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Is there a finite index torsion-free subgroup $G$ of $GL_n(\mathbb Z)$, where $n\ge 3$, such that the coinvariants group $\mathbb Z^n_G$ is finite?

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 $\{gz-z: g\in G, z\in\mathbb Z^n\}$.

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.

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2 Answers 2

up vote 7 down vote accepted

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 $G$.

Edit (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 $m$.

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Thank you! In fact, it is enough to consider the diagonal matrices $E_i(m)$ which have $m$ at $(i,i)$ entry and $1$ on all other diagonal entries. I am still in the process of proving that the group generated by $E_1(m), \dots, E_n(m)$ sits in any given finite index subgroup for large $m$ but it does sound believable. –  Igor Belegradek Jul 27 '12 at 14:57
    
That will not work. They do not lie in $GL_n(\mathbb Z)$. –  Wilberd van der Kallen Jul 27 '12 at 15:19
    
Oops. So how does one prove the claim in the first sentence of your answer? –  Igor Belegradek Jul 27 '12 at 15:22
    
@Igor: $e_{i,j}(m)=e_{i,j}(1)^m$. –  Mark Sapir Jul 27 '12 at 16:43
    
thank you, Mark, now it is all clear. –  Igor Belegradek Jul 27 '12 at 16:53

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).

LEMMA : Let $V$ be a nontrivial finite-dimensional irreducible representation of $\text{SL}_n(\mathbb{R})$ over $\mathbb{R}$ and let $\Gamma$ be any lattice in $\text{SL}_n(\mathbb{R})$. Then $V_{\Gamma} = 0$.

To see this, observe that using the Borel density theorem (which says that $\Gamma$ is Zariski dense in $\text{SL}_n(\mathbb{R})$), we can get that $V$ is also a nontrivial irreducible $\Gamma$-representation. Now, $V_{\Gamma} = V/K$ where $K$ is spanned by the set $\{x-g(x)\text{ $|$ }x \in V, g \in \Gamma\}$. Clearly $K$ is a nontrivial $\Gamma$-subrepresentation of $V$, so by the irreducibility of $V$ we must have $K=V$.

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Thanks, Andy. This is very nice. –  Igor Belegradek Jul 27 '12 at 17:50

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