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 $k$?



How complicated can a finite double complex over a field be?
It seems that [Jonas Stelzig](https://arxiv.org/abs/1812.00865) has answered your question completely, for the sake of applications to non-Kähler geometry.
$\require{AMScd}$


I will quote here Theorem A, which enumerates all of the isomorphism classes of double complexes as direct sums of certain standard double complexes. Because it is easy to write down the set of maps between any two of the standard double complexes, I am hopeful this description is enough to determine any categorical facts you might wonder about.



> 
> For any bounded double complex over a field $K$, there exist unique cardinal numbers $\text{mult}\_S(A)$ and a non-functorial isomorphism $A \cong \bigoplus\_S S^{\oplus \text{mult}\_S(A)}$, where $S$ runs over squares and zig-zags:
> 
> 
> $$\begin{CD}K @>-1>>K\\@A1AA @A1AA\\K @>1>> K\end{CD}, \;\;\;K,\;\;\; K \xrightarrow{1} K, \;\;\; \begin{CD}K \\ @A1AA \\ K\end{CD}, \;\;\; \begin{CD}K \\ @A1AA \\ K @>1>> K\end{CD}, \;\;\; \cdots$$
> 
> 
> 


That is, ignoring bigradings, $S$ consists of the square, the single-element term, and two zig-zags of each length $L > 0$ (depending on orientation of the 'initial' arrow); there are $\Bbb Z^2$ each of these generators, depending on the bigrading of the initial element.


