From a result Obtained by O. Schreier and B. L. van der Waerden [Math. Sem. Univ. Hamburg 6, 303- 322 (1928)], one can show that for two fields $\mathbb F$ and $\mathbb G$, and integers $n>m>2$, the only group homomorphism $\mathrm{SL}\_n(\mathbb F) \to \mathrm{SL}\_m(\mathbb G)$ is the trivial homomorphism.


My question is: Is there a non trivial group homomorphism $\mathrm{SL}\_n(\mathbb Z) \to \mathrm{SL}\_m(\mathbb Z)$ with $n>m>2$?


Thanks.



Morphisms $\mathrm{SL}_n(\mathbb Z) \to \mathrm{SL}_m(\mathbb Z)$
I'll try to flesh out the first comment to give an answer: for all $n\ge 4$ there is a representation $\rho:A\_{n+1}\to \rm SL\_n(\bf Z)$ whose image normally generates $\rm SL\_n(\bf Z)$ (because it injects into all $\mathrm{PSL}\_{n-1}(\mathbf{Z} / N)$ for $N\ge 1$ and $\rm SL\_n(\bf Z)$ has CSP). But for $n\ge 5$ there is no nontrivial complex representation of $A\_{n+1}$ in dimension less than $n$, so the image of $\rho$ must be contained in the kernel of any morphism $\rm SL\_n(\bf Z)\to\rm SL\_m(\bf Z)$ if $m < n$, and so this morphism must be trivial. 


This leaves the case $n=4$ when there is a real representation of dimension 3 ; however (as pointed in the comment below) it is not rational, so the argument above works as well.


