The recent striking progress on the chromatic number of the plane by de Grey arises from the interesting fact that certain Cayley graphs have large chromatic number; namely, the graph whose vertices are the ring of integers of a certain number field K endowed with a complex absolute value ||, and in which x and y are adjacent if and only if |x-y| = 1.

This made me realize I know very little about the chromatic numbers of infinite Cayley graphs, and too my surprise, I wasn't able to find much in the literature!

So here's a sample question. Let A be a finitely generated free abelian group, let S be a finite subset of A closed under negation, and let G be the Cayley graph whose vertices are the elements of A and where a,b are adjacent if and only if |a-b| lies in S.

For every integer N, G has a quotient G/N, a finite Cayley graph whose vertices are A/NA and whose edges are given by the images of S in A/NA. (Maybe better to take N large enough so that no element of S lies in NA, so G/N has no loops.)

Evidently a coloring of G/N pulls back to G, so we get an inequality of chromatic numbers

$\chi(G) \leq \chi(G/N)$.

My question is: is it always the case that

$\chi(G) = \inf_N \chi(G/N)$?

In other words: if G has a k-coloring, does it necessarily have a periodic k-coloring?

(You might think of $\inf_N \chi(G/N)$ as the chromatic number of a profinite Cayley graph.)

I don't even know how to do this for A=Z!

  • $\begingroup$ Do you want $S$ to be finite? $\endgroup$
    – verret
    Jul 10, 2018 at 21:09
  • $\begingroup$ Totes. I have edited. $\endgroup$
    – JSE
    Jul 10, 2018 at 21:20
  • $\begingroup$ This has a lot of the feel of asking whether there's a single shape that tiles the plane only aperiodically, although clearly the two problems aren't strictly equivalent. $\endgroup$ Jul 10, 2018 at 23:18
  • 2
    $\begingroup$ I asked a particular case of this question here, where $A$ is a Euclidean lattice and $S$ the set of Voronoi neighbors of the origin, I obtained a number of results in this are last year with Mathieu Dutour and Frank Vallentin which we are still in the process of writing down, but I hope to upload them to the arXiv soon. All of our optimal colorings happen to be periodic but we don't know if there's a general reason. $\endgroup$
    – Gro-Tsen
    Jul 11, 2018 at 0:12
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    $\begingroup$ In the case of two-colorings, the answer is yes, and the proof is fairly straightforward. Here's the idea. Suppose c is a proper two-coloring, where the codomain of c is {0,1}. For simplicity, assume that S generates A. Without loss of generality, assume that c(0)=0. (For if not, then reverse the colors.) Then c is itself a periodic two-coloring. If S does not generate A, then first restrict c to the subgroup generated by S, and then extend to all of A. For > 2 colors, this logic no longer works, and the problem seems to be much more difficult. $\endgroup$
    – Mike Krebs
    Oct 13, 2020 at 0:08

1 Answer 1


Here's an answer for $A = \mathbb Z$ (assuming that $S$ is finite):

Let $c$ be an optimal (i.e. using the minimal number of colours) colouring of $\mathrm{Cay}(\mathbb Z,S)$ and let $k > \max S$. The degrees in $\mathrm{Cay}(\mathbb Z,S)$ are bounded, hence $c$ uses a finite number of colours.

This means there are only finitely many possibilities for the sequence $(c(n+i))_{0 \leq i \leq k}$ and hence there are $n_1,n_2 \in \mathbb Z$ such that $n_2 > n_1 + k$ and $c(n_1+i) = c(n_2 + i)$ for $0 \leq i \leq k$. Without loss of generality assume that $n_1 =0$, $n_2=l$

Define a colouring $d$ by $d(n) = c(n \mod l)$. This clearly is periodic. Since $k > \max S$ and $c(i) = c(l + i)$ for $0 \leq i \leq k$, the neighbours of any vertex $n$ have the same colours in $d$ as the neighbours of $(n \mod l)$ have in $c$, whence the colouring is proper.

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    $\begingroup$ This is also the same proof given by Eggleton, Erdos, and Skilton in "Coloring Prime Distance Graphs", Graphs and Combinatorics 6, 17-32 1990, theorem 2. $\endgroup$ Jul 11, 2018 at 2:43

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