[Math] Are the underlying undirected graphs of two mutation-equivalent acylic quivers isomorphic

co.combinatoricsderived-categoriesgraph theoryquiversrt.representation-theory

Quiver mutation, defined by Fomin and Zelevinsky, is a combinatorial process. It is important in the representation theory of quivers, in the theory of cluster algebras, and in physics.

We consider a finite directed graph Q (aka a quiver) without loops and without 2-cycles which may contain parallel edges. Then Fomin-Zelevinsky associated with every vertex k of Q a new quiver μk(Q) which has again neither loops nor 2-cycles. Every μk is involutive in the sense that μkk(Q)=Q for every quiver Q and every vertex k. The precise definition may be found in Fomin-Zelevinsky's works on cluster algebras. In the theory of cluster algebras the quivers encode exchange relations.

You might check Keller's java program to see quiver mutation in action.

Although quiver mutation has an easy combinatorial definition, many natural questions are either open or rely on sophisticated techniques. For example, there is no recipe (simple algorithm) to decide whether two given quivers can be obtained from each other by a sequence of mutations.

Question: Suppose I start with an acyclic quiver Q. (That is, Q contains no oriented cycles.) After a sequence of mutations I get another acyclic quiver Q'. Does it follow that the underlying undirected graphs of Q and Q' obtained by ignoring the orientation of the edges are isomorphic as (undirected) graphs?

Best Answer

Two acyclic quivers which are mutation equivalent are always related by a sequence of mutations at sinks and sources. This is proved in Caldero-Keller's 2006 paper.

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