Timeline for Surfaces that can be rolled by a ball
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 23, 2012 at 13:33 | vote | accept | Joseph O'Rourke | ||
| Sep 22, 2012 at 23:09 | comment | added | Joseph O'Rourke | @Gerhard: I had not considered disconnected surfaces, but that is an interesting expansion of the topic... | |
| Sep 22, 2012 at 21:48 | comment | added | Gerhard Paseman | Actually, by definition it is unclear. One assumes "image of the interval (0,1) under a continuous map" when one sees the words "path" and "connecting", but this is not the same as "path-connected". Joseph may intend the surface to be connected, but unless he clarifies, I see enough ambiguity in the wording to admit disconnected surfaces. A certain form of connecting may involve the surface of the ball as well. Even so, I don't think you need to change your characterization much, if at all, for this situation. Gerhard "Ask Me About System Design" Paseman, 2012.09.22 | |
| Sep 22, 2012 at 21:33 | comment | added | Joseph O'Rourke | @Anton: Thank you! @Agol: Nice, succinct formulation! | |
| Sep 22, 2012 at 18:37 | comment | added | Anton Petrunin | @Gerhard, no, by definition the rollable surfaces are connected. | |
| Sep 22, 2012 at 18:35 | comment | added | Anton Petrunin | @Agol, yes, sure. | |
| Sep 22, 2012 at 16:16 | comment | added | Gerhard Paseman | Note that there are disconnected surfaces which are rollable. You might consider incorporating them. Gerhard "Ask Me About System Design" Paseman, 2012.09.22 | |
| Sep 22, 2012 at 15:53 | comment | added | Ian Agol | I think your condition is equivalent to saying that the tubular neighborhood of radius $r$ is embedded, when parameterized by Fermi coordinates. | |
| Sep 22, 2012 at 15:21 | comment | added | Deane Yang | Anton, thanks for answering such an easy question. | |
| Sep 22, 2012 at 15:12 | comment | added | Anton Petrunin | @Deane, take a smooth simple curve with curvature $<1$ which comes $(2\cdot r)$-close to itself. The body is its $\varepsilon$-neighborhood (smoothed as needed). | |
| Sep 22, 2012 at 15:01 | comment | added | Deane Yang | This sounds right, but what's the easiest example showing that the global condition is necessary? | |
| Sep 22, 2012 at 14:49 | history | answered | Anton Petrunin | CC BY-SA 3.0 |