Are there some nontrivial group homomorphisms from $SL\_n(\mathbb{Z})$ to $GL\_{n-1}(\mathbb{Z})$ for $n\geq3$ except the determinant? This should be a natural question and any references are welcomed.


PS. A similar question has the answer 'NO' for a finite field $F\_p$ instead of $\mathbb{Z}$ as explained below. 



Are there some nontrivial group homomorphisms from $SL_n(\mathbb{Z})$ to $GL_{n-1}(\mathbb{Z})$ for $n\geq3$?
Se [Does $SL\_3(R)$ embed in $SL\_2(R)$?](https://mathoverflow.net/questions/47407/does-sl-3r-embed-in-sl-2r/) for a related discussion. 


That any homomorphism $\varphi\colon SL\_n(\mathbb{Z}) \to GL\_{n-1}(\mathbb{Z})$ is trivial can be seen as follows. 


By Margulis' normal subgroup theorem, either the kernel of $\varphi$ is finite (in which it is trivial or the center of $SL\_n(\mathbb Z)$, which is $\pm I\_n$) or the image of $\varphi$ is finite. 


In the second case, the kernel of $\varphi$ contains a congruence subgroup, so $\varphi$ factors through a group $SL\_n(\mathbb Z/m)$ for some $m \in \mathbb Z$. But then $\varphi$ gives rise to a nontrivial representation of this group of degree $n-1$, which is impossible.


In the first case, note first that the subgroup $U\_n(\mathbb{Z}) \leq SL\_n(\mathbb{Z})$ of upper triangular matrices is nilpotent of class $n-1$, and it is a nice exercise to see that any subgroup $U'\leq U$ of finite index is also nilpotent of class $n-1$. 


Now if $\varphi$ were nontrivial, it would inject $U$ into $GL\_{n-1}(\mathbb{C})$ since $U$ intersects the center of $SL\_n(\mathbb{Z})$ trivially. Then $U$ contains a finite index subgroup $U'$ such that the Zariski closure of $\varphi(U')$ is connected: take $U'$ to be the preimage under $\varphi$ of the connected component of the identity of $\overline{\varphi(U)}$. 


By the Lie-Kolchin theorem, it follows that $\varphi(U')$ is conjugate to a subgroup of the upper triangular matrices, which is a contradiction since any such subgroup has nilpotency class at most $n-2$.


