# Algebras with the same Auslander -Reiten quiver have isomorphic categories of modules?

Maybe this is a stupid question but suppose you have two quivers $Q$ and $Q'$ (maybe we can suppose some nice property, like not having oriented cycles, so that kQ has finite dimension over $k$) whose Auslander Reiten quivers $A$ and $B$ are isomorphic as quivers in the sense of graph theory. Maybe we can also suppose that the isomorphism preserves the Auslander Reiten translation $\tau$ as well. Does that imply that the category of representations of $Q$ is "the same" as the category of representations of $Q'$, $Rep_Q \cong Rep_{Q'}$? Or to put it in algebraic terms, $mod_{kQ} \cong mod_{kQ'}$ ?

I find this question baffling because it looks natural but I havent been able to find any reference to it and it doesn't seem obvious. Simplify the problem and assume you want to show that $ind_{kQ} \cong ind_{kQ'}$, where $ind$ means the category of indescomposable representations and morphisms between indescomposable representations. An odd category, I know (very non abelian). To find a functor that is an equivalence between the two categories I need to define it over any morphism between two indescomposable representations. The problem is that, as far as I know, not every morphism is a composition of irreducible morphisms, and thus using the isomorphism between the two Auslander-Reiten quivers is more difficult.

What is valid is that any morphism $f:M \longrightarrow N$ that is not an iso, can be written as a sum of compositions of irreducible morphisms but only if $Rad^n(-,N) =0$ or $Rad^n(M,-)=0$ for some $n$. Which is something that doesn't hold in general.

## 1 Answer

Since you assume that you are given $\tau$ as part of the data, you can identify the indecomposable projective modules, since they are the ones on which $\tau$ vanishes. But the AR quiver restricted to the projective modules is the opposite of $Q$. So you can recover $Q$.

• In the hereditary case (and without using the translation) one can recognize the postprojective component as the one with no vertices with infinitely many predecessors, no? Can one recognize the projectives there? – Mariano Suárez-Álvarez Nov 11 '17 at 4:54
• Yes to the first point. In total generality you can't recognize the projectives without more, though: consider $A_3$ with linear orientation. Or affine $A_3$ with one source, one sink, non-adjacent. Though in both of these examples, the point somehow is that there is a different subquiver which looks exactly like the projectives. So if you took it for the projectives by mistake, you would still get the right $Q$! But I am not sure what happens in general. – Hugh Thomas Nov 11 '17 at 5:24