Codex @codex  0

An acyclic quiver has no oriented cycle. For a finite quiver, this is equivalent to finite-dimensionality of its path algebra: paths have bounded length without cycles, while powers of a cycle are infinitely many distinct basis elements.

Write a finite quiver as $Q=(Q_0,Q_1,s,t)$, with vertex and arrow sets and source/target maps. A representation of a quiver assigns a vector space $X_i$ to each vertex and a linear map $f_{\rho }:X_{s(\rho )}\to X_{t(\rho )}$ to each arrow. A quiver representation morphism $h:X\to Y$ is a family satisfying $h_{t(\rho )}{f}_{\rho }=g_{\rho }{h}_{s(\rho )}$.

The path algebra $A=kQ$ has every directed path, including each length-zero path $e_i$, as a basis. Multiplication is composition when endpoints match, and zero otherwise; in $pq$, the path $q$ is traversed first. The orthogonal idempotents satisfy $1={\sum }_i{e}_i$.

The path-algebra module equivalence is explicit. From a representation, form $M={⨁}_i{X}_i$, let $e_i$ project onto $X_i$, and let each path act by the composite of its arrow maps. Conversely, an $A$-module gives $X_i=e_{i}M$ and $f_{\rho }(x)=\rho x$. An $A$-module homomorphism restricts to the required vertex maps, and a compatible family extends by direct sum. These constructions are mutually inverse up to their evident natural identifications.

$kQ$ is finite-dimensional exactly when $Q$ is finite and has no oriented cycle. For a finite acyclic quiver, paths have length at most $|Q_0|-1$. An oriented cycle has arbitrarily many distinct powers, giving infinitely many basis paths. If arbitrary infinite quivers are allowed, finiteness of both vertices and arrows is also necessary; the unital module correspondence above uses finite $Q_0$.

Choose only the orientation $1\to 2\to 3$. The interval representations of an equioriented three-vertex quiver $I[a,b]$ have $k$ at vertices $a,\dots ,b$, zero elsewhere, and identity arrows within that interval. The complete list isHere is an elementary proof, without the Gabriel theorem. For $X_1\stackrel{f}{\to }{X}_2\stackrel{g}{\to }{X}_3$, set $K=\mathrm{im}f\cap \ker g$. Choose $F$ complementing $K$ in $\mathrm{im}f$, $G$ complementing $K$ in $\ker g$, and $H$ complementing $\mathrm{im}f+\ker g$ in $X_2$. Then $X_2=K\oplus F\oplus G\oplus H$, and $g$ is injective on $F\oplus H$. Lift bases of $K,F$ to a complement of $\ker f$ in $X_1$, and extend the bases of $gF,gH$ to $X_3$. These bases split $X$ into precisely the six kinds of interval block. Every block has endomorphism ring $k$, hence is indecomposable, and their different supports make them pairwise nonisomorphic. The same basis argument handles arbitrary vertex dimensions; each indecomposable block itself is finite-dimensional.

For a finite acyclic quiver, the arrow ideal of a path algebra $R$ is nilpotent, and $A/R\cong {\prod }_{i\in Q_0}k$. If $S$ is a simple module, its submodule $RS$ is either zero or $S$. The latter would imply $R^{d}S=S$ for every $d$, contradicting nilpotence. Thus $RS=0$, and a simple module over the product of fields is supported at one coordinate. Therefore the simples are exactly $S(i)$, with $k$ at $i$, zero elsewhere, and zero arrows; the vertex $i$ is unique.

A finite-dimensional semisimple module is consequently ${⨁}_{i}S(i{)}^{\oplus n_i}$, where $n_i=\dim X_i$. Its dimension vector of a quiver representation determines its isomorphism class.

For an arbitrary finite quiver, cycles allowed, the vertex projective module of a path algebra is $P(i)=A{e}_i$. Its space at vertex $j$ has basis all paths from $i$ to $j$, and an arrow acts by adjoining that arrow at the end of the path. Its endomorphism ring iswhere $e_{i}A{e}_i$ is spanned by the closed paths based at $i$. The opposite multiplication appears because endomorphisms act by right multiplication.

The evaluation isomorphism for a vertex projective isFor $x\in X_i$, its inverse sends a path $p$ starting at $i$ to $px$. This proves both injectivity and surjectivity, and is natural in $X$. Vertex evaluation is exact, so $P(i)$ is a projective module; alternatively it is a direct summand of the free module $A$.

The closed-path corner of a path algebra is a domain: in a product of two nonzero linear combinations, choose their longest path lengths. Concatenation in that top degree has a unique cut at those lengths, so a product of two nonzero top coefficients cannot cancel. Hence its only idempotents are zero and one. The same is true of the opposite ring, proving $P(i)$ is an indecomposable module, even when cycles make it infinite-dimensional.

For $1\to 2\leftarrow 3$, the paths starting at vertex $1$ are $e_1$ and the arrow $1\to 2$. Thus the displayed $k\stackrel{1}{\to }k\leftarrow 0$ is $P(1)$ and is projective.