TITLE: Chromatic numbers of infinite abelian Cayley graphs
QUESTION [15 upvotes]: 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!

REPLY [7 votes]: 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.