3
$\begingroup$

Unbounded flat Euclidean spaces can be either infinite (e.g., an infinite plane) or finite--e.g., a flat torus, constructed by starting with a square and identifying opposite edges.

Meanwhile, the most obvious example of a space with constant positive curvature is the sphere, which is finite. And there doesn't seem to be an obvious way to cut a piece out and tile it like you could to go from a closed torus to an open plane. So, are constant-positive-curvature spaces inherently an unavoidably finite, or is it possible to construct an infinite space with constant positive curvature?

$\endgroup$
3
  • $\begingroup$ What do you mean by "infinite" and "finite"? Do you mean max distance is finite? Volume is finite? Something else? $\endgroup$ Oct 8, 2021 at 18:32
  • $\begingroup$ @JasonDeVito It did not occur to me that those could be different things. I suppose I mean area or volume. Can you take an open plane and give it constant positive curvature? $\endgroup$ Oct 8, 2021 at 18:38
  • 1
    $\begingroup$ If you view the plane as, say, the northern hemisphere of a sphere, does that answer your question? For the distance vs volume issue, if your graph is something like the graph of $e^{-x}$ for $x\geq 1$ rotated around the $x$-axis, then volume (which means surface area in this context) is finite, while max distance is infinite. I don't have an example where surface area is infinite but max distance is finite. Maybe that can't happen? $\endgroup$ Oct 8, 2021 at 18:44

1 Answer 1

4
$\begingroup$

I think the following may be what you want. By the Killing-Hopf Theorem, every complete connected Riemannian manifold of constant positive sectional curvature is isometric to the quotient of a sphere by a group acting freely and properly discontinuously. Quotient spaces of compact spaces are also compact, so it follows that all complete connected Riemannian manifolds of constant positive sectional curvature are compact. There's a nice proof of the Killing-Hopf Theorem in chapter 11 of John M Lee's Riemannian Manifolds: An Introduction to Curvature.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.