It is well known that any unipotent algebraic group (over a field) can be embedded as a closed subgroup of $U\_n$ for some $n$, where $U\_n$ denotes the set of all $n \times n$ upper triangular matrices with $1$'s on the diagonal. 


An algebraic subgroup $H$ of an algebraic group $G$ is called $\textbf{observable}$ if every (rational) linear representation of $H$ arises as the restriction of a (rational) linear representation of $G$. My question:


Under what conditions, if any, is a unipotent algebraic group observable with respect to its overgroup $U\_n$?


Background: I have in hand a certain result concerning the characteristic $p>0$ representation theory of the groups $U\_n$, when $p$ is sufficiently large when compared to $n$ and the dimension of a representation. It essentially says that such representation look functorially identical to representations of $U\_n^\infty$ (countable direct product of copies of $U\_n$) in characteristic zero.


If I can pass representations of a unipotent algebraic group $U$ to representations of its overgroup $U\_n$, then I get the theorem for free for $U$. Of course, I have no reason to believe when and if this is true, it’s just a shot in the dark that gives me a wonderful shortcut.


Ideas?



Observable Unipotent Algebraic Subgroups of the Unipotent Upper Triangular Groups
To answer your comment, take the representation 


$\left(\begin{array}{ccc} 1 & a & b \\ 0 & 1 & a \\ 0 & 0 & 1\end{array}\right) \to \left(\begin{array}{cc} 1 & b-a^2/2 \\ 0 & 1 \end{array}\right)$


If this were a pullback of a representation of the additive group, it would be a pull-back of a two-dimensional unipotent representation, thus an abelian representation, so its kernel would include the commutator subgroup:


$\left(\begin{array}{ccc} 1 & 0 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$


while this representation's kernel does not include that subgroup.


