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The go-to example of elliptic space is a sphere where geodesics turn into great circles of finite length. But is it possible to have an elliptic space which doesn't 'merge' with itself once it's made a full turn? ie. infinite, unbounded, simply-connected, but with constant positive curvature everywhere.

I can't find anything like it online, but then maybe I simply don't know what word to search for?

Edit: due to user427327's remarks I thought I'd elaborate with a 1D example, nevermind that curves don't have intrinsic curvature.

Circle 'space' vs spiral 'space'.

The above image shows two 'spaces', both of constant positive curvature. In the left space if you travel 2pi you end up back where you started. In the right space you end up somewhere else entirely. You can keep on travelling and keep on getting further and further away from where you started, despite travelling inside a 'space' of positive curvature. Is the same not possible for 2D spaces with constant positive curvature?

David Rutten
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  • @Narasimham, "[...] From this however the Infinity in no way follows. Rather would space, if one presumes bodies independent of place, that is ascribed to it a constant curvature, necessarily be finite so soon as this curvature had ever so small a positive value." (page 48, Bonola) Seems to suggest that positive curvature must mean the volume of a space with positive curvature is finite. But I still don't understand why. I can't find the the infinite vs. indefinite quote though. – David Rutten Mar 26 '16 at 15:17
  • It's worth noting that going from a circle to a spiral as you describe is exactly what happens when you lift to a covering space - you separate points into various parts based on different paths between them. The sphere is simply connected, which in particular means that there is no covering space for the sphere (other than itself), since if any point on the sphere lifted to two in some covering space, the path between those two points wouldn't be null-homotopic (on the sphere), which is impossible. – Milo Brandt Mar 26 '16 at 15:53
  • @MiloBrandt, of course my example is lacking, since it assumes the circle and helix are embedded into a higher space. But if that weren't the case, if the helix was all there is, does the same argument from a covering space still apply? – David Rutten Mar 26 '16 at 15:57
  • Please check in the comments by Prof. Carlsaw. Google has a limited pages preview. – Narasimham Mar 26 '16 at 18:04
  • the spiral space picture looks more like an cillindrical than an elliptic space. the curvature of a cillinder is zero (not positive) – Willemien Mar 31 '16 at 08:12

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I came across this question while trawling the internet, and on the off-chance that you're still interested in this question and haven't seen an answer, I'll write a few words - it is possible to be infinite, non-repeating and positively curved. What you must lose, however, is completeness. (This is forced by the Bonnet-Myers theorem.) There are a couple of ways of viewing this.

Firstly, one can use longitude/latitude coordinates $\theta, \phi$ for $S^2$, so that the round metric is $ds^2 = d\phi^2 + \cos^2 \phi d \theta^2.$ (These are not the usual polar coordinates, but I think are more useful here). This metric breaks down in these coordinates at $\phi = \pm \pi/2.$ However, there is no reason why we cannot now allow $\theta$ to take any real value! That is, $$ds^2 = d \phi^2 + \cos^2 d \theta^2 $$ defines a metric on $(- \pi/2, \pi/2)_\phi \times \mathbb{R}_\theta$ (which as promised, is not complete - all 'vertical' lines $\theta =$ const are maximal geodesics of length $\pi$). Locally, this is the same as the sphere, and so it has constant curvature 1. However, the line $\phi = 0$ is a geodesic of infinite length.

Secondly, one can see the above construction as effectively removing the north and south poles from the sphere, and then 'unwrapping' the the rest of the sphere, by saying that when we travel around the equator once, we do not in fact return to the starting point, but to a 'new' point in the next part of the orange peel. This is just like your example of 'unwinding' a loop to create a helix - we 'unwrap' the sphere to create this thing.

Thirdly, the above thing may be seen as the Riemannian universal cover of $S^2 \backslash$ poles.

Latimer Leviosa
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Any surface with constant positive curvature, $\kappa$, is a sphere with radius $\frac{1}{\kappa}$. Surely that's not what you are asking? An "ellipse" has positive but varying curvature at every point.

user247327
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  • Why does it need to be a sphere, why can't it consist of an infinite layering of itself? The 1D case would be the difference between a circle and a spiral whose radius is fixed. Two points on the spiral may well look like they are coincident to us who are observing it embedded in our own space, but in spiral coordinates those two points could be miles apart. Or is this not possible? – David Rutten Mar 26 '16 at 15:05
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    Bernhard Riemann made distinction between infinite and unbounded(not indefinite), a comment in epilogue of this subject book by Roberto Bonola by Carlsaw? I think it is the basic difference between the two types of non-euclidean geometries – Narasimham Mar 26 '16 at 18:10
  • @Narasimham, thanks, I'll see if I can lay my hands on that. – David Rutten Mar 26 '16 at 18:35
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One view.

If you go around a constant negative Gauss curvature surface either way, you encounter cuspidal edges at two locations.(David Hilbert). There are 3 types, progressive, regressive and non-return asymptotics at lowest distances from symmetry lines.

As for non constant K positive surfaces of revolution only under special conditions do geodesics trace their own path after 1,2,3,.. circuits. Generally they are non-repeating.

Narasimham
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