Timeline for Geodesic current supported on a pencil?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 1, 2018 at 21:25 | history | edited | Dylan Thurston | CC BY-SA 4.0 |
TeX formatting
|
| Oct 1, 2018 at 14:43 | history | edited | DMG | CC BY-SA 4.0 |
fixed the argument, including a missing step, and reformatted the answer to improve legibility.
|
| Oct 1, 2018 at 14:08 | comment | added | DMG | Thanks to both of you: yes, I was considering a proper subset of a pencil, but in the answer I wrongly stated I was considering the whole pencil. I will edit the answer. My apologies for the confusion. | |
| Sep 29, 2018 at 18:03 | comment | added | Dylan Thurston | I think you should be looking at not the whole pencil $P(a)$ (i.e., the set of all geodesics with an endpoint at $a$), but rather some subset of it: pick a window $w$ disjoint from $a$, and consider the set $G(a,w)$ of geodesics with one endpoint at $a$ and the other in $w$. Then I believe local finiteness impless that $\mu(G(a,w)) < \infty$, while the intersection $\cap_{n > 0} g^n G(a,w)$ is a single geodesic and the union $\cup_{n < 0} g^n G(a,w) = P(a)$. | |
| Sep 29, 2018 at 1:19 | comment | added | R W | Why is $\cap_n g^n P(a)=\gamma'$? Even if this were true why would this imply $\mu(g^n P(a)) \to \mu(\gamma')=0$? (as the measure $\mu$ is infinite). | |
| Sep 28, 2018 at 14:05 | review | First posts | |||
| Sep 28, 2018 at 14:13 | |||||
| Sep 28, 2018 at 14:00 | history | answered | DMG | CC BY-SA 4.0 |