Intuition behind the canonical projective resolution of a quiver representation

Question

Let $Q$ be a finite acyclic quiver, and $X$ some representation of $Q$. For $i \in Q_0$ define the $kQ$-modules $P_i = kQe_i$, and $X(i) = e_i X$. The representation $X$ has a canonical projective resolution, called the Ringel resolution, given by $$ 0 \longrightarrow \bigoplus_{a \in Q_1} P_{h(a)} \otimes X(t(a)) \xrightarrow{\quad d \quad} \bigoplus_{i \in Q_0} P_i \otimes X(i) \xrightarrow{\quad f \quad} X \longrightarrow 0 $$ where in each direct summand the maps are given by $$ d(p \otimes x) = p \otimes (a \cdot x) - pa \otimes x \qquad\quad f(p \otimes x) = p \cdot x $$

Does anyone have some intuition for this resolution that they could share? Can the modules and the maps in this resolution be nicely interpreted in terms of the paths in $Q$? And can this interpretation be extended to get us projective resolutions for quivers with relations? Or for non-acylic quivers?

Answer 1

I suggest to think about it in terms of an analogy. If $G$ is a group, and $M$ is a module over $G$, you have a resolution of $M$ known as the bar complex. The way you produce this resolution can be thought of as follows. You freely generate a module over $G$ by elements of $M$, namely consider $\mathbb{Z}[G] \otimes M$, with the action only on the first coordinate. This maps to $M$ by $g \otimes m \mapsto gm$. Then, there are many relations, and the only uniform way to encode all of them in general is to note that the defining relation for a module is $g(hm) = (gh)m$ so you encode this relation using $\mathbb{Z}[G \times G] \otimes M$. Then essentially the only way which is uniform in $M$ and $G$ to encode the relation among those is to use $G \times G \times G$, e.t.c.

In general, if $C$ is a category, you can consider modules over $C$, and to construct a very similar resolution, encoding the full uniform collection of relations you can impose on a module without any assumptions on $C$ or the module $M$. The resulting complex is the bar construction of $C$ with coefficients in $M$, and a typical element in it is a tuple of composable maps, tensored with a module element concentrated at the source of the first map. The details are not important for our discussion, just that it is similar to the bar resolution for group cohomology.

Now, if $G$ turn out to be a $\textbf{free}$ group, you can write a much shorter resolution for $M$ over $G$. Namely, choose a set of generators $S$ for $G$ and then you can resolve $M$ by $$\mathbb{Z}[G]^S\otimes M \to \mathbb{Z}[G] \otimes M \to M$$ where the first map is $g \otimes m \mapsto gm$ and the second is $(x)_i \otimes m \mapsto xs_i \otimes m - x \otimes s_im$. Again we implement the relations defining $M$ as a module, but this time we restrict only to the relations on the free generators. The fact that they generate $G$ imply that all relations follows from these, and the fact that they freely generate $G$ imply that there are no relations-among-relations, so we get a very short resolution.

As we saw, a group can be generalized to a category, and the analogous notion of a free group is a quiver: A category freely generated by some objects and morphisms. The resolution you asked about is the analogous resolution to the one I discussed for the free group. You write down only the "module relations" coming from the generating morphisms, and the fact that they freely generate the whole collection of morphisms, imply that there are no relations-among-relations, so we get a very short resolution.

Answer 2

The answer to all your questions is "yes". I'm not sure I could do a better job than the intuition in Derksen's notes, though: think of the projective modules in terms of paths as described there, and start by working out the resolution for the simple modules. In the case of a quiver with relations, the resolution extends with more terms describing the relations, the relations between relations, etc. I'm not sure of a reference for this -- I think it was folklore back when I was working on this stuff -- but it's not too hard to work out by hand. I did the first step (with no claim to originality) in my long ago paper hep-th/0609225, eq (5.3). As one example, Calabi-Yau quivers, which have loops and relations, etc., there's a nice self-dual resolution from Bocklandt.

Answer 3

There's some exposition along these lines in Theorem 2.15 in Schiffler's "Quiver Representations".

Using your notation, let $Q = (Q_0,Q_1)$ be a finite acyclic quiver and $X$ be some representation of $Q$. Let $d_i$ denote the dimension of $X_i$. Then the standard projective resolution of $X$ is $$0 \rightarrow P_1 \rightarrow P_0 \rightarrow X \rightarrow 0$$ where $$P_1 = \bigoplus_{\alpha \in Q_1} d_{s(\alpha)} P(t(\alpha)), \qquad P_0 = \bigoplus_{i \in Q_0} d_i P(i).$$ Looking at $P_0$, it contains $d_i$ copies of $P(i)$ for each vertex $i$. We want $P_0 \rightarrow X$ to be surjective; this choice of $P_0$ is definitely "big enough" to make that possible, since $(P_0)_i$ has at least dimension $d_i$. Basically what we've done is the "cheapest" way to find a projective representation that can surject onto $X$. Then the natural thing to do for $P_0 \rightarrow X$ is to have the map act like the identity on $d_i$ copies of $(P_0)_i$ and like the zero map on any "extra" copies.

So then we need to understand what $P_1 = \textrm{ker}(P_0 \rightarrow X)$ needs to look like (intuitively, $P_1$ is all the "extra" stuff in $P_0$ that's not in $X$). Essentially we want to take $P_0$ and reduce the dimension of each $(P_0)_i$ by $d_i$. This is equivalent to replacing $d_i P(i)$ with $\oplus_{\alpha: i \rightarrow j} d_i P(j)$ (i.e., you replace each copy of $P(i)$ with the direct sum of the projective representations at each vertex $j$ such that there's an arrow $\alpha: i \rightarrow j$ in $Q$).

The same reasoning should carry through for bound quivers, with the relations showing up in the way $P(i)$ is defined for a bound quiver.

I hope this was somewhat useful and not just a repetition of information in the previous answers and/or something that you already knew.