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Patrick Le Meur
Patrick Le Meur
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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$$\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.

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.

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.

Source Link
Patrick Le Meur
Patrick Le Meur
  • 61
  • 3

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.