Is there a name for this acyclic quiver? Sorry for the trivial question, but I don't know much about the subject and don't seem to be able to come up with much by Googling.  Is there an established name for quivers of the form
$$\require{AMScd}
\begin{CD}
\cdot @>>> \cdot @>>> \cdot \\
@VVV @VVV @VVV \\
\cdot @>>> \cdot @>>> \cdot \\
@VVV @VVV @VVV \\
\cdot @>>> \cdot @>>> \cdot
\end{CD}$$
(where there could be an arbitrary number of rows and columns)?  Actually I'm only interested in the case where it's a square, but surely that's not important.
 A: The quiver in your example could be called a tensor product of two copies of the quiver
$$\require{AMScd}
\begin{CD}
\cdot @>>> \cdot @>>> \cdot \\
\end{CD}$$
(which is a quiver of type $A_3$, linearly oriented).
This terminology appears, for instance, in section 3.3 of the paper The periodicity conjecture for pairs of Dynkin diagrams, by Bernhard Keller (link to the arXiv version of the paper).  It is quite likely that it has appeared before in the literature, although I could not find (and did not look too hard for) an earlier reference.
In the paper cited above, the tensor product of any two quivers $Q$ and $Q'$ is defined as follows:


*

*its vertices are pairs $(i,i')$, where $i$ is a vertex of $Q$ and $i'$ is a vertex of $Q'$;

*the number of arrows from $(i,i')$ to $(j,j')$ is given by these rules:


*

*if $i\neq j$ and $i'\neq j'$, then it is equal to $0$;

*if $i=j$, then it is equal to the number of arrows from $i'$ to $j'$;

*if $i'=j'$, then it is equal to the number of arrows from $i$ to $j$.



Thus, to get a quiver as in your post with $m$ rows and $n$ columns, one should take the tensor product of a quiver of type $A_m$ with a quiver of type $A_n$, both linearly oriented.
