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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.

An algebraic group is connected when its underlying Zariski topology is a connected space. An unipotent algebraic group is a linear algebraic group whose elements, in a faithful matrix realization, satisfy that $u-I$ is nilpotent. A reductive algebraic group, in the connected convention, is a connected linear algebraic group whose unipotent radical is trivial: it has no nontrivial connected normal unipotent algebraic subgroup.

Let $E={\mathrm{End}}_Q(X)$. It is a finite-dimensional associative algebra, and ${\mathrm{Aut}}_Q(X)=E^{\times }$. The latter is the nonempty open subset where the determinant on ${⨁}_i{X}_i$ is nonzero. Since the affine space $E$ is irreducible, this open subset is irreducible and hence connected. It is a linear algebraic group: within $\mathrm{GL}({⨁}_i{X}_i)$ it is cut out by the linear equations of preserving the vertex summands and commuting with the arrow maps. Thus the automorphism group is connected.

By the Krull–Schmidt theorem, write $X\cong {⨁}_{a=1}^s{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposables $M_a$. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local algebra. Its residue division algebra is $k$: every element of a finite-dimensional division algebra over the algebraically closed field has an eigenvalue for left multiplication, and subtracting that scalar gives a noninvertible element, hence zero.

The standard radical description of the endomorphism algebra of a Krull-Schmidt decomposition therefore givesConcretely, maps between nonisomorphic summands lie in the radical, while on each isotypic block one reduces all entries modulo the local radical. The off-diagonal rule follows because a composition $M_a\to M_b\to M_a$ cannot be a unit for $a\ne b$: it would split $M_a$ off the indecomposable $M_b$, forcing an isomorphism. The finite-dimensional radical description then identifies the kernel of this block reduction with $J(E)$. Put $U=1+J(E)$. The Jacobson radical of the finite-dimensional algebra is nilpotent, so every $1+j$ is a unit and is unipotent on ${⨁}_i{X}_i$. Also $U$ is closed, being the translate of the linear subspace $J(E)$, and normal, being the kernel ofThis map has an explicit algebraic section: after fixing the direct-sum decomposition, a matrix $B_a$ acts on the $m_a$ copies by $B_a\otimes 1_{M_a}$. These block actions give a subgroup $L\cong {\prod }_a{\mathrm{GL}}_{m_a}(k)$ with trivial intersection with $U$. Every unit is uniquely a product of an element of $U$ and one of $L$. HenceThis Levi decomposition of a quiver automorphism group holds in every characteristic; the section is constructed directly from the multiplicity spaces.

For $G={\mathbb{G}}_a$, write $k[G]=k[t]$. Under the left-right regular representation of an affine algebraic group,so the action factors through the difference map ${\mathbb{G}}_a\times {\mathbb{G}}_a\to {\mathbb{G}}_a$. The degree filtrationis stable and has trivial one-dimensional successive quotients. The module is indecomposable; every nonzero submodule contains a nonzero translation difference of lower degree and, on iteration using suitable translations, meets the unique invariant line $k\cdot 1$.

For every finite-dimensional rational $G$-module $V$, the matrix-coefficient construction givesOn the other hand,Indeed, a nonzero finite-dimensional quotient of $k[G]$ would have a simple quotient. Every simple rational representation of the unipotent algebraic group ${\mathbb{G}}_a$ is trivial, so this would give a nonzero translation-invariant functional on $k[t]$. No such functional exists: if $c_j$ is its first nonzero value on a monomial, translating a sufficiently high monomial produces a nonzero coefficient times $c_j$, contradicting invariance.

The derived subgroup of an affine algebraic group $[G,G]$ is the closed subgroup generated by the commutators $xy{x}^{-1}{y}^{-1}$. If $G$ is connected, the image of every finite product of commutator maps is connected and contains the identity. The closures of these images form an increasing sequence; once their dimensions stabilize, the stable member is closed under products and inverses and equals $[G,G]$. Hence $[G,G]$ is connected.

Now suppose the connected group $G$ is solvable. The Lie-Kolchin theorem conjugates a faithful representation of $G$ into the upper triangular matrices. Every commutator then has all diagonal entries equal to one, so every element of $[G,G]$ is unipotent. Moreover $[G,G]$ lies in the upper unitriangular group, whose superdiagonal filtration is a central series. It is therefore a nilpotent group.

A diagonalizable algebraic group is a closed subgroup of a product of copies of ${\mathbb{G}}_m$. A unipotent algebraic group has only unipotent elements, while a semisimple algebraic group here means one all of whose elements are semisimple. A reductive algebraic group is smooth, connected, affine, and has trivial connected normal unipotent radical.

The Kolchin theorem conjugates a faithful representation of a unipotent algebraic group $U$ into the upper unitriangular group $U_n$. Let $U_n^{(r)}$ consist of matrices whose first $r-1$ superdiagonals vanish. Matrix multiplication givesSince $U_n^{(n)}=1$, this filtration is a finite central series, so $U_n$ and every subgroup of it are nilpotent groups. Hence $U$ is nilpotent.

The converse fails: ${\mathbb{G}}_m$ is abelian, hence nilpotent, but its nonidentity points are semisimple rather than unipotent.

A reductive algebraic group is a smooth connected affine algebraic group whose largest connected normal unipotent algebraic group is trivial.

An element $u$ of a ring is unipotent when $u-1$ is nilpotent. For a matrix, this says that every eigenvalue is one. A unipotent algebraic group consists of unipotent elements in a faithful linear representation.

A unipotent algebraic group is built from unipotent elements. A unipotent algebraic group is an affine algebraic group all of whose elements are unipotent. The Kolchin theorem embeds it into an upper unitriangular group.