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In this question, all representations are finite-dimensional representations over $\mathbb{C}$.

Fix some $n \geq 3$. Assume that $V$ is a representation of $\text{SL}(n,\mathbb{Z})$. Also, assume that $W$ is a subrepresentation of $V$ and set $V' = V/W$, so we have a short exact sequence

$$0 \longrightarrow W \longrightarrow V \longrightarrow V' \longrightarrow 0$$

of $\text{SL}(n,\mathbb{Z})$ representations. Assume that the actions of $\text{SL}(n,\mathbb{Z})$ on $W$ and $V'$ extend to actions of $\text{SL}(n,\mathbb{C})$. Question : Must the action of $\text{SL}(n,\mathbb{Z})$ on $V$ extend to an action of $\text{SL}(n,\mathbb{C})$?

Margulis superrigidity says that the action virtually extends, but I can't construct representations where it doesn't extend on the nose. Of course, if it extends then $V = V' \oplus W$ as $\text{SL}(n,\mathbb{C})$-representations.

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What does "virtually extends" mean? If it means "can be extended to a subgroup of finite index of $SL(n,\mathbf C)$", the question becomes trivial as $SL(n,\mathbf C)$ does not have any proper subgroups of finite index as it is connected in the Zariski topology. (Note that the map $SL(n,\mathbf C) \to PSL(n,\mathbf C)$ has finite kernel, but $PSL(n,\mathbf C)$ doesn't lift to a subgroup.) –  Guntram Mar 21 '11 at 7:37
"Virtually extends" means that there is a representation of $SL(n,\mathbf{C})$ that coincides with the given representation of $SL(n,\mathbb{Z})$ on a subgroup of finite index of $SL(n,\mathbb{Z}).$ –  Victor Protsak Mar 21 '11 at 12:30
This is intriguing, but maybe the assumption that actions of the discrete group extend to "actions" of the Lie/algebraic group could be clarified: are the latter actions taken in the sense of an abstract group or as a Lie/algebraic group? (Borel-Tits work on abstract homomorphisms of algebraic groups might or might not be relevant.) Also, the set-up seems meaningful for more arbitrary simple Lie/algebraic groups of rank at least 2. –  Jim Humphreys Mar 21 '11 at 14:10
@Jim : Extends to a representation of the Lie/algebraic group. And it is definitely interesting for general lattices in higher rank Lie groups! I just chose the first nontrivial case to make the statement accessible to the widest range of people. –  Tim O Mar 21 '11 at 14:54
You probably mean $SL(n,R)$ in place of $SL(n,C)$, if you want $SL(n,Z)$ to be a lattice. –  Ben Wieland Mar 21 '11 at 22:51

1 Answer 1

up vote 9 down vote accepted

Consider the surjective map of $SL(n,\Bbb Z)$-modules $Hom_{\Bbb C}(V',V)\to Hom_{\Bbb C}(V',V')$. Tim tells us that the identity map from $V'$ to $V'$ lifts to an $f:V'\to V$ which is invariant under a finite index subgroup $\Gamma $ of $SL(n,\Bbb Z)$. Then by averageing one can make it invariant under $SL(n,\Bbb Z)$.

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+1: But you might want to make this more explicit. I had to dig through the comments and then think for a while to decode your answer. –  Daniel Litt Mar 21 '11 at 18:16
Yes, this is a case where the rule "more is more" works better than "more is less". –  Jim Humphreys Mar 21 '11 at 22:06

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