Is this quiver with relations of finite representation type

Question

Let $Q=(Q_0,Q_1)$ be the following quiver, $Q_0$ consist of 2 vertices, denoted by 1,2. $Q_1$ consist a loop at 1 called $\gamma$, an arrow $\alpha$ from 1 to 2 and an arrow $\beta$ from 2 to 1. The relation $\rho$ is $\{\beta\alpha, \beta\gamma, \gamma\alpha, \gamma^m\}$ for some integer $m\geq 2$. Given a field $k$, is the quiver algebra $kQ$ always of finite representation type?

Answer 1

An easy way to see that the algebra is of infinite representation type (for any $m \geq 2$) is to observe that it is a string algebra, and that you have strings of arbitrary length, each corresponding to an indecomposable module.

For instance the strings $(\alpha \beta \gamma^{-1})^t$, $t \geq 1$, correspond to indecomposable modules of vector space dimension $3t+1$.

Answer 2

The algebra $A$ is not representation finite for $m=2$, where the algebra has vector space dimension 6 (thus it is also not representation-finite for $m >3$). The first simple module $S_1$ has that the indecomposable module $\tau^7(S_1)$ has vector space dimension $32>max(2 dim A, 30)$, which implies for a quiver algebra that it is representation-finite by a result of Bongartz (see the final comment in 3.4 in https://arxiv.org/pdf/0904.4609.pdf )

Here the QPA (see https://folk.ntnu.no/oyvinso/QPA/) code:

m:=2;;Q:=Quiver(2,[[1,1,"y"],[1,2,"a"],[2,1,"b"]]);KQ:=PathAlgebra(GF(3),Q);AssignGeneratorVariables(KQ);rel:=[b*a,b*y,y*a,y^m];A:=KQ/rel;Dimension(A);S:=SimpleModules(A)[1];DTr(S,7);

If you want to prove this by hand, then one way is to show that all the indecomposable modules $\tau^i(S_1)$ are different for $i \geq 0$, when $\tau$ is the Auslander-Reiten translate.