MT 855: Graduate Combinatorics

Transcription

1 MT 855: Graduate Combinatorics Lecture 24: News Flash: the chromatic number of the plane is at least 5 We interrupt this regularly scheduled lecture series in order to bring the latest news: the chromatic number of the plane is at least five! Let s describe what the problem is and its history a little bit; we ll also tie it into some of our recent discussion using the linear algebra method. Historical progress. In 1950, Edward Nelson, who was then a graduate student, raised an innocuous-sounding problem: suppose that we color the points of the plane subject to the rule that points at distance one apart get different colors. How many colors are required to do so? The presence of an equilateral triangle of side-length 1 immediately shows that at least three colors are required, but further thought is required in order to see whether more are required, or that a finite number suffice. Before more ideas, here is some more formal language. The unit distance graph is the graph G = (V, E) with vertex set V = R 2 and edge set E consisting of pairs of points at a unit distance apart. Nelson s question is to determine the chromatic number of this graph. This value is often called the chromatic number of the plane and denoted χ(r 2 ). The same problem was taken up by Hugo Hadwiger in Switzerland at around the same time, and so this problem is often dubbed the Hadwiger-Nelson problem. The best bounds on χ(r 2 ) were determined in the same year that Nelson raised the problem until just a few weeks ago! Let s first survey the early progress on this problem. The brothers Moser did a little better than the equilateral triangle to show that χ(r 2 ) 4. Here is what they did. Take an equilateral triangle, and glue another one onto it that shares a common edge. In any three-coloring of this 4-vertex graph, the points A and B at distance 3 apart have to get the same color. Now rotate this configuration about A until B swings out to a new point C at distance 1 from where it began. The union of these two configurations is now a 7-vertex subgraph of the unit distance graph called the Moser spindle. By what we have argued, in any 3-coloring of the Moser spindle, A and B have to get the same color, and A and C have to get the same color; but then B and C have the same color and lie at a unit distance apart, violating χ(r 2 ) = 3. Therefore, χ(r 2 ) 4, and we have said a little bit more: namely, that R 2 contains a finite subgraph of chromatic number 4. We will return to this point in a little while. To see that a finite number of colors suffice to color the plane, we begin with a square tiling of the plane of side length d to be determined. We will give all of the points of a given square the same color. So long as the diameter of a square is < 1, we can do this without creating a pair of points the same color at a unit distance in a given square. However, we must take care in how we color different squares. We color a 3 3 arrangement of squares by painting all of the points of each square one color, using nine different colors total (and giving points on the shared boundary of two different squares one of the colors of an adjacent square). We then copy this 3 3 coloring template onto the plane. Note that the 9 color classes are translates of one another. For a fixed color class, a pair of points of that color in different squares lie at distance 2d apart. Therefore, so long as we can arrange that d 2 < 1 and 2d > 1, then we get a 9-coloring of the plane. Taking d = 0.6 will do. In fact, using a tiling by hexagons, we can do even better and show that χ(r 2 ) 7. 1

2 2 The news. Thus, we have argued that 4 χ(r 2 ) 7, and that is where things stood until the following breakthrough: Theorem 0.1 (de Grey (2018)). There exists a finite subgraph N of the unit distance graph for which χ(n) 5. Thus, χ(r 2 ) 5. This result has generated quite a bit of publicity recently, including on Tao s blog and in a Quanta magazine article. On the human interest side, Aubrey D.N.J. de Grey is 55 years old and is an amateur mathematician. He is spearheading an effort to extend human longevity through gene therapy. He contends that the science exists to extend the human lifespan by decades and potentially indefinitely. He has also thought seriously about the philosophical questions about whether this should be done. de Grey s original construction of an N as in Theorem 0.1 requires 20, 245 vertices and a computer to check that it is not 4-colorable. He managed to reduce the size of his construction. Along with Dustin Mixon, he has initiated a PolyMath project to produce even smaller examples. This is an open project that anyone can jump onto to contribute their ideas. The record from a couple of days ago (as of the time of this writing) has 826 vertices, which still takes a computer to check is not 4-colorable. It would be fascinating to give N of any size with a short (human-readable) proof that χ(n) 5. We will describe de Grey s approach in the next lecture. It is easy to understand, based on a very natural idea, and, we may say, it succeeds with a fair dose of luck. Before that, let us address an important feature of the problem. This has to do with χ(r 2 ) versus the chromatic number of its finite subgraphs, an issue that arose earlier when we discussed Borsuk s graphs. Coloring infinite graphs. Suppose that G is an infinite graph, and that χ(g) = t, for some positive integer t. Does it follow that G contains a finite subgraph H such that χ(h) = t? Equivalently, suppose that all of the finite subgraphs of G are k-colorable; does it follow that G itself is k-colorable? Indeed, this is the case, with a caveat that we shall discuss after the proof see whether you can figure out what it is. Theorem 0.2 (Erdős - de Bruijn (1951)). A graph is k-colorable if all of its finite subgraphs are. The proof harkens back to material you learned in point-set topology. Proof. Let G = (V, E) be a graph. We are interested in the set Col G,k of k-colorings of G; in particular, we want to show that this set is non-empty. A k-coloring of G is, firstly, a special kind of mapping from V to a color set of size k, which we may take to be {1,..., k} for concreteness. We can denote the set of all such by X G,k = {f : V {1,..., k}} = {1,..., k} V = v V {1,..., k}. The set of functions is naturally a topological space, and this will be critical in arguing that the subset Col G,k X G,k is non-empty. To describe the topology, we give the set {1,..., k}

3 the discrete topology and X G,k the product topology. Thus, X G,k has a basis of open sets given by v V U v, where U v is a subset of the copy of {1,..., k} indexed by v. How does the condition on a k-coloring manifest? We wish to enforce the condition that f(x) f(y) whenever x y. To that end, for each edge e = (x, y) E, we introduce a set F e = {f X G,k f(x) f(y)}. This set is a closed set. In fact, its complement is the union, from j = 1,..., k, of U j = {f X G,k f(x) = f(y) = j}. Each U j is a basic open set, because it is the product of {1,..., k} for each v x, y and {j} for v = x, y. Thus, F e is closed. The set Col G,k is the intersection of all of the F e, e E. We need to show that this intersection is non-empty, assuming that all of the finite subgraphs of G are k-colorable. If we select finitely many e E, then their union spans a finite subgraph H G. By the assumption, this subgraph is k-colorable. A k-coloring of it (extended arbitrarily to the vertices of G not appearing in H) therefore defines an element of the intersection of these finitely many F e. Consequently, this family of closed sets F e, e E, has the finite intersection property: any finite intersection of them is non-empty. It follows that the intersection of all of the F e, e E, is non-empty, and a point in the intersection is a k-coloring of G. There is an important ingredient in the proof that went unmentioned. It is not true that every collection of closed sets in a topological space that has the finite intersection property always has a non-empty intersection. For instance, in X = (0, 1], if we let F n = (0, 1/n], then these sets are closed and have the finite intersection property, but their total intersection is empty. The issue is that X is not compact. However, the set X G,k used above is compact... isn t it? Each productand {1,..., k} is compact, and the Tychonoff theorem therefore implies that X G,k is assuming the Axiom of Choice. Therefore, the above proof holds, subject to the caveat that we assume the Axiom of Choice. This feature of infinite graph coloring has led some people to speculate that the value of χ(r 2 ) could depend on the logical axioms. It could be that, without the Axiom of Choice, the value of χ(r 2 ) is different from the maximum value of χ(h) taken over all finite subgraphs H R 2. A cautionary example. In fact, the following example, due to Saharon Shelah and Alexander Soifer, is a fairly innocuous-looking example of a graph whose chromatic number depends on the axiom system that gets used. I learned about it from Jacob Fox when I was in graduate school. The relevant graph G has vertex set equal to the real line and edge set consisting of pairs of points that differ by an amount of the form a + b 2, where a and b are integers of opposite parity. This graph does not contain any odd cycles. For suppose that x 1,..., x k were a cycle in the graph. Write each difference x j+1 x j = a j + b j 2 for j = 1,..., k, indices (mod k), where a j and b j are integers of opposite parity. The sum of all of these differences is 0, since the x i form a cycle, and on the other hand it adds to a + b 2, where a is the sum of the a j s and b is the sum of the b j s. The only way we could have 0 = a + b 2 is for a = b = 0, since 2 is irrational. In particular, 0 a + b k j=1 (a j + b j ) k 1 k (mod 2) shows that k is even, establishing the claim about cycle lengths. It follows that all 3

4 4 of the finite subgraphs of this graph are 2-colorable. Assuming the axiom of choice, it follows that G itself is 2-colorable. However, there is another way to complete ZFC to an axiomatic system in which the axiom of choice does not hold, but for for which every subset of the real line has a positive Lebesgue measure. (You should recall that one of the first amazing facts from measure theory is that, assuming the axiom of choice, there exist non-measurable subsets of R.) Under these axioms, it is not even possible to color G using countably many colors! For suppose we countably-color G. Then one of the color classes C has a positive Lebesgue measure. A lemma in measure theory implies that there exists a small ɛ > 0 with the property that for all δ < ɛ, there exists a pair of points in C at distance δ apart. Since the set a + b 2 with a and b integers of opposite parity is dense in R, it follows that C contains a pair of endpoints of an edge in G. Actually, it is even easier to describe a graph whose chromatic number depends on whether or not one assumes the Axiom of Choice. Consider the graph consisting of a disjoint union of uncountably many edges {(x α, y α )}, where α ranges over an uncountable index set A. Twocoloring this graph is equivalent to selecting a point from the product α A {x α, y α }, which in turn is equivalent to the Axiom of Choice. Thus, the graph has chromatic number two iff the Axiom of Choice holds. The first example mainly serves to indicate that a graph whose description is not so far off from the unit distance graph is known to has a chromatic number which behaves very differently depending on the axioms used; so perhaps the same could hold for the unit distance graph. Therefore, it is particularly satisfying that de Grey s theorem asserts the existence of a finite subgraph of R 2 that has chromatic number 5. However, for all we know still, if we accept an axiomatic system in which all subsets of R 2 have a Lebesgue measure, then it could well be that all finite subgraphs of R 2 are 5-colorable, yet R 2 is not. Along these lines, there is an interesting complement. Consider the variation on coloring the unit distance graph, where the color classes are required to be Lebesgue measurable subsets of R 2 and where we furthermore accept the Axiom of Choice. This variation leads us to definite the measurable chromatic number of the plane, χ meas (R 2 ). For instance, our 9- and 7-colorings from earlier (can be made to) use measurable color classes, and so χ meas (R 2 ) 7. Falconer proved in 1981 that χ meas (R 2 ) 5. The proof uses some basic measure theory having to do with density points. Thus, de Grey s result can be read as an improvement on Falconer s. de Grey s work has generated some speculation that perhaps we can improve the lower bound on χ meas (R 2 ) to 6, although it is not clear how to adapt de Grey s ideas to this setting quite yet. The chromatic number of Euclidean space. To conclude, let us consider another natural generalization of the problem under consideration to n-dimensional Euclidean space. We can define the unit distance graph in this setting just as before, replacing R 2 by R n, and write χ(r n ) for the chromatic number of n-dimensional Euclidean space. How does this function grow with n? A naïve construction based on a tiling by cubes shows that χ(r n ) n n. A somewhat smarter approach yields χ(r n ) 9 n, which you can consider for homework. Is the growth of χ(r n ) truly exponential in n? Indeed it is:

5 Theorem 0.3 (Frankl - Wilson (1981)). χ(r n ) > (1.1) n ; moreover, there exists a finite subgraph X R n with χ(x) > (1.1) n. This is not the first time that we have seen 1.1 and in fact, the construction will not surprise you at all, as it ties in beautifully with our recent work. Proof. We just prove the theorem for some special values of n; the extension to all n follows from some properties of prime numbers, like the Prime Number Theorem, or even just Bertrand s postulate. Thus, suppose that n = 4p, where p is a prime number. Let X be the set of indicator vectors of subsets A {1,..., n} of cardinality 2p 1, scaled by 1/ 2p, for good luck; so elements of X are vectors with entries 0 and 1/ 2p, with 2p + 1 of the former and 2p 1 of the latter. Suppose that we color R n with (1.1) n or fewer colors. Then one of the color classes meets X in a subset of size at least X /(1.1) n = ( n 2p 1) /(1.1) n. By an earlier estimate, this value is larger than ( ( n 0) + n ) ( n p 1). Hence, the Ray-Chaudhuri Wilson theorem the idea that keeps giving guarantees that this subset of X contains a pair of vectors (1/ 2p)v A and (1/ 2p)v B for which A B = p 1. The squared-distance between them equals dist( 1 2p v A, 1 2p v B ) 2 = (1/2p)( v A 2 2v A v B + v B 2 ) = (1/2p)(2p 1 2(p 1)+2p 1) = 1. Hence, if we color R n with (1.1) n or fewer colors, then there exists a pair of points (in X) of the same color at a unit distance apart. Notice the remarkable similarity between this construction and the one used a dozen years later by Kahn and Kalai to disprove Borsuk s conjecture. Perhaps the overarching theme of the lecture is that sometimes simple and natural ideas take a long time to develop! 5