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.



Graded quivers vs "ordinary" quivers and derived categories
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.


