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I understand that via the Borel density theorem given a finite dimensional (polynomial) representation of the simple non-compact Lie groups $SL_n \mathbb R$ or $Sp_n \mathbb R$, I get an irreducible representation when I restrict to $SL_n \mathbb Z$ or $Sp_n \mathbb Z$, respectively.

I wonder if the restriction is injective on the set of isomorphism classes of irreducible polynomial representations. I.e. is it true that if $V$ and $W$ are irreducible polynomial $SL_n \mathbb R$-representations and $$Res_{SL_n \mathbb Z} V \cong Res_{SL_n \mathbb Z} W$$ then $V\cong W$?

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    $\begingroup$ I wouldn't call the fact that $\text{SL}_n(\mathbb{Z})<\text{SL}_n(\mathbb{R})$ is Z-dense a result of the Borel density theorem (even though this fact indeed follows from the BDT). This is almost a triviality. $\endgroup$ – Uri Bader Aug 5 '16 at 13:54
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(This is too long for a comment.) This follows from the fact that $\text{SL}_n\mathbb{Z}$ is Zariski dense in $\text{SL}_n\mathbb{R}$. Let $\phi:V\to W$ be a linear isomorphism such that for every $g\in \text{SL}_n\mathbb{Z}$, the "conjugate" $\phi^g = g^{-1}\cdot\phi(g\cdot -)$ equals $\phi$. The subset of $\text{SL}_n\mathbb{R}$ of all $g$ such that $\phi^g$ equals $\phi$ is the simultaneous zero locus of finitely many polynomial functions of the matrix entries of $g$: just look at the matrix entries of $\phi^g - \phi$ as functions of $g$. Thus, since these polynomials vanish on the Zariski dense subset $\text{SL}_n\mathbb{Z}$, they vanish on all of $\text{SL}_n\mathbb{R}$.

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  • $\begingroup$ This is exactly what I hoped for. $\endgroup$ – Peter Patzt Aug 5 '16 at 13:32

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