Timeline for Does every hyperbolic, almost-transitive, triangulation of $\mathbb{R}^n$ have boundary homeomorphic with $\mathbb{S}^{n-1}$?
Current License: CC BY-SA 4.0
8 events
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| May 20, 2022 at 18:00 | history | edited | Sam Nead | CC BY-SA 4.0 |
Added the discussion of complex hyperbolic spaces, at Igor's direction.
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| May 20, 2022 at 17:34 | comment | added | Igor Belegradek | Examples of uniform complex hyperbolic lattices? It is in Borel's Compact Clifford-Klein forms of symmetric spaces, Topology 1963, sciencedirect.com/science/article/pii/0040938363900260. And of course one needs qi rigidity for the complex hyperbolic space. | |
| May 20, 2022 at 16:44 | comment | added | Sam Nead | I could not find a reference for the existence of uniform examples! I’ll add it back. | |
| May 20, 2022 at 15:52 | comment | added | Igor Belegradek | As you wrote before the edit, uniform complex hyperbolic lattices aren't qi to real hyperbolic ones so the answer to Q3 is no. This follows from qi rigidity of rank one symmetric spaces. There are other examples, e.g. fundamental groups of some Gromov-Thurston closed negatively curved 4-manifolds are not qi to the real hyperbolic space. | |
| May 20, 2022 at 15:39 | comment | added | Agelos | I added the asphericity condition. Suggestions for additional conditions are welcome, as long as the answer may be helpful towards Question 1. | |
| May 20, 2022 at 15:35 | history | edited | Sam Nead | CC BY-SA 4.0 |
deleted 80 characters in body
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| May 20, 2022 at 15:25 | history | edited | Sam Nead | CC BY-SA 4.0 |
added 95 characters in body
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| May 20, 2022 at 15:16 | history | answered | Sam Nead | CC BY-SA 4.0 |