Timeline for Two geodesics with angle $\pi$ in Alexandrov space
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 25, 2015 at 20:22 | comment | added | Richard Montgomery | @AntonPetrunin. Thanks for the correction! Yes. Of course! I didn't take my own advice and draw lines on paper of a plane minus a sector before folding the paper to make a cone. Any small sliver of a sector deleted shows you that as you say any $\beta > 0$ does the trick | |
| Jul 25, 2015 at 10:49 | vote | accept | asv | ||
| Jul 24, 2015 at 14:54 | comment | added | Anton Petrunin | @RichardMontgomery, it works for any $\beta>0$, but the angle in this case is $<\pi$... | |
| Jul 24, 2015 at 14:51 | comment | added | Anton Petrunin | For $\phi(t)=|t|$ the angle is $<\pi$, something like $\phi(t)=|t|^{1.00001}$ should work. | |
| Jul 24, 2015 at 14:34 | comment | added | Richard Montgomery | No, but it works for $\phi(t) = \beta |t|$ when $\beta$ is large enough, I believe any $\beta > \sqrt{3}$ works. (Make cones out of paper. Lines drawn before folding are geodesics.) The border line case is a ``cone over a circle of circumference $\pi$''. The metric of a cone over a circle of circumf $2 \pi \lambda$ is $ds^2 = dr^2 + \lambda^2 r^2 d \theta ^2$ in polar coord based at the cone point. The border-line case is the cone made from a standard rectangular piece of paper (fold a half-plane), so $\lambda =1/2$, and I believe corresponds to $\beta = \sqrt 3$. | |
| Jul 24, 2015 at 11:48 | comment | added | asv | If one takes $\phi(t)=|t|$, will it work? In other words, one takes the graph of $z=\sqrt{x^2+y^2}$. | |
| Jul 24, 2015 at 11:15 | history | answered | Anton Petrunin | CC BY-SA 3.0 |