Quiver representations

Last time  I introduced the idea of quivers. Basically the idea is that if you want to study algebras and their modules, then picking a nice presentation of those algebras may be useful.  Of course, when stated like this, this is obvious.  No one wants difficult presentations of their abstract objects, but at this point it remains to be seen if quivers actually are nice presentations.

Last time I left out one detail that I should include here so that the following statements are actually true.  I said that we can find a quiver presentation for all* algebras.  Well it turns out that there is some important fine print here.  Primarily, if your algebras is basic and connected; then, there is a unique quiver with relations that we can associate to it, I will come back to this business about relations in a moment.  Why do we want these conditions on the algebra?  I will work backwards.  If your algebra is not connected, then the quiver you would build will not be connected.  This doesn't seem like a big deal, but most of the time we like to work with connected quivers and generalize from there.  The connected condition is mostly a technical condition, at least so far. The basic condition is more important but there is a convenient work around: given an algebra $A$, there is a basic algebra $A^b$ such that the category of $A$ modules and equivalent to the category of $A^b$ modules. This is the concept of Morita equivalence.  I really only want to study modules, so as far as considering all the possible module categories is concerned, quivers will produce everything that I want to consider.

What are these relations on the quiver?  Basically, this is how we introduce additional zero-divisors.  I call the pair $(Q,I)$ a bound quiver where $I$ is some ideal in $\mathbb CQ$ such that $I$ does not contain any arrows (or sums of arrows).  For instance, what if we wanted to only consider the algebra generated by the polynomials with degree 1?  We would consider $\mathbb C[X]/(X^2)$.  In terms of quivers this is $\mathbb CQ/(a^2)$ where we use the identification of $\mathbb C[X]$ with

Figure 1

In this example $a^2 = 0$.  Similarly, we can consider the quiver

Figure 2

and the ideal $I = \langle a_2a_3\rangle$.  In this case, we would consider any paths that contain $a_2a_3$ to be zero, so $a_1a_2a_3 = 0$ in $\mathbb CQ/I$.

The short story of why quivers are a good way to present your algebras is that it immediately gives us a way to present the corresponding modules as quiver representations.  Given a quiver $Q$, a quiver representation is a collection $M$ of vector spaces and linear maps $\{ V_i, \phi_a\}$ where $i\in Q_0$ is a vertex and $a\in Q_1$ is an arrow such that $\phi_a \colon V_{s(a)} \to V_{t(a)}$, that is the linear map associated to $a$ makes sense with the vector spaces at the source and target of the arrow.

If $(Q,I)$ is the bound quiver associated to an algebra, then we require the that the maps associated to paths in $I$ should compose to zero.  Going back to Figure 2 and $I = \langle a_2a_3\rangle$, a quiver representation would be $V_i = \mathbb C$ for $i\in Q_0$ and $\phi_1 = \phi_2 = 1$ and $\phi_3 = 0$. We do not have much choice in which linear maps we can pick for the arrows.  Since each vector space in this representation is one-dimensional, each of the linear maps is multiplication by a constant from $\mathbb C$. We need $\phi_3 \circ \phi_2 = 0$, so either $\phi_3 = 0$ or $\phi_2 = 0$.  So, for instance, $V_i = \mathbb C$ for $i\in Q_0$ and $\phi_a = 1$ for $a\in Q_1$ is not a representation of $(Q,I)$, but, it would be a representation of $Q$ (here we can think of setting $I=0$).

We can of course present quiver representations as diagrams. For instance here two representation of $(Q,I)$ as given above:

Figure 3

In the same way maps between representations are easy to describe as well:

Figure 4

Morphisms of quiver representations $M = \{ V_i, \phi_a\}$ and $N= \{ W_i, \xi_a\}$ are collections of linear maps $\psi_i$ for $i\in Q_0$ such that $\psi_i \colon V_i \to W_i$ and so that all of the obvious squares commute, that is $\psi_i\xi_a = \phi_a\psi_{t(a)}$.

Throughout the rest of these posts, I will always assume that my algebra is finite-dimensional and that the vector spaces in a quiver representation are also finite-dimensional.  The benefit of doing this is that when it gets down to the gritty details of proving stuff or working with concrete representations  everything is vector spaces and matricies.

What we have done is generalize the example of $\mathbb C[X]$ in which all of its modules can be described by a vector space and a particular matrix.  Next time we will talk about some special quivers that deserve a distinguished place in this theory.