Timeline for Equidistant hypersurfaces in symmetric space via exponentiation?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| S Aug 1, 2013 at 10:50 | history | bounty ended | CommunityBot | ||
| S Aug 1, 2013 at 10:50 | history | notice removed | CommunityBot | ||
| Jul 24, 2013 at 13:31 | history | edited | A. Pascal | CC BY-SA 3.0 |
Original question too naive. Revised question based on negative answers.
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| Jul 24, 2013 at 13:00 | answer | added | Mikhail Katz | timeline score: 1 | |
| Jul 24, 2013 at 12:29 | comment | added | Mikhail Katz | As Benoît Kloeckner pointed out, I misunderstood the question. The OP is apparently not exponentiating from the midpoint. My answer is related to showing that the equidistant hypersurface is the image of the exponential map from the midpoint of the segment. | |
| Jul 24, 2013 at 12:25 | comment | added | Benoît Kloeckner | @Misha, katz: I think we did not understand the question in the same way. Do you talk about exponentiating a linear subspace from $m$, or an affine subspace from $p_0$? | |
| Jul 24, 2013 at 12:18 | answer | added | Benoît Kloeckner | timeline score: 2 | |
| Jul 24, 2013 at 10:20 | comment | added | Mikhail Katz | In the complex projective case this follows from the theorem of cosines that can be found in my paper here. Later I noticed that the formula had already been discovered by Shirokov in the 1950s. There is a dual formula in the complex hyperbolic case. | |
| Jul 24, 2013 at 10:17 | answer | added | Mikhail Katz | timeline score: 1 | |
| S Jul 24, 2013 at 9:37 | history | bounty started | A. Pascal | ||
| S Jul 24, 2013 at 9:37 | history | notice added | A. Pascal | Draw attention | |
| Jul 20, 2013 at 13:31 | comment | added | Misha | This is clearly true in real-hyperbolic case; you should check complex-hyperbolic case, where bisectors are explicitly computed (see e.g. Goldman's book). | |
| Jul 20, 2013 at 11:22 | comment | added | Benoît Kloeckner | I do not see why this should be the case. Did you at least check the case of the hyperbolic plane, where everything can be computed easily? | |
| Jul 20, 2013 at 9:47 | history | asked | A. Pascal | CC BY-SA 3.0 |