Quiver

A quiver is a directed multigraph, allowing loops and multiple arrows.

The Kronecker quiver has two vertices and two parallel arrows from the first to the second. For a representation with arrows $I,N$, an endomorphism consists of the same matrix at both vertices and must commute with $N$. If $N$ is a single Nilpotent Jordan block, this centralizer is $k[N]$, yielding an indecomposable non-brick when the block size exceeds one.

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.

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.

The four-subspace quiver has four sources with arrows into one sink. Four one-dimensional sources mapped to the lines $k{e}_1$, $k{e}_2$, $k(e_1+e_2)$ and $k(\lambda e_1+e_2)$ in $k^2$ give a brick module for every $\lambda$. Preserving the first three lines forces scalar endomorphisms. Its Tits form of a quiver is zero, so the Ringel form gives one-dimensional self-extensions.

The subspace quiver with three sources reduces to the problem of three subspaces in one vector space. The three-subspace decomposition gives eleven indecomposables with vertex dimensions at most one, and one exceptional indecomposable of the three-subspace quiver.

Every triple of subspaces is a direct sum of one-dimensional membership blocks and two-dimensional three-line blocks. Split off the triple intersection, then the pairwise intersections using projections killing the third subspace. Next split complements to $U_i\cap (U_j+U_{\ell })$, using projections killing the other two subspaces. The remaining subspaces are pairwise disjoint and each lies in the sum of the other two. Their sum is $U_1\oplus U_2$, with $U_3$ the graph of an isomorphism $U_1\to U_2$. A graph basis splits this into two-dimensional blocks. A complement to the total span supplies zero-membership blocks. This elementary argument classifies the three-subspace quiver.

Take sources $k,k,k$, sink $k^2$, and arrows sending $1$ to $e_1,e_2,e_1+e_2$. An endomorphism preserving these three lines must be scalar, so the representation has endomorphism ring $k$ and is a brick module. It is the only indecomposable of the three-subspace quiver with a two-dimensional sink.

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.

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.

Its path algebra is $k[t]$. Finite-dimensional representations are vector spaces with a single endomorphism. Thus oriented cycles can give an infinite-dimensional algebra even when the representations studied are finite-dimensional.

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.

A representation of a quiver over a field assigns a vector space to each vertex and a linear map to each arrow. A morphism is a compatible family of vertexwise linear maps.

For $1\to 2\to 3$, $I[a,b]$ has $k$ on the interval of vertices $a,\dots ,b$, zero elsewhere and identity maps along its internal arrows. These six representations are exactly the indecomposables. To decompose $V_1\stackrel{f}{\to }{V}_2\stackrel{g}{\to }{V}_3$, split $\mathrm{im}f\cap \ker g$, its complements inside each of $\mathrm{im}f$ and $\ker g$, and a complement to their sum. Lift the image basis through $f$ and transport the complementary nonkernel basis through $g$. This yields interval blocks directly, without the Gabriel theorem.

For fixed dimensions $(d_1,d_2,d_3)$, a pair $V_1\stackrel{f}{\to }{V}_2\stackrel{g}{\to }{V}_3$ is classified up to vertexwise change of basis by its three ranks $r,s,t$. The multiplicities of its interval representations of an equioriented three-vertex quiver are $m_{13}=t$, $m_{12}=r-t$, $m_{23}=s-t$, $m_{11}=d_1-r$, $m_{22}=d_2-r-s+t$ and $m_{33}=d_3-s$. Nonnegativity is precisely the existence condition. For dimensions $(2,2,2)$ there are ten orbits, of which $(2,2,2)$ is open; its boundary has nine orbits. The two cases $(1,1,0)$ and $(1,1,1)$ show why the individual arrow ranks are insufficient.

At fixed dimension vector, ${\mathrm{Rep}}_Q(\mathbf{n})={\prod }_{\rho :i\to j}{\mathrm{Hom}}_k(k^{n_i},k^{n_j})$ is an irreducible affine space of dimension ${\sum }_{\rho }{n}_{s(\rho )}{n}_{t(\rho )}$. Its points are arrow-matrix collections, and its base-change orbits are isomorphism classes.

The group $\mathrm{GL}(\mathbf{n})={\prod }_i{\mathrm{GL}}_{n_i}(k)$ acts by $(g\cdot f{)}_{\rho }=g_{t(\rho )}{f}_{\rho }{g}_{s(\rho )}^{-1}$. Inverse entries are regular on a general linear group, so this is an algebraic group action. Its stabilizer is the automorphism group of a quiver representation.

The infinitesimal base change action on quiver representations is ${\xi }_x(u{)}_{\rho }=u_{t(\rho )}{x}_{\rho }-x_{\rho }{u}_{s(\rho )}$. Its image is the Zariski tangent space to the orbit, because the stabilizer is a smooth open subset of the endomorphism ring. The extension complex of quiver representations identifies the quotient of the ambient tangent space by this image with ${\mathrm{Ext}}_Q^1(X,X)$. Consequently rigid quiver representations have open orbits.

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.

For a nonzero dimension vector, quotient ${\prod }_i{\mathrm{GL}}_{n_i}(k)$ by the common scalar subgroup $(\lambda I_{n_i}{)}_i$. It acts trivially, so the quotient acts algebraically on the quiver representation space. This is one scalar quotient, rather than a product of the individual projective general linear groups. At dimension vector zero the action is trivial.

This is ${\mathrm{End}}_Q(X{)}^{\times }$, a connected linear algebraic group by the units of a finite-dimensional algebra argument. It is the stabilizer under base change. For $X={⨁}_a{M}_a^{\oplus m_a}$, reduction modulo the radical gives quotient ${\prod }_a{\mathrm{GL}}_{m_a}(k)$.

Fixing a Krull-Schmidt decomposition embeds $L={\prod }_a{\mathrm{GL}}_{m_a}(k)$ by its actions on multiplicity spaces. This splits reduction modulo the Jacobson radical. Hence ${\mathrm{Aut}}_Q(X)=(1+J({\mathrm{End}}_Q(X)))⋊L$, with a closed normal unipotent first factor. The explicit section works in every characteristic.

A finite-dimensional quiver representation is rigid when ${\mathrm{Ext}}_Q^1(X,X)=0$. Its extension complex of quiver representations then has surjective differential. Equivalently, it has an open orbit under base change. Rigidity constrains arrow ranks and path identities.

The stabilizer is the automorphism group of a quiver representation, an open set in its endomorphism algebra. Thus $\dim {\mathcal{O}}_X={\sum }_i{n}_i^2-\dim {\mathrm{End}}_Q(X)$. The Ringel form gives orbit codimension $\dim {\mathrm{Ext}}_Q^1(X,X)$. Since orbits are locally closed and the representation space is irreducible, rigidity is equivalent to an open dense orbit.

A base-change invariant polynomial identity holding on a rigid representation holds on its dense orbit and hence throughout the representation space. A path through nonzero vertex spaces cannot vanish identically there: assign every involved arrow a rank-one map carrying a chosen source vector to the chosen target vector. This works for repeated arrows. Consequently a nonzero path image in a rigid representation cannot be killed by an outgoing arrow into a nonzero vertex space.

This bilinear form is $⟨\mathbf{n},\mathbf{m}{⟩}_Q={\sum }_i{n}_i{m}_i-{\sum }_{\rho :i\to j}{n}_i{m}_j$. The standard projective resolution of a quiver representation gives its value as $\dim {\mathrm{Hom}}_Q(X,Y)-\dim {\mathrm{Ext}}_Q^1(X,Y)$ for the dimension vectors of $X,Y$. Its diagonal is the Tits form of a quiver.

The quadratic form is $q_Q(\mathbf{n})=⟨\mathbf{n},\mathbf{n}{⟩}_Q={\sum }_i{n}_i^2-{\sum }_{\rho :i\to j}{n}_i{n}_j$. It depends on the underlying graph rather than its orientation. For a representation it equals $\dim {\mathrm{End}}_Q(X)-\dim {\mathrm{Ext}}_Q^1(X,X)$. Its positive definiteness forces indecomposables to be rigid brick modules, using the Ringel lemma on bricks.

In the positive definite case this is a nonnegative integer vector with $q_Q(\mathbf{n})=1$, so it is nonzero. The positive definite Tits form indecomposable classification gives one indecomposable isomorphism class per positive root.

For the chain, with $n_0=n_{m+1}=0$, the quadratic form is $\frac{1}{2}{\sum }_{i=0}^m(n_{i+1}-n_i{)}^2$. Its value is one precisely for interval vectors: one $+1$ difference followed by one $-1$, all other differences zero. Thus there are $m(m+1)/2$ positive roots, independently of orientation.

The Ringel lemma on bricks makes every indecomposable a brick module; positive definiteness and the Ringel form then force $q_Q=1$ and rigidity. Conversely, a maximal-orbit representation at a positive root cannot split: nonzero cross extensions increase orbit dimension, while vanishing cross extensions make the quadratic form of a split sum at least two. Open dense orbits give uniqueness. A compact bound on the integer roots gives finite representation type.

The dimension vector of a finite-dimensional quiver representation is $({\dim }_k{X}_i{)}_{i\in Q_0}$. Direct sums add dimension vectors. Fixing the vector gives an affine space of possible arrow matrices, the quiver representation space.

For quiver representations $X=(X_i,f_{\rho })$ and $Y=(Y_i,g_{\rho })$, a morphism consists of maps $h_i:X_i\to Y_i$ satisfying $h_{t(\rho )}{f}_{\rho }=g_{\rho }{h}_{s(\rho )}$ for every arrow. Composition is vertexwise. These families are the kernel of the extension complex of quiver representations.

A quiver representation morphism is an isomorphism exactly when all its vertex maps are invertible. Choosing bases turns this into the base change action on quiver representations, whose orbits coincide with isomorphism classes.

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

A connected quiver has finite representation type exactly when its underlying unoriented graph is a Dynkin diagram of type $A$, $D$, or $E$. In that case, dimension vectors give a bijection from indecomposable representations to positive roots.

At a sink $i$, the Bernstein–Gelfand–Ponomarev reflection functor reverses every arrow into $i$ and replaces the vector space at $i$ by the kernel of the sum of its incoming maps. Away from the simple representation supported at $i$, reflection at the resulting source is an inverse functor.

The path algebra $kQ$ has the directed paths in a quiver $Q$ as a basis; multiplication concatenates composable paths and is zero otherwise. Representations of $Q$ are naturally modules over $kQ$.

Closed paths at vertex $i$ form a basis of $e_{i}kQ{e}_i$. For nonzero finite linear combinations $f,g$, take their largest path lengths $a,b$. In the length-$a+b$ part of $fg$, a concatenated path has a unique cut into lengths $a,b$. Thus a product of nonzero top-degree coefficients cannot cancel, and $fg\ne 0$. The corner is a noncommutative domain and has no idempotents except zero and its identity $e_i$. This argument works with oriented cycles and proves indecomposability of the vertex projective module of a path algebra without falsely assuming that its endomorphism ring is local.

The vertex projective module of a path algebra $P(i)=kQ{e}_i$ has at vertex $j$ the span of paths from $i$ to $j$. Each arrow appends itself to such paths. It is a projective module because it is a summand of the free module $kQ$. Its endomorphism ring is $(e_{i}kQ{e}_i{)}^{\mathrm{op}}$, acting by right multiplication. The closed-path corner of a path algebra is a domain, so this ring has no nontrivial idempotents and $P(i)$ is indecomposable. Directed cycles may make these projectives infinite-dimensional.

Evaluation at $e_i$ identifies a module homomorphism $h:kQ{e}_i\to X$ with $h(e_i)\in e_{i}X$. Given $x\in e_{i}X$, the inverse sends $p$ to $px$. The isomorphism is natural and shows that Hom functor evaluation on a vertex projective is exact.

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 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.