I have heard the "slogan" that graded quivers are (derived) equivalent to ordinary quivers (with this "result" being attributed to Keller) and am looking for a precise statement and a reference.

By a "graded quiver" I would understand the same as an ordinary quiver, except that arrows come with a grading, making the path algebra a graded algebra.

I am envisioning a derived equivalence between suitable module categories of the path algebra of a graded quiver, and the path algebra of an ordinary quiver.


1 Answer 1


I have not heard the slogan and perhaps do not understand the context, but it seems to me that this has nothing to do with the derived categories. For any graded quiver (with or without relations) there exists an ordinary quiver (respectively, with or without relations) such that the abelian category of representations of the graded quiver is equivalent to the abelian category of representations of the associated ordinary quiver.

Indeed, let us presume that a "graded quiver" means a set of vertices $V$, a set of edges $E$ with a map $(v',v'')\colon E\to V\times V$, and a grading function $g\colon E\to\mathbb Z$. A representation of a graded quiver $(V,E,g)$ is a collection of graded vector spaces $X_v$, $v\in V$, and homogeneous linear maps $x_e\colon X_{v'(e)}\to X_{v''(e)}$ of degree $g(e)$ defined for all $e\in E$.

Then I'd define the associated ungraded quiver $(\widetilde V,\widetilde E)$ as follows. The idea is to have $\mathbb Z$ vertices of the ungraded quiver for each vertex of the graded one, and $\mathbb Z$ edges of the ungraded quiver for each edge of the graded one. Given a collection of graded vector spaces sitting at the vertices of the graded quiver, the grading components of these graded vector spaces are placed at the vertices of the ungraded quiver.

Set $\widetilde V = V\times\mathbb Z$ and $\widetilde E = E\times\mathbb Z$. Given an edge $(e,n)\in\widetilde E$ of the ungraded quiver, the two vertices that it joins are defined by the rules $v'(e,n)=(v'(e),n)$ and $v''(e,n)=(v''(e),\,n+g(e))$. Given a representation $X$ of the graded quiver, the related representation $\widetilde X$ of the ungraded quiver is defined by the rule that the vector space $\widetilde X_{(v,n)}$ is the $n$-th grading component of the graded vector space $X_v$. Given an edge $(e,n)$ of the undgraded quiver, the linear map $\tilde x_{(e,n)}\colon\widetilde X_{(v'(e),n)}\to\widetilde X_{(v''(e),n+g(e))}$ is then defined as the restriction of the homogeneous linear map $x_e\colon X_{v'(e)}\to X_{v''(e)}$ to the component of degree $n$ in the graded vector space $X_{v'(e)}$ (which lands in the component of degree $n+g(e)$ in the graded vector space $X_{v''(e)}$, as $x_e$ is a map of degree $g(e)$).

When there is a system of homogeneous multiplicative relations imposed on the homogeneous linear maps $x_e$, one can easily transform it into an equivalent system of multiplicative relations imposed on the linear maps $\tilde x_{(e,n)}$.

An equivalence of the abelian categories then, of course, induces an equivalence of their (bounded or unbounded) derived categories.

  • $\begingroup$ Thank you very much. The context is the following (even if my lack of understanding prevents me from making it precise). Consider the subcategory of Dᵇ(X) generated by some finite set of sheaves on (say, sm. proj.) X. Consider a graded quiver, with one vertex for each sheaf, dim Hom(Fᵢ,Fⱼ) arrows of deg. 0, dim Ext¹(Fᵢ,Fⱼ) arrows of deg. 1, etc. There should be a derived equivalence of some module category (ordinary graded modules?) of the graded quiver with the Δ'ated category generated by the sheaves. The pair Fᵢ,Fⱼ generates the same sheaves as Fᵢ,Eᵢⱼ, where Eᵢⱼ is a non-trivial extension $\endgroup$
    – Earthliŋ
    Sep 13, 2017 at 6:30
  • $\begingroup$ of Fᵢ by Fⱼ, as Fᵢ can be recovered as Eᵢⱼ/Fⱼ, but there may no longer be extensions between Fⱼ and Eᵢⱼ. Again, one can draw a corresp. quiver, replacing the vertex Fᵢ by a vertex Eᵢⱼ (introducing a relation). Now there may be no more deg. 1 arrows between Fⱼ and Eᵢⱼ. My understanding was that there is some such process for (a large class of) arbitrary graded quivers, producing an ordinary quiver from a graded one. If this process yields an ordinary quiver, and if the graded quiver was finite, the ordinary quiver would also be finite. $\endgroup$
    – Earthliŋ
    Sep 13, 2017 at 6:32
  • 1
    $\begingroup$ Oh, these are cohomologically graded quivers. This is a different story. For one thing, a DG-algebra does not need to be formal, so replacing the complexes RHom with the cohomologically graded vector spaces Ext is not always possible. Also, I never heard of a possibility to transform an arbitrary DG-algebra into an algebra sitting in the cohomological degree 0 by mutations. So, I don't know. What will you do if you have a single sheaf, so your quiver has only one vertex? $\endgroup$ Sep 13, 2017 at 9:04
  • $\begingroup$ Indeed, a single sheaf with self-extensions looks like a hopeless case, so I must have misunderstood the statement (or necessary hypotheses). I guess when considering direct summands of a tilting sheaf (or better, exceptional collections) one is already in some "best case scenario". Could you explain the "D" of "DG-algebra" in this setup, or is it only there to indicate the type of grading (which I guess are largely sign rules)? Is the correct module category in this setting the category of DG-modules? $\endgroup$
    – Earthliŋ
    Sep 13, 2017 at 12:03
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    $\begingroup$ In these cohomologically graded quivers, you do not have such a grading. Rather, your complexes are endowed with a single grading which is raised both by the quiver operators and by the differential. Such complexes are generally called "DG-modules". $\endgroup$ Sep 13, 2017 at 12:32

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