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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.

flag	asked Apr 15 2010 at 16:13 tkf 127●2

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If there is a uniformly continuous map onto a complete space, then the domain is complete. And completeness implies compactness, see Ulrich Pennig's answer. – Sergei Ivanov Apr 15 2010 at 17:07

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Sergei Ivanov (in a comment to my deleted answer) makes us notice that a metric on $\mathbb{R}^2$ with constant curvature $+1$ is not unique. One example with bounded diameter is constructed identifying $\mathbb{R}^2$ with, say, the complement of the North pole in the unit sphere (or pulling back the metric from any open subset of the unit sphere diffeomorphic to $\mathbb{R}^2$, with any area between $0$ and $4\pi$, as noted by Robin Chapman). On the other hand, the universal cover of a tubular neighbourhood of the equator in the sphere provides an example with infinite diameter. – Qfwfq Apr 15 2010 at 17:32

Ulrich Pennig · Accepted Answer · 2010-04-16 17:22:06Z UTC

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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 $k$.

edited Apr 16 2010 at 17:22

answered Apr 15 2010 at 16:41

Ulrich Pennig
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er... I think sectional curvature has more information than Ricci curvature, in general? Sectional curvature tells you about transversal geodesic propagation of arcs; Ricci only tells you about volumes. – some guy on the street Apr 15 2010 at 17:55

Of course, you are right. As stated on the Wikipedia page, the result also applies if the sectional curvature is bounded by $k$ from below. – Ulrich Pennig Apr 15 2010 at 20:23

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I edited the answer according to the above comment by some guy on the street (okay... that sounds really awkward). – Ulrich Pennig Apr 15 2010 at 20:43

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Sectional curvature bound is stronger than Ricci curvature bound, not weaker. The Ricci curvature at a unit vector is just a sum of $n-1$ sectional curvatures, so $Sec\ge k$ implies $Ric\ge (n-1)k$. – Sergei Ivanov Apr 15 2010 at 22:44

Whoops, sorry again, I edited the comment according to your comments. Thank you. – Ulrich Pennig Apr 16 2010 at 17:24

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Miguel · Answer 2 · 2010-04-15 17:02:25Z UTC

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

Gunnar Magnusson · Answer 3 · 2010-04-15 16:43:53Z UTC

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I don't think so:

A surface of constant positive curvature is diffeomorphic to a sphere, so it would be compact. If the identity map were continuous, then by composition the image of the surface would be a compact subset of $\mathbb R^2$ and thus not equal to the whole plane.

At least this works if the surface is orientable. If not I'm not really sure what happens (I'm a complex geometer; everything is of even dimension and orientable.)

answered Apr 15 2010 at 16:43

Gunnar Magnusson
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	@Gunnar: a non-complete surface of constant positive curvature is not necessarily diffeomorphic to a sphere. – Qfwfq Apr 15 2010 at 17:14
	Really? But it needs to be non-orientable, right? My understanding is that if not you can put some almost complex structure compatible with the metric on the surface. This structure will automatically be integrable, so we have a Riemann surface with a metric of constant positive curvature. Thus its universal cover is $\mathbb P^1$, and the only thing covered by the sphere is the sphere itself. Does this line of reasoning go wrong somewhere? – Gunnar Magnusson Apr 16 2010 at 8:55

Can you have a spherical plane?

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