TITLE: Intuition behind the canonical projective resolution of a quiver representation
QUESTION [8 upvotes]: 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?

REPLY [3 votes]: 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.