Let $Q$ be the quiver:

$$\begin{array}{ccccc} & & \\ 1 & \rightrightarrows & 2 & \rightrightarrows &3 & \end{array}$$

I want to find all simple modules over the complex path algebra of $Q$.

In class we defined the category $Q\textbf{-Mod}$ of modules $M$ over a quiver $Q$. Where $Q\textbf{-Mod} \ni M$ consists of vector spaces $M_i$ for every vertex $i \in Q_0$ and $k$-linear maps $M(\alpha) \colon M(i) \to M(j)$ for every $\alpha \in Q_1$ with morphisms between any two such modules $M,N$: $\phi \colon M\to N$ is a collection $\{\phi_i\}_{i\in Q_0}, \phi_i \colon M(i) \to N(i)$ such that the obvious square commutes. We also proved that we have a bijection between $Q\textbf{-Mod}$ and $K[Q]\textbf{-Mod}$, the category of modules over the path algebra of $Q$ over $K$.

Since we have this bijection (which I believe actually gives us equivalence of categories?) we can work with $Q\textbf{-Mod}$ instead of $K[Q]\textbf{-Mod}$, right?

Then $M\in Q\textbf{-Mod}$ is said to be simple if its only submodules are $M$ and $0$. So if the dimension of the direct sum $\bigoplus_{i\in Q_0} M(i)$ is equal $1$ then $M$ must be simple so I think that

$$\begin{array}{ccccc} & & \\ \mathbb{C} & \rightrightarrows & 0 & \rightrightarrows &0 & \end{array}$$

$$\begin{array}{ccccc} & & \\ 0 & \rightrightarrows & \mathbb{C} & \rightrightarrows &0 & \end{array}$$

$$\begin{array}{ccccc} & & \\ 0 & \rightrightarrows & 0 & \rightrightarrows & \mathbb{C} & \end{array}$$

are all simple. But are there any other such simple modules? Or rather is this even correct? It looks too simple. To be honest I find this very hard and confusing. Is it better to work directly with modules over $\mathbb{C}[Q]$ instead?


If your quiver $Q$ has no oriented cycles (and only in this case), then it is true that the simple representations of $Q$ are exactly the one-dimensional representations of $Q$ (so there is exactly one for every vertex of $Q$).

One way to see this is to use Nakayama's lemma: Let $R$ be a unital (not necessarily commutative) ring, and let $J$ be its Jacobson radical. If $M$ is a finitely-generated right $R$-module, then $M\cdot J$ is a proper submodule of $M$.

Now, let $M$ be a simple $\mathbb{C}[Q]$-module (this implies that $M$ is finitely-generated). The Jacobson radical $J$ of $\mathbb{C}[Q]$ is the two-sided ideal generated by the arrows of $Q$. Since $M\cdot J$ is a proper submodule of $M$ (by Nakayama's lemma) and $M$ is simple, we must have $M\cdot J = 0$. In other words, the action of all the arrows of $Q$ on $M$ is the zero action.

This means that in the $Q$-representation corresponding to $M$, all the linear maps attached to arrows of $Q$ are zero maps. In this case, the only way for $M$ to be simple is to be one-dimensional.

Some comments:

  • Since we have this bijection (which I believe actually gives us equivalence of categories?)

Indeed, it is an equivalence of categories.

  • Is it better to work directly with modules over $\mathbb{C}[Q]$ instead?

It depends on the situation. The best is to know both categories.


By Gabriel's theorem, the indecompsable module are classified by positive roots. For your case A_3, they are like a consecutive string of dimension 1 vector spaces. As a result, the only simple modules are 1 dimension at 1 vertex and 0 otherwise. You assertion is correct. The three are the only simple module for A_3.

  • $\begingroup$ But is my example really $A_3$? My reason for asking is that when trying to research quivers a bit my example looks more like a "extented " Kronecker quiver or doesn't the extra arrows matter when searching for simple modules? Sorry if this is naive. $\endgroup$ – Martin Garb Dec 7 '16 at 9:43
  • $\begingroup$ Indeed, the OP's quiver is not of type $A_3$. $\endgroup$ – Hugh Thomas Dec 24 '16 at 2:12

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