I am studying representation theory at the moment and I was wondering the following (Since I found no source discussing it):
Let $Q$ be a quiver and let $KQ$ be its path algebra. Let Rep($Q$) be the category of representations of $Q$ and Mod($KQ$) be the category of modules over $KQ$.
Now in representation theory one finds that there is an equivalence between Rep($Q$) and Mod($KQ$).
Now my question is: Given a representation, can you determine by looking at its dimension vector the length of the module that is associated to it?
Here is an example:
Let $Q$ be the quiver $1 \rightarrow 2 \rightarrow 3$ and $K \stackrel{id}{\longrightarrow} K \longrightarrow 0$ a representation of $Q$. Can one say that the module associated to it has length $2$?
When the quiver has no oriented cycles, then the simple modules of $KQ$ are all one dimensional, and are indexed by the vertices. (In fact, the Jacobson radical is the ideal generated by all the arrows, and the semisimple quotient $KQ/J(KQ)$ is isomorphic to $K^n$, where $n$ is the number of vertices.) Thus the length of a module equals its dimension as a $K$-vector space, which in turn equals the dimension of the corresponding representation.
This breaks down when the quiver has oriented cycles, since then there are simple modules which are not one dimensional. For example, when we have one vertex and one loop, so the path algebra is $K[t]$, then simple modules correspond to monic irreducible polynomials, and the dimension equals the degree of the polynomial. This is always one if and only if the field $K$ is algebraically closed.
If the quiver has two vertices $1,2$ and arrows $1\to2$ and $2\to1$, then there is always a 2-dimensional simple module where both vector spaces are $K$ and both maps are non-zero.