As mentioned in the title, I would like to know a proof of the "well known" fact that the A6 preprojective algebra is of wild representation type.


Ideally, I would like to see an explicit two-parameter family of indecomposable modules.


If you happen to prefer type D to type A and want to give an answer there, that would be appreciated too.



Why is the A6 preprojective algebra of wild representation type?
Inspired by the reference Julian gave in his comment, here's an explicit example of a two-parameter family of indecomposable representations.


First, I'll describe a one-parameter family of indecomposable representations for the preprojective algebra of type $A\_5$.


Let $k$ be the coefficient field, and $\alpha\in k$. Then let $M(\alpha)$ be the representation
$$\begin{array}{ccccccccc}
&\begin{pmatrix}0\\1\end{pmatrix}&&\begin{pmatrix}1&0\\0&0\end{pmatrix}
&&\begin{pmatrix}0&0\\1&\alpha\end{pmatrix}
&&\begin{pmatrix}1&0\end{pmatrix}\\
k&\rightleftarrows&k^2&\rightleftarrows&k^2&\rightleftarrows&k^2&
\rightleftarrows&k\\
&\begin{pmatrix}1&0\end{pmatrix}&&\begin{pmatrix}0&0\\1&1\end{pmatrix}
&&\begin{pmatrix}0&0\\1&0\end{pmatrix}
&&\begin{pmatrix}0\\\alpha\end{pmatrix}\\
\end{array}$$


It's a straightforward calculation to check that $M(\alpha)$ is indecomposable and that if $\alpha\neq\beta$ then any homomorphism $M(\alpha)\to M(\beta)$ is zero at the right hand vertex, so that a direct sum decomposition of $M(\alpha)\oplus M(\beta)$ into two non-zero summands is uniquely determined at the right hand vertex.


Now extend $M(\alpha)\oplus M(\beta)$ to a representation $N\left(\{\alpha,\beta\}\right)$ of the preprojective algebra of type $A\_6$ by adding a vertex on the right hand end:
$$\begin{array}{ccccccccccc}
&&&&&&&&&\begin{pmatrix}0&0\end{pmatrix}\\
k\oplus k&\rightleftarrows&k^2\oplus k^2&\rightleftarrows&k^2\oplus k^2&\rightleftarrows&k^2\oplus k^2&
\rightleftarrows&k\oplus k&\rightleftarrows&k\\
&&&&&&&&&\begin{pmatrix}1\\1\end{pmatrix}\\
\end{array}$$


Then it follows easily from the properties stated for the representations $M(\alpha)$ that $N\left(\{\alpha,\beta\}\right)$ is indecomposable and $N\left(\{\alpha,\beta\}\right)\not\cong N\left(\{\alpha',\beta'\}\right)$ unless $\{\alpha,\beta\}=\{\alpha',\beta'\}$.


