# Representation-finite quivers over dual numbers

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?

• What is the relationship between $k$ and $K$? – LSpice Jul 10 at 16:51
• @LSpice Thanks, they are the same. I corrected it. – 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 $$n$$.
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 $$n=2$$.