TITLE: Can Hom_gp(G,H) fail to be representable for affine algebraic groups?
QUESTION [16 upvotes]: Let $G$ and $H$ be affine algebraic groups over a scheme $S$ of characteristic 0 and let $\textbf{Hom}_{S,gp}(G,H)$ be the functor $T \mapsto \text{Hom}\_{T,gp}(G,H)$

Theorem (SGA 3, expose XXIV, 7.3.1(a)): Suppose G is reductive. Then 
  $\textbf{Hom}_{S,gp}(G,H)$ is representable by a scheme.

Can this fail if $G$ is not reductive? I worked out a few example with $G = \mathbb{G}_a$, but they were representable.

REPLY [17 votes]: There is a reasonable salvage, at least if the base scheme is a field: Hom(G,H) is a direct limit of representable subfunctors.  See Lemma A.8.13 in the book "pseudo-reductive groups" (where it is used to prove that the scheme-theoretic fixed locus for a linearly reductive group acting on a connected reductive group is always reductive (possibly disconnected) provided the base is a field.