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?

  • $\begingroup$ In your definition of bound quiver algebra, is $I$ supposed to be contained in the square of the arrow ideal? $\endgroup$ – Dag Oskar Madsen Dec 18 '15 at 9:14
  • $\begingroup$ @DagOskarMadsen - Since the ideal $I$ is just the ideal generated by the commutative relations from the arrows in the quiver, then it is admissible, yes. $\endgroup$ – Rachel Dec 21 '15 at 10:59
  • $\begingroup$ This means you don't allow commutative triangles in your quiver then (path of length 2 never equal to path of length 1). $\endgroup$ – Dag Oskar Madsen Dec 21 '15 at 11:23
  • $\begingroup$ @DagOskarMadsen - Oh wow, I didn't realize that's what that meant! I see it in the definitions now. I was thinking that the length of the path was counting the vertices, not the number of arrows. Thanks for the clarification. $\endgroup$ – Rachel Dec 23 '15 at 1:35

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 $\tilde{D}_4$.


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