2
$\begingroup$

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)?

$\endgroup$

1 Answer 1

2
$\begingroup$

(I think you are also assuming $M$ indecomposable? If not, there are easy counterexamples with the Kronecker algebra.)

It is enough to prove the claim for $M$ indecomposable projective for the following reasons: Every other indecomposable preprojective module is of the form $\tau^{-n}P$ for some $n \geq 0$ and $P$ indecomposable projective. As long as we stay away from the preinjective component, the translate $\tau^{-1}$ preserves exact sequences [Auslander, Reiten, Smalø, "Representation Theory of Artin Algebras", Lemma VIII.4.4]. $\tau^{-1}$ also sends indecomposables to indecomposables and regular simples to regular simples.

In the $\tilde {\mathbb A}_n$ case the claim follows from string algebra combinatorics. One reference for string algebras is [Butler, Ringel, "Auslander-Reiten sequences with few middle terms and applications to string algebras"]. You can construct an irreducible monomorphism from $M$ into a larger string module by adding a "right hook" (or "left hook"). The cokernel is a uniserial string module which is a regular simple module.

I don't have a general proof for the $\tilde {\mathbb D}_n$ case, but your example sequence together with the sequences similar to $$0 \longrightarrow \begin{smallmatrix} & 1 & & \cr 1 & 0 & 0 & 0 \end{smallmatrix} \longrightarrow \begin{smallmatrix} & 2 & & \cr 1 & 1 & 1 & 0 \end{smallmatrix} \longrightarrow\begin{smallmatrix} & 1 & & \cr 0 & 1 & 1 & 0 \end{smallmatrix} \longrightarrow 0$$ shows that it is true for $\tilde {\mathbb D}_4$ with your chosen orientation.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .