Can we infer an isomorphism of quivers from an isomorphism of their corresponding path algebras?

Question

Given a pair $\Delta, \Gamma$ of quivers and a field $K$ one can construct the corresponding path algebras $K\Delta, K\Gamma$. I came upon a paper claiming (section 3, 2nd paragraph) that an isomorphism of $K\Delta \cong K\Gamma$ of the path algebras induces an isomorphism of the quivers $\Delta \cong \Gamma$. However, no proof is given for this claim, and so I am looking for a reference giving a proof and stating explicitly the required conditions on the quivers or the exact meaning of a quiver isomorphism in this context.

Answer 1

If your quiver is finite and acyclic then you can recover it up to isomorphism of digraphs by taking as vertices the isomorphism classes of simple modules and taking as directed edges in bijection with a basis for Ext^1 between the simples. I'll have to think about the general case.

Answer 2

Benjamin and Theo's answers are already very good, but just to add a little; it is in fact true that any basic finite-dimensional algebra $A$ over a field $K$ determines (up to non-canonical isomorphism) a quiver $Q$ such that $A\cong KQ/I$ for some ideal $I$ contained in that generated by paths of length $2$. (Any finite-dimensional algebra is Morita equivalent to a basic one, and all path algebras are basic.) This is proved in Assem–Simson–Skowroński's Elements of the representation theory of associative algebras, Vol. 1, and implies the result you want when $\Delta$ and $\Gamma$ are acyclic.

The construction (which is essentially the same as Benjamin and Theo's) is to first pick a maximal set of pairwise orthogonal idempotents in $A$ (equivalent to a decomposition of $A$, as a module on one side, into indecomposable projective modules, which are pairwise non-isomorphic by basicness). Let $R$ be the radical, the minimal ideal such that $A/R$ is semi-simple (i.e. isomorphic to $K^n$). Then the vertices of $Q$ correspond to the idempotents, and the number of arrows $i\to j$ is the dimension of $e_j(R/R^2)e_i$ for $e_i$, $e_j$ the corresponding idempotents. (Or possibly of $e_i(R/R^2)e_j$ depending on how you like to compose paths.) Choosing elements of $R$ descending to a basis of $R/R^2$ determines a surjection $KQ\to A$, and one can show that the kernel has the required property.

I am fairly certain that one can get $A\cong KQ/I$ as above, with $A$ determining $Q$, whenever $A$ is semi-perfect (meaning every finitely generated $A$-module has a projective cover), but I don't have a reference for this. If true, then $\widehat{K\Delta}\cong \widehat{K\Gamma} \implies \Delta\cong\Gamma$, where $\widehat{K\Delta}$ denotes the completion of $K\Delta$ with respect to the arrow ideal (so that one allows formal linear combinations of paths). After the discussion in the comments, I am now uncertain whether the same might be true without completion: since an isomorphism $K\Delta\to K\Gamma$ need not preserve the arrow ideals, one cannot deduce an uncompleted result from the completed one.

The fact that quivers of algebras are determined only up to non-canonical isomorphism is going to prevent you from writing down a specific isomorphism $\Delta\to\Gamma$ given an isomorphism $K\Delta\to K\Gamma$, however.

Answer 3

Here's one way to recover $Q$ from $KQ$ when $Q$ is finite but not necessarily acyclic.

Consider the one-dimensional $KQ$-modules. Each is associated to some vertex of $Q$, and two such modules $M,N$ are associated to the same vertex if and only if there is some free algebra quotient $F=K\langle x_1,\dots,x_n\rangle$ of $KQ$ such that the actions of $KQ$ on $M$ and $N$ both factor through $F$.

So the vertex set of $Q$ can be recovered from $KQ$. Picking one one-dimensional module $M_v$ for each vertex $v$, the number of arrows from vertex $v$ to vertex $w$ is the dimension of $\text{Ext}^1(M_v,M_w)$.

Answer 4

If your quiver is finite (but not necessarily acyclic), then you should be able to do the following. Take as vertices the set of isomorphism classes of irreducible projective modules for the path algebra. Given projective modules $M,N$, call a morphism $f:M \to N$ factorizable if it can be written as $f = pq$ with neither $p$ nor $q$ invertible; call the quotient of $\hom(M,N)$ by the vector space of spanned by factorizable morphisms the space of "unfactorizable morphisms". For each $M,N$, choose a basis for the space of unfactorizable morphisms. I think that basis should give the arrows in the quiver, but there could be some subtleties I have not considered.

Note that both my answer and Benjamin Steinberg's imply something stronger than your request: the isomorphism type of the quiver can be recovered from the Morita equivalence class of the path algebra, not just the isomorphism class.