Timeline for Is the cut locus of a generic point in a hyperbolic manifold a generic polyhedron?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 29, 2010 at 21:17 | vote | accept | Bruno Martelli | ||
| Oct 28, 2010 at 9:01 | answer | added | Roberto Frigerio | timeline score: 7 | |
| Oct 27, 2010 at 10:31 | answer | added | Bruno Martelli | timeline score: 2 | |
| Oct 27, 2010 at 9:40 | comment | added | Bruno Martelli | Thanks Mohan, I have quickly looked at Buchner's papers. He has defined a notion of "cut-stable" metric; as far as I have understood, these metrics form an open (dense?) set, and the cut locus on a generic point for these "cut-stable" metrics has links of some prescribed types. He described these types in dimension 2 and 3. They are slightly more general than simple, but this cannot be avoided: for instance 1-valent vertices are necessary. However, 1-valent vertices cannot occur on a hyperbolic surface. | |
| Oct 27, 2010 at 2:25 | answer | added | Anton Petrunin | timeline score: 1 | |
| Oct 26, 2010 at 22:45 | comment | added | Dylan Thurston | @Joseph: This doesn't address Bruno's main question, since nothing says the simplicial complex must be simple. | |
| Oct 26, 2010 at 20:21 | comment | added | Joseph O'Rourke | Following Mohan's lead, the paper by Buchner entitled "Simplicial structure of the real analytic cut locus" [Proc. Amer. Math. Soc. 64 (1977), no. 1, 118–121.] "establishes that the cut locus of a compact real analytic Riemannian manifold of dimension $n$ is homeomorphic to a finite $(n-1)$-dimensional simplicial complex." | |
| Oct 26, 2010 at 19:18 | comment | added | Mohan Ramachandran | Bruno:Have you looked at the papers of Michael Buchner.He showed the cut locus of any compact real analytic manifold with real analytic metric is subanalytic. He also classified the cut loci in the generic situation upto dimension six. | |
| Oct 26, 2010 at 19:12 | comment | added | Bruno Martelli | A 1-dimensional polyhedron is simple if it is a 3-valent graph (or a circle). Vertices have valence 2 or 3. I was trying unsuccessfully to embed a picture in the text... I will try again later (probably tomorrow morning) | |
| Oct 26, 2010 at 19:07 | history | edited | Bruno Martelli | CC BY-SA 2.5 |
No, it didn't work. I explain with words
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| Oct 26, 2010 at 18:20 | comment | added | Joseph O'Rourke | @Bruno: IF $C(p)$ is the cut locus from $p$, $M \setminus C(p)$ is homeomorphic to an open ball. So on a 2-manifold, $C(p)$ is a tree. Would you call a tree a 1-dimensional polyhedron? Just trying to undertand your terminology... | |
| Oct 26, 2010 at 18:05 | history | asked | Bruno Martelli | CC BY-SA 2.5 |