Timeline for distance regular metric spaces
Current License: CC BY-SA 2.5
19 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2010 at 13:58 | vote | accept | Dima Fon-Der-Flaass | ||
| Jan 5, 2010 at 13:58 | vote | accept | Dima Fon-Der-Flaass | ||
| Jan 5, 2010 at 13:58 | |||||
| Jan 5, 2010 at 13:58 | vote | accept | Dima Fon-Der-Flaass | ||
| Jan 5, 2010 at 13:58 | |||||
| Jan 5, 2010 at 7:03 | history | edited | Dima Fon-Der-Flaass | CC BY-SA 2.5 |
summary of the week's activity. Open problem.
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| Dec 29, 2009 at 18:49 | answer | added | fedja | timeline score: 4 | |
| Dec 29, 2009 at 6:38 | answer | added | Tom LaGatta | timeline score: 0 | |
| Dec 28, 2009 at 20:45 | history | edited | Dima Fon-Der-Flaass | CC BY-SA 2.5 |
added 250 characters in body
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| Dec 28, 2009 at 20:32 | comment | added | Dima Fon-Der-Flaass | I should have mentioned that the values of the distance are all non-negative reals. Ohterwise, all distance regular graphs (and they are many) would serve as examples. | |
| Dec 28, 2009 at 20:01 | comment | added | Nurdin Takenov | May be it should be mentioned that V is not finite, because we can take the vertices of a octahedron as V, and then p(\sqrt(2),1,1)=4. Also p(1,1,\sqrt(2))=2, i.e. not equal to (\sqrt(2),1,1). | |
| Dec 28, 2009 at 19:08 | answer | added | Anton Petrunin | timeline score: 8 | |
| Dec 28, 2009 at 18:34 | comment | added | Anton Petrunin | Yet an other example is Minkowski plane (i.e. $\mathbb R^2$ with metric induced by norm). | |
| Dec 28, 2009 at 18:20 | history | edited | Dima Fon-Der-Flaass | CC BY-SA 2.5 |
clarification added
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| Dec 28, 2009 at 18:19 | comment | added | macbeth | Oh, I see, it's the function p that you care about. | |
| Dec 28, 2009 at 18:16 | comment | added | Dima Fon-Der-Flaass | Heather: yes, this is another example. But the parameters p(a,b,c) are the same. | |
| Dec 28, 2009 at 17:57 | comment | added | macbeth | What about the hyperbolic plane? It's symmetrical enough for distance-regularity. As for your condition that any triangle-inequality-satisfying triple (a, b, c) be realized as a triangle: I think it holds, since we can achieve c = b-a and c = b+a, and thus, since distances vary continously with pairs of points, everything in between. | |
| Dec 28, 2009 at 16:18 | comment | added | Martin M. W. | Oh, OK. Good point. | |
| Dec 28, 2009 at 16:16 | comment | added | Dima Fon-Der-Flaass | Does not work. The number p(1, 1/2,1/2) is not defined: if we take points B,C at distance 1 in the same plane, there is a midpoint, if in different planes, there is none | |
| Dec 28, 2009 at 16:13 | comment | added | Martin M. W. | If I understand correctly, there exist other examples. E.g., as a subset of Euclidean 3-space, take the union of the planes where z=0 and z=1. Maybe you could add the hypothesis that the space is connected? | |
| Dec 28, 2009 at 15:22 | history | asked | Dima Fon-Der-Flaass | CC BY-SA 2.5 |