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Oct 22, 2016 at 2:15 history edited Venkataramana CC BY-SA 3.0
$I had written $\lambda-1$ ; this should read $\lambda =-1$
Oct 13, 2016 at 10:03 vote accept Honing
Oct 13, 2016 at 9:02 comment added Venkataramana @Honig: not quite. If $G$ is a Chevalley group which is simply connected, then yes. Otherwise, there are conditions. There are results by Adler about these matters (there is an article in the Algebraic Groups and Discontinuous subgroups edited by Borel, where these issues are discussed).
Oct 13, 2016 at 8:13 comment added Honing @Venkataramana Thank you for your answer. Is it a general fact of life that, if $G$ is a connected reductive group over $\mathbb Z$, then $G(\mathbb Z)$ is a maximal arithmetic subgroup of $G(\mathbb Q)$?
Oct 13, 2016 at 7:49 history edited Uri Bader CC BY-SA 3.0
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Oct 13, 2016 at 7:02 history edited Friedrich Knop CC BY-SA 3.0
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Oct 13, 2016 at 6:45 history edited Venkataramana CC BY-SA 3.0
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Oct 13, 2016 at 6:36 history edited Venkataramana CC BY-SA 3.0
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Oct 13, 2016 at 6:34 comment added Venkataramana Oops, you are right. I meant to write every intermediate subgroup contains the integral subgroup of infinite index
Oct 13, 2016 at 6:30 comment added Venkataramana That is not the claim: every intermediate subgroup contains the integral symp group as an infinite index subgroup. Pls read carefully
Oct 13, 2016 at 4:56 comment added YCor $Sp_{2g}(\mathbf{Z})$ is not maximal in $Sp_{2g}(\mathbf{Q})$ (an intermediate subgroup is $Sp_{2g}(\mathbf{Z}[1/2])$). Actually, a finitely generated group cannot be maximal in an infinitely generated group since any group is generated by any of its maximal subgroups and one further element.
Oct 13, 2016 at 3:30 history edited Venkataramana CC BY-SA 3.0
fixed notation
Oct 13, 2016 at 2:00 history edited Venkataramana CC BY-SA 3.0
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Oct 13, 2016 at 1:41 history answered Venkataramana CC BY-SA 3.0