Timeline for Unbounded sectional curvature implies infinite diameter?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 19, 2022 at 23:48 | vote | accept | L.F. Cavenaghi | ||
| Dec 19, 2022 at 23:38 | answer | added | Nate Eldredge | timeline score: 7 | |
| Dec 18, 2022 at 17:28 | comment | added | Anton Petrunin | @NateEldredge could you make an answer from your comment (so the question would disappear from unanswered). | |
| Dec 1, 2022 at 1:56 | comment | added | Willie Wong | I’m voting to close this question because it was answered in the comments. | |
| Nov 30, 2022 at 19:16 | review | Close votes | |||
| Dec 13, 2022 at 3:02 | |||||
| Dec 20, 2019 at 11:24 | history | edited | Ali Taghavi |
I add a tag.
|
|
| Dec 20, 2019 at 4:23 | history | edited | Nate Eldredge |
edited tags
|
|
| Dec 20, 2019 at 4:22 | comment | added | L.F. Cavenaghi | thank you @NateEldredge, I forgot the completeness assumption, I do appreciate your answer, that is what I was looking for. | |
| Dec 20, 2019 at 4:21 | comment | added | Nate Eldredge | Since you did not ask for $(M,g)$ to be complete, can't you just take something like $(0,1)^2$ and add a bunch of "dimples" that get very sharp as you approach the boundary? If $(M,g)$ is complete, then the statement is true, simply because a manifold with unbounded sectional curvature must be non-compact and therefore by Hopf-Rinow it has infinite diameter. | |
| Dec 20, 2019 at 4:08 | history | asked | L.F. Cavenaghi | CC BY-SA 4.0 |