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Let $V$ be a finite-dimensional irreducible representation of the Lie group $\text{SL}_n(\mathbb{R})$. Must $V$ remain irreducible when you restrict the action to $\text{SL}_n(\mathbb{Z})$? More generally, when you restrict it to other lattices in $\text{SL}_n(\mathbb{R})$?

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Yes, I agree. I was actually answering a different question. – Igor Rivin Sep 10 '12 at 2:42
Plus the fact that all finite-dimensional representations of semisimple groups are algebraic. – Misha Sep 10 '12 at 2:45
@Misha: @Harpo had abbreviated his comment, which initially remarked on the fact you are alluding to. The OP (and I) would probably appreciate a reference to where the fact should be found. – Igor Rivin Sep 10 '12 at 2:56
@Misha: Over $\mathbf{C}$ what you write is correct, but over $\mathbf{R}$ you need to impose simply connectedness in additional to semi-simplicity. The usual counterexample is the inverse of the analytic isomorphism ${\rm{SL}}_n(\mathbf{R}) \rightarrow {\rm{PGL}}_n(\mathbf{R})$ for odd $n > 1$. The "problem" is that an $\mathbf{R}$-isogeny between connected linear algebraic $\mathbf{R}$-groups can have finite kernel with no nontrivial $\mathbf{R}$-points; over $\mathbf{C}$ this isogeny issue (invisible to Lie algebras) cannot be missed at the level of $\mathbf{C}$-points. – grp Sep 10 '12 at 5:05
@Misha: A more basic kind of counterexample is that if the connected semisimple $G$ isn't simply connected then $G(\mathbf{R})$ may be disconnected (e.g., ${\rm{PGL}}_{2n}$ and various other $G$ with "fundamental group" of even order). In such cases the projection from $G(\mathbf{R})$ onto its nontrivial finite component group leads to rather non-algebraic representations. This highlights the significance of the connectedness of $G(\mathbf{R})$ when $G$ is simply connected (see my comments on the answer below). – grp Sep 10 '12 at 6:04
up vote 8 down vote accepted

The result is essentially the statement of Borel's stability theorem for $\mathrm{SL}_n(\mathbf{R})$, see for example Theorem 4.39 of the following:

Sometimes Borel's stability theorem is phrased in terms of one of the corollaries, which in this case would be that any lattice in $\mathrm{SL}_n(\mathbf{R})$ is Zariski dense. Then to deduce the result one would also have to note that all finite dimensional representations of $\mathrm{SL}_n(\mathbf{R})$ are algebraic.

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Thanks! I was previously vaguely aware of the Borel density theorem, but I hadn't connected it to this question and was not aware that it was deduced from something that is almost exactly what I wanted. – Sue Sep 10 '12 at 3:14
This still lacks a pointer to the algebricity fact, though the Witte-Morris book is most excellent. – Igor Rivin Sep 10 '12 at 3:57
@Igor: Let's show for linear algebraic $\mathbf{R}$-groups $G'$ and $G$ with $G'$ connected semisimple and simply connected, analytic $f:G'(\mathbf{R}) \rightarrow G(\mathbf{R})$ are algebraic. The Lie algebra of $\Gamma_f$ is a perfect subalgebra of $\mathbf{g}' \times \mathfrak{g}$, so by 7.9 in Borel's LAG it arises from a connected closed $\mathbf{R}$-subgroup $H$ of $G' \times G$. Clearly $H$ is semisimple, and $H \rightarrow G'$ is an isogeny, so an isomorphism as $G'$ is simply connected. This gives $G' \rightarrow G$ recovering $f$ on $G'(\mathbf{R})^0 = G'(\mathbf{R})$ (Steinberg!) – grp Sep 10 '12 at 5:22
@Igor (cont'd): At the final step above I am hauling out the big guns: if $G'$ is a connected semisimple $\mathbf{R}$-group that is simply connected then $G'(\mathbf{R})$ is connected. The only proof I know is to deduce it from considerations with maximal compact subgroups, Cartan involutions, and a theorem of Steinberg (or topological precursor by Cartan) on Zariski-connectedness of centralizers of semisimple automorphisms of simply connected semisimple groups, way too much to explain here. I would love to know a literature reference (beyond the "classical" anisotropic case). – grp Sep 10 '12 at 5:31
The Zariski-density of $G(\mathbf{Z})$ in $G(\mathbf{R})$ for any "split semisimple" group $G$ over $\mathbf{Z}$ (e.g., ${\rm{SL}}_n$) admits a direct proof more elementary than the proof of Borel's theorem. Indeed, by considering pairs of opposite root groups and the open cell structure relative to a positive system of roots for a split maximal $\mathbf{Z}$-torus, we get $\mathbf{Z}$-subgroups $U \simeq \mathbf{G}_a$ such that the subgroups $U_{\mathbf{Q}}$ generate $G$ over $\mathbf{Q}$. Thus, we just need to note that $\mathbf{Z}$ is Zariski-dense in $\mathbf{G}_a$ over $\mathbf{Q}$! – grp Sep 10 '12 at 7:15

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