Lecture 1. Introduction and first examples.(adsbygoogle = window.adsbygoogle || []).push({});

I thought that this forum was looking a little sparse recently and decided to try to write something interesting for people to read, think about, possibly do some work on. One aspect of algebra that is not taught anywhere at undergraduate level that I'm aware of, but that is incredibly important (and reasonably simple) are quivers. In some sense absolutely everything in algebra is to do with quivers. So here goes.

Definition. A quiver is a directed graph (with a finite number of vertices and arrows).

Examples. (Let's see how much we can typeset in here.) Well, it's kinda obvious what they are but here are some of the more important ones we'll talk about:

[tex] \bullet \rightrightarrows \bullet[/tex]

is the so called kronecker quiver.

[tex]\bullet \to \bullet[/tex]

doesn't have a name but is important.

So, how on earth is a purely combinatorial thing like a graph actually algebraic?

Here's the simplest way.

Definition. Let Q be a quiver. A representation of Q is the following data: a finite dimensional complex vector space for each vertex (we may talk about other fields later, and in general any algebraically closed field will do for this part), and for each arrow between vertices a linear map between the corresponding vector spaces. Two such sets of data are defined to be equivalent/isomorphic if they differ by change of bases in the vector spaces. The collection vector spaces is denoted [itex]\{W_v\}[/itex] where v are the vertices, and the linear maps are [itex]\{f_a\}[/itex] labelled by the arrows a. It is useful to have a way of talking about where the arrows go from an to. The arrow starts at the source, and ends at the target space. We use s(a) and t(a) to represent these vertices in the graph. Thus in a representation we are assigning to each arrow, a, an element of [itex]{\rm Hom}(W_{s(a)},W_{t(a)})[/itex]

Example. Consider

[tex] \stackrel{\bullet}{\circlearrowleft}[/tex]

A representation of this is just a vector space and a linear map i.e. a pair (W,f). (W',g) is equivalent if W' is isomorphic to W (i.e. the dimensions are the same) and, identifying End(W') with End(W), if f is conjugate to g. Hence equivalence classes of representations of this quiver are parametrized by Jordan Normal form. This parametrization is relatively simple, especially if we make a further definition.

Definition. Given a representation [itex](\{W_v\},\{f_a\})[/itex] of Q a subrepresentation is a choice of subspace of each W_v which are all preserved by the linear maps f_a.

A representation is simple if it has no subrepresentations.

Example. For

[tex] \stackrel{\bullet}{\circlearrowleft}[/tex]

and a representation (W,f), a subrepresentation is just a subspace that f maps into itself. Since we are working over the complex numbers every linear map has an eigenvalue and eigenvector, thus the only simple representations of this quiver are those where W is 1-dimensional (and f is nonzero) or when W is the zero vector space.

If we just think about this quiver alone for a while, there are two different kinds of subrepresentation. Let W be of dimension 2. Consider the case when f has just one jordan block compared to when f has two jordan blocks. In the either case, the representation is not simple, but the former is a very different beast from the latter: in the second case f has two eigenvectors, and we can write W as W'+W'' where both W' and W'' are subrepresentations. But in the former case there is no complementary space we can pick that is preserved by f.

Thus we are led to the following definition.

Definition: a representation [itex](\{W_v\},\{f_a\}[/itex] is decomposable if we can find a proper subrepresentation (i.e. one where not all subspaces are zero or W_v) and a choice of complementary subspaces that is also a representation. If a representation is not decomposable we call it indecomposable. Thus we can talk about sums of representations and summands of representations.

Indecomposable is *not* the same as simple. A two dimensional W and a linear map with one jordan block above is indecomposable but not simple.

Example.

[tex] \bullet \to \bullet[/tex]

A representation is then a pair of vector spaces X and Y and a linear map f:X--->Y.

Let us classify all indecomposable representations where X and Y are nonzero vector spaces and f is nonzero. We can write X as ker(f)+X' for some complementary subspace X' of the kernel, and We can write Y as Im(f)+Y' for some complementary subspace of the Image of f called acokernel. Thus if the representation is indecomposable, we must have ker(f)=Im(f)=0 and f is an isomorphism (invertible linear map). Further, since we are completely free to choose our basis we may assume that if {x_i} is a basis of X and {y_i} a basis of Y then f sends x_i to y_i. Then the only indecomposable is if X and Y are 1 dimensional spanned by x and y resp and f maps x to y, call this P_1

Now, let us look at the simple representations. If ({X,Y},{f}) is any representation there is the following subrepresentation ({0,Y},{f=0}), and hence there is a simple (nonzero) subrepresentation ({0,Y},{0}) where Y is one dimensional, call this S_2. Similarly if f is the zero map from X to Y the representation ({X,0},{0}) is a subrep, hence we find the only other simple is ({X,0},{0}) where X is one dimensional, call this S_1.

Summary: given [itex] \bullet \to \bullet[/itex] there are exactly two nonzero simple representations, and one more indecomposable representation. Further every representation is a direct sum of these three representations. Explicitly if f is a map between X and Y we can choose a basis of X and Y so that the representation of the quiver = [itex] (S_1)^p \oplus (P_1)^q \oplus (S_2)^r[/itex] where p is the dimension of the kernel of f, and r is the dimension of the cokernel.

This is an example of so-called finite representation type.

Exercise. Repeat this for [itex]\bullet \to \bullet \to \bullet[/itex]

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# Quivers and representations

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