Timeline for The action of the mapping class group of a punctured disk on the boundary at infinity of the universal cover
Current License: CC BY-SA 3.0
21 events
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| Sep 11, 2017 at 7:43 | vote | accept | azureai | ||
| Sep 9, 2017 at 7:38 | answer | added | Sam Nead | timeline score: 2 | |
| Sep 9, 2017 at 5:27 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 9, 2017 at 0:46 | comment | added | azureai | How about what I proposed in my second edit. Do the deleted object and $\mathbb{D}_n$ have the same mapping class group? | |
| Sep 8, 2017 at 21:30 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 8, 2017 at 20:42 | history | edited | azureai |
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| Sep 8, 2017 at 18:38 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 8, 2017 at 18:23 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 8, 2017 at 8:14 | comment | added | YCor | $\pi_1(D_n)$ is not QI to $\tilde{D}_n$ because the former is a free group (so has 0, 2 or $\infty$ ends), while the second one is the hyperbolic space and hence has 1 end. | |
| Sep 8, 2017 at 1:36 | comment | added | Lee Mosher | 2) is mis-copied from earlier in the question where it asks whether $\pi_1(\mathbb{D}_n)$ is quasi-isometric to $\widetilde{\mathbb{D}}_n$. However, that's not really a sensible question either. It is certainly sensible to ask whether the orbit map $\pi_1(\mathbb{D}_n) \to \widetilde{\mathbb{D}}_n$ of the action is a quasi-isometry, but the answer is "no" because the action is not cobounded. | |
| Sep 7, 2017 at 23:25 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 7, 2017 at 23:19 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 7, 2017 at 23:13 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 7, 2017 at 23:08 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 7, 2017 at 23:02 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 7, 2017 at 22:56 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 7, 2017 at 22:22 | comment | added | YCor | $\pi_1(\mathbb{D}_n)$ is free on $n$ generators, so has infinitely many ends for $n\ge 2$, while $\mathbb{D}_n$ has finitely many ends ($n+1$). So they are not quasi-isometric. For $n=1$, $\mathbb{D}_1$ (punctured disk) can't be endowed with a hyperbolic metric for which both ends are cusps. Anyway, for any complete hyperbolic metric, it will have exponential growth, while the fundamental group $\mathbf{Z}$ has linear growth, so again they are not QI for $n=1$ as well (a similar argument works for $n=0$). | |
| Sep 7, 2017 at 22:17 | history | edited | azureai | CC BY-SA 3.0 |
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| Sep 7, 2017 at 22:16 | history | edited | YCor |
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| Sep 7, 2017 at 22:00 | review | First posts | |||
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| Sep 7, 2017 at 21:56 | history | asked | azureai | CC BY-SA 3.0 |