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.

Sound problems

Yesterday I finally decided to fix a problem  with my laptop that I had been ignoring for at least a year. Essentially, when I used the volume buttons on my keyboard the volume would change by approximately 25% of its max loudness, so it got loud or quiet very fast.  I could change the volume in smaller increments using the volume slider but I like using the keyboard buttons, so my laptop had essentially 3 volume levels: loud, audible, and mute.  The source of the problem seems to be this bug  and unfortunately upstream (Gnome I think) did not want to spend much, or any, time on fixing this.

Fortunately, after much sleuthing, I found this post on ask Ubuntu, which in turn uses information found on Ubuntu's wiki which unfortunately has removed that page.    The fix boils down to adjusting two lines in a config file:

Fix / Workaround

There are a few variables which control how PulseAudio controls the volume. You can either edit /etc/pulse/default.pa (you'll have to be root to do that) to change the behavior for all users, or copy that file to ~/.pulse/default.pa and then edit that file, to change behavior for the current user only.

Open the file mentioned above. Find the row saying "load-module module-udev-detect" and change it to:

load-module module-udev-detect ignore_dB=1

and I also needed

Find the row saying "#load-module module-alsa-sink" and change it to:

load-module module-alsa-sink control=PCM

Of course you may be wondering why I am blogging about something like this.  Well, simply put this is mostly intended to be a personal trail of breadcrumbs.  More math will be coming later this afternoon.

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.