
Lately I've been having fun in this series discussing some things that I don't really know much about, like lattice packings of spheres. Next week I'll get back to subjects that I actually know something about, but today I want to talk about the 4color theorem, the golden mean, the silver root, knots and quantum field theory. I know a bit about some of these subjects, but I've only become interested in the 4color theorem recently, thanks to my friend Bruce Smith, who has a hobby of trying to prove it, and Louis Kauffman's recent work connecting it to knot theory. The sources for what follows are:
1) The FourColor Problem: Assault and Conquest, by Thomas L. Saaty and Paul C. Kainen, McGrawHill, 1977.
and
2) Map coloring and the vector cross product, by Louis Kauffman, J. Comb. Theory B, 48 (1990) 45.
Map coloring, 1deformed spin networks, and TuraevViro invariants for 3manifolds, by Louis Kauffman, Int. Jour. of Mod. Phys. B, 6 (1992) 1765  1794.
An algebraic approach to the planar colouring problem, by Louis Kauffman and H. Saleur, Yale University preprint YCTPP2791, November 8, 1991.
(I discussed this work of Kauffman already in "week8," where I described a way to reformulate the 4color theorem as a property of the vector cross product.)
Where to start? Well, probably back in October, 1852. When Francis Guthrie was coloring a map of England, he wondered whether it was always possible to color maps with only 4 colors in such a way that no two countries (or counties!) touching with a common stretch of boundary were given the same color. Guthrie's brother passed the question on to De Morgan, who passed it on to students and other mathematicians, and in 1878 Cayley publicized it in the Proceedings of the London Mathematical Society.
In just one year, Kempe was able to prove it. Whoops! In 1890 Heawood found an error in Kempe's proof. And then the real fun starts....
But I don't want to tell the whole story leading up to how Appel and Haken proved it in 1976 (with the help of a computer calculation involving 10^10 operations and taking 1200 hours). I don't even understand the structure of the AppelHaken proof  for that, one should probably try:
3) Every Planar Map is Four Colorable, by Kenneth Appel and Wolfgang Haken, Contemporary Mathematics (American Mathematical Society), v. 98, 1989.
Instead, I'd like to talk about some tantalizing hints of relationships between the 4color problem and physics!
First, note that to prove the 4color theorem, it suffices to consider the case where only three countries meet at any "corner," since if more meet, say four:
      
we can stick in a little country at each corner:
  / \   \ /  
so that now only three meet at each corner. If we can color the resulting map, it's easy to check that the same coloring with the little countries deleted gives a coloring of the original map.
Let us talk in the language of graph theory, calling the map a "graph," the countries "faces," their borders "edges," and the corners "vertices." What we've basically shown is it suffices to consider trivalent planar graphs without loops  that is, graphs on the plane that have three edges meeting at any vertex, and never have both ends of the same edge incident to the same vertex.
Now, it's easy to see that 4coloring the faces of such a graph is equivalent to 3coloring the edges in such a way that no two edges incident to the same vertex have the same color. For suppose we have a 4coloring of faces with colors 1, i, j, and k. Wait  you say  those don't look like colors, they look like the quaternions. True! Now color each edge either i, j, or k according to product of the the colors of the two faces it is incident to, where we define products by:
1i = i1 = i 1j = j1 = j 1k = k1 = k ij = ji = k jk = kj = i ki = ik = j.
These are almost the rules for multiplying quaternions, but with some minus signs missing. Since today (October 16th, 1993) is the 150th birthday of the quaternions, I suppose I should remind the reader what the right signs are:
ij = ji = k, jk = kj = i, ki = ik = j, i^2 = j^2 = k^2 = 1.
Anyway, I leave it to the reader to check that this trick really gives us a 3coloring of the edges, and conversely that a 3coloring of the edges gives a 4coloring of the faces.
So, we see that the edgecoloring formulation of the 4color problem points to some relation with the quaternions, or, pretty much the same thing, the group SU(2)! (For what SU(2) has to do with quaternions, see "week5".) Those wrong signs look distressing, but in the following paper Penrose showed they weren't really so bad:
4) Applications of negative dimensional tensors, by Roger Penrose, in Combinatorial Mathematics and its Applications, ed. D. J. A. Welsh, Academic Press, 1971.
Namely, he showed one could count the number of ways to 3color the edges of a planar graph as follows. Consider all ways of labelling the edges with the quaternions i, j, and k. For each vertex, take the product of the quaternions at the three incident edges in counterclockwise order and then multiply by i, getting either i or i. Take the product of these plusorminusi's over all vertices of the graph. And THEN sum over all labellings!
This recipe may sound complicated, but only if you haven't ever studied statistical mechanics of lattice systems. It's exactly the same as how one computes the "partition function" of such a system  the partition function being the philosopher's stone of statistical mechanics, since one can squeeze out so much information from it. (If we could compute the partition function of water we could derive its melting point.) To compute a partition one sums over states (labellings of edges) the product of the exponentials of interaction energies (corresponding to vertices). The statistical mechanics of 2dimensional systems is closely connected to all sorts of nice subjects like knot theory and quantum groups, so we should suspect already that something interesting is going on here. It's especially nice that Penrose's formula makes sense for arbitrary trivalent graphs (although it does not count their 3colorings unless they're planar), and satisfies some juicy "skein relations" reminiscent of those satisfied by the quantum group knot invariants. Namely, we can recursively calculate Penrose's number for any trivalent graph using the following three rules:
A. Wherever you see
\ / \ / \ /    / \ / \ / \
you can replace it with
  \ /   \ /   \ /    \   / \   / \   / \
In other words, replace the problem of computing Penrose's number for the original graph by the problem computing the difference of the Penrose numbers for the two graphs with the above changes made. For knot theory fans I should emphasize that we are talking about abstract graphs here, not graphs in 3d space, so there's no real difference between an "overcrossing" and an "undercrossing"  i.e., we could have said
\ / \ / \ / / / \ / \ / \
instead of
\ / \ / \ / \ / \ / \ / \
above, and it wouldn't matter.
B. If you do this you will start getting weird loops that have NO vertices on them. You are allowed to dispose of such a loop if you correct for that by multiplying by 3. (This is not magic, this is just because there were 3 ways to color that loop!)
C. Finally, when you are down to the empty graph, use the rule that the empty graph equals 1.
Greg Kuperberg pointed out to me that this is a case of the quantum group knot invariant called the Yamada polynomal. This is associated to the spin1 representation of the quantum group SU(2), and it is a polynomial in a variable q that represents e^h, where h is Planck's constant. But the "Penrose number" is just the value at q = 1 of the Yamada polynomial  the "classical case" when h = 0. This makes perfect sense if one knows about quantum group knot invariants: the factor of 3 in rule B above comes from the fact that the spin1 representation of SU(2) is 3dimensional; this representation is really just another way of talking about the vector space spanned by the quaternions i, j, and k. Also, quantum group knot invariants fail to distinguish between overcrossings and undercrossings when h = 0.
Now let me turn to a different but related issue. Consider the problem of trying to color the vertices of a graph with n colors in such a way that no two vertices at opposite ends of any given edge have the same color. Let P(n) denote the number of such ncolorings. This turns out to be a polynomial in n  it's not hard to see using recursion relations similar to the skein relations above. It also turns out that the 4color theorem is equivalent to saying that the vertices of any planar graph can be 4colored. (To see this, just use the idea of the "dual graph" of a graph  the vertices of the one being in 11 correspondence with the edges of the other.) So another way to state the 4color theorem is that for no planar graph does the polynomial P(n) have a root at n = 4.
P(n) is called the "chromatic polynomial" and has been intensively investigated. One very curious thing is this. Remember the golden mean
G = (sqrt(5) + 1)/2 = 1.61803398874989484820458683437...?
Well, G + 1 is never a root of the chromatic polynomial of a graph! (Unless the polynomial vanishes identically, which happens just when the graph has loops.) The proof is not all that hard, and it's in Saaty and Kainen's book. However  and here's where things get really interesting  in 1965, Hall, Siry and Vanderslice figured out the chromatic polynomial of a truncated icosahedron. (This looks like a soccer ball or buckyball.) They found that of the four real roots that weren't integers, one agreed with G + 1 up to 8 decimal places! Of course, here one might think the 5fold symmetry of the situation was secretly playing a role. But in 1966 Barri tabulated a bunch of chromatic polynomials in her thesis, and in 1969 Berman and Tutte noticed that most of them had a root that agreed with G + 1 up to at least 5 decimal places.
This curious situation was at least partially explained by Tutte in 1970. He showed that for a triangular planar graph (that is, one all of whose faces are triangles) with n vertices one has
P(G + 1) <= G^{5n}
(that little thingie is a "less then or equals" sign). This is apparently not a complete explanation, though, because the truncated icosahedron is not triangular.
This is not an isolated freak curiosity, either! In 1974 Beraha suggested checking out the behavior of chromatic polynomials at what are now called the "Beraha numbers"
B(n) = 4 cos^2(π/n).
These are
B(1) = 4 B(2) = 0 B(3) = 1 B(4) = 2 B(5) = G + 1 B(6) = 3 B(7) = S
etc.. Note by the way that B(n) approaces 4 as n approaches ∞. (What's S, you ask? Well, folks call B(7) the "silver root," a term I find most poetic and eagerly want to spread!
S = 3.246979603717467061050009768008479621265....
If anyone knows charming properties of the silver root, I'd be interested.) Anyway, it turns out that the roots of chromatic polynomials seem to cluster near Beraha numbers. For example, the four nonintegral real roots of the chromatic polynomial of the truncated icosahedron are awfully close to B(5), B(7), B(8) and B(9). Beraha made the following conjecture: let P_i be a sequence of chromatic polynomials of graphs such whose number of vertices approaches ∞ as i → ∞. Suppose r_i is a real root of P_i and suppose the r_i approach some number x. Then x is a Beraha number.
In work in the late 60's and early 70's, Tutte proved some results showing that there really was a deep connection between chromatic polynomials and the Beraha numbers.
Well, to make a long story short (I'm getting tired), the Beraha numbers also have a lot to do with the quantum group SU(2). This actually goes back to some important work of Jones right before he discovered the first of the quantum group knot polynomials, the Jones polynomial. He found that  pardon the jargon burst  the Markov trace on the TemperleyLieb algebra is only nonnegative when the Markov parameter is the reciprocal of a Beraha number or less than 1/4. When the relationship of all this stuff to quantum groups became clear, people realized that this was due to the special natural of quantum groups when q is an nth root of unity (this winds up corresponding to the Beraha number B(n).)
This all leads up to a paper that, unfortunately, I have not yet read, in part because our library doesn't get this journal!
5) Zeroes of chromatic polynomials: a new approach to the Beraha conjecture using quantum groups, by H. Saleur, Comm. Math. Phys. 132 (1990) 657.
This apparently gives a "physicist's proof" of the Beraha conjecture, and makes use of conformal field theory, that is, quantum field theory in 2 dimensions that is invariant under conformal transformations.
I should say more: about what quantum groups have to do with conformal field theory and knot polynomials, about the Kauffman/Saleur translation of the 4color theorem into a statement about the TemperleyLieb algebra, etc.. But I won't! It's time for dinner. Next week, if all goes according to plan, I'll move on to another puzzle in 2dimensional topology  the AndrewCurtis conjecture  and Frank Quinn's ideas on tackling that using quantum field theory.
© 1993 John Baez
baez@math.removethis.ucr.andthis.edu
