What is the quiver representation of an algebra, and how is it obtained from a decomposition into spaces of morphisms (the opposite algebra)?

Question

I will make everything concrete so I have something to work with. Let $A$ be an algebra, free, and 4-dimensional, and let $A = P_1 \oplus P_2$ be a decomposition into two 2-dimensional free modules. I can then write $$A = \operatorname{End}_A(A) = \operatorname{End}_A(P_1) \oplus \operatorname{End}_A(P_2) \oplus \operatorname{Hom}_A(P_1,P_2) \oplus \operatorname{Hom}_A(P_2, P_1).$$ Multiplication in $A$ is then just matrix multiplication on the right in each $2 \times 2$ block, so I can form the opposite algebra $A^{op}$ by transposing the blocks corresponding to $\operatorname{End}_A(P_1) \leftrightarrow \operatorname{End}_A(P_2)$ and then switching the blocks corresponding to the $\operatorname{Hom}_A(P_i,P_j)$.

Now I have read somewhere, but I cannot find it anymore, that the "quiver representation of the algebra" can be read off this geometric picture on the right. I can't entirely make sense of it, though.

My first thought is that we treat this as a quiver of four vertices with no directed edges, unless $P_1 \cong P_2$ in which case it is a quiver of four vertices with directed edges between all vertices in all directions. Even so, my understanding of quiver representations is quite basic, and I only know to form the path algebra here but am not sure where else to go. I don't know how Gabriel's theorem could apply here to classify the connected quiver in the second case - maybe it 'collapses' into one vertex with 12 loops? Any ideas or clarification here would be wonderful.

Answer 1

Let $X: P_1 \to P_2$ be a linear map and $Y: P_2 \to P_1$ be some other linear map (not necessarily $X^{_1}$, if it exists). Then the graph has two vertices, one for each $P_i$, and two directed edges representing $X$ and $Y$. By doing some matrix multiplication (possible here since we're working with free modules), we can obtain relations among the $X$ and $Y$ - if they're two dimensional, then that's even better. Finally from this quiver and relations we can form the path algebra, and see that this is isomorphic to the algebra itself.

Answer 2

I am assuming that when the OP says $A$ is free and four-dimensional, they mean that $A$ is a rank four free module over the ground ring, while $P_1$ and $P_2$ are rank two free modules over the ground ring. I am going to assume that the ground ring is an algebraically closed field.

My reference for this material is "Elements of the representation theory of associative algebras" by Assem, Simson, and Skowroński. They only associate a quiver to an algebra under the assumption that it is basic, i.e. that the projective summands of the algebra are non-isomorphic. Their approach for non-basic algebras would be to pass first to the corresponding basic algebra. Let's assume, then, that $P_1$ and $P_2$ are not isomorphic as $A$-modules.

The algebra $A$ is isomorphic to a quotient of the path algebra of a certain quiver. The quiver has a vertex for each projective indecomposable summand of $A$, so in this case there would be two vertices. Let's call the quiver $Q$, and write $KQ$ for its path algebra, and $KQ/I$ for a quotient of the path algebra isomorphic to $A$.

If there were a vertex in $Q$ with no arrow pointing out of it, the corresponding simple representation would be projective. But that contradicts the hypothesis that the two projective modules are simple. This implies that there must be an arrow leaving each vertex. By the assumption that the projective modules are only two-dimensional, there can't be more than one arrow leaving either vertex. The two vertices and the two arrows already account for a four-dimensional algebra, so any compositions of the arrows must be zero.

There are two possible configurations of the quiver: an arrow from 1 to 2 and an arrow from 2 to 1, or else a loop at 1 and a loop at 2.

Edited to add the following: To pass from an algebra to the associated basic algebra, we take a set of complete orthogonal idempotents, $e_1,\dots,e_n$, and then we take a subset $\{e_{i_1},\dots,e_{i_r}\}$ such that each indecomposable projective module shows up exactly once in the list $Ae_{i_j}$. (These are the projective summands of $A$, and saying it is non-basic says exactly that this list contains repetition, so we will wind up choosing a proper subset of the set of complete orthogonal idempotents.)

Then, let $e=e_{i_1}+\dots +e_{i_r}$, and let $A'=eAe$. It is shown in section I.6 of Assem-Simson-Skowroński that $A'$ is basic, does not depend on the choice of idempotents, and that the module category of $A'$ is equivalent to that of $A$.