Consider the infinite graph G defined as follows:

The vertex set of G is the set of points of n-dimensional Euclidean space R^n. Two vertices are adjacent if the corresponding points are unit distance apart. The question is, what is the chromatic number of G?

First note that a simple compactness argument shows that this is equivalent to finding the maximum chromatic number of a unit distance graph in R^n; i.e., if we can show that any finite subgraph of G is k-chromatic, then G is k-chromatic.

For n = 2, this is a fairly famous problem. But for higher dimensions, although a lot of work has been done, the known results are (as often happens in combinatorics!) sometimes woefully poor.

Just as one example: The best known general upper bound for the chromatic number of R^n is (3+o(1))^n. But consider the c x c x ... x c hypercube, where c < 3, subdivided into 3^n hypercubes of edge length c/3. If we color each of these hypercubes distinctly, we may tessellate n-space with them, so \Chi(R^n) \leq 3^n is practically trivial!

This is one of the many examples in combinatorics where the best known results are only marginally better than easy, or even trivial, results. (Another good one: upper bounds for the Ramsey numbers R(n,n).) The interesting thing here is that it derives from a geometric fact: namely, that the only regular polytope that tiles n-space for n \geq 5 is the hypercube. The best known bound (7) for 2-space derives from a hexagonal tiling; the best known bound for 3-space is 16 -- I don't know where that comes from. The best bound for 4-space is 49 -- does this arise directly from the tessellation of hyperspace by 24-cells? I'd be greatly indebted to anyone who can shed light on that.

Incidentally, the best known lower bound for \Chi(R^n) is something like (1.239+o(1))^n. I don't see any obvious way to construct even a superlinear unit distance graph in R^n. Anyone know the details of the proof of that bound?

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