Timeline for Nilpotent subgroups of uniform finite index
Current License: CC BY-SA 3.0
8 events
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| Mar 17, 2013 at 18:44 | comment | added | Misha | @Yves: You are right: I will add more details, but the proof still works. For the Auslender's result, one needs to assume that $K$ embeds in $Aut(N)$; if not, then one first splits off the compact kernel $K_2$: $H\cong (N\rtimes K_1)\times K_2$, where $K_1$ embeds in $Aut(N)$. Projection of a lattice in H to $N\rtimes K_1$ is still a lattice. | |
| Mar 17, 2013 at 18:11 | comment | added | YCor | @Misha: you mean "for every compact Lie group $K$". Also I don't see why your argument is enough to conclude: assuming your argument works, you reduce to the case when the projection of $\Gamma$ on $K$ is abelian. This does not imply $\Gamma$ nilpotent, so some argument should be used, in the spirit of the fact that finite subgroups of $SL_n(\mathbf{Z})$ have a bounded order. Besides, your projection statement is suspicious: if $N\rtimes K$ is a direct product and $N=\mathbf{R}$, $K$ is the circle, you can pick $\Gamma$ to be an infinite cyclic lattice with a dense image in $K$. | |
| Mar 17, 2013 at 18:00 | comment | added | Misha | I will when I get to my copy of the book; for now, I added some online references. | |
| Mar 17, 2013 at 17:59 | history | edited | Misha | CC BY-SA 3.0 |
Added references
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| Mar 17, 2013 at 16:05 | comment | added | Davis | Do you recall where in Raghunathan's book these results are? I find that book particularly hard to navigate through. | |
| Mar 17, 2013 at 15:08 | vote | accept | Davis | ||
| Mar 16, 2013 at 23:18 | history | edited | Misha | CC BY-SA 3.0 |
added 29 characters in body
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| Mar 16, 2013 at 23:11 | history | answered | Misha | CC BY-SA 3.0 |