The String Coffee Table

A quiver is just a directed graph. It is called ‘quiver’ in order to indicate the context, namely algebra. Algebras have quivers associated to them and quivers have algebras associated to them.

Given any quiver $Q$ we can identify it with its graph category, also to be called $Q$ here, which is simply the category with objects the vertices of the quiver and morphisms the set of morphisms generated by the edges of the quiver under composition.

The path algebra $kQ$ of the quiver over the field $k$ is the algebra generated by all the morphisms in $Q$ (which are paths of edges in $Q$) with the product between two generators defined to be equal to their composition as morphisms if composable, and zero otherwise.

There are two important categories associated to any quiver $Q$, namely the functor category

of functors from $Q$ to the category $Vect$ of vector spaces,…

(I am typing on a public internet terinal which does not even have curly brackts on its keyboard. Therefore the somewhat odd look of some formulas currently.)

as well as the category

of left (say) $kQ$-modules.

One simple but very nice fact is that these two categories are really the same.

One simple way to see this is the following: $kQ$ contains the algebra of $k$-valued functions on the vertex set of $Q$ as a subalgebra (namely that generated by the identity morphisms). The modules of this subalgebra are of course nothing but collections of $k$-vector spaces over each vertex.

Now pick an element of this module which is nonzero only in the vector space over some vertex $x$. Acting now with a path $x \overset \gamma \to y$ must transport this object $k$-linearly to $y$. Hence edges must be associated with linear operators. This is just what elements in $Rep(Q)$ do.

The above fact is useful because many problems become obvious when quiver representations are regarded as $kQ-modules$ while other become obvious when these are regarded as functors $Q \to Vect$.

For instanc to $kQ-mod$ it is relatively easy to associate the Picard group, which I hope to talk about later, while $(Vect)^Q$ plays a role in an argument that I want to give below for how the derived category of quiver representations is related to 2-bundles.

Representations of quivers appear in string theory when one studies 0-branes (3-branes) on orbifolds of the form $C^3/G$ ($R^4 \times C^3/G$) for some finite group $G$. The vacuum of such a theory is described by an element in $D(Rep(Q))$, the derived category of $Rep(Q)$ for some quiver $Q$. By using some stringy magic these systems have alternative descriptions where the quiver turns from an abstract object to a lattice description of spacetime dimensions. It is this latter aspect which makes a connection between quiver representations and categorified geometry, as I have mentioned before and will indicate from a somewhat new perspective below.

So what is $D(Rep(Q)) = D(kQ-mod) = D((Vect)^Q)$ like?

An object in this category is not just a functor $Q \to Vect$ but actually a cochain complex of such functors. A morphisms in $D(Rep(Q))$ is a cochain map between such cochains modulo quasi-isomorphisms.

At this point one should really draw a diagram, which I cannot do at the moment. But I can describe it.

In order to get rid of the annoying issue of quasi-isomorphisms for a moment consider instead of $D(Rep(Q))$ its ‘pre-image’ $Ch(Rep(Q))$, the category of chain complexes in $Rep(Q)$.

Also, for ease of visualization consider the quiver which has just a single edge going between its only two vertices.

Now a functor from that to $Vect$ is an image of this edge in $Vect$, an edge in $Vect$. A chain complex of such functors is a commuting rectangle in $Vect$. A morphism of such chain complexes is a 3-dimensional commuting diagram in $Vect$, with square base in this case.

By looking at this diagram I believe it is obvious that we can reformulate the situation as follows:

A chain complex of functors $Q \to Vect$ is the same as a functor $Q \to Ch(Vect)$. In formulas

(I believe this is true, but since that’s just my idea it wouldn’t hurt to check.)

The point of all this is that this way it becomes clear that the objects of $D(Rep(Q))$ can be regarded as 2-sections of a (trivial) 2-bundle

with typical 2-fiber $Ch(Vect)$.

I’ll end this entry with a brief discussion of what this seems to suggest:

Consider a quiver regarded as a latticized space which is not simply connected. The functors $Rep(Q) \ni F : Q \to Ch(Vect)$ are like a categorification of fields from the point space to smething like complex numbers. We know that it is generally neccesary to view such fields not as functions but as sections of some possibly nontrivial bundle. If that’s true for ordinary fields it should be all the more relevant for quiver reps.

Let me just assume this to be true. This would mean that for a not simply-connected quiver we should refine the notion of what the elements of $D(Rep(Q))$ should be. We should first restrict attention to simply conncted sub-quivers $Q_i$ of $Q$, consider their reps $F_i : Q_i \to Ch(Vect)$ and then ‘glue them together’ appropriately on double overlaps $Q_i \cap Q_j$.

There should be ‘transition 2-maps’ relating the restriction of $F_i$ and $F_j$ to $Q_i \cap Q_j$.

All this tells us that more generally an element of $D(Rep(Q))$ should be considered a 2-section of a 2-bundle over $Q$ with typical fiber $Ch(Vect)$. There is an obvious choice for the transition 2-functions: These should be elements of the ‘weak’ Picard (2-)group of $Q$.

It should be obvious how the details would have to be spelled out here. This would give a good notion of line 2-bundle, I think. But I’ll stop here for the moment.