TITLE: Chip-firing clocks
QUESTION [6 upvotes]: Let $G$ be some outdegree-regular directed graph with $n$ vertices and let $H$ be the Laplacian of $G$, so that the rows of $H$ correspond to chip-firing moves. I’m interested in linear functions $f$ from $\mathbb{Z}^n$ to $\mathbb{Z}/k\mathbb{Z}$ with the property that performing a chip-firing move on a vector in $\mathbb{Z}^n$ increases the value of $f$ by 1 mod $k$ (I call such a function $f$ a chip-firing “clock”). Such an $f$ will exist only for certain values of $k$. This feels like it’s related to the structure of the cokernel of the Laplacian but I’m having trouble seeing exactly what's going on. I’m interested in knowing about all such functions $f$ for a given $G$.

REPLY [2 votes]: This is just to record the observation of lambda from the comments.
I'll keep your convention that the rows of $H$ determine the chip-firing moves, so that for a sequence of firings $v\in\mathbb{Z}^n$ the result of caring out these firings (starting from the zero configuration) is $vH$. (But this means that we should really be talking about the kernel and cokernel of $H^t$ everywhere...)
For $x\in \mathbb{Z}^n$, the linear function $f(v)=\langle v, x\rangle \mod k$ is a clock if and only if $Hx=\mathbf{1} \mod k$. (Here $\langle \cdot , \cdot \rangle$ is the usual inner product, and $\mathbf{1}$ is the all ones vector.) This is because the requirement we have to satisfy to be a clock is that $f(vH)=\langle v, \mathbf{1}\rangle \mod k$ for all $v\in\mathbb{Z}^n$, so we need $f(vH)=\langle vH, x\rangle =\langle v,Hx\rangle$ to be equal moulo $k$ to $\langle v, \mathbf{1}\rangle$ for all $v\in \mathbb{Z}^n$, which happens if and only if $Hx=\mathbf{1} \mod k$.
It still is not totally clear how given a graph $G$ to find the (finite!) list of $k$ for which a clock exists, but for a fixed $k$ at least this makes it clear that the question of whether such a $k$ exists is a "linear algebra" problem (if $k$ is prime then indeed we're talking about linear algebra over a field).
[ By the way, here is the argument that the set of such $k$ is finite. $H^t$ is a singular M-matrix, so there is some vector $v_{*}\in\mathrm{ker}(H^t)$ with $v_{*}\neq 0$ but all entries of $v_{*}$ nonnegative. Hence in particular $\langle v_{*}, \mathbf{1} \rangle > 0$. But, as mentioned, if there is a $k$-clock we need that $\langle v, \mathbf{1} \rangle = 0 \mod k$ for any $v \in \mathrm{ker}(H^t)$, so as long as $k > \langle v_{*}, \mathbf{1}\rangle$ then there cannot be a $k$-clock. ]