Unipotent algebraic groups and unipotent representations

Let $k$ be a field and let $G$ be an affine algebraic group over $k$, that is, an affine group scheme of finite type over $k$ (Affine schemes and their coordinate rings, Group schemes of finite type over a field).

(a) Unipotent representations. A finite-dimensional rational representation $(V,r,\rho )$ of $G$ (Rational representations and comodules of an affine group scheme, Vector space over a field) is unipotent if there is a $k$-basis of $V$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis) such that, for every commutative $k$-algebra $R$ and every $g\in G(R)$, the operator $r_R(g)$ on $V{\otimes }_{k}R$ is upper triangular with all diagonal entries equal to $1$ in the scalar-extended basis. Equivalently, by the comodule dictionary, $V$ is unipotent if and only if $V$ has a complete $G$-stable flag $V=V_m\supseteq V_{m-1}\supseteq \cdots \supseteq V_0=0$ with $G$ acting trivially on each quotient $V_i\mathrm{/}{V}_{i-1}$; the equivalence uses that a complete flag with trivial successive quotients is exactly a chain of subspaces in a basis as above, and conversely.

(b) Unipotent groups. The group $G$ is unipotent if every nonzero rational representation of $G$ has a nonzero $G$-fixed vector, equivalently if every simple rational representation of $G$ is one-dimensional with trivial action. Because every rational representation is a union of finite-dimensional subrepresentations (Every element of a comodule lies in a finite-dimensional subcomodule), it suffices to test finite-dimensional representations: $G$ is unipotent if and only if every nonzero finite-dimensional rational representation has a nonzero fixed vector.

No smoothness, reducedness or connectedness of $G$ is imposed; the matrix formulation in (a) is written out because the group scheme $U_n$ of upper unitriangular matrices is introduced separately and is not used in the definition.