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.