Oct 13, 2015 · 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 curve which is null-homologous is the longitude. Therefore, the longitude is null-homotopic and by the disk theorem bounds an embedded disk. Thus the knot is the unknot.

Sep 8, 2015 · Any knot has crossings, and changing enough of them from over to under or vice versa will give us the unknot. (The minimum number of changes is known as the unknotting number .) So to show that we can always get the unknot, all we have to do is show that we can change any crossing we want.

Oct 26, 2023 · 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 and shown they are equivalent to the unknot, but I am unsure how to proceed.

Oct 30, 2019 · In a real world sense, this would probably be time spent unknotting an unknot. In the math sense, this is usually done with number of crossings. There are unknots diagrams which require you to increase the number of crossings before you can simplify it to the standard unknot.

Apr 3, 2020 · Unknot: embedded vs immersed bounding disk. Ask Question Asked 4 years, 11 months ago. Modified 4 years ...

You can blow up and then do handle slides to change the crossings of any component of your link without changing the resulting $3$-manifold. It is a classical fact that every knot can be turned into the unknot by changing some of its crossings.

Nov 17, 2020 · Joel Hass gave some nice lectures about these things here.The main ideas are (1) taking a disk and manipulating it to get a new disk that is a normal surface with respect to the triangulation, therefore there is a spanning disk among the vectors, (2) finding a collection of fundamental vectors whose positive linear combinations span all spanning surfaces, and (3) one of these fundamental ...

Aug 19, 2019 · 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 reparameterization of the Alexander-Conway polynomial $\nabla$ , which is defined by a sort of "recurrence relation" whose base case is $\nabla(\textrm{unknot})=1$ .

Apr 30, 2020 · In fact, I think maybe one could twist the unknot into a string and join the ends together to make a circle. And then form any knot with this circle and it will have unknotting number=1. So it seems like the classification of knots with unknotting number 1 is paradoxically just as difficulat as the classification of all knots!

Apr 25, 2020 · $\begingroup$ Note that Khovanov homology (which is a refinement of the Jones polynomial), does detect the unknot. $\endgroup$ – Moishe Kohan Commented Apr 25, 2020 at 15:32