I'm trying to show that if $G$ is a Chevalley group, then every finite index subgroup of $G(\Bbb Z)$ is Zariski dense in $G(\Bbb C)$. ($G(\Bbb Z)$ is the Chevalley group over $\Bbb Z$ and similarly for $G(\Bbb C)$)
But I'm struggling to understand some basic stuff,
It seems to me that there can't be finite index subgroup in $G(\Bbb C)$ how can I show that?
also is there a good source that covers Chevalley groups over $\Bbb Z$?
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3$\begingroup$ For Chevalley groups over the integers, you might look at the more recent article by Luszttig, or else go back to Steinberg's lectures as republished by the AMS recently with corrections and typesetting. $\endgroup$– Jim HumphreysCommented Jun 22, 2019 at 23:29
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$\begingroup$ It is true that $G = {\rm GL}(n,\mathbb{C})$ has no proper subgroup of finite index. Here is an outline when $n >1.$ . If there were such a subgroup, say $N$, it could be assumed normal (using Cayley's theorem). $S \cap N$ would have finite index in $S = {\rm SL}(n,\mathbb{C}).$ All proper normal subgroups of $S$ consist of scalar matrices, and these all have infinite index in $S$. Hence $N$ contains $S.$ It follows that $D = \mathbb{C}^{\times}$ has a proper subgroup of finite index, whereas $D$ is divisible, a contradiction. Note that ${\rm GL}(n,\mathbb{R})$ has a subgroup of index $2.$ $\endgroup$– Geoff RobinsonCommented Jun 23, 2019 at 9:56
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2$\begingroup$ It's just Borel-Harish-Chandra + Borel's density theorem (+ definition of Chevalley group). More precisely, $G$ is a $\mathbf{Q}$-defined group without nontrivial rational characters (since it has no nontrivial character at all), so Borel-Harish-Chandra implies $G(\mathbf{Z})$ is a lattice in $G(\mathbf{R})$, and hence so are its finite index subgroups. Then Borel's density theorem says that $G(\mathbf{Z})$ is Zariski-dense in $G(\mathbf{R})$ (and hence in $G(\mathbf{C})$ by Rosenlicht and connectedness of $G$). $\endgroup$– YCorCommented Jun 23, 2019 at 14:27
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2$\begingroup$ @GeoffRobinson Any connected real Lie group has no proper finite index subgroup, i.e., has no nontrivial quotient (of abstract group), this follows from being generated by 1-parameter subgroups. $\endgroup$– YCorCommented Jun 23, 2019 at 14:31
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(Essentially copied from comments)
It's just Borel-Harish-Chandra + Borel's density theorem (+ definition of Chevalley group). More precisely, $G$ is a $\mathbf{Q}$-defined group without nontrivial rational characters (since it has no nontrivial character at all), so Borel-Harish-Chandra implies $G(\mathbf{Z})$ is a lattice in $G(\mathbf{R})$, and hence so are its finite index subgroups. Then Borel's density theorem says that $G(\mathbf{Z})$ is Zariski-dense in $G(\mathbf{R})$ (and hence in $G(\mathbf{C})$ by Rosenlicht and connectedness of $G$).
As regards the question about finite index subgroups: this argument probably appears several times on this site: any connected real Lie group has no proper finite index subgroup, i.e., each homomorphism to a finite group is trivial: this follows from being generated by 1-parameter subgroups (which satisfy the given property, by divisibility).
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$\begingroup$ To use the Borel's density theorem I need to show that $G(\Bbb C)$ have no compact factors. (I'm looking at $\Bbb C$ because by some definitions $G(\Bbb Z)$ is already a lattice in $G(\Bbb C)$ ) Is there an easy way to see that? $\endgroup$– AmiCommented Jun 23, 2019 at 21:54
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1$\begingroup$ Well, Borel's density uses the definition of lattice: discrete subgroup of finite covolume, for which $G(\mathbf{Z})$ is not a lattice in $G(\mathbf{C})$. We have $G$ split over $\mathbf{Q}$ and hence over $\mathbf{R}$, so $G(\mathbf{R})$ has no compact factor. $\endgroup$– YCorCommented Jun 24, 2019 at 5:37
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$\begingroup$ Can you please elaborate? $G(\mathbb R)$ is simple so how do we show it's not compact? $\endgroup$– AmiCommented Jul 5, 2019 at 1:03
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$\begingroup$ Because $G$ is split over $\mathbf{R}$, $G(\mathbf{R})$ is not compact. $\endgroup$– YCorCommented Jul 5, 2019 at 3:15
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$\begingroup$ Yes, but can you please explain why $G(\mathbb R)$ is split over $\mathbb Q, \mathbb R$? I did not found a good source for this... $\endgroup$– AmiCommented Jul 5, 2019 at 21:41