Finding quivers of finite-dimensional algebras

637 Views Asked by At

What is the most streamlined way to calculate the quiver of a finite-dimensional algebra over an algebraically closed field of characteristic zero? In particular how can I find the quiver of a group algebra?

I know that the vertices of the quiver correspond to primitive idempotents $e_i$ and I know that the number of arrows from $i$ to $j$ corresponds to the number of basis elements of $e_i(\text{Rad} A/ \text{Rad}^2 A)e_j$. Caculating these things can be quite a task. Finding the correct admissible ideal to satisfy the proper relations seems to be difficult in general.

For the class of algebras I'm looking at, I somehow know the projective indecomposable modules and the $\text{Hom}$-sets between them, thus I'm able to construct the quiver using that. But I imagine there must be other more streamlined and basic methods to find the quiver.

Thank you.

1

There are 1 best solutions below

3
On BEST ANSWER

If your main interest lies in group algebras, then there is one thing you need to remember: the quiver of a finite-dimensional algebra $A$ represents a set of generators for this algebra if the algebra is basic, that is, if $A$ (as an $A$-module) is a direct sum of pairwise non-isomorphic projective modules.

The general statement is that $A$ is only Morita-equivalent to a basic algebra $A_{basic}$, which is itself isomorphic to the path algebra of some quiver, modulo an admissible ideal. In particular, your second paragraph only applies to basic algebras.

For group algebras: a group algebra $A$ is basic if and only if the group is abelian. In any case, group algebras are semi-simple, so the quiver of $A_{basic}$ will contain only vertices (one per irreducible character of the group) and no arrows.

Added: As for the general question of computing the quiver of a given basic algebra, the answer will depend heavily on the specifics of the given algebra. I don't think a general method can be developed.