Property of lattices in Lie groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:35:32Z http://mathoverflow.net/feeds/question/122776 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122776/property-of-lattices-in-lie-groups Property of lattices in Lie groups anton 2013-02-24T04:50:41Z 2013-03-04T14:35:56Z <p>Let $\Gamma$ be a lattice in a (real or p-adic) Lie group. Is it true that for a given natural number $n$ there exists a finite index subgroup $\Sigma\subset\Gamma$ such that each $\sigma\in\Sigma$ is an $n$-th power of some element of $\Gamma$?</p> <p>In other words, is it true that for given $\sigma\in\Sigma$ there exists $\gamma\in\Gamma$ such that $\sigma=\gamma^n$?</p> <p>If not true for general lattices, are there some restrictions under which this holds (cocompactness, arithmeticity,...)?</p> http://mathoverflow.net/questions/122776/property-of-lattices-in-lie-groups/122785#122785 Answer by Ian Agol for Property of lattices in Lie groups Ian Agol 2013-02-24T07:16:35Z 2013-02-24T21:39:08Z <p>This is false for uniform lattices in rank one semi simple Lie groups and large $n$ by a result of <a href="http://www.ams.org/journals/tran/1996-348-06/S0002-9947-96-01510-3/home.html" rel="nofollow">Ivanov and Olshanskii</a>, which implies that the normal subgroup generated by $n$th powers is infinite index for certain large $n$. </p> http://mathoverflow.net/questions/122776/property-of-lattices-in-lie-groups/122791#122791 Answer by Aakumadula for Property of lattices in Lie groups Aakumadula 2013-02-24T09:05:30Z 2013-03-04T14:35:56Z <p>The answer is no, even in higher rank groups. For example, take $\Gamma = SL_3({\mathbb Z})$. If such a $\Sigma$ existed for any $n$, then its completion in the profinite (same as congruence) completion of $\Gamma$ would have this property that every element in $\Sigma$ would be an $n$-th power. But the completion of $\Sigma$, by strong approximation, is open. So for almost all primes, $SL_3({\mathbb Z}_p)$ would have this property. But you can easily cook up a prime $p$ and elements in $SL_3({\mathbb F}_p)$ which are not $n$-th powers. </p> <p>To see this, get large primes such that $p^3 \equiv 1\quad (mod\quad n)$ but $p-1$ is not divisible by $n$. Choose a cubic extension of ${\mathbb F}_p$; the group of norm one elements in the cubic extension is cyclic of order $\frac{p^3-1}{p-1}$. Hence the $n$-th power map on this subgroup of $SL_3({\mathbb F}_p)$ is not surjective; but any $n$-th root of a generator for this subgroup already lies in this subgroup. </p> <p> To summarise: for any lattice in a semi-simple group, and for any integer $n\geq 2$, the image of the $n$-th power map never contains a finite index subgroup. The argument using strong approximation works if the lattice has number field entries (this holds for all lattices except those in $SL(2,{\mathbb R})$) . I think that it is true for any Zariski dense finitely generated subgroup of a semi-simple group, using strong approximation due to Nori (and Weisfeiler). </p>