2 corrected one 'obvious' oversight

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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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.

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').