Timeline for Equidistant hypersurfaces in symmetric space via exponentiation?
Current License: CC BY-SA 3.0
12 events
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| Aug 1, 2013 at 10:50 | history | bounty ended | CommunityBot | ||
| Jul 24, 2013 at 14:20 | comment | added | Benoît Kloeckner | You should try the case of $\mathbb{H}^2\times \mathbb{R}$: it is the simplest higher-rank case, and appears anyway inside every other ones. Moreover it is simple enough that all computation can be done plainly (which does not mean without some effort,). | |
| Jul 24, 2013 at 14:19 | comment | added | A. Pascal | OK. Thank you for the clarification. Katz below suggests another argument for the revised claim using Leuzinger's trigonometry. It seems to make sense, but perhaps it too needs tempering. | |
| Jul 24, 2013 at 14:14 | history | edited | Benoît Kloeckner | CC BY-SA 3.0 |
removed unfunded claims
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| Jul 24, 2013 at 14:11 | comment | added | Benoît Kloeckner | (continued) The claim in the second paragraph should hold in rank 1 symmetric spaces, but I am not sure for higher rank ones except in very specific cases. You should try the case of $\mathbb{H}^2\times \mathbb{R}$: it is the simplest higher-rank case, and appears anyway inside every other ones. | |
| Jul 24, 2013 at 14:10 | comment | added | Benoît Kloeckner | @A.Pascal: I have probably been quite optimistic, so I tempered my claim a bit. I will try to think, but in Hadamard manifolds this seems too bold as the bissector is a global object and should not in general be constructed in such a local way. | |
| Jul 24, 2013 at 14:08 | history | edited | Benoît Kloeckner | CC BY-SA 3.0 |
deleted 4 characters in body
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| Jul 24, 2013 at 13:10 | comment | added | A. Pascal | line through the tangent space at $p_{0}$ in the direction of the geodesic. This is the intuition I had for my original question, but I guess that translation in the tangent space giving an affine hyperplane is simply not compatible with translation in the symmetric space after exponentiating from the tangent space. | |
| Jul 24, 2013 at 13:07 | comment | added | A. Pascal | Thanks. Could you expand on the argument in the 2nd paragraph, or provide a reference? I'm just a novice with Riemannian geometry and although I find this intuitive, I don't see the argument. I would also like to know if the following works: extend the geodesic $[p_{0},p]$ to $[-p,p]$ with midpoint $p_{0}$. Compute the bisector at $p_{0}$ in the way you described and then translate it along the geodesic $[-p,p]$ until you get to the midpoint of $[p_{0},p]$. Is that the bisector between $[p_{0},p]$? By translate, I think I mean using a 1 parameter subgroup gotten by exponentiating the... | |
| Jul 24, 2013 at 12:27 | comment | added | Mikhail Katz | You are right, I misunderstood the question (didn't pay enough attention to the "affine" part). | |
| Jul 24, 2013 at 12:23 | history | edited | Benoît Kloeckner | CC BY-SA 3.0 |
Changed slightly the concluding argument.
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| Jul 24, 2013 at 12:18 | history | answered | Benoît Kloeckner | CC BY-SA 3.0 |