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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.

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  • $\begingroup$ For $n$ at least 7 it has to be trivial since such homomorphism has to kill the alternating group $A_n$. Something similar probably also works for the smaller numbers. $\endgroup$
    – Misha
    Mar 30, 2013 at 2:41
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    $\begingroup$ See mathoverflow.net/questions/70311/… for a related question. $\endgroup$
    – Guntram
    Mar 30, 2013 at 4:36

2 Answers 2

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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.

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    $\begingroup$ For $n=4$, there are complex representations, just not rational, of dim $3$, because $A_5$ is the symmetry group of an oriented dodecahedron. $\endgroup$
    – Will Sawin
    Mar 30, 2013 at 16:17
  • $\begingroup$ Right. I have edited accordingly, thanks for pointing that out. $\endgroup$ Mar 31, 2013 at 3:24
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$SL_n(\mathbb Z)$ is generated by elementary matrices. In general let $E_n(R)$ be the subgroup of $SL_n(R)$ generated by elementary matrices. Let's show that if $m\lt n$ and $n\ge 3$ then the only homomorphism $E_n(R)\to GL_m(F)$ is the trivial one, if $F$ is a field. Wlog $F$ is algebraically closed.

Let $T_n(R)\subset E_n(R)$ be the group of strictly upper-triangular matrices. It is generated by elementary matrices $e_{ij}(a)$ where $1\le i\lt j\le n$ and $a\in R$. Let's first show that every homomorphism $h:T_n(R)\to GL_m(R)$ takes $e_{1n}(a)$ to a scalar matrix. This is by induction on $n$. The element $e_{1n}(a)$ is central in $T_n(R)$, so the image of $h$ is contained in the centralizer of $h(e_{1n}(a))$. If $h(e_{1n}(a))$ is not a scalar, then by conjugating we can assume the image of $h$ is contained in $GL_{m_1}(F)\times GL_{m_2}(F)$ for some positive $m_1$ and $m_2$ adding up to $m$. (EDIT: I have changed the next few sentences to correct an error.) Or rather, that the image of $h$ is contained in the subgroup of $GL_m(F)$ preserving $F^{m_1}\times 0\subset F^{m_1}\times F^{m_2}=F^m$. By induction on $n$ the images of $e_{1,n-1}(a)$ under the resulting homomorphisms to $GL_{m_1}(F)$ and $GL_{m_2}(F)$ are central. (To begin the induction, if $n=3$ then $m_1$ and $m_2$ are both $1$ so this step still is OK.) So $h$ takes $T_n(R)$ into some conjugate of $T_m(F)$. Since $n>m$, it follows that $e_{1n}(a)$, being an $(n-1)$-fold commutator in $T_n(R)$, to the identity after all.

In particular a homomorphism $E_n(R)\to GL_m(F)$ must take $e_{1n}(a)$ to a scalar matrix. Of course more generally it must take $e_{ij}(a)$ to a scalar matrix. Since $n\ge 3$, we can express $e_{ij}(a)$ as the commutator of $e_{ik}(a)$ and $e_{kj}(1)$, and it follows that $e_{ij}(a)$ goes to the identity.

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  • $\begingroup$ That's a great argument, both more elementary and more easily generalized than competing arguments. However, to nitpick: you seem to assume that all matrices are diagonalizable when you talk of a product of $GL$. $\endgroup$ Apr 9, 2013 at 20:17
  • $\begingroup$ Good point. I'm sure that I never in my life thought that every matrix over an algebraically closed field is diagonalizable, but I did mess up somehow. I've corrected some typos, and will try to corret the thinko soon. $\endgroup$ Apr 9, 2013 at 23:28
  • $\begingroup$ I don't follow the step "If h(e1n(a)) is not a scalar, then by conjugating we can assume the image of h is contained in GLm1(F)×GLm2(F) for some positive m1 and m2 adding up to m." What about the case that n=3, m=2 and h(e13(a)) = [[1,a],[0,1]]? $\endgroup$
    – Guntram
    Apr 10, 2013 at 4:41
  • $\begingroup$ That step is wrong. I'll fix the proof. $\endgroup$ Apr 10, 2013 at 13:55

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