A complete metric on $\mathbb{R}^2 \setminus \{p_1, p_2\}$ with zero curvature

Question

For two distinct arbitrary points $p_1, p_2 \in \mathbb{R}^2$, is there a complete flat Riemannian metric on $\mathbb{R}^2 \setminus \{p_1, p_2\}$?

Answer 1

No. Set $M = \mathbb{R}^2 \setminus \{ p_1, p_2 \}$, let $g$ be a complete flat metric on $M$ and choose some $q \in M$. The results leading to the Cartan-Hadamard theorem tell you that $\exp_q \colon T_q M \rightarrow M$ is a local diffeomorphism and a covering map (see Do Carmo, pages 149-150). Thus we can pull back $g$ from $M$ to $T_q M$ and make the map $\exp_q$ a local isometry. Since the pullback metric on $T_q M$ is also complete and flat and $T_q M \approx \mathbb{R}^2$, we have that $(T_q M, \exp_q^{*}(g))$ is isometric to $\mathbb{R}^2$ with the standard flat metric.

This implies that $M$ is a quotient of $\mathbb{R}^2$ by a subgroup of the group of isometries of $\mathbb{R}^2$ which acts totally discontinuously on $\mathbb{R}^2$. All such possible groups have been classified explicitly and result in quotients which are cylinders, infinite Mobious strips, tori or Klein bottles so we have a contradiction.