Timeline for Is the canonical map from isometry group of a Gromov hyperbolic space to homeomorphisms of its Gromov boundary injective?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2023 at 14:16 | answer | added | YCor | timeline score: 2 | |
| Aug 2, 2023 at 14:09 | comment | added | YCor | Yes, sorry, this is indeed another counterexample. No need to add hair: just the graph $\mathbf{Z}$ is already one. In the other direction, a bounded normal subgroup need not act trivially on the boundary (e.g., modify a bit the metric on the hyperbolic plane so that the isometry group is reduced to $\mathrm{SO}(2)$). I'll post an answer with details. | |
| Aug 2, 2023 at 13:30 | comment | added | YCor | Yes, sorry, this is the other counterexample. No need to add hair: just the graph $\mathbf{Z}$ is already one. I'll post an answer with details. | |
| Aug 2, 2023 at 8:26 | comment | added | John Depp | @YCor : You are saying that if we exclude the case when the boundary of $X$ is a singleton and the action is unbounded , then the map $\Phi$ is injective. Am I correct? This is both for proper and non-proper case, right? Could you explain the proof? However, AGenevois provided a counter-example below in the case when the boundary has exactly two points ! | |
| Aug 2, 2023 at 5:54 | comment | added | YCor | Unless the boundary is a aingleton: if $X$ is proper, the kernel is the "compact radical", i.e., largest compact normal subgroup of $\mathrm{Isom}(X)$ (which exists in this case). In the non-proper case one can still define an obvious natural notion of boundedness in $\mathrm{Isom}(X)$ and I think we get the largest normal bounded subgroup. The exception is when the boundary is a singleton and the action is unbounded. For instance, when $X$ is a horodisc in the hyperbolic plane. This exception can't occur when there are loxodromics / when the action is cobounded. | |
| Aug 2, 2023 at 4:54 | answer | added | AGenevois | timeline score: 2 | |
| Aug 2, 2023 at 2:39 | comment | added | Moishe Kohan | Consider $X$ which is bounded... How about $X={\mathbb R}$? What answer do you get in these examples? | |
| Aug 1, 2023 at 23:30 | history | asked | John Depp | CC BY-SA 4.0 |