Timeline for In hyperbolic 3-orbifold with totally geodesic boundary case, is it true: rank(the fundamental group of boundary M)< or equal 2 rank(fundmental group of M)?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 6, 2011 at 16:53 | vote | accept | Lin Jianfeng | ||
| Oct 6, 2011 at 5:59 | history | edited | Lin Jianfeng | CC BY-SA 3.0 |
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| Oct 6, 2011 at 5:58 | comment | added | Lin Jianfeng | I am very sorry, my question is a little misleading. I should not use the term "hyperbolic oblifold". But a "hyperbolic obrifold with totally geodesic boundary". That means the boundary is a totally geodeic 2-hyperbolic orbifold. we can double it to get a really closed orbifold. We can assume that the boundary is connected. | |
| Oct 6, 2011 at 5:17 | answer | added | Ian Agol | timeline score: 3 | |
| Oct 6, 2011 at 3:41 | comment | added | Ian Agol | What if the boundary is disconnected? Do you want to take the sum of the ranks of the boundary components? This would work in the torsion-free case by your observation. | |
| Oct 5, 2011 at 19:30 | answer | added | Igor Rivin | timeline score: 1 | |
| Oct 5, 2011 at 17:12 | comment | added | Lin Jianfeng | rank(G) is the least number of elements in G that can generate G. | |
| Oct 5, 2011 at 17:05 | comment | added | Igor Rivin | What do you mean by "rank"? | |
| Oct 5, 2011 at 16:59 | history | edited | Lin Jianfeng | CC BY-SA 3.0 |
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| Oct 5, 2011 at 16:44 | history | asked | Lin Jianfeng | CC BY-SA 3.0 |