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The closure of a rational $4$-tangle is a rational knot. But is the converse true? We could tangle up even the unknot to a hopeless mess before cutting it up, and we could cut it were it "hurts most" (where we need to do a Reidemeister-$3$ to unknot).

Note that the analogous statement already for algebraic knots is "No", as it seems to me - $8_{16}$ is algebraic but has a crossing (the leftmost on KnotAtlas) that, if cut out, yields a non-algebraic $4$-tangle.

Thus: Can the closure of a non-rational tangle yield a rational knot?

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The closure of a non-rational four-tangle can yield an unknot.

Here is one family of examples. Start with a nontrivial two-tangle (that is, an arc embedded in a three-ball, which is not boundary-parallel). Double it to obtain a "ribbon four-tangle". One closure gives the double of a non-trivial knot. (So the ribbon four-tangle is not rational.) The other closure (capping off parallel ends) yields an unknot.

Here is a picture of the simplest example:

two-tangle and its double

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  • $\begingroup$ Mind to draw the simplest example? (I'm quite the amateur, also I just believe what I can see :-) $\endgroup$ Mar 3 at 20:33
  • $\begingroup$ I’ve now added a figure. I hope this makes the construction clear. $\endgroup$
    – Sam Nead
    Mar 3 at 21:49
  • $\begingroup$ Yup, THX. I just had to think why the ribbon closure can't be rational (e.g. the nonalternating crossings might crosscancel, so you can't argue "it's nonalternating, so it's nonrational) - that a knot double isn't rational is a deep theorem for me). $\endgroup$ Mar 4 at 8:55

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