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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.

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 an algebraic group action, the orbit dimension is the dimension of the group minus that of its stabilizer. For a finite-dimensional representation, the stabilizer is an automorphism group, open in its endomorphism ring, so their dimensions agree. In a quiver representation space, the Ringel form turns the orbit codimension into the dimension of the self-extension group.

A unipotent algebraic group admits a faithful linear representation in which every group element is a unipotent matrix. For $E={\mathrm{End}}_Q(X)$, invertibility is the nonvanishing condition ${\prod }_i\det h_i\ne 0$. Therefore ${\mathrm{Aut}}_Q(X)=E^{\times }$ is a nonempty Zariski-open subset of the vector space $E$.

Take a Krull-Schmidt decomposition $X\cong {⨁}_{a=1}^r{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposable modules. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local endomorphism ring. Its residue division algebra is $k$: over an algebraically closed field, every element of a finite-dimensional division algebra has an eigenvalue and hence must be scalar. The semisimple quotient of a module endomorphism algebra consequently gives, for the Jacobson radical $J=J(E)$,is surjective with kernel $1+J$. The nilpotence of $J$ makes $(1+j{)}^{-1}=1-j+j^2-\cdots$ finite and each $1+j$ unipotent. The kernel is closed and normal. Acting on the multiplicity spaces embeds the product of general linear groups back into $E^{\times }$ and splits this quotient. This proves the Levi decomposition of a quiver automorphism groupSince $1+J$ is the unipotent radical, a nonzero $X$ is indecomposable exactly when ${\mathrm{Aut}}_Q(X)/R_u({\mathrm{Aut}}_Q(X))\cong {\mathbb{G}}_m$: the product has a single factor of size one.

For the base change action on quiver representations, the orbit map is $g\mapsto (g_j{x}_{\rho }{g}_i^{-1}{)}_{\rho :i\to j}$. Substituting $g_i=I+ϵu_i$, with ${ϵ}^2=0$, shows its differential isIts kernel is ${\mathrm{End}}_Q(X)$. The stabilizer is smooth because it is open in that vector space. Hence the differential has rank $\dim \mathrm{GL}(\mathbf{n})-\dim {\mathrm{Aut}}_Q(X)=\dim {\mathcal{O}}_X$, and its image is the Zariski tangent space $T_x{\mathcal{O}}_X$. The normal space to a quiver orbit is thereforeThe ambient quiver representation space is an irreducible affine space, and orbits are locally closed. An orbit is open exactly when its dimension equals that ambient dimension, equivalently when ${\mathrm{Ext}}_Q^1(X,X)=0$. This proves that rigid quiver representations have open orbits. Such an orbit is dense and unique, since two nonempty open subsets of an irreducible space intersect.

The dimension of a topological space by irreducible chains is the supremum of the integers $d$ for which there is a chain $Z_0⊊Z_1⊊\cdots ⊊Z_d$ of nonempty irreducible closed subsets of the space. Closedness here is relative to the given locally closed space. For varieties this is their Krull dimension.

An algebraic group is a group whose underlying space is an algebraic variety and whose multiplication $G\times G\to G$ and inversion $G\to G$ are morphisms. An algebraic group action on a variety $V$ is a morphism $G\times V\to V$ satisfying the identity and associativity axioms of a group action.

For the homomorphism $\phi$, the kernel is ${\phi }^{-1}(1_H)$, so the kernel is closed. To prove closedness of the image, use the Chevalley constructibility theorem: the image of a morphism of varieties is constructible. Thus $S=\phi (G)$ is a constructible subset of a variety and an abstract subgroup. Its closure $C$ is also a subgroup: translation by elements of $S$ preserves $C$, and continuity then extends multiplication and inversion to $C$.

A dense constructible subset contains a dense open subset $U$ of its closure. For $c\in C$, both $U$ and $cU$ are dense open subsets of $C$, so their intersection is nonempty. If $u=cv$ with $u,v\in U\subseteq S$, then $c=u{v}^{-1}\in S$. Hence $S=C$, proving that the image is closed. This is the principle that a constructible subgroup is closed.

Every nonempty fiber of $G\to S$ is a translate of $\ker \phi$ and has that same dimension. The fiber dimension theorem therefore givesThis dimension formula for an algebraic group homomorphism is a dimension statement, so it does not require separability of $\phi$.

For a dimension vector of a quiver representation $\mathbf{n}$, setThe base change action on quiver representations isThe entries are regular functions on the product of the general linear groups and the quiver representation space, because inverse entries are cofactors divided by the invertible determinant. Thus this is an algebraic group action.

For nonzero $\mathbf{n}$, let $\Delta k^{\times }$ be the common scalar subgroup $(\lambda I_{n_i}{)}_i$ and define $\mathrm{PGL}(\mathbf{n})=\mathrm{GL}(\mathbf{n})/\Delta k^{\times }$. Common scalars act trivially, so the formula descends to an algebraic group action of this projective base change group of a quiver. This is a quotient by one common scalar, not a product of the individual projective groups. If every $n_i=0$, the representation space is a point and both actions are taken to be trivial.

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.

An indecomposable of dimension vector $\mathbf{n}$ is a brick module and has $q_Q(\mathbf{n})=1$ by the previous parts. Its automorphism group of a quiver representation has dimension one. ThereforeIt follows that its orbit is open and dense in the irreducible quiver representation space. Two different indecomposables of the same dimension vector would give two disjoint nonempty open orbits. Nonempty open subsets of an irreducible space must intersect, so this is impossible. Hence an indecomposable is uniquely determined by its dimension vector, up to isomorphism. This is the positive definite Tits form indecomposable classification.

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.