Timeline for The fundamental group of a $3$-manifold with a boundary of genus $>0$
Current License: CC BY-SA 3.0
9 events
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| Dec 17, 2017 at 20:56 | comment | added | Ian Agol | @AnubhavMukherjee: I'm guessing what I meant was that if $M$ is reducible, then a connect summand has non-trivial fundamental group by the Poincaré conjecture. Hence, one may assume that the manifold is irreducible, and thus aspherical. (If there's a non-separating 2-sphere, then there's and $S^2\times S^1$ connect summand.) | |
| Dec 17, 2017 at 19:33 | comment | added | Anubhav Mukherjee | @IanAgol I have a question regarding your first comment. I can't see why $g>1$ implies $M$ is aspherical? For example if we start with $S^3$ and now consider a two sided sphere and scoop out two solid $S_g$ from both side, then the resulting manifold is not aspherical. Am I making any mistake? Or did I get your comment wrong? Thanks. | |
| Jul 23, 2013 at 16:46 | vote | accept | aglearner | ||
| Jul 23, 2013 at 15:59 | history | edited | Misha | CC BY-SA 3.0 |
added 212 characters in body
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| Jul 23, 2013 at 15:57 | comment | added | Misha | @IanAgol: Right, I should have thought about it, I will modify the answer. | |
| Jul 23, 2013 at 14:38 | comment | added | Ian Agol | Also, any finite-index subgroup of $\mathbb{Z}^2$ will induce an isomorphism of $H_1(;\mathbb{Q})$, giving an immediate contradiction. | |
| Jul 23, 2013 at 14:33 | comment | added | Ian Agol | In the case that $g>1$, one may use the fact (by the sphere theorem) that $M$ is aspherical, and $\chi(M)=\frac12 \chi(\partial M)$. | |
| Jul 23, 2013 at 13:09 | comment | added | aglearner | Misha, thank you very much for writing the detailed answer! | |
| Jul 23, 2013 at 12:59 | history | answered | Misha | CC BY-SA 3.0 |