Prove that the shadow of any link diagram can be checkerboarded (2 colored).
Is there an elementary proof for this result (not requiring advanced theory)
2 Answers
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An argument I rather like for this as follows. Replace each crossing with an arbitrary smoothing:
After these smoothings, the shadow diagram is a collection of non-intersecting simple closed curves, possibly nested. Such a situation is $2$-colorable: it follows from the Jordan curve theorem that, if you start in a region and count how many times you cross a curve to get to the outermost region, no matter the path you took it will always be the same modulo $2$. Color the regions according to the modulus. (I'm just saying a reason why it is you can start coloring regions without worry.)
Now, unsmooth the crossings.
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The shadow of an $n$-component link diagram can be obtained from the standard, zero-crossing diagram of the $n$-component unlink via a sequence of shadows of Reidemeister moves. This diagram of the $n$-component unlink is checkerboard colorable (color the insides of the components black and the unbounded face white).
One can show that a checkerboard coloring before a shadow of a Reidemeister move induces a checkerboard coloring after the shadow of a Reidemeister move. Therefore, any link diagram is checkerboard colorable.
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$\begingroup$ is there any argument that works without involving Reidemeister moves? $\endgroup$J. Allen– J. Allen2018-03-01 01:51:48 +00:00Commented Mar 1, 2018 at 1:51
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$\begingroup$ Here is an argument that does not require Reidemeister moves. Try to start coloring the faces of the link diagram in a checkerboard fashion by picking one face to color black. Then color all the adjacent faces white, and then all their adjacent faces black, and so on. If you end up with a coloring that is not proper (e.g. you've forced a black face to be adjacent to another black face), then there is a loop on your link diagram that intersects the diagram only in transverse double points in an odd number of points. The Jordan curve theorem can be used to show this is impossible. $\endgroup$Adam Lowrance– Adam Lowrance2018-03-01 01:57:29 +00:00Commented Mar 1, 2018 at 1:57
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$\begingroup$ What do you mean by "there is a loop on your link diagram that intersects the diagram only in transverse double points in an odd number of points"? $\endgroup$J. Allen– J. Allen2018-03-01 02:08:06 +00:00Commented Mar 1, 2018 at 2:08
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$\begingroup$ Suppose you are forced to color two adjacent faces black. Take a point in one of the black faces, draw a path to the adjacent black face, then continue that path through other faces, alternating black, white, black, white, etc. until you get back to the starting point. So the colors of the faces you are in are: black, black, white, black, white, ... ,white, black. Every time the path goes to a new face, you have an intersection point. So there is an odd number of intersection points of this path and the link diagram (which can't happen). $\endgroup$Adam Lowrance– Adam Lowrance2018-03-01 02:13:09 +00:00Commented Mar 1, 2018 at 2:13
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$\begingroup$ thank you so much I understand now $\endgroup$J. Allen– J. Allen2018-03-01 02:14:36 +00:00Commented Mar 1, 2018 at 2:14