Are the underlying undirected graphs of two mutation-equivalent acylic quivers isomorphic?

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?

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If two acyclic quivers are mutation equivalent, their Ginzburg dg-algebras (whith trivial potential, because there are no cycles to construct a potential!) are derived equivalent. I'm pretty sure this implies the quivers are obtained from each other by a sequence of APR tiltings... –  Mariano Suárez-Alvarez Jun 4 '10 at 13:53
(APR tilting = mutation at a sink or a source, by the way) –  Mariano Suárez-Alvarez Jun 4 '10 at 13:53
Mariano -- could you find a reference or proof for that? I would like to see it. –  David Speyer Jun 4 '10 at 14:36
David, this is proved in arxiv.org/abs/0906.0761. –  Mariano Suárez-Alvarez Jun 4 '10 at 14:54
Mariano, I don't see where 0906.0761 proves the second part of your statement, about mutation at source or sink. Can you be more precise? –  Dylan Thurston Jun 5 '10 at 13:19

I guess that Mariano is right:

Assume that $Q$ and $Q'$ are acyclic and mutation-equivalent quivers. Following the result of Keller and Yang, this implies, as Mariano noticed, that their Ginzburg dg algebras, say $A$ and $A'$, are derived equivalent. This equivalence induces an equivalence between the generalised cluster category of $A$ and that of $A'$ (the generalised cluster category of $A$ is defined as the Verdier quotient of the perfect derived category of $A$ by the bounded derived category of $A$, it is well-defined in the present situation). Since we started with acyclic quivers, the generalised cluster category of $A$ coincides with the usual cluster category $\mathcal C_Q$ (and the same holds true for $Q'$).

Thus, the cluster categories $\mathcal C_Q$ and $\mathcal C_{Q'}$ are equivalent. Therefore, their Auslander-Reiten quivers are isomorphic. The link between $Q$ and $Q'$ then follows from that fact. More precisely:

The paths algebras of $Q$ and $Q'$ have the same representation type, hence $Q$ is of Dynkin type if and only if so is $Q'$, in which case the Auslander-Reiten quiver of $\mathcal C_Q$ is isomorphic to $\mathbb{Z}Q/\langle\sigma\rangle$ for some automorphism $\sigma$ of $\mathbb{Z}Q$. Since the translation quivers $\mathbb{Z}Q$ and $\mathbb{Z}Q'$ are universal covers of $\mathbb{Z}Q/\langle\sigma\rangle$ and $\mathbb{Z}Q'/\langle\sigma'\rangle$, respectively, there is an isomorphism $\mathbb{Z}Q\simeq\mathbb{Z}Q'$ (the covering is understood in the sense of "Covering spaces in representation theory" by Bongartz and Gabriel, Invent. Math. 65 (1982) n°3, 3331--378).

Now assume that neither $Q$ nor $Q'$ is of Dynkin type. Then the Auslander-Reiten quiver of $\mathcal C_Q$ has a unique connected component with only finitely many $\tau$-orbits, it is called the transjective component, and it is isomorphic to $\mathbb{Z}Q$. Therefore $\mathbb{Z}Q\simeq\mathbb{Z}Q'$ in any case.

The isomorphism $\mathbb{Z}Q\simeq\mathbb{Z}Q'$ implies that the paths algebras of $Q$ and $Q'$ have equivalent bounded derived categories and, therefore, that $Q$ and $Q'$ are related by a sequence of reflections (or APR-tilts, see 4.8 in the article "On the derived category of a finite dimensional algebra" of Happel, Coment. Math. Helv., 62 (1987) 339--389).

There should be a shorter arugment, though.

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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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Why, Doc? I started Keller's applet and in 5 minutes mutated D5 into another D5, i.e., with different connections on vertices. I guess this does not quite answer the question as they are isomorphic but the isomorphism is not identity on vertices.

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