On page 6 of this article, the author defined a new quiver.
Fix positive integers $n$ and $s$. We define the quiver $Q:=Q^{(n, s)}$ with the set $Q_{0}$ of vertices and the set $Q_{1}$ of arrows by $$ \begin{aligned} &Q_{0}:=\left\{x=\left(x_{1}, x_{2}, \ldots, x_{n+1}\right) \in \mathbf{Z}_{\geq 0}^{n+1} \mid \sum_{i=1}^{n+1} x_{i}=s-1\right\}, \\ &Q_{1}:=\left\{x \stackrel{i}{\rightarrow} x+f_{i} \mid 1 \leq i \leq n+1, x, x+f_{i} \in Q_{0}\right\} \end{aligned} $$ where $f_{i}$ denotes the vector $f_{i}:=(0, \ldots, 0,\stackrel{i}{-1},\stackrel{i+1}{1},0, \ldots, 0)$ for $1 \leq i \leq n$ and $f_{n+1}:=(\stackrel{1}{1},0, \ldots, 0,\stackrel{n+1}{-1})$.
For example, $Q^{(1,5)}$ is
$$40 \underset{2}{\stackrel{1}{\rightleftarrows}} 31 \underset{2}{\stackrel{1}{\rightleftarrows}} 22 \underset{2}{\stackrel{1}{\rightleftarrows}} 13 \underset{2}{\stackrel{1}{\rightleftarrows}} 04,$$
$Q^{(2,4)}$ is
$Q^{(3,3)}$ is
Now we can think of $Q^{(1,s)}$ as a line segment and $Q^{(2,s)}$ as the face of a tetrahedron. How about $Q^{(3,s)}$ and $Q^{(n,s)}$ for larger $n$? What is the intuitive explanation of them? Thank you:)
$Q^{(3,s)}$ is a tetrahedron. The picture given for $Q^{(3,3)}$ has four vertices and six additional points which are the midpoints of each of the six edges.
If you like (and if this helps), you can view it as the 1-skeleton of a part of the affine type A hyperplane arrangement. (If the hyperplane arrangement story is unfamiliar, I'm not sure it's worth going into.)
$Q^{n,s}$ is an n-dimensional simplex.