 Solution

Set $A=kQ$, $X_i=e_{i}X$, and let $f_{\rho }$ be the action of the arrow $\rho$ on $X_{s(\rho )}$. Write $P_i=A{e}_i$, a left projective module. The standard projective resolution of a quiver representation isThe algebra acts on the first tensor factor. The augmentation is $\epsilon (p\otimes x)=px$. On the summand for $\rho :i\to j$, the differential sends $p\otimes x$ to $p\rho \otimes x$ in the $i$-summand minus $p\otimes f_{\rho }(x)$ in the $j$-summand. Here $p\in A{e}_j$, so $p\rho \in A{e}_i$. These formulas specify every term and map, and $\epsilon d=0$.

Exactness is the standard path resolution fact: the relations identify a path acting on a vector with successively applying its arrows; uniqueness of the first traversed arrow supplies injectivity of the relation map. Each $A{e}_i$ is a direct summand of $A$ because $e_i$ is an idempotent, and tensoring with a $k$-vector space gives a direct sum of copies of $A{e}_i$. Both terms preceding $X$ are consequently projective modules. Thus the displayed sequence is a projective resolution of length at most one.

The same path resolution works for arbitrary left modules, with possibly infinite-dimensional $X_i$ and arbitrary direct sums of projective modules. Hence every left $A$-module has projective dimension at most one. Equivalently, $A$ is a left hereditary ring: if $N\subseteq P$ with $P$ projective, dimension shifting gives ${\mathrm{Ext}}_A^1(N,M)\cong {\mathrm{Ext}}_A^2(P/N,M)=0$ for every $M$, so $N$ is projective. This argument also covers quivers with oriented cycles.

For dimension vectors $\mathbf{n},\mathbf{m}\in {\mathbb{R}}^{Q_0}$, define the Ringel formApply ${\mathrm{Hom}}_A(-,Y)$ to the projective resolution. Evaluation at $e_i$ identifies ${\mathrm{Hom}}_A(P_i{\otimes }_{k}V,Y)$ with ${\mathrm{Hom}}_k(V,Y_i)$. Consequently there is an exact sequencewhereThis is the extension complex of quiver representations. Taking its alternating dimension sum yields