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Good Morning from Belgium,

I'm no stranger to the mantra that quiver-algebras are an extremely powerful tool (see for example the representation theory of finite dimensional algebras). But what is a bit unclear to me is what is known about what kind of algebras are quiver algebras. The first case I can prove is for graded rings with finite dimensional semisimple $A_0$. But what about the ungraded case (I'm pretty sure that it is also true for finite-dimensional algebras over an algebraically closed field for example). Is there a larger generality possible ?

Answers (and references) are as always much appreciated !

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Louis, do you mean by a 'quiver-algebra' the path algebra of a quiver, or do you mean a quotient of a path algebra?

If the first, then I do not understand your comments. k[x,y] is graded with semi-simple part of degree zero but not a path algebra.

If you mean by quiver-algebra a quotient of a path algebra then the answer is simple : any finitely generated C-algebra will do as they are quotients of free algebras (path algebra of one vertex multiple loop quiver).

If you mean by quiver-algebra really the path algebra of a quiver, the answer is trickier.

If your algebra is finite dimensional (I'm always working over C) then the classification is : hereditary and basic (that is, all simples are one-dimensional). In that case, any hereditary is Morita equivalent to a path algebra. All this goes back to Gabriel.

If your algebra is infinite dimensional one has to be careful. Surely it must be formally smooth (that is, it has the lifting property for algebra maps through nilpotent ideals) and have a finite number of isoclasses of one-dimensional representations.

EDIT : Oops, if one has loops then there are of course infinitely many 1-dmls. I should have said that there are only finitely many components of 1-dml representations, all parametrized by affine spaces and such that one can pick one rep in each component to perform the trick with the structural morphism described below.

But that is not enough, take e.g. the groupalgebra of the modular group PSL(2,Z). It has 6 one-dimensionals but is not isomorphic to the 'obvious' quiver one would construct out of these 6 (arrows corresponding to extensions between the simples).

What one needs is that the structural morphism A --> End(SS) (where SS is the semi-simple on the finite number of one-dimls) splits so that A becomes a C^k-algebra. Then one can use formal smoothness of A and the path algebra with semi-simple part End(SS) and arrow part determined by M/M^2 (where M is the kernel of the structural morphism above) to prove that they are isomorphic. An argument like this appears in the paper by Cuntz and Quillen on noncommutative smoothness (they call a formally smooth algebra 'quasi-free').

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  • $\begingroup$ Professor, I guess my terminology wasn't very clear (I thought there was a standard way to talk about these things, but as MO proves, there isn't) What a quiver algebra is to me is a quotient of a path algebra by an ideal generated by relations $I \subset k[Q]^2$. This restriction might be trivial, but I don't think we can just simply apply the free-algebra trick. PS. thanks for the related material on path-algebras, I wasn't aware that the situation was so intricate and subtle ! $\endgroup$
    – louis de Thanhoffer de Völcsey
    Commented May 19, 2010 at 15:56
  • $\begingroup$ The reason for this restriction $(I \subset k[Q]^2)$ being that the quiver is uniquely determined in this case. $\endgroup$
    – louis de Thanhoffer de Völcsey
    Commented May 19, 2010 at 16:53
  • $\begingroup$ Louis, the answer to your question is : always, provided you allow for Morita equivalence. In the 'basic'-case you use free C^k-algebras if A has a complete set of orthogonal set of idempotents of length k. Morita equivalence is important because in general these idempotents can get higher ranks. So even in your graded case with degree zero semi-simple, I think you need to take the simple components all to be C rather than matrix-algebras. $\endgroup$
    – lieven lebruyn
    Commented May 19, 2010 at 19:14
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I do not know the full answer, but I believe that in order for an algebra to be a path algebra of a quiver it needs to be a formally smooth algebra (see here for a definition). I think this is something I read on Lieven Le Bruyn's blog at some point, so it could be there and there is probably a nice exposition in his noncommutative geometry book (which is available on his homepage).

Another thing, you ask about the ungraded case. Well a path algebra is always graded by path length so in order to see some ungraded algebras you will have to look at quivers with relations.

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  • $\begingroup$ By the ungraded case, I meant a quotient-algebra of a path algebra by relations which need not be homogeneous,a construction in which you lose the grading $\endgroup$
    – louis de Thanhoffer de Völcsey
    Commented May 19, 2010 at 12:52
  • $\begingroup$ Ahh, yes of course. I meant relations not potentials. $\endgroup$
    – GMRA
    Commented May 19, 2010 at 13:02

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