I am a beginner studying knot theory, and we covered the Reidemeister moves on link diagrams in class today. The question in the title is the one I am struggling with now. I attached an image of the solution to a similar problem (for diagrams with only one crossing) that was provided in class.
(I originally only had the first two diagrams drawn but was told in class that the correct answer included the additional two.)
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. The logic given for the solution in the image was that a ‘rotation is not a Reidemeister move,’ so by this logic, should I be including four additional diagrams? Any help is greatly appreciated. Thank you!
It looks like your are suppose to approach this problem purely from a combinatorial viewpoint. That means, in my opinion, treat crossings like 4-valent vertices of a graph. Then you have to connect the edges up without adding more vertices. Then once you have all the graphs, turn them into knot diagrams by converting the vertices to crossings, which there are two ways for each vertex, of course.
Now, you have found all the ways to do this for a single crossing. But you can recreate your proof by first finding the two ways to connect your graph's edges up and then turning each of those into the two different knot diagrams. I mention this because it will be easier for you to draw all graphs with two crossing and then turn them into the different knot diagrams afterwards. Thus, your problem turns into creating all the ways to connect up two vertices like the ones below. Then deal with your Reidemeister moves. You can solve this problem while ignoring all the things like planar rotations and things, but it will be a little less elegant. But it simplifies the arguments you need to make.