Timeline for Parametric Seifert surfaces for parametric families of knots in $\mathbb{R}^3$
Current License: CC BY-SA 4.0
7 events
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| Jan 11, 2020 at 0:24 | comment | added | X1921 | @Ryan Budney: could i please ask you one more question? I was wandering if the result (existence of 1-parametric Seifert surfaces for 1-parametric families of knots) may be true for cablings of torus or hyperbollic knots? Or for certain satellites with additional restrictions on the companions and patterns? Thank you very much. | |
| Nov 2, 2019 at 9:51 | comment | added | X1921 | Aha, yes, I see. Thank you for your answer. | |
| Oct 24, 2019 at 17:58 | comment | added | Ryan Budney | I've been meaning to write a paper on this issue, as it's a bit subtle. The answer depends both on the geometrization of the knot exteriors and your choice of Seifert surface. But at its most basic, when you perform isotopy extension on your 1-parameter family, it gives you a diffeomorphism of the knot exterior. Compatibility means that diffeomorphism is isotopic to one that preserves your Seifert surface. For a fibred hyperbolic or torus knot, this can be done. But for more complicated knots (and seifert surfaces) the answer is nuanced. | |
| Oct 24, 2019 at 11:37 | comment | added | X1921 | Yes, I see the non-trivial case is exactly when the 1-dimensional parameter space is a circle rather than the interval. That was the exact case I was concerned. Thank you for your answer, Ryan. By the way, I would appreciate if you could provide a more explicit description of the notion of compatibility with the symmetry of the knot. Thanks again! | |
| Oct 17, 2019 at 21:04 | comment | added | Ryan Budney | This phenomenon is ultimately another way to see the difference in the homotopy-type between the embedding space of a circle, and the embedding space of a surface. | |
| Oct 17, 2019 at 20:53 | comment | added | Ryan Budney | The answer will depend on your precise formulation, but if your 1-parameter family of knots is literally parametrized by an interval $[0,1]$ then by isotopy extension there is a compatible 1-parameter family of Seifurt surfaces. Your question becomes more interesting if your parameter is a circle, or some higher-dimensional manifold. For the circle, your Seifert surface would have to be compatible with the symmetry of your knot (i.e. diffeomorphism of the knot exterior). There are symmetries that do not have compatible Seifert surfaces. | |
| Oct 17, 2019 at 17:15 | history | asked | X1921 | CC BY-SA 4.0 |