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.

The original problem asking to "prove that every knot diagram with one crossing is equivalent to the unknot." Along with the student's solution where two infinity shaped knots and two figure eight shaped knots are equivalent to the unknot with a single R1 move.

(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!

  • 2
    $\begingroup$ Your instructor is not correct (although so are you). If you want to use only the first two diagrams in the 1-crossing example, you need to show that there is a PL-homeomorphism of $S^3$ that gives you the equivalent rotation of the diagram. If you have proved that then your solution is complete. $\endgroup$ Commented Oct 24, 2023 at 22:54
  • $\begingroup$ For the 2-crossing version, same thing. Your instructor is presumably looking for a solution that use R2 to bring the two intersection to some "standard" position (say at $(\pm 1,0)$ with the usual diagonal directions) because he/she does not believe in PL-homeomorphism of the plane. After you have done that you just exhaust all possibilities of joining up the 4 segments. $\endgroup$ Commented Oct 24, 2023 at 22:59
  • $\begingroup$ Hi, welcome to Mathematics Stack Exchange! Before posting a question, kindly read how to ask a good question and how to use LaTeX. I hope you enjoy your time being a part of this community! $\endgroup$ Commented Oct 24, 2023 at 23:37

1 Answer 1


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.

one X shape being turned into two different crossings for knot diagrams.

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.

enter image description here


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