Timeline for Gradient of distance function at cut points on Alexandrov spaces
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Jul 1, 2015 at 8:20 | comment | added | John Harvey | First off, the sector angle only has to be $2 \theta < 2 \pi$. Your calculation of the gradient at $q$ is correct. I have edited my answer a little to clarify -- if you allow $x$ to vary along the ray through $q$, then the gradient at $x$ varies through all possible values. Move $x$ so that it is $\left| p \right| \cos \theta$ from the vertex. Then the angle is $\pi/2$. Move it to infinity and the angle becomes $\pi$. | |
| Jul 1, 2015 at 8:15 | history | edited | John Harvey | CC BY-SA 3.0 |
In paragraph four, the gradient should be measured at $q$, not $x$. Paragraph 5 is edited to clarify where $x$ is allowed vary.
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| Jul 1, 2015 at 3:00 | comment | added | mafan | Maybe I misuderstand you, I don't know how to get every possible value between $0$ and $1$ for a fixed cone. Cut the cone along the ray through p, we get a sector with angle $2\theta<\pi$. Then the angel between $p$ and $q$ is $\theta$. So the angle formed by $qp$ and the ray from the vertex through $q$ is $\alpha=(\pi+\theta)/2$. So $|\nabla_q d_p|=-\cos \alpha$. | |
| Jun 29, 2015 at 11:51 | history | answered | John Harvey | CC BY-SA 3.0 |