When does the Auslander-Reiten translation operate linearly on the Grothendieck group? -- Part I

Carlo Klapproth

Published 2022-03-11Version 1

For a hereditary, finite dimensional $\Bbbk$-algebra $A$ the Coxeter transformation is a linear endomorphism $\Phi_A \colon \mathsf{K}_0(\mathsf{mod} A) \to \mathsf{K}_0(\mathsf{mod} A)$ of the Grothendieck group $\mathsf{K}_0(\mathsf{mod} A)$ of $\mathsf{mod} A$ such that $\Phi_A[M] = [\tau_A M]$ for every non-projective indecomposable $A$-module $M$, that is the Auslander-Reiten translation $\tau_A$ induces a linear endomorphism of $\mathsf{K}_0(\mathsf{mod} A)$. It is natural to ask whether there are other algebras $A$ having a linear endomorphism $\Phi_A \in \operatorname{End}_\mathbb{Z}(\mathsf{K}_0(\mathsf{mod} A))$ with $\Phi_A [M] = [\tau_A M]$ for all non-projective $M \in \mathsf{ind} A$. We will show that this is the case for all Nakayama algebras. Conversely, we will show that if an algebra $A = \Bbbk Q / I$, where $Q$ is a non-acyclic quiver and $I \lhd \Bbbk Q$ is an admissible ideal, admits such a linear endomorphism then $A$ is already a cyclic Nakayama algebra.