TITLE: How complicated can a finite double complex over a field be?
QUESTION [16 upvotes]: A finite complex over a field $k$ is pretty simple: it's the direct sum of its homology with a split-exact complex. How complicated can a finite double complex be? Does it make a difference if $k$ is algebraically closed?
It seems as though quiver representation theory may be relevant here, but since a double complex is a representation of a quiver with relations, I'm not really sure where to start looking. So my question is:
Question: Let $k$ be a field. How complicated is the category of finite-(total) dimensional double complexes over $k$?

REPLY [4 votes]: The representation theory of those representations concentrated in the $n \times m$-grid are just representations of the tensor product of the radical square zero algbras $A_n$ and $A_m$ with underlying quiver a linear oriented line.
For $n=m$, they have been also classified in https://arxiv.org/pdf/1703.08377.pdf (see proposition 2)and this then also gives the classificaiton for arbitrary $n$ and $m$ when looking at the indecomposables which lie only in the $n \times m$ grid for $m \leq n$.
Here another proof using some theorems. Let $C=A_n \otimes_K A_m$.
By Theorem 4.5. a) in chapter IX. of the book "Elements of the representation theory of associative algebras" by Assem, Simson and Skowronski the algebra $C$ has a postprojective component. Thus one just have to show that for every indecomposable projective $C$-module $P$ one has $\tau^{-l}(P)=0$ for $l$ large enough to conclude that there are only finitely many indecomposable modules and all are isomorphic to $\tau^{-r}(P)$ for some $r$ and some indecomposable projective module $P$. (starting from this one can get the Auslander-Reiten quiver of $C$ easily)
Seeing those complexes as tensor products also shows that in any dimension higher than 2, there are infinitely many indecomposable representations in general as already $A_2 \otimes_K A_2 \otimes_K A_2$ is representation-infinite.