Complete Riemannian metric on ${\mathbb R}^2\setminus\{0\}$. It seems to me that the Riemannian metric $g_{ij}=\delta_{ij}/|x|^2$ on the punctured plane is complete, but I don't find a proof not involving explicit computations of the geodesic equation. Does anyone know one?
 A: As user8268 noted, this space is isometric to $\mathbb{R}\times S^1$. Using the polar coordinates $(r,\phi)$ on $\mathbb{R}^2\setminus \{0\}$, the isometry is 
$$F(r,\phi) = (\log r,\phi)\in \mathbb{R}\times S^1$$
Indeed, this map is  smooth and a bijection. So to check that it's an isometry it remains to consider the action on tangent vectors. Conveniently, the space $(\mathbb{R}^2\setminus \{0\},g_{ij})$ has two vector fields that form an orthonormal basis of every tangent plane:
$$
r\,\frac{\partial}{\partial r} \quad\text{and}\quad \frac{\partial}{\partial\phi}
$$
The map $F$ pushes them forward to $$
 \frac{\partial}{\partial z} \quad\text{and}\quad \frac{\partial}{\partial\phi}
$$
which are orthonormal vector fields on $\mathbb{R}\times S^1$. The push-forward is basically the chain rule: since $r=e^{z}$, 
$$
\frac{\partial f}{\partial z} = \frac{\partial r}{\partial z} \frac{\partial f}{\partial r} = e^z \frac{\partial f}{\partial r} = r\,\frac{\partial f}{\partial r} 
$$
Since $dF$ maps one orthonormal basis to another, it is an isometry.
