Suppose that $A = kQ/I$ is a bound quiver algebra for $k$ an algebraically closed field, $Q=(Q\_0, Q\_1)$ a finite connected quiver with no oriented cycles with no multiple edges or self-loops, and $I$ the ideal of commutative relations when viewing $Q$ as a commutative diagram (say of vector spaces and linear maps over the field $k$).


Suppose that I modify $Q$ by adding a single arrow $\alpha$, without introducing any oriented cycles, self-loops, or multiple edges. Call the new bound quiver algebra $A' = kQ'/I'$, where $I'$ is now the new ideal of commutative relations when viewing $Q'=(Q\_0, Q\_1\cup{\alpha})$ as a commutative diagram.


My question is this: If $A$ is representation-finite, is it necessarily the case that $A'$ is representation-finite?


It seems to me (inutitively) that it should be true, but doing some preliminary computations using Auslander-Reiten quivers, this type of modification alters the Auslander-Reiten quiver $\Gamma(A)$ quite a bit, and the indecomposables may be quite different. I was considering looking at using Tits quadratic form, but unfortunately there does not seem to be an if and only if statement concerning bound quiver algebras and the weak positivity of the form (the implication goes the wrong way for trying to use it here).


Any ideas from the community?



Is this modified bound quiver algebra necessarily representation-finite?
The quiver
$$\begin{array}{ccccc}
&&1&\to&2\\
&&\uparrow&&\uparrow&\\
3&\to&4&&5\\
&&\uparrow&&\\
&&6&&
\end{array}$$
is a Dynkin quiver of type $D\_6$, so has finite representation type. But if you add an arrow from vertex $4$ to vertex $5$, together with commutation relations, it has infinite representation type, since the representations with the zero vector space at vertex $2$ are just representations of the tame quiver $\tilde{D}\_4$.


