Vaughan Jones often quips at the beginning of talks on Planar Algebras (see these lectures, for example) that the worst thing you can say about Planar Algebras is that they have not yet yielded a proof of the 4-color theorem. In this post I’ll sketch how a common “evaluation algorithm” (used by Greg Kuperberg and by Emily Peters, for example) almost proves the 4-color theorem. I believe this (failed) argument is due to Penrose, though I’m taking it from an article of Chmutov, Duzhin, and Kaishev and some notes of John Baez’s. There are some more elaborate attacks (by Kauffman, Saleur, Bar Natan, and probably others) that I won’t discuss at all. This is the second of what hopefully will be a short series of posts on “evaluation algorithms” (the first was on the Jellyfish algorithm).
The outline of the post is as follows. First I’ll explain a standard reduction of the 4-color theorem to a question about 3-coloring edges of trivalent graphs. Second I’ll explain why 3-colorings of edges is a question about finding a positive evaluation algorithm for a certain planar algebra. Third, I’ll discuss “Euler characteristic” evaluation algorithms. Fourth I’ll explain how this technique almost answers the 4-color theorem.
Suppose you have a planar graph and you want to 4-color the faces (note, I said faces, not vertices). First notice that you can reduce to the case of 3-valent planar graphs by “adding a new country at every higher valency vertex” (see Week 22 in Mathematical Physics for a delightful ASCII drawing). This observation is from the early days of the 4-color problem and due to Cayley and Kempe. The next reduction also comes from the 19th century, and is due to Tait. There is a bijection between “4-colorings of faces” and “4-coloring one distinguished face and then 3-coloring the edges.” The way to understand this bijection is to label the edges with “rules for changing face colorings.” Namely there are 3 ways to assign a pairing of the four colors. If you’ve chosen such a rule for every edge then you can propogate your single face coloring to the whole graph. It’s a simple exercise to check that this gives a bijection between valid colorings.
The invariant of planar graphs “number of 3-colorings of edges” extends to a planar algebra/TQFT/tensor category by the recipe similar to that Ben outlined here. Any planar graph with boundary can be thought of as a functional on the space Span{labelings of the boundary}. Extending by linearity any linear combination of planar graphs with boundary can also be thought of as such a functional. Mod out by the equivalence setting two diagrams equal to each other if they give the same functional. The key fact is that these relations play well with gluing together diagrams and hence gives a planar algebra (or tensor category together with a chosen generating object).
What explicitly do these relations look like? You can remove a bigon for a multiplicative cost of 2, a triangle for a multiplicative cost of 1, and a circle with no trivalent vertices for a multiplicative cost of 3. Furthermore there is a version of the I = H relation (I strongly encourage you to check yourself that this relation between number of 3-colorings of edges holds for any coloring of the boundary):
So what do we need to do to prove the 4-color theorem? We need to give a manifestly positive algorithm for evaluating closed diagrams in this planar algebra! That’s it.
One common technique for evaluating closed planar diagrams is to concentrate on the faces. Notice above we know that we can remove all bigons and triangles. What about bigger faces? Well the beautiful thing is that by Euler characteristic arguments all you need to do is prove that you can remove squares and pentagons because any planar diagram has a pentagonal or smaller face. In fact this technique and small variations on it have been remarkably successful at answer planar algebraic questions (once for quantum groups and once for an exotic subfactor).
So let’s try that. Here’s the square (finding this formula using the above relations is kinda fun):
Now there’s just one case left, the pentagon. Just find a positive formula there and you’d be done!