TITLE: Can you have a spherical plane?
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: Intrinsically, no, because one way to understand positive curvature is as a "force" that pulls diverging geodesics closer together. If the curvature is bounded from below away from zero, then any two geodesic arcs leaving a given point will intersect at a finite distance as Ulrich comments above.
But extrinsically, the situation is a lot more interesting and fun. You can embed an intrinsically flat plane in a hyperbolic 3-space and, from within the hyperbolic 3-space, it will look like it has constant positive curvature with respect to geodesic planes (intrinsically, hyperbolic planes) of the hyperbolic 3-space. There are models of the hyperbolic 3-space where the 3-space is represented by the interior of a ball of radius 1, and then one of this embedded flat planes is represented by a 2-sphere internally tangent to the surface of the 3-ball.
See http://en.wikipedia.org/wiki/Horoball, and (for a nice picture) http://en.wikipedia.org/wiki/Horocycle