 Solution

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.