Timeline for Equidistant hypersurfaces in symmetric space via exponentiation?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2013 at 14:28 | comment | added | Benoît Kloeckner | The case of a singular direction for $[p_0 p]$ is probably not an obstruction; rather than an approximation argument, you can also use the structure theory for symmetric spaces: the set of geodesics parallel to $[p_0,p]$ is a totally geodesic subspace, isometric to the product of lower-rank space with a Euclidean space. This should be of helo, you can look at Eberlein's book. | |
| Jul 24, 2013 at 14:21 | comment | added | A. Pascal | Thanks. About a simpler argument than using Leuzinger. Benoit above at first thought it was simple but then was not quite certain. So a simple argument is still desired. | |
| Jul 24, 2013 at 13:45 | comment | added | Mikhail Katz | Yes. For example, in the complex projective case, one gets a law of cosines that's identical with the spherical case except for one additional term involving the angle in the tangent space between the complex lines spanned by the two directions. | |
| Jul 24, 2013 at 13:37 | comment | added | A. Pascal | OK. Thanks. Although the trigonometry is hard to set up, it is intuitive to use, no? | |
| Jul 24, 2013 at 13:32 | comment | added | Mikhail Katz | A generic one is regular, so for the purposes of merely showing that the exponential map gives the equidistant surface that's enough by a suitable passage to the limit. But perhaps there is a simpler argument for showing this that does not require trigonometry on a symmetric space. | |
| Jul 24, 2013 at 13:24 | comment | added | A. Pascal | Thanks. One would like to argue just like this. Leuzinger's result applies to geodesic triangles whose geodesic sides are regular. Is that clearly the case here? What if the geodesic $[p_{0},p]$ is singular? | |
| Jul 24, 2013 at 12:30 | comment | added | Mikhail Katz | For those interested in exponentiating from the midpoint, I would be interested in seeing a proof less involved than the one using Leuzinger's detailed results. | |
| Jul 24, 2013 at 10:17 | history | answered | Mikhail Katz | CC BY-SA 3.0 |