Timeline for Towards a metric characterization of Euclidean spaces
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 16, 2011 at 19:33 | comment | added | Sergei Ivanov | I assume uniqueness of a line through a pair of points (this implies uniqueness of segments). This assumption is used virtually everywhere in the argument. I don't know how to weaken it. | |
| May 16, 2011 at 4:22 | comment | added | Marcos Cossarini | Is the non branching of geodesics required apart from the existance of a unique segment joining each pair of points? I think that you are assuming that. If not, e. g. how do you get rid of the case $c=1$, $\lambda=0$? It's not that I imagine a counterexample along these lines, since it would look like a tree in which the branchings occur everywhere. | |
| May 16, 2011 at 4:13 | comment | added | Marcos Cossarini | 1. Yes, I meant similarities, not isometries. The emphasis was in the finite vs. infinite subspace. | |
| May 14, 2011 at 16:49 | comment | added | Marcos Cossarini | Note that the Busemann functions here are defined as minus the Busemann functions of Springer's EoM. | |
| May 14, 2011 at 5:55 | history | edited | Sergei Ivanov | CC BY-SA 3.0 |
simplified a part of the argument
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| May 14, 2011 at 5:54 | comment | added | Sergei Ivanov | @Marcos: 1. If you allow only isometries, how do you get rid of spheres and hyperbolic spaces? 2. Yes it is easier to consider only isosceles triangles, I'll edit the text accordingly. 3. Removing local compactness could be possible if one also removes finiteness in the transitivity assumption. The most essential use of local compactness is the existence of asymptotic rays. | |
| May 14, 2011 at 0:29 | comment | added | Marcos Cossarini | I don't see that the triple $(p,x,y)$ realizes all similarity types, but I see that it realizes all types in which $d(p,x)=d(,p,y)$. This is enough to prove that all pairs (oriented line, point outside) are similar. In particular, $(\mathcal l,p)~(\mathcal l^{op},p)$. | |
| May 14, 2011 at 0:09 | comment | added | Marcos Cossarini | Some remarks in my first attempt of reading: From the motivation paragraph, it is perhaps more natural to require that the automorphism group is transitive on each class of isometric finite subspaces, since the experimenter would not know more than this. Also, the Hilbert spaces would be counterexamples. However, in this answer, this possibility is eliminated by the additional hypothesis of local compactness. | |
| May 13, 2011 at 21:09 | history | answered | Sergei Ivanov | CC BY-SA 3.0 |