Timeline for Unbounded curvature implies infinite diameter on complete metric spaces
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 11, 2023 at 18:48 | answer | added | Ethan Dlugie | timeline score: 0 | |
| Dec 20, 2019 at 20:25 | vote | accept | L.F. Cavenaghi | ||
| Dec 20, 2019 at 20:09 | comment | added | YCor | In en.wikipedia.org/wiki/Alexandrov_space to have Alexandrov curvature $\le\kappa$ is defined, but I don't see how this makes "Alexandrov curvature bounded from above" meaningful. It seems that the relevant definition is rather being CAT($\kappa$), and the question seems to already be answered by the boundary of a 2-sphere (quarague's answer), which is clearly a GH-limit of smooth Riemannian surfaces. | |
| Dec 20, 2019 at 20:04 | history | edited | YCor | CC BY-SA 4.0 |
fixed typos
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| Dec 20, 2019 at 18:14 | comment | added | L.F. Cavenaghi | I have updated the question. | |
| Dec 20, 2019 at 18:12 | history | edited | L.F. Cavenaghi | CC BY-SA 4.0 |
added 352 characters in body
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| Dec 20, 2019 at 15:05 | comment | added | Alexandre Eremenko | If the space (with a Riemannian metric) is complete but not compact, it must have infinite diameter. | |
| Dec 20, 2019 at 15:02 | comment | added | Andy Sanders | The curvature of a (complete) metric space has no, at least obvious, meaning. Within certain classes, you'll find Alexandrov spaces as limits of Riemannian manifolds where certain curvature bounds make sense, but you need to formulate a more precise question to get a reasonable answer. | |
| Dec 20, 2019 at 15:01 | comment | added | HJRW | Could you say exactly what you mean by unbounded curvature of a metric space? For instance, do you mean it doesn't satisfy the $CAT(\kappa)$ condition for any $\kappa$? | |
| Dec 20, 2019 at 14:56 | comment | added | Alexandre Eremenko | Your condition on curvature is irrelevant. If you have a complete non-compact surface of finite diameter, you can always make a sequence of very little holes in it and glue in spheres of very large curvature. So the question is equivalent to: "does there exist a complete Riemannian manifold of finite diameter?" | |
| Dec 20, 2019 at 14:53 | answer | added | quarague | timeline score: 2 | |
| Dec 20, 2019 at 14:10 | history | asked | L.F. Cavenaghi | CC BY-SA 4.0 |