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I am studying the text An introduction to Quiver Representations by Derksen. Exercise 1.5.4 asks to prove that if the path algebras $\mathbb{C}Q$ and $\mathbb{C}Q'$ are isomorphic $\mathbb{C}$-algebras then $Q$ and $Q'$ are isomorphic directed graphs if they are acyclic. This I am able to prove(the proof is wrong).

However, will the above result hold if the quivers have cycles? I think the answer should be NO but I am unable to find a counterexample. In any case if the path algebras are isomorphic then $Q$ and $Q'$ will have the same number of edges. This is because they form the free generators and the path algebras are finitely generated algebras. As vector spaces they will now be infinite dimensional which is the reason why I think the underlying quivers might not be isomorphic since it is "easier" for infinite dimensional vectors spaces to be isomorphic and the basis here are the paths in the quivers. But still I don't have a counterexample.

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  • $\begingroup$ When you say "isomorphic", do you mean as vector spaces or as algebras? For example, the polynomial algebras in two and three variables give vector spaces of countable dimension, which are isomorphic to each other, but they are not isomorphic as algebras. $\endgroup$
    – Pedro
    May 6, 2021 at 17:55
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    $\begingroup$ @PedroTamaroff As algebras. As vector spaces any two path algebras with underlying quiver being cyclic are isomorphic, again because of countably infinite dimension. $\endgroup$ May 6, 2021 at 17:59
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    $\begingroup$ Can you quickly remind me how you proved the claim for acyclic quivers? I suppose you can look at say the collection of simple projective modules (corresponding to sinks) and try to "remove" one of them? Perhaps it is even more straightforward. $\endgroup$
    – Pedro
    May 6, 2021 at 19:14
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    $\begingroup$ @SubhamJaiswal It’s not true that in the path algebra of an acyclic quiver, the only idempotents are the $e_x$. For example, if $a$ is an arrow with $a=e_yae_x$, then $e_y+a$ is idempotent. $\endgroup$ May 7, 2021 at 3:35
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    $\begingroup$ I suspect the counterexample will be a wild algebra, since those basically all embed into each other. Like, I bet a counterexample comes from how utterly misbehaved those algebras are, rather than simply being acyclic. $\endgroup$ May 9, 2021 at 23:21

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I'll assume that the quiver $Q$ has finitely many vertices and arrows.

Then even if $Q$ has oriented cycles, it is it is still true that the path algebra $\mathbb{C}Q$ determines the quiver $Q$.

(1) In the case of a quiver with no oriented cycles, a quick way to recover the quiver is as follows:

There is a simple module $S_i$ associated with each vertex $i$, and the number of arrows from vertex $i$ to vertex $j$ is $\dim_{\mathbb{C}}\operatorname{Ext}^1_{\mathbb{C}Q}(S_i,S_j)$. [I'm not sure that this is the proof that Derksen was asking for, as the book doesn't seem to assume knowledge of $\operatorname{Ext}$.]

So knowing the simple $\mathbb{C}Q$-modules and the extensions between them lets you recover the quiver.

(2) If the quiver has oriented cycles but no loops (arrows with the target equal to source) then there are more simple modules, but the obvious simples associated to the vertices are the only $1$-dimensional simples, and the same method of recovering the quiver works, if you only consider the $1$-dimensional simples.

(3) If the quiver has loops, then there are more $1$-dimensional simples (consider the representation with $\mathbb{C}$ at vertex $i$, zero at every other vertex, with the loops at vertex $i$ acting by multiplication by arbitrary scalars). But the same method work if we can pick out one $1$-dimensional simple module for each vertex.

There may be a simpler method, but one way to do this is to consider the abelianization of $\mathbb{C}Q$. This is a product of polynomial algebras $\mathbb{C}[x_1,\dots,x_{r_i}]$, one for each vertex $i$, where $r_i$ is the number of loops at vertex $i$. So it has one primitive idempotent for each vertex, and for each of these idempotents we can choose any $1$-dimensional simple module (it doesn't matter which) that is not annihilated by that idempotent. As before, the quiver can then be recovered by considering $\operatorname{Ext}^1_{\mathbb{C}Q}$ between these simple modules.

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    $\begingroup$ What if we allow infinite arrows and vertices? There might be cardinality problems (like adding a single point to something infinite), but beside such things it might be still true? $\endgroup$
    – Mare
    May 25, 2021 at 16:37
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    $\begingroup$ I think it's probably still true (although with infinitely many vertices, the path algebra will not be unital). But if $S$ and $T$ are simples at two vertices, then $\text{Ext}^1(S,T)$ is the dual of the space spanned by arrows from $S$ to $T$, and this may not be enough to recover the cardinality of the set of arrows if it's infinite. $\endgroup$ May 26, 2021 at 7:33
  • $\begingroup$ @JeremyRickard sorry for bringing it up so late...I understand your answer and yes, derksen has Ext in the 2nd chapter... But my question is, is there a 'conceptual' reason why a Path Algebra determines a unique quiver? Sorry if I'm question doesn't make sense... $\endgroup$ Jan 23, 2022 at 15:42
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I am answering the question for the case of acyclic quivers. The main reason is that I have done a solution that doesn't require use of Ext but only projective modules and some matrix algebra.

I will show that given a finite dimensional path algebra the underlying acyclic direct graph can be uniquely recovered solely from the algebra.

Say the path algebra is $A$.

First the vertices can be recovered since the simple modules are in one one correspondance with the vertices. One can also do this by considering maximal ideals (all but $e_i$) for some $i$.

So, we can now label the vertices $1, \cdots n$

We know the projective modules!

The indecomposable projective corresponding to the vertex $i$ is clearly $P_i = A e_i$. One doesnt need to know about projectives. Simply take the projection of the path algebra using the idempotent corresponding to vertex $i$.

Now using the module $P_i$ we know the number of paths from the $i$th vertex to any other vertex $j$.

So, overall we have determined the no. of directed paths from any vertex $i$ to any vertex $j$.

So, now the question is does this determine the graph uniquely? The answer is YES as I show.

Claim: Knowing the number of directed paths(same as walks due to acyclicty) from $i$ to $j$ are known for all pairs $(i, j)$ then the graph is determined.

Let $A$ be a matrix corresponding to a directed graph such that $A_{ij}$ is the number of edges from $i$ to $j$.

Note that $A^{k}$ is the number of paths(directed obviously) from $i$ to $j$.

NOte that $A$ is nilpotent due to acyclicity. $A^{n+1}=0$ in fact.

What we are given is $A+ A^2+ \cdots A^n = B$. We are given the $B$

Now we simplify as $I = I -A^{n+1} = (A+ A^2+ \cdots A^n)(I-A) = (B+I)(I-A)$

Note that above shows that $(B+I)$ is invertible so we get $(B+I)^{-1} - I = -A$

So, $A$ can be solved in terms of $B$. So, there is a unique $A$ which means there is a unique graph.

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