Take a clock-wise 3-flip mobius strip,

  1. Cut it down the middle and then let the ribbon cross itself 6 times.
  2. This forms a framed knot of which there are many.
  3. Get the knot diagram.

I've found that there is only one knot, which is a perfect braid, but the rest of the group are either figure of 8 knots with two loops, or trefoil knots with 3 twists.

Seeing the all knots have the same etiology, how many are there and prove it. I've found about 80 with paper models.

See my video

link text

  • 3
    $\begingroup$ The tag Mobius-inversion means something else. $\endgroup$ – Jim Conant Oct 21 '12 at 15:33

Let's try to make your question a little more precise. I am going to restate your questions with terminology slightly more friendly to mathematicians, if that's okay. You start with a Mobius band with 3-twists in it. If you mark the center of your band, it is embedded in space as an unknotted circle. Now consider the boundary of your band. As you note in the video, the band has one component so the boundary is also an embedded knot. Drawing a projection of the knot to the plane you can see it is an alternating knot with 3 crossings and therefore must be the trefoil. (That last observation relies on starting with the core of the band being an unknotted curve. If the core of your band is embedded in space as an non-trivial knot then you will get more interesting knots in this projection.)

You then cut along the center of the band. Notice this creates two boundary components. Your band is now two sided as you note. Also, the two boundary components of the band and the core of the band are all isotopic in space, that is you can move either boundary component to the core without having the knot cross itself. Also, the original boundary component is isotopic to one of the one of the boundary components you now have. The side that is coming from the core is not isotopic to the trefoil because the cutting operation creates two boundary points from one point on the core and isotopy is a one to one mapping.

After cutting, you observe that your band is now a trefoil. (In your video, I think you got a right handed trefoil. You will get a trefoil of the opposite handedness if you switched your 3 twists and the start of this process.)

Your final step is that you allow the band to cross itself up to 6 times. Strictly speaking the core your band is no longer an embedded circle instead it is a embedded graph in space with 6 degree 4 vertices. I am going to assume that you choose a resolution here so the core of your band froms a positive or negative crossing. (At least, it looks like you do this in the video.) So the result is knot with up to 9 crossings. 3 from the initial trefoil, then 6 more by your assumption. What you now try to do is classify knots up to ambient isotopy. This problem is well understood. There are exactly 100 knots with 9 or fewer crossings. You can see all of them here: http://katlas.math.toronto.edu/wiki/The_Rolfsen_Knot_Table

So there is a definite upper bound of 100 knot types (including the unknot) you should see. Depending on how you do your resolutions you may also be affecting the writhe of your knot projection. Again, this step is unclear to me from the video. However, if you are making a projection where the writhe is restricted in some way you may cut down the number of possible knots. The key question is how may Type 1 Reidemeister moves do you allow while adding the 6 crossings. (Writhe and Reidemeister moves have wikipedia entries.)

Hope that helps!


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