ is the surface (hyper)area of the unit sphere. Since we know $k>n,$ this integral converges; so we know $\int_M E$ converges. Thus the ADM mass integral converges if and only if the integral of the scalar curvature does.
