Timeline for Virtually large groups of small rank (related to 3-manifolds)
Current License: CC BY-SA 4.0
8 events
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| May 27, 2021 at 16:09 | comment | added | Moishe Kohan | @lemon314: There is more than one way to compute $\pi_1$ of the Dehn filling in this situation. If you do this naively, then, indeed, you get an extra generator. But, as Sam Need says, if you do the computation in two steps then you only add an extra relator and no extra generators. I do not understand the meaning of "will probably be related to how the inclusion of a boundary is somewhat constrained algebraically." | |
| May 27, 2021 at 15:26 | comment | added | lemon314 | @MoisheKohan, I did mean Seifert-van Kampen in the first sentence, I apologize for the confusion; I will head your advice, but I stand by the initial impression: since S-vK gives a presentation with +1 generators and +1 relations, that this doesn't change the rank of the group will need a reason that will probably be related to how the inclusion of a boundary is somewhat constrained algebraically | |
| May 27, 2021 at 14:48 | comment | added | Moishe Kohan | @lemon314: I think you are quite confused, mixing up homology, fundamental group and who knows what else. The fact that you somehow came with a 3-generator set for a group (more precisely, its abelianization given by the first homology) does not imply that there is no 2-generator set. I suggest, you start by going through a proof that each torus knot group is 2-generated (meaning admits a 2-generator set). | |
| May 25, 2021 at 18:35 | comment | added | Sam Nead | You can think of Dehn filling as consisting of two steps. First attach a (three-dimensional) two-handle and second attach a three-ball (a three-handle). In terms of the fundamental group, the first adds a relation and the second does nothing. Thus Dehn filling causes a quotient in $\pi_1$. | |
| May 25, 2021 at 15:28 | comment | added | lemon314 | If you don't mind me asking half a year later, how is $\pi_1(M)$ obviously 2-generated? Well, how is it obviously a quotient, actually - on the face of it, Mayer-Vietoris will give us three generators, two from $N$ and one from the filling - is there any obvious reason why the rank doesn't rise? Is that related to how the inclusion of the boundary in a 3-manifold obeys the "half lives half dies" rule in $H_1$? | |
| Jan 17, 2021 at 16:39 | comment | added | lemon314 | Thank you for this answer and clarifications. You are right, somehow I convinced myself that $\mathbb{H}^2\times\mathbb{R}$ geometry forces the base orbifold to be of higher genus. I will unconvince myself now, then accept your answer and edit the question accordingly:) | |
| Jan 15, 2021 at 18:17 | history | edited | Moishe Kohan | CC BY-SA 4.0 |
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| Jan 15, 2021 at 18:06 | history | answered | Moishe Kohan | CC BY-SA 4.0 |