Chromatic numbers of infinite abelian Cayley graphs

Question

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!

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.