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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.

Choose integer vertex weights increasing along every arrow of a finite acyclic quiver. Acting at vertex $i$ by $t^{w_i}I$ multiplies each arrow by a positive power of $t$. At $t=0$ all arrows vanish. Thus every orbit closure contains the origin of the quiver representation space.

This two-sided ideal is spanned by paths of positive length; its $m$th power is spanned by paths of length at least $m$. The quotient is a product of copies of the ground field indexed by vertices. For a finite acyclic quiver, the ideal is nilpotent and equals the Jacobson radical. A loop shows that the equality can fail in general.

A quiver has finite representation type over a field when it has only finitely many isomorphism classes of indecomposable finite-dimensional representations.

The quiver has two vertices and one arrow. Splitting a representation's kernel, image and complementary target decomposes it into source-only and sink-only simples and copies of the identity $k\to k$. These are its three indecomposables. Fixed-dimension isomorphism classes are rank orbits of a matrix under left-right multiplication.

An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

For a finite-dimensional algebra over an algebraically closed field, basic algebra means $A/J(A)\cong k^n$, so its simple modules are one-dimensional and occur once in the semisimple quotient of the regular module. A hereditary ring has every submodule of a projective module projective; equivalently its global dimension is at most one.

Choose representatives $S_1,\dots ,S_n$ of the simple modules. Our Ext quiver convention gives one vertex $i$ for $S_i$ andWith primitive orthogonal idempotents $e_i$, put $P_i=A{e}_i$ and $J=J(A)$. Applying ${\mathrm{Hom}}_A(-,S_j)$ to $0\to J{P}_i\to P_i\to S_i\to 0$ givesThe restriction map from $\mathrm{Hom}(P_i,S_j)$ is zero since every map to a simple kills $J{P}_i$, and $P_i$ is projective. Thus the arrow multiplicity is $\dim e_j(J/J^2)e_i$. This specifies the orientation rather than silently replacing the quiver by its opposite.

Because $A$ is hereditary, $J{P}_i$ is projective. Its semisimple top has $a_{ij}$ copies of $S_j$, where $a_{ij}$ is the arrow multiplicity. Its projective cover therefore supplies an epimorphismThis splits by projectivity of $J{P}_i$. Its kernel is a summand contained in the radical of the source; projection onto that summand and Nakayama lemma force the kernel to be zero. HenceEvery arrow $i\to j$ forces $\dim P_i>\dim P_j$, so $Q$ has no directed cycles.

Choose lifts in $e_{j}J{e}_i$ of bases of the arrow spaces. Send the vertices and arrows of the path algebra to these idempotents and lifts, multiplying paths in the order $\beta \alpha$ for consecutive arrows $\alpha :i\to j$, $\beta :j\to l$. This defines an algebra homomorphism $kQ\to A$. It is surjective: vertices span $A/J$, arrows span $J/J^2$, and products show that paths of length $r$ span $J^r/J^{r+1}$. Nilpotence of $J$ then gives all of $A$.

Let $N_i$ be the number of directed paths starting at $i$, including the length-zero path. Acyclicity gives the same recursion $N_i=1+{\sum }_j{a}_{ij}{N}_j$, with value one at a sink. Induction from sinks gives $N_i=\dim P_i$. ThereforeThe surjective homomorphism is an isomorphism. This proves the basic hereditary algebra path-algebra theorem:

For each connected component $C$ of the underlying unoriented graph, $e_C={\sum }_{i\in C}{e}_i$ is a central idempotent. Paths never connect distinct components, so $kQ={⨁}_C{e}_{C}kQ$ as a direct sum of two-sided ideals. To prove each is a block, let $z$ be any central idempotent. Modulo the arrow ideal it has coordinates $c_i\in \{0,1\}$. Commuting $z$ with an arrow $i\to j$ and reducing modulo the square of that ideal gives $c_i=c_j$. Thus these coordinates are constant on each component. If they are zero on a component, the restriction of $z$ there is a nilpotent idempotent, hence zero; if they are one, apply that argument to $1-z$. Consequently

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.

Use the following Jacobson radical facts for a unital algebra: a nilpotent ideal is contained in the Jacobson radical; under a surjective algebra homomorphism the image of the Jacobson radical is contained in the radical of the quotient; and a finite product of fields has zero Jacobson radical. For completeness, if $I^N=0$ and $x\in I$, then $ax\in I$ for every $a\in A$, and $1-ax$ is invertible with inverse ${\sum }_{j=0}^{N-1}(ax{)}^j$. The usual unit criterion for the Jacobson radical therefore gives $I\subseteq J(A)$. The quotient assumption gives $J(A)\subseteq I$. HenceThis is the nilpotent ideal with semisimple quotient radical criterion.

For the finite quiver under consideration, let $R$ be the arrow ideal of a path algebra, spanned by paths of positive length. If $Q$ has $r$ vertices and no oriented cycle, a path cannot repeat a vertex, so $R^r=0$. Meanwhile $kQ/R\cong {\prod }_{i\in Q_0}k$, with the constant paths giving the coordinate idempotents. The criterion just proved yields $J(kQ)=R$.

The condition on cycles is necessary. A quiver with one loop has path algebra $k[t]$, whose arrow ideal is $(t)$. But $J(k[t])=0\ne (t)$: the maximal ideals $(t-a)$, $a\in k$, have intersection zero, since the algebraically closed field $k$ is infinite and a nonzero polynomial has only finitely many roots. Thus the arrow ideal need not be the Jacobson radical when oriented cycles are present.

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

We first record why rigid quiver representations have open orbits. Let $\mathbf{n}=\dim X$, $G={\prod }_i\mathrm{GL}(X_i)$ and $R={\mathrm{Rep}}_Q(\mathbf{n})$. The stabilizer is ${\mathrm{Aut}}_Q(X)$, the nonempty open set of units in ${\mathrm{End}}_Q(X)$, so it has dimension $\dim {\mathrm{End}}_Q(X)$. The dimension formula for an algebraic group homomorphism, or the same constant-fiber argument for the orbit map, givesBy the Ringel form identity and ${\mathrm{Ext}}_Q^1(X,X)=0$, this equals ${\sum }_{\sigma }{n}_{s(\sigma )}{n}_{t(\sigma )}=\dim R$. An algebraic-group orbit is locally closed; since $R$ is an irreducible affine space, this full-dimensional orbit is open and dense.

Write $q=\rho \pi$ for the composed path. Each matrix entry of $q$ is a polynomial function on $R$. Since $qX=0$, change of basis makes it zero on all of ${\mathcal{O}}_X$, hence on all of $R$ by density. Suppose instead that $X_{t(\rho )}\ne 0$. The condition $\pi X\ne 0$ ensures that every vertex space visited by $\pi$ is nonzero. Choose a vector $v_a\ne 0$ and a functional ${\ell }_a$ with ${\ell }_a(v_a)=1$ at every vertex visited by $q$. Assign to every arrow $\sigma$ occurring in $q$ the map $v_{t(\sigma )}{\ell }_{s(\sigma )}$, and assign arbitrary maps, say zero, to the other arrows. At this representation, the path $q$ carries its initial chosen vector to its final chosen vector, so $q$ is nonzero, a contradiction.

The construction uses a single assigned map per arrow, so it still works if an arrow or vertex occurs repeatedly in the path. ThereforeThis proves the path identities in a rigid quiver representation claim for an arbitrary quiver. Maximal rank of the individual arrows alone would not justify the conclusion about their composition.

Let $K=\ker f_{\rho }$ and $C=Y_j/\mathrm{im}{g}_{\rho }$. Consider the map from the arrow term of the extension complex of quiver representations to ${\mathrm{Hom}}_k(K,C)$ that restricts its $\rho$-component to $K$ and then takes the quotient in $Y_j$. This map is surjective: a map $K\to C$ can be lifted to $Y_j$ and extended from $K$ to $X_i$.

Every coboundary is killed by this map, since for $x\in K$,It therefore induces a surjectionBoth vector spaces in the final Hom functor are nonzero, so its dimension is positive. Hence ${\mathrm{Ext}}_Q^1(X,Y)\ne 0$. This is the kernel-cokernel obstruction to splitting a quiver extension and remains valid for loops and repeated arrows elsewhere in the quiver.

An acyclic finite quiver has integer vertex weights $w_i$ such that $w_j>w_i$ for every arrow $i\to j$; for example, let $w_i$ be the maximum length of a path ending at $i$. For $t\ne 0$, act by $g_i(t)=t^{w_i}{I}_{n_i}$. ThenEvery exponent is positive. The arrow matrices therefore extend polynomially to $t=0$ with all arrows zero, which is the origin of the quiver representation space. The values for $t\ne 0$ lie in ${\mathcal{O}}_X$, soThis proves that acyclic quiver representations degenerate to zero, also when some vertex spaces are zero.

A vertex $i$ of a quiver is a sink when no arrow starts at $i$. A representation of a quiver assigns a vector space $V_j$ to every vertex and a linear map $V_a:V_j\to V_{\ell }$ to every arrow $a:j\to \ell$. A morphism $f:V\to W$ is a family of linear maps $f_j:V_j\to W_j$ such that $f_{\ell }{V}_a=W_a{f}_j$ for every arrow. The quiver has finite representation type when it has only finitely many isomorphism classes of indecomposable finite-dimensional representations.

At a sink $i$, the Bernstein–Gelfand–Ponomarev reflection functor replacesand reverses the arrows ending at $i$; the new arrow maps are the kernel inclusion followed by the coordinate projections. Every representation is a direct sum of copies of the simple representation $S_i$ and a representation for which the displayed incoming map is surjective. On the latter representations, reflection at the resulting source, using the corresponding cokernel, is inverse up to natural isomorphism. Thus reflection gives a bijection between indecomposable representations other than $S_i$ on the two sides. Adding the one omitted simple representation on each side proves that reversing all arrows into a sink preserves finite representation type.

For the four-arrow star $Q_1$, take $V_1=\cdots =V_4=k$, $V_5=k^2$, and let the four arrows have imagesAn endomorphism must preserve the first two lines, so its map on $V_5$ is diagonal. Preserving the third forces its diagonal entries to agree, and then all vertex maps are multiplication by that same scalar. The endomorphism ring is therefore $k$, so this representation is a brick module and hence indecomposable. Any isomorphism between parameters preserves the first three labelled lines; the induced projective linear transformation is therefore the identity, and the fourth line gives $\lambda =\mu$. Since $k$ is infinite, this is an infinite family of pairwise nonisomorphic indecomposables. Hence $Q_1$ is not of finite representation type. Repeatedly applying the reflection result to sinks or, dually, to sources shows that every orientation obtained by reversing some of its arrows also has infinite representation type.

No. Let $Q$ be the quiver $1\to 2$ over a field $k$, and take the representation of a quiverAn endomorphism is a pair of scalar maps $(a,b)$ satisfying $b=a$, so its endomorphism ring is $k$. Every nonzero endomorphism is therefore an isomorphism, making this representation a brick module. It nevertheless has the proper nonzero subrepresentation $0\to k$, so it is not an irreducible module.

Equivalently, this is a nonsimple module over the path algebra $kQ$ whose endomorphism ring is a division ring.

For a quiver with finitely many vertices, send a representation of a quiver $(X_i,f_{\rho })$ to ${⨁}_i{X}_i$. Vertex idempotents act as projections, and paths act as composites of the arrow maps. Conversely, a unital module $M$ gives vertex spaces $e_{i}M$ and arrow maps $x\mapsto \rho x$. Compatible vertex maps are exactly module homomorphisms. These constructions give inverse equivalences of categories.

This quiver has one vertex and $r$ loops. Paths are words in the loops, so its path algebra is a free associative algebra. A representation assigns an arbitrary endomorphism to each loop.

A vertex of a quiver is a sink when no arrow has that vertex as its source. Dually, a source is a vertex at which no arrow ends.

This quiver has $r$ source vertices, each with one arrow to a common sink. Its path algebra has the vertex idempotents and the arrows as basis. After extracting source kernels from a representation, its remaining arrows identify the source spaces with subspaces of the sink.