Representation of a quiver

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.