Let $(N, g)$ be a complete Riemannian $n$-manifold such that there is a compact set $K$ such that $N\setminus K$ is diffeomorphic to $\mathbb{R}^n \setminus D^n$.
Suppose there exist $n$ parallel vector fields $V_1, …, V_n$ that are orthonormal at a point.
I want to prove that the manifold is globally isometric to $\mathbb{R}^n$ with the euclidean metric.
First part of the proof:
As $\forall X \in \mathfrak{X}(N)$
$$
\nabla_X \langle V_i, V_j \rangle = \langle \nabla_X V_i, V_j \rangle +\langle V_i, \nabla_X V_j \rangle = 0
$$
$\langle V_i, V_j \rangle$ is constant.
Moreover, as it is equal to $\delta_{ij}$ a a point, $\{V_i\}$ are orthogonal everywhere by continuity.
Now, as the Levi-Civita connection is torsion-free
$$
[V_i, V_j] = \nabla_{V_i} V_j – \nabla_{V_j} V_i = 0
$$
and hence they are integrable locally to a chart which is an isometry because of the equation $\langle V_i, V_j \rangle = \delta_{ij}$.
Missing part:
The goal is to use Killing-Hopf theorem to prove the following lemma
Lemma:
Let $(M,g)$ be a complete flat Riemannian manifold such that there is a compact set $K \subset M$ with $M\setminus K$ diffeomeorphic to $\mathbb{R}^n \setminus D^n$. Then $(M,g)$ is isometric to Euclidean space.
but I am not really sure how to do this. Does anybody have a hint? (In case someone is wondering I need this to give a rigorous proof of the positive mass rigidity in the positive mass theorem).
Best Answer
Lemma that you are trying to prove is false in dimension 2. Consider $(M,g)$ which is the open Moebius band with a complete flat metric and take $K$ to be a simple 1-sided loop in $M$. Then $M-K$ is diffeomorphic to the annulus, i.e. to $E^2-D^2$, where $D^2$ is a closed 2-disk.
On the other hand, the open Moebius band is the only counter-example. Indeed, all noncompact complete connected flat surfaces are diffeomorphic either to the open Moebius band or to the open annulus or to the Euclidean plane.
If $n\ge 3$ then the boundary sphere of $D^n$ corresponds to a n embedded $n-1$-dimensional sphere $\Sigma\subset M$. If $M$ is not simply-connected then the preimage of $\Sigma$ in the universal cover $\tilde M$ of $M$ consists of at least two components, each homeomorphic to $\Sigma$ (since $\Sigma$ is simply connected); each of these components does not bound a compact submanifold in $\tilde M$. Therefore, each of these components determines a nontrivial element of $H_{n-1}(\tilde M)$, which implies that $\pi_{n-1}(\tilde M)\ne 0$, contradicting contractibility of $\tilde M$. Therefore, $M$ is simply connected, hence, Hopf-Killing, is isometric to $E^n$.