Given a Dynkin quiver $Q$ and a field $K$.

Question 1: For which such $Q$ are there only finitely many indecomposable representations over the dual numbers $K[x]/(x^2)$?

Note that those representations are exactly those of $KQ \otimes_K K[x]/(x^2)$. This is for example true for $Q$ being of type $A_1, A_2$ or $A_3$.

Question 2: For which combinations of $Q$ and $n$ is $KQ \otimes_K K[x]/(x^n)$ representation-finite?

I think question two has an easy answer for $n \geq 3$, namely only when $Q$ is of type $A_1$ for arbitrary $n$ or $A_2$ for $n=3$, but is there an elementary argument?

  • 1
    $\begingroup$ What is the relationship between $k$ and $K$? $\endgroup$ – LSpice Jul 10 at 16:51
  • 1
    $\begingroup$ @LSpice Thanks, they are the same. I corrected it. $\endgroup$ – Mare Jul 10 at 16:52

You can find this result in [Geiss, Leclerc, Schröer: Quivers with relations for symmetrizable Cartan matrices I: Foundations] as Proposition 13.1. They consider a more general class of algebras. The class of algebras you are considering is obtained by setting their parameters $(c_1,\dots,c_m)$ to all the same number, which you call $n$.

Translating to your setup, their result reads:

The algebra $KQ\otimes_K K[x]/(x^n)$ for $Q$ a Dynkin quiver is representation-finite if and only if you are in one of the following cases:

  1. $n=1$ and $Q$ arbitrary, in which case you just recover $KQ$.
  2. $Q=A_1$, the quiver with only one vertex, in this case $n$ is arbitrary.
  3. $Q=A_2$ and $n=2$ or $n=3$.
  4. $Q=A_3$ and $n=2$.

The proof they give (unfortunately) uses the Bongartz-Gabriel and the Happel-Vossieck list. The question for a more elementary argument thus remains open. For this one should try to construct an infinite family of modules.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.