Timeline for Curvature $\geq-1$ but not $\geq1$
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
|
|
| Aug 14, 2017 at 17:47 | history | edited | stefaNo | CC BY-SA 3.0 |
added information about the spaces of directions of a cone-manifold
|
| Aug 12, 2017 at 21:27 | history | edited | stefaNo |
added tag "curvature" (because "big list" was removed by an expert user)
|
|
| Aug 12, 2017 at 15:15 | history | edited | Henry.L |
edited tags
|
|
| Aug 12, 2017 at 13:07 | history | edited | stefaNo | CC BY-SA 3.0 |
changed < vith \leq
|
| Aug 12, 2017 at 10:34 | history | edited | stefaNo | CC BY-SA 3.0 |
added assumption on the dimension
|
| Aug 12, 2017 at 9:12 | history | edited | stefaNo | CC BY-SA 3.0 |
improved formatting
|
| Aug 12, 2017 at 8:33 | comment | added | stefaNo | I am sorry. I edited again by reformulating and by adding an example of answer to clarify better. | |
| Aug 12, 2017 at 8:32 | history | edited | stefaNo | CC BY-SA 3.0 |
to clarify better: reformulated, added trivial example of answer
|
| Aug 11, 2017 at 13:05 | comment | added | Benoît Kloeckner | I don't understand what "Let us assume that $\kappa=\kappa_{max}$ is the maximum possible" means. Also, I don't see how Fact 1 could hold for any $\mathcal{P}$, since having curvature bounded below by $1$ is stronger than having it bounded by $-1$: if the latter implies something, then certainly the former does. I must have misunderstood your question. | |
| Aug 11, 2017 at 9:00 | history | edited | stefaNo | CC BY-SA 3.0 |
added tags 'reference request' and 'big list', rephrased, added motivations
|
| Aug 8, 2017 at 20:29 | comment | added | Igor Belegradek | If you provide some motivation (i.e. why do you care), the question would be easier to answer. | |
| Aug 8, 2017 at 17:35 | comment | added | StefaNo | And if I edit by adding the tag 'reference request', do you still vote for the closure? Sorry but I'm not an expert in metric geometry. If you know a book with some propositions answering my question, it would be greatly appreciated. | |
| Aug 8, 2017 at 13:26 | review | Close votes | |||
| Aug 8, 2017 at 16:38 | |||||
| Aug 8, 2017 at 13:10 | comment | added | Igor Belegradek | The question is overly broad (many results in comparison geometry can serve as an answer). Voted to close. | |
| Aug 8, 2017 at 12:18 | comment | added | stefaNo | Probably what you say is true for curvature bounds from above. For example, by doubling a geodesic hyperbolic triangle you get a simply connected (homeomorphic to $S^2$) Aleksandrov space with curvature bounded below that is not contractible. | |
| Aug 8, 2017 at 12:14 | history | edited | stefaNo | CC BY-SA 3.0 |
added an optional hypothesis: cone-manifold case
|
| Aug 8, 2017 at 8:04 | comment | added | Dmitrii Korshunov | By analogy with the Riemannian picture that would be natural to use another global result: namely Cartan-Hadamard theorem. So the property would be: the universal cover is contractable. | |
| Aug 8, 2017 at 6:30 | history | edited | Ben McKay | CC BY-SA 3.0 |
spelling
|
| Aug 8, 2017 at 4:29 | review | First posts | |||
| Aug 8, 2017 at 4:48 | |||||
| Aug 8, 2017 at 4:28 | history | asked | stefaNo | CC BY-SA 3.0 |