ADE classification

Last time I introduced the idea of quiver representations, basically, collections of vector spaces and linear maps that are compatible with a given quiver with relations $(Q, I)$.  It turns out that the collection of representations for a fixed quiver  is a Krull-Schmidt category.  This means that we have a notion of direct sums of quiver representations.  It is constructed in the most obvious way, by taking direct sums at each vertex and along the arrows.  For example, it we consider the simple quiver $1 \longrightarrow 2$, then we have the following direct sum of representations

Here I have taken the direct sum at each vertex, producing $\mathbb C^3$, and then identified the last basis element with the the representation $\mathbb C \xrightarrow{1}\mathbb C$.

In practical terms the Krull-Schmidt property means that we can restrict our focus to only the indecomposable representations of $(Q,I)$, those representations that can not be broken into a non-trivial direct sum.   The obvious question is how do I distinguish which representations are indecomposable? In the example given above only the representation $\mathbb C\xrightarrow{1}\mathbb C$.   The larger representation can be broken up into the following direct sum

Where the isomorphism is realized by the pair of matrices $\left(\begin{smallmatrix} 1& i\\0&1\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix} 1& 0\\0&1\end{smallmatrix}\right)$.
(Here I am starting with the original representation and mapping it to the representation with the identity matrix along the arrow.)

The follow up question might be, how many indecomposable representations are there? This question of course requires the question : are there only finitely many indecomposable representations?   Especially as computers get more powerful, the knowledge that there are only finitely many representations would be nice, to identify the indecomposables we would simply need to (get a computer) to enumerate them.   Those algebras that have only finitely many indecomposables are said to be of finite type.

To simplify things I will only consider quivers without relations, that is $I=0$.  With this restriction we get a satisfying though short answer the number questions thanks to P. Gabriel:

Theorem. The path algebra of a quiver has only finitely many indecomposable representations if and only if the underlying graph is a Dynkin diagram of type $A$, $D$, or $E$. 

You may have heard of the Dynkin diagrams before; as the this discussion on MathOverflow shows, they show up in many places across mathematics. The unfortunate result of this theorem is that most quivers will have infinitely many indecomposables.  Of course, the not so unfortunate part is that there is plenty of more mathematics to be done.

So for this restricted case of finite type quivers without relations, how many indecomposables are there?  For type $A_n$ where $n$ is the number of vertices, the indecomposable representations are counted by the $\frac12 n(n+1)$.  For type $D_n$, the number of indecomposables is $n(n-1)$. For types $E_6$, $E_7$, and $E_8$m, the number of indecomposables is 36, 69, and 120 respectively.   In the previous example, that is, $Q$ being $1\longrightarrow2$, there are exactly 3 indecomposables which we can list as
\[\mathbb C\longrightarrow 0,\qquad  \mathbb C\xrightarrow{\ 1\ } \mathbb C,\quad \text{ and } 0 \longrightarrow \mathbb C.\]
I have not said anything about how I generated this  list, that will be a post for a future date.

For quivers with relations the story is much more complicated.  In fact, I don't believe that there is any concise method for determining if an arbitrary quiver with relations is finite representation type. Most work seems to be done in special cases or classes of quivers. Next time I will introduce a class of quivers that are constructed from surface triangulations.  These quivers will not always be finite type, but as a whole have an elegant theory for describing their representations.