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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?

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    $\begingroup$ Any block-diagonal embedding of $U_{a_1} \times U_{a_2} \times \dots \times U_{a_k}$ into $U_{a_1+a_2+\dots+a_k}$ is observable for the rather trivial reason that it is also a quotient group. Any subgroup $G$ such that $[G,G] \neq G \cap [U_n,U_n]$ is not observable because it has two-dimensional representations that are not pullbacks of two-dimensional, therefore abelian, representations of $U_n$. For an example, take any $2$-dimensional subgroup of $U_3$ or $5$-dimensional subgroup of $U_6$. $\endgroup$ – Will Sawin Jul 1 '12 at 6:50
  • $\begingroup$ A typo: $U_6$ should be $U_4$. $\endgroup$ – Will Sawin Jul 1 '12 at 17:47
  • $\begingroup$ Thanks for the answer, a big help. Another question: Let, for example, $G$ be the subgroup of $U_3$ consisting of matrices where the (1,2) entry and (2,3) entry are equal. Do you happen to have in mind a representation of $G$ that does not lift to one of $U_3$? $\endgroup$ – Mike Crumley Jul 2 '12 at 23:37
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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.

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