When I was at school I wondered if a surface could locally appear to be a unit sphere, yet `carry on forever'. More formally, my question is:


Can you place a metric of constant curvature +1 on ${\mathbb R}^2$, such that the identity map to ${\mathbb R}^2$ (with standard Euclidean metric) is uniformly continuous?


It is possible to induce such a metric on ${\mathbb R}^2 - {\mathbb Z}^2$, by identifying each unit square with integer vertices, with a hemisphere on the unit sphere.



Can you have a spherical plane?
By Myers's theorem (see <http://en.wikipedia.org/wiki/Myers%27s_theorem>) you have that if the Ricci curvature of a complete $n$-manifold $M$ is bounded below by $(n − 1)k > 0$, then its diameter is bounded by some constant depending on $k$. In particular, it is compact. Therefore if there is such a metric, it cannot be complete.


As pointed out by Sergei Ivanov in the comments below, the bound on the Ricci curvature holds (and therefore compactness follows) if the sectional curvature is bounded below by $k$. 


