Chromatic Number of the Odd-Distance Graph Jacob Steinhardt Chromatic Number of the Odd-Distance Graph

Odd-Distance Graph Make a graph where the vertices consist of the points in the plane, and two points are connected if their distance is an odd-integer, i.e. if sqrt((x_1-x_2)^2+(y_1-y_2)^2) = 2k+1 for some k. Question: Is the chromatic number of this graph finite?

Chromatic Number The chromatic number is the minimum number of colors we need to color the vertices of a graph so that no two adjacent vertices have the same color. E.g.: The four-color theorem asserts that the chromatic number of all planar graphs is at most four. The problem is equivalent to asking if we can color the points in the plane with finitely many colors so that no two points of the same color are at an odd integral distance from each other.

Ideas Let A be the adjacency matrix of a graph, and let lambda_max and lambda_min be the largest and smallest eigenvalues. Then: Chi >= 1-lambda_max/lambda_min Chi is chromatic number For infinite graphs, only applies to measurable colorings Idea: Find a sequence of weightings for our graph such that lambda_max/lambda_min -> - infinity

Weighting If we weight vertices exponentially by their distance (i.e. as alpha^-dist), then we can prove that lambda_max/lambda_min -> -infinity as alpha -> 1 This shows that no finite measurable coloring exists

Non-measurable colorings If we can find a sequence of finite subgraphs whose adjacency matrices in some sense “converge” to that for our infinite graph, we may be able to get our result to hold for non- measurable colorings as well, as all colorings on a finite graph are measurable. Unsure how to do this...maybe use the coherent topology on our graph, but it doesn’t have all the properties one might hope for (and I also don’t know enough topology)‏