Quivers

The next couple of weeks will be an experiment to help get me blogging.  This coming Friday I am giving a talk on "Representation theory via surfaces"

We will discuss the representation theory of finite dimensional algebras by using quiver theory and surface triangulations.  These methods allow us to give concrete descriptions of modules over finite dimensional algebras and open the theory to rich combinatorial methods.  We will focus on the algebras of finite representation type and the related tilted algebras of global dimension two.   

Today I want to start with some background and my perspective on what representation theory is and why you might want to use quivers to study it.

Of course deciding where to start is always troublesome.  We will start with Jordan normal form.  If you have done any linear algebra past a basic first semester course, you will have run into Jordan normal form. What you may not have run into is the fact this completely describes the representation theory of the polynomial ring $\mathbb C[X]$, that is using Jordan normal form we can describe all of the modules over $\mathbb C[X]$.  You can actually find the "proof" of this in Dummit and Foote so it is likely that you have seen this but didn't realize it. Basically the idea goes like this, given an $n$-dimensional  ($\mathbb C$) vector space $V$ and an $n\times n$ matrix $A$, we can have a polynomial
    $$p(x) = a_0 + a_1x + \cdots a_m x^m$$
act on $V$ by defining $$p(x) \cdot v = a_0 v + a_1 A\cdot v + \cdots a_m A^m \cdot v.$$   The standard exercise is to show that this is in-fact the only way for $\mathbb C[X]$ to act on a vector space.   Putting this together with Jordan normal form means that we have a complete description of how $\mathbb C[X]$ acts on vectors spaces and further we have a method to determine when two seemingly different actions are the same: the actions are distinct (up to an isomorphism) if and only if Jordan normal form of the corresponding matricies are the same.

The lesson to take from the Jordan normal form business is that by picking an appropriate basis  on $V$ we can read important information about the action of $\mathbb C[X]$.   Quivers are a method for choosing a special presentation of the algebra we want to study and in turn (as we will see later) this will dictate a special presentation for the corresponding module we want to study. For $\mathbb C[X]$ we have the following presentation

Figure 1

We need some terminology and notation before we can formalize the connection between the above picture and $\mathbb C[X]$.  A quiver $Q$ is simply a directed graph, so $Q = (Q_0,Q_1)$ where $Q_0$ is a set of vertices (my quivers are always finite, so I will generally use natural numbers) and $Q_1$ is the collection of arrows.   When needed we also have the maps $s, t\colon Q_1 \to Q_0$ that read off the source and target of an arrow. A path in a quiver is simply a sequence of arrows $a_1a_2\cdots a_t$ in $Q$ such that $t(a_i) = s(a_{i+1})$.  For example consider

Figure 2

Then $a_1a_2$ is a path in $Q$, so is $a_1a_3^5a_2$, where $a_3^5$ simply means to repeat $a_3$ five times.   We also define "stationary paths" at each vertex in $Q$, denoted $e_i$. These are paths with $s(e_i) = t(e_i) = i$ and containing no arrows.  With this notation, we can define the path algebra $\mathbb CQ$ of a quiver to be the

1. the $\mathbb C$ vector space spanned by the paths in $Q$ (really any field can be used, but we might as well use a nice field),
2. with multiplication given by concatenation of paths when it makes sense or zero otherwise,
3. the identity $1$ is given by the sum of the stationary paths,  $e_1+\cdots + e_n$.

 This is of course a good time for an example.

Figure 3

 Let $Q$ be the quiver in Figure 3, then some elements in $\mathbb CQ$ would be 

\[\begin{aligned}
a_1,&\\
2a_2,&\\
i a_1 + a_1a_2,&
\end{aligned}\]

and we could consider the multiplication by $a_1$ on each of these elements
\[\begin{aligned}
a_1 \cdot a_1 &= 0\\
a_1 \cdot 2a_2 &= 2a_1a_2\\
a_1 \cdot (i a_1 + a_1a_2) &= 0
\end{aligned}\]
With this machinery now built we can be explicit about how  $\mathbb C[X]$  is isomorphic to path algebra $\mathbb CQ$ of

.

Let $p(X) = c_0 + c_1 X + \cdots c_m X^m\in \mathbb C[X]$, then isomoprhism $\mathbb C[X] \cong \mathbb CQ$ is given by
\[p(x) \mapsto c_0 1+ c_1 a + \cdots c_m a^m.\]

In general, we can do this for all* algebras.  A proof of this claim can be found at the end of chapter 2 in Elements of the Representation Theory of Associative Algebras I by Assem, Simson, and Skowronski.  Next time I will talk about how to present modules in terms of quivers.