Timeline for Can distinct meridians commute in a knot group?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 26, 2023 at 17:30 | comment | added | Calvin McPhail-Snyder | I was hopeful that, say, writing $K$ as a braid closure would be enough but it looks like that isn't the case. | |
| Apr 26, 2023 at 17:21 | comment | added | Andy Putman | Yeah, I don't think there is any simple condition on the knot diagram that ensures that two particular generators in the Wirtinger Presentation are distinct. As far as the general condition about 3-manifold groups, what it follows from is the classification of abelian subgroups of 3-manifold groups, which can be found in Hempel's book. | |
| Apr 26, 2023 at 17:20 | comment | added | Calvin McPhail-Snyder | The goal of the distinct arcs condition was to try to guarantee that $x_1$ and $x_2$ are actually different elements of $\pi(K)$ but I think it might be a little harder to ensure this. | |
| Apr 26, 2023 at 17:19 | comment | added | Calvin McPhail-Snyder | Thanks! That's quite helpful. I thought there was some fact like this about 3-manifold groups but didn't know it precisely. | |
| Apr 26, 2023 at 13:42 | comment | added | Andy Putman | (it is easy to find examples where they are the same — take a diagram and do a bunch of silly Reidemeister moves to complicate it unnecessarily) | |
| Apr 26, 2023 at 13:38 | comment | added | Andy Putman | They will always commute if it’s the diagram of a trivial knot. But I think this points to the general phenomenon: in a 3-manifold group, two infinite-order elements commute if and only if they either lie in a common cyclic subgroup or they are in the image of a pi_1-injective torus. Since your loops are meridians, thinking about the fact that they both map to the same generator of H_1 implies that they will commute if and only of they are identical elements of pi_1. | |
| Apr 26, 2023 at 13:02 | history | asked | Calvin McPhail-Snyder | CC BY-SA 4.0 |