Complete Riemannian metric on ${\mathbb R}^2\setminus\{0\}$.

Question

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?

Answer 1

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.