Given four points in general position on $\mathbb{P}^2$ there exists a projection to $\mathbb{P}^1$ collapsing these four pairwise to two points. Its kernel is some fifth point on $\mathbb{P}^2$.

In terms of representations of the Euclidean quiver $\tilde{\mathbb{D}}_4$ with the four subspaces orientation, this corresponds to the nonsplit extension of the direct sum of two regular simple representations, say $\begin{smallmatrix} & 1 & & \cr 1 & 1 & 0 & 0 \end{smallmatrix} \oplus\begin{smallmatrix}& 1 & & \cr 0 & 0 & 1 & 1 \end{smallmatrix}$ by the (pre)projective $\begin{smallmatrix} & 1 & & \cr 0 & 0 & 0 & 0 \end{smallmatrix}$. That is,

$0 \longrightarrow \begin{smallmatrix} & 1 & & \cr 0 & 0 & 0 & 0 \end{smallmatrix} \longrightarrow \begin{smallmatrix} & 3 & & \cr 1 & 1 & 1 & 1 \end{smallmatrix} \longrightarrow\begin{smallmatrix} & 1 & & \cr 1 & 1 & 0 & 0 \end{smallmatrix} \oplus\begin{smallmatrix}& 1 & & \cr 0 & 0 & 1 & 1 \end{smallmatrix} \longrightarrow 0,$

where the middle term is indecomposable preprojective.

(By way of analogy, if one thinks of preprojective modules as somehow like vector bundles on a $\mathbb{P}^1$ then the passage from $\begin{smallmatrix} & 1 & & \cr 0 & 0 & 0 & 0 \end{smallmatrix}$ to $\begin{smallmatrix} & 3 & & \cr 1 & 1 & 1 & 1 \end{smallmatrix}$ is like twisting by a divisor.)

Can this be done in general? Given a preprojective representation $M$ of a Eucilidean quiver, does there always exist an $indecomposable$ nonsplit extension of some sum of regular simples by that $M$? If $M$ is rank one, then yes. But what about higher rank (as in the example above)?

Given four points in general position on $\mathbb{P}^2$ there exists a projection to $\mathbb{P}^1$ collapsing these four pairwise to two points. Its kernel is some fifth point on $\mathbb{P}^2$.

In terms of representations of the Euclidean quiver $\tilde{\mathbb{D}}_4$ with the four subspaces orientation, this corresponds to the nonsplit extension of the direct sum of two regular simple representations, say $\begin{smallmatrix} & 1 & & \cr 1 & 1 & 0 & 0 \end{smallmatrix} \oplus\begin{smallmatrix}& 1 & & \cr 0 & 0 & 1 & 1 \end{smallmatrix}$ by the (pre)projective $\begin{smallmatrix} & 1 & & \cr 0 & 0 & 0 & 0 \end{smallmatrix}$. That is,

$0 \longrightarrow \begin{smallmatrix} & 1 & & \cr 0 & 0 & 0 & 0 \end{smallmatrix} \longrightarrow \begin{smallmatrix} & 3 & & \cr 1 & 1 & 1 & 1 \end{smallmatrix} \longrightarrow\begin{smallmatrix} & 1 & & \cr 1 & 1 & 0 & 0 \end{smallmatrix} \oplus\begin{smallmatrix}& 1 & & \cr 0 & 0 & 1 & 1 \end{smallmatrix} \longrightarrow 0,$

where the middle term is indecomposable preprojective.

(By way of analogy, if one thinks of preprojective modules as somehow like vector bundles on a $\mathbb{P}^1$ then the passage from $\begin{smallmatrix} & 1 & & \cr 0 & 0 & 0 & 0 \end{smallmatrix}$ to $\begin{smallmatrix} & 3 & & \cr 1 & 1 & 1 & 1 \end{smallmatrix}$ is like twisting by a divisor.)

Can this be done in general? Given a preprojective representation $M$ of a Eucilidean quiver, does there always exist an $indecomposable$ and $preprojective$ nonsplit extension of some sum of regular simples by that $M$? If $M$ is rank one, then yes. But what about higher rank (as in the example above)?