Timeline for Can one use Brownian motion to prove that two manifolds are not conformally equivalent?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 21, 2013 at 1:29 | comment | added | Ritwik | Yes, that is correct. I am thinking of the two objects as Riemann surfaces with boundary. | |
| Jul 20, 2013 at 23:34 | comment | added | Igor Rivin | I think (s)he thinks of the two objects as marked Riemann surfaces (with boundary), as is common in in the subject. | |
| Jul 20, 2013 at 8:49 | answer | added | Alexander Shamov | timeline score: 10 | |
| Jul 20, 2013 at 8:42 | comment | added | Alexander Shamov | ... And the Riemannian mapping doesn't take vertices to vertices, so you don't want to call it conformal. Is that what you mean? | |
| Jul 20, 2013 at 8:35 | history | asked | Ritwik | CC BY-SA 3.0 |