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This question - so far as I know - has no broader mathematical significance, but it occurred to me a while ago and I haven't been able to make any headway.

Any knot diagram $D$ splits the plane into a finite number of pieces. For example, in a standard diagram for a trefoil (e.g., http://en.wikipedia.org/wiki/File:Trefoil_knot_left.svg), the plane is split into five pieces: one lying "outside the diagram," three "lobes" of the knot, and a central region around which the three "lobes" are arrayed. The outer region has three crossings on its boundary, as does the inner region; each of the lobes only have two crossings each on their boundaries.

A diagram for a more complicated knot may split the plane into many more regions, and one of these regions may have many crossings on its boundary. My question is: Does there exist some $n$ such that for any knot $K$, there is a diagram $D$ of $K$ such that any region created by $D$ has at most $n$ crossings on its boundary?

If the answer is no, a (very soft) question that then arises is: given a knot $K$, how hard is it to find the least $n$ ($=n_K$) such that for some diagram $D$ of $K$, every region created by $D$ has at most $n$ crossings on its boundary? For example, for any knot $K$, we can easily see that $n_K>2$.

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I think you can always ensure there are at most $4$ vertices in all the regions that the knot diagram divides the plane into -- think of grabbing a bit of the knot, stretching it out and making a very dense vertical and horizontal "grid" of lines, overlayed on top of the diagram. If you have the grid fine enough then you get at most 4 vertices on the boundary of any region. –  Ryan Budney Oct 3 '10 at 1:00
Ryan: Wouldn't there also be 5-gons in your construction? e.g. a small square with one of its corners clipped off. Also, an Euler characteristic argument shows that if the graph is large and some regions are 3-gons (or 2-gons or 1-gons), then other regions must be (>4)-gons. (I'm working in S^2 and counting the "region at infinity".) –  Kevin Walker Oct 3 '10 at 13:33
Kevin, yes I seem to have mis-counted. –  Ryan Budney Oct 3 '10 at 15:30
Or maybe even 6-gons, if a grid square contains one of the crossings of the original diagram and the edges incident to the crossing enter and leave the square on adjacent sides. –  David Eppstein Oct 3 '10 at 15:54
Kevin, I used the Euler characteristic in my answer, but all I get out of it is that you can have as many as 8 3-gons with all the other regions being 4-gons. I like to think of it as a 4-gon conclusion. –  Gerry Myerson Oct 3 '10 at 22:31

3 Answers 3

up vote 8 down vote accepted

Colin Adams, Reiko Shinjo and Kokoro Tanaka have a paper (http://arxiv.org/abs/0812.2558) that shows that for any knot you can find a diagram which has only regions with 2, 4 and 5 sides.

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They say it's an open problem as to whether you can always get by with just 3 and 4 sides. They don't mention the possibility that you can always do with 2, 3, and 4 sides; presumably, that too is open (unless someone can make rigorous the hand-waving argument in the answer I posted). –  Gerry Myerson Oct 3 '10 at 22:56
I'm still interested in whether $n$ can be reduced to 4, but this certainly answers my question. Thanks a lot! –  Noah S Oct 4 '10 at 23:16

I'm pretty sure something like the following construction can be used to transform any knot diagram into another diagram for the same knot in which each region has bounded complexity. I didn't draw the over-under relationships at the crossings but that's not a problem. alt text

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I don't understand the construction. It seems to me that you have turned a diagram in which no region has more than 5 vertices to a diagram in which there is a region (the very topmost one) with 8 vertices. –  Gerry Myerson Oct 3 '10 at 4:16
The construction is: isolate each critical point of the drawing by a sequence of vertical line segments, pull off a loop of the diagram and connect one end of the pulled-off loop in a spiral pattern through all the vertical segments and connect the other end of the loop to the center of the spiral. Each region in the resulting diagram has at most two critical points (the part of the loop crossing the spiral prevents more than that) so the eight-sided region in the top of the diagram is the worst case. –  David Eppstein Oct 3 '10 at 5:01
So you are saying that for any graph you can find a diagram in which every region has at most 8 sides? That's good, though it's well short of the alleged best possible value of 4. –  Gerry Myerson Oct 3 '10 at 12:47
I suspect that a more careful version of this argument should get it down to 6. –  David Eppstein Oct 3 '10 at 15:51

The consensus seems to be that the answer is 4. Here's a simple proof that it can't be anything less (that is, that 3 can't always work). It can be arranged that all vertices have degree 4. Let the diagram have $n$ vertices. By Euler, it has $n+2$ regions (including the outside region). Each vertex takes part in 4 regions, so the average number of vertices per region is $4n/(n+2)$. For $n\ge7$, this exceeds 3. If the average region has over 3 vertices, some region must have at least 4.

Now here's a hand-waving argument that 4 is always possible (presented because I don't understand Ryan Budney's argument in the comment on the question). Find a region that has more than 4 vertices. Take two more-or-less opposite strands bounding that region, and do a Type 2 Reidemeister move on them. That will split the offending region into two regions (and a digon), each with fewer vertices than the original. Unfortunately, it will also introduce two new vertices into the region on the other side of each of the two strands, which may move one or both of them over 4. The hand-waving is that if you persist with doing Type 2 moves you will eventually get to all regions having 4 vertices or fewer.

EDIT: I have lost faith in the hand-waving argument. If you start with a pentagram, which as a knot diagram represents a knot with 10 crossings, 10 triangles and 2 pentagons, then, persist as I may, I keep producing at least as many regions with 5 or more sides as I remove whenever I do a Reidemeister.

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