{
  "id": "2203.06238",
  "version": "v1",
  "published": "2022-03-11T20:44:29.000Z",
  "updated": "2022-03-11T20:44:29.000Z",
  "title": "When does the Auslander-Reiten translation operate linearly on the Grothendieck group? -- Part I",
  "authors": [
    "Carlo Klapproth"
  ],
  "categories": [
    "math.RT"
  ],
  "abstract": "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.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2022-03-11T20:44:29.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "16G70",
      "16E20"
    ],
    "keywords": [
      "auslander-reiten translation operate",
      "grothendieck group",
      "linear endomorphism",
      "cyclic nakayama algebra",
      "coxeter transformation"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 0,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}