This is a minor curiosity that I've been wondering about. Suppose that we draw a closed curve in the plane and that this curve intersects itself several times, but never twice in one spot. We can knot the curve by traveling along the curve and assigning over- and under-crossings alternately as we reach each intersection. It seems, as I have been told, that this will never cause a contradiction, i.e. under and over crossings will always match up. Why is this case?

I am not too familiar with know nothing about knot theory. If there is a non-technical proof of this result, or perhaps a proof involving graph theory, that would be wonderful.

Below are some diagrams to illustrate what I mean


A trefoil knot formed by assigning alternate over and under crossings.

alternating knot

A more complicated alternating curve.

  • $\begingroup$ I take the question to mean, how do you know that the second time you come to a crossing you're guaranteed to be due for an over if you previously did an under there, and for an under if you did an over? $\endgroup$ – Gerry Myerson Oct 8 '12 at 2:57
  • $\begingroup$ @Gerry Myerson Precisely. Sorry if the question wasn't formulated well. $\endgroup$ – EuYu Oct 8 '12 at 2:58
  • 1
    $\begingroup$ A proof is given in the answer to this very similar question: math.stackexchange.com/questions/182971/… $\endgroup$ – Jack Schmidt Oct 8 '12 at 3:29
  • $\begingroup$ @JackSchmidt Wow, that's quite the beautiful proof. Thank you for directing me to that question. $\endgroup$ – EuYu Oct 8 '12 at 3:34
  • $\begingroup$ @Gerry: oh. Somehow I did not understand the significance of the word "alternating." $\endgroup$ – Qiaochu Yuan Oct 8 '12 at 3:44

Consider the loop formed by starting at some crossing and following a strand until you get back to that crossing. The loop may cross itself, and it may be crossed by other strands. Each time it crosses itself, it does so twice (once over, once under). Each starnd that crosses the loop does so twice. So all told, the loop does an even number of crossings between visits to the starting crossing. Thus, it always winds up being due for the correct kind of crossing when it gets back to the start.

  • $\begingroup$ A very nice parity argument. Thank you Gerry. $\endgroup$ – EuYu Oct 8 '12 at 5:06

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