2015年10月13日 · This means that the boundary torus is not incompressible. So there is a simple closed curve on the torus that bounds a disk into the complement. Furthermore, the only such …

2023年10月26日 · The question I am looking to answer now: Prove that every knot diagram with two crossings is equivalent to the unknot. So far, I have drawn four possible knot diagrams …

2019年8月19日 · The obvious presentation of the unknot has no crossings, so you're taking the determinant of a $0\times0$ matrix, which equals $1$. The Alexander polynomial $\Delta$ is a …

2020年4月3日 · Unknot: embedded vs immersed bounding disk. Ask Question Asked 4 years, 10 months ago. Modified 4 years ...

2019年9月19日 · If you really want to think about this in the standard unknot in $\Bbb R^3$, it starts with the obvious disc, then the disc moves upward to look like a spherical cap; …

2018年1月14日 · When I was reading in introduction chapter in composite knot subchapter, I came up with idea that if we somewhat composite 2 or more factor knots each another under …

2020年4月25日 · $\begingroup$ Note that Khovanov homology (which is a refinement of the Jones polynomial), does detect the unknot. $\endgroup$ – Moishe Kohan Commented Apr 25, …

2012年4月28日 · Choose a clasper decomposition of the knot or link (replacing it by the unknot, with some tangled web of claspers inside it), identify moves on claspers induced by moves on …

2019年12月8日 · I've come across the fact that the only knot whose knot group is isomorphic to $\mathbb{Z}$ is the unknot; ...

2022年12月15日 · A contender for a counter example was conjectured to have to fail a disc at each point thus a contender would be the Bing Sling. However this condition is not sufficient …