Timeline for Are there CAT(-1) spaces which are not trees whose Gromov boundary is disconnected?
Current License: CC BY-SA 3.0
11 events
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| Mar 19, 2015 at 3:01 | review | Close votes | |||
| Mar 19, 2015 at 9:27 | |||||
| Mar 15, 2015 at 2:21 | comment | added | Sam Nead | This is homework; the question should be moved. As a hint - you should think of examples of CAT(-1) spaces and their boundaries. Then think about how you can cut spaces into pieces (or glue spaces together) and how the boundary changes under those operations. | |
| Mar 14, 2015 at 23:52 | review | Close votes | |||
| Mar 15, 2015 at 1:22 | |||||
| Mar 14, 2015 at 23:31 | comment | added | YCor | oh, I saw "totally disconnected". Otherwise it's even much easier, just take a wedge of two copies of $H^2$, then the Gromov boundary is a disjoint union of 2 circles. | |
| Mar 14, 2015 at 17:19 | answer | added | Anton Petrunin | timeline score: 0 | |
| Mar 14, 2015 at 17:03 | answer | added | Igor Rivin | timeline score: 2 | |
| Mar 14, 2015 at 16:59 | comment | added | Igor Rivin | @YCor How is a point disconnected? | |
| Mar 14, 2015 at 15:16 | review | Low quality posts | |||
| Mar 14, 2015 at 15:55 | |||||
| Mar 14, 2015 at 15:08 | comment | added | YCor | Second example (if you don't want something quasi-isometric to a tree): consider a horodisc in the hyperbolic plane. Then it's CAT($-1$) and its Gromov boundary is reduced to a point. | |
| Mar 14, 2015 at 15:05 | comment | added | YCor | Yes: let $T$ be (the 2-skeleton of) an equilateral triangle in the hyperbolic plane $H^2$. Consider two copies of $T$ glued on their vertices, and take the universal covering, with the length metric. Then it is obviously QI to a tree and CAT($-1$), but not isometric to a tree. | |
| Mar 14, 2015 at 15:00 | history | asked | Yellow Pig | CC BY-SA 3.0 |