Timeline for (1-Lipschitz) + (length-preserving) = isometry
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Aug 21, 2018 at 7:40 | answer | added | Wlod AA | timeline score: 0 | |
| Aug 18, 2018 at 18:31 | answer | added | Hee Kwon Lee | timeline score: 1 | |
| Aug 9, 2011 at 21:04 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Aug 9, 2011 at 21:03 | comment | added | Anton Petrunin | @Pietro: I forget to say "closed"; now it is corrected | |
| Aug 9, 2011 at 20:58 | comment | added | Pietro Majer | I still miss something. What e.g. if $\alpha=[0,\pi]$, $\beta$ is the upper half-circle of radius 1 and $f(t)=e^{it}$ ? This $f$ is certainly a 1-Lipschitz bijection between two simple convex curves of length $\pi$, but it is not an isometry (wrto the Euclidean distance). | |
| Jul 31, 2011 at 9:02 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Jul 30, 2011 at 7:35 | comment | added | Anton Petrunin | @Zarathustra, for higher dimension one has to find right formulation (simple change length to area makes the statement trivial). I need only 2-dimensional case in the proof of rigidity case of Reshetnyak Majorization Theorem. | |
| Jul 30, 2011 at 3:04 | comment | added | Zarathustra | Does higher dim. analog holds true? | |
| Jul 29, 2011 at 18:04 | answer | added | Arseniy Akopyan | timeline score: 5 | |
| Jul 29, 2011 at 17:47 | answer | added | Gjergji Zaimi | timeline score: 4 | |
| Jul 29, 2011 at 12:49 | comment | added | Deane Yang | So "1-Lipschitz" here means $d(f(x),f(y)) \le d(x,y)$ for any $x, y \in \alpha$, where $d$ is the Euclidean distance function? I always think of "Lipschitz" as a local property, but I guess in metric geometry it is really a global condition. | |
| Jul 29, 2011 at 12:35 | comment | added | Tapio Rajala | Deane, when using the Euclidean metric for example an ellipse and a circle with the same length do not satisfy the assumptions because the diameter of the ellipse is larger than the diameter of the circle and hence there is no 1-Lipschitz bijection. | |
| Jul 29, 2011 at 12:02 | comment | added | Deane Yang | Tapio, if the map $f$ is defined only as a map from $\alpha$ to $\beta$, then the theorem is trivially true using the length metric and false using the Euclidean metric because any two simple convex curves with the same length satisfy the assumptions of the theorem. So the theorem makes sense to me only if $f$ more than just a map between the curves. Or am I missing something here? | |
| Jul 29, 2011 at 10:31 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Jul 29, 2011 at 9:34 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Jul 29, 2011 at 6:56 | comment | added | Tapio Rajala | @Deane: they are isometric in the length-metric, but not in the original Euclidean metric of $\mathbb{R}^2$. Also, if I understand this correctly the formulation given by Anton is the correct one: the bijection is required between $\alpha$ and $\beta$. The convexity assumption is the key point, as otherwise the claim would be trivially false (you could for example "turn bumps from the outside to the inside" with a non-convex curve). | |
| Jul 28, 2011 at 15:55 | comment | added | Deane Yang | I'm rather confused. Aren't any two curves of the same length always isometric? Do you really mean that $f$ is a 1-Lipschitz bijection of $\mathbb{R}^2$ to itself and not just of $\alpha$ to $\beta$? Or something else? | |
| Jul 28, 2011 at 11:44 | answer | added | Benoît Kloeckner | timeline score: 3 | |
| Jul 28, 2011 at 11:17 | history | asked | Anton Petrunin | CC BY-SA 3.0 |