Timeline for Stationary curves on homogeneous spaces
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 27, 2014 at 16:28 | vote | accept | Benjamin | ||
| Jan 27, 2014 at 14:31 | answer | added | Robert Bryant | timeline score: 2 | |
| Jan 27, 2014 at 1:41 | comment | added | Benjamin | And yes, I would like to assume non-degeneracy. Sorry for lack of clarity. | |
| Jan 26, 2014 at 11:03 | history | edited | Benjamin | CC BY-SA 3.0 |
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| Jan 25, 2014 at 22:03 | comment | added | Benjamin | Yes, I mean only the images of the curves coincide | |
| Jan 25, 2014 at 22:02 | history | edited | Benjamin | CC BY-SA 3.0 |
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| Jan 25, 2014 at 20:52 | comment | added | Robert Bryant | Are you at least going to assume that $\mathcal{L}^2:TM\to\mathbb{R}$ is nondegenerate? Also, I suppose that you mean only to ask that the stationary curves will be of the desired form up to reparametrization, since, otherwise, your desired form could never hold for all stationary curves. | |
| Jan 25, 2014 at 18:09 | comment | added | Benjamin | I should have said $\lambda > 0$. | |
| Jan 25, 2014 at 17:53 | comment | added | Benjamin | Let's say initially that absolute homogeneity is not guaranteed. I guess I'm asking what the needed invariance properties are. | |
| Jan 25, 2014 at 17:02 | comment | added | Robert Bryant | Are you assuming any invariance properties of the Lagrangian, e.g., that $\mathcal{L}$ is invariant under the induced action of $G$ on $TM$? Also, would you rather have $\mathcal{L}(\lambda v) = |\lambda|\mathcal{L}(v)$ for all $\lambda\in\mathbb{R}$ and $v\in TM$? (Without the absolute value sign, the value of the action will depend on the orientation of the curve.) | |
| Jan 25, 2014 at 15:18 | history | asked | Benjamin | CC BY-SA 3.0 |