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Let $\pi_0$ be an irreducible representation of a discrete subgroup $\Gamma$ of the reductive group $G$. My question in general is

What can be said of representations $\pi$ of $G$ extending $\pi_0$, in the sense that $\pi$ acts on $\Gamma$ as $\pi_0$?

I would like to know technical possibilities in this direction, even assuming freely further properties on $\pi_0$. What if in particular if $\pi_0$ is finite dimensional?

The base case $\Gamma = SL(2, \mathbb{Z})$ and $G=SL(2, \mathbb{Q})$ is already of interest for me.

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    $\begingroup$ The word "lifting" in the title should be "extending" (as in the question). "lift" refers to passing from a representation from a quotient to a given group. $\endgroup$ – YCor Apr 4 '18 at 15:02
  • $\begingroup$ Anything more precise than "what can be said"? An answer is "we can say that they extend $\pi_0$", and can have an endless list of reasonable answers, from superrigidity to induction etc. The question is far too broad. $\endgroup$ – YCor Apr 4 '18 at 15:04
  • $\begingroup$ @YCor Thanks for your comments, I am mainly interested in the specific cases of low rank, for instance the base case I mentioned. $\endgroup$ – Automorphic Apr 5 '18 at 12:54
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If $\Gamma$ is a lattice in a connected semisimple $\mathbb{R}$-algebraic group $G$ without compact factors, then the Borel density theorem says that $\Gamma$ is Zariski dense in $G$. This implies that there can be at most one extension of a representation of $\Gamma$ to a rational representation of $G$.

Of course, extensions need not exist. For finite-dimensional representations, however, you can often get information using the Margulis superrigidity theorem. This says that if $\Gamma$ is a lattice in a semisimple Lie group $G$ satisfying some technical assumptions and if $V$ is a finite-dimensional representation of $\Gamma$, then there exists a finite-index subgroup $\Gamma'$ of $\Gamma$ such that the action of $\Gamma'$ on $V$ extends to a rational representation of $G$.

Passing to a finite-index subgroup is necessary here. For instance, if the action of $\Gamma$ on $V$ factors through a finite group, then you first have to pass to a subgroup $\Gamma'$ that acts trivially on $V$ (and the extension to $G$ is then the trivial action).

The most important technical assumption in the superrigidity theorem is that $G$ is of higher rank. It applies to $\text{SL}(n,\mathbb{Z})$ in $\text{SL}(n,\mathbb{R})$ for $n \geq 3$, but not for $n=2$.

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    $\begingroup$ Superrigidity works in some rank 1 cases ($Sp(n,1)$ and $F_{4(-20)}$). But obviously it fails for $SL_2(Z)\subset SL_2(R)$ (or $SL_2(Z)\subset SL_2(Q)$). $\endgroup$ – YCor Apr 4 '18 at 14:56
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    $\begingroup$ By the way, the OP talks of arbitrary discrete subgroups, and superrigidity dramatically fails even for Zariski-dense discrete subgroups in higher rank. $\endgroup$ – YCor Apr 4 '18 at 15:03
  • $\begingroup$ @YCor: Of course. I can't imagine any general results about representations of arbitrary Zariski-dense subgroups and the one example the OP listed was a lattice, so I thought it was reasonable to just focus on those. $\endgroup$ – Andy Putman Apr 4 '18 at 16:04
  • $\begingroup$ Thank you very much for your answer, this is exactly the kind of result I would like to explore. Is there any standard reference for these matters? And also, is there anything that can be said for $SL(2, \mathbb{Z})$ or, typically, its congruence subgroups? $\endgroup$ – Automorphic Apr 5 '18 at 12:55
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    $\begingroup$ @Automorphic: The group $\text{SL}(2,\mathbb{Z})$ is very close to a free group (and its congruence subgroups are actually free groups except for level $2$), so its representation theory is completely unstructured: if $F_n$ is a free group, then you can make each of the $n$ generators act by any invertible linear transformations you want. So general representations of $\text{SL}(2,\mathbb{Z})$ and its congruence subgroups have nothing to do with representations of $\text{SL}(2,\mathbb{Q})$. $\endgroup$ – Andy Putman Apr 5 '18 at 15:24

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