I've recently started studying quivers and their representations. Currently, I'm trying to build some handiness with path algebras and to do so I was trying to prove the following fact (which sounds pretty true, to me at least): given the quiver $Q$ of type $A_3$ with arrows pointing out of the central vertex and the quiver $Q'$ of type $A_3$ with arrows pointing towards the central vertex, prove that their path algebras $KQ$ and $KQ'$ are not isomorphic. What I know is that $KQ' = KQ^{op}$ the opposite algebra, but that doesn't necessarily mean that they can't be isomorphic (take $M_n(K)$ and $M_n(K)^{op}$). Any hint? Thank you in advance
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Both path algebras have the same idempotents. The obvious ones are the vertices.
Exercise 1. Find the non-obvious primitive idempotents. (Two $k^\times$-indexed families per edge.)
Exercise 2. Show the central vertex is the only primitive idempotent with exactly three orthogonal primitive idempotents (namely, the other three vertices).
Exericse 3. In $kQ$ versus $kQ'$: How many primitive idempotents does the central vertex left-annhilate? What about right-annihilate? Conclude $kQ\not\cong kQ'$.
coiso
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Tysm! I'll work through these steps you gave me and hopefully make it to the end. Out of curiosity, is there some sort of classification of path algebras? Is looking for idempotents and their left multiplication action a generally worth tool or is this a lucky example? – Pickman02 May 25 '25 at 19:35
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I am not an expert. This answer says every finite dim associative algebra is Morita equivalent to a quotient of a path algebra, suggesting there's no nice explicit parameterization of all path algebras up to iso. A recent paper claims they are "classified" (i.e. determined) by their "GK dimension" and "algebraic entropy" invariants. Anyway, idempotents and left/right actions are very important for understanding algebras in general, and I assume path algebras are no different. Idempotents seem important for representations in particular. – coiso May 25 '25 at 20:03