Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:20:07Z http://mathoverflow.net/feeds/question/103301 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103301/subgroups-of-gl-n-mathbb-z-with-finite-coinvariants Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants Igor Belegradek 2012-07-27T12:17:43Z 2012-07-30T18:30:15Z <p>Is there a finite index <b>torsion-free</b> subgroup $G$ of $GL_n(\mathbb Z)$, where $n\ge 3$, such that the coinvariants group $\mathbb Z^n_G$ is finite? </p> <p>Here $G$ acts on $\mathbb Z^n$ in the standard way, and $\mathbb Z^n_G$ by definition is the quotient of $\mathbb Z^n$ by the subgroup generated by the set <code>$\{gz-z: g\in G, z\in\mathbb Z^n\}$</code>. </p> <p>If $G=GL_n(\mathbb Z)$, then $\mathbb Z^n_G$ is finite because G contains $-I_n$, but then $GL_n(\mathbb Z)$ isn't torsion-free.</p> http://mathoverflow.net/questions/103301/subgroups-of-gl-n-mathbb-z-with-finite-coinvariants/103309#103309 Answer by Wilberd van der Kallen for Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants Wilberd van der Kallen 2012-07-27T14:19:20Z 2012-07-30T18:30:15Z <p>For any finite index subgroup $G$ there is a nonzero integer $m$ so that $G$ contains the elementary matrices $e_{ij}(m)$ that have ones on the diagonal, $m$ at the $(i,j)$ entry and zeroes elsewhere. So the span of $\{gz-z: g\in G, z\in\mathbb Z^n\}$ contains all multiples of $m$ and ${\mathbb Z}^n_G$ is finite. Now just take a torsion-free finite index subgroup $G$.</p> <p><b>Edit</b> (by Igor Belegradek): I add some detail for my records. If $e_r$ is a vector in the standard basis, then it is easy to write $me_r$ in the form $v-e_{ij}(m)v$ for some $i\neq j$ and $v\in\mathbb Z^n$. It remains to show that for any finite index subgroup $G$ of $GL_n(\mathbb Z)$ all such $e_{ij}(m)$ lie in $G$ for some $m$. By making $G$ smaller we can assume it is normal and of index $k$. Since $e_{ij}(m)=e_{ij}(1)^m$, it suffices to assume that $k$ divides $m$.</p> http://mathoverflow.net/questions/103301/subgroups-of-gl-n-mathbb-z-with-finite-coinvariants/103332#103332 Answer by Andy Putman for Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants Andy Putman 2012-07-27T17:27:23Z 2012-07-27T17:27:23Z <p>van der Kallen gave a nice answer for the situation at hand, but I thought I'd give a somewhat more general one. The question is equivalent to showing that $(\mathbb{R}^n)_G=0$ for some finite-index torsion-free subgroup $G$ of $\text{GL}_n(\mathbb{Z})$. The representation $\mathbb{R}^n$ is a nontrivial irreducible representation of $\text{SL}_n(\mathbb{R})$, so this follows from the following more general result (just take $\Gamma$ to be a torsion-free lattice, for instance the level $3$ principle congruence subgroup).</p> <p>LEMMA : Let $V$ be a nontrivial finite-dimensional irreducible representation of <code>$\text{SL}_n(\mathbb{R})$</code> over $\mathbb{R}$ and let $\Gamma$ be any lattice in <code>$\text{SL}_n(\mathbb{R})$</code>. Then $V_{\Gamma} = 0$.</p> <p>To see this, observe that using the Borel density theorem (which says that $\Gamma$ is Zariski dense in <code>$\text{SL}_n(\mathbb{R})$</code>), we can get that $V$ is also a nontrivial irreducible $\Gamma$-representation. Now, $V_{\Gamma} = V/K$ where $K$ is spanned by the set <code>$\{x-g(x)\text{$|$}x \in V, g \in \Gamma\}$</code>. Clearly $K$ is a nontrivial $\Gamma$-subrepresentation of $V$, so by the irreducibility of $V$ we must have $K=V$.</p>