Let $A=KQ/I$ be a finite dimensional quiver algebra with an admissible ideal $I$.


Is it true that in case $A$ is representation-finite, $Q$ has to be planar?


In case it is true a possible approach would be to use Kuratowski's criterion which says that a graph is planar iff it does not contain a subgraph homeomorphic to $K\_{3,3}$ or $K\_5$. Then one might prove that for all possible orientations and $I=J^2$ for $Q=K\_{3,3}$ or $Q=K\_5$ one only gets representation-infinite algebras, but probably the "homeomorphic part" says that there might be more cases to consider than those two (but hopefully only finitely many cases)?


Here one uses that in case $A$ is representation-finite then also $eAe$ for any idempotent (corresponding to a subgraph) and also $KQ/I'$ when $I'=J^2$ is generated by all paths of length at least two.



Representation-finite implies planar for quiver algebras?
Consider path algebras $KQ$ modulo "radical square zero" relations (i.e., paths of length two are zero). It is well known that these have finite representation type if and only if the separated quiver of $Q$ is a disjoint union of Dynkin quivers.


If $Q$ is the complete graph $K\_5$ on five vertices (nonplanar), oriented such that there are arrows from vertex $i$ to vertices $i+1$ and $i+2$ (modulo $5$), then the separated quiver is of Euclidean type $\tilde{A}\_9$, so we don't have an algebra of finite representation type.


However, if we put an extra vertex in the middle of one edge, replacing
$$\bullet\longrightarrow\bullet$$
with
$$\bullet\to\bullet\to\bullet$$
then the separated quiver becomes of Dynkin type $A\_{12}$, so we do have an algebra of finite representation type.


