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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.

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(Question in a similar flavour:… ) – Qfwfq Jul 14 '11 at 11:33
This poster and the one: are both for injective homomorphisms. Here the question is more general. I guess the answer is No. But I don't know how to prove it. – yeshengkui Jul 14 '11 at 12:13
up vote 7 down vote accepted

Se Does $SL_3(R)$ embed in $SL_2(R)$? 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 $n-2$.

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Why is the $U$ injected into $GL_{n-1}(\mathbb{C})$ by a nontrivial map? – yeshengkui Jul 14 '11 at 11:31
Thank you for your comment, I forgot about the second case. – Guntram Jul 14 '11 at 13:55
In the case of $SL_n(\mathbb{Z})$, completely elementary proofs of the Margulis subgroup theorem and of the congruence subgroup property, are available: see R. Steinberg, Some consequences of the elementary relations in ${\rm SL}_n$. Finite groups—coming of age (Montreal, Que., 1982), 335–350, Contemp. Math., 45, Amer. Math. Soc., 1985. To prove that $SL_n(\mathbb{Z})$ does not embed into $GL_{n-1}(\mathbb{Z})$, one can also appeal to virtual cohomological dimension. @Guntram: How do you see that every non-trivial representation of $SL_n(\mathbb{Z}/m)$ has degree at least $n$? – Alain Valette Jul 15 '11 at 19:04

Write $G=SL_n(\mathbb{Z}/p)$ and $H=GL_{n-1}(\mathbb{Z}/p)$. By standard formulae for the orders we have $|G|=|H|p^{n-1}(p^n-1)/(p-1)$. Let $Z$ be the centre of $G$. It is known that $G/Z$ is simple, and it follows easily that all normal subgroups are contained in $Z$. Moreover, the order of $Z$ divides $p-1$. It follows that the image of any nontrivial quotient of $G$ has order at least $|G|/(p-1)$, which is too big for it to be a subgroup of $H$. Thus, there are no nontrivial homomorphisms from $G$ to $H$.

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good point! Thanks for your comments. – yeshengkui Jul 14 '11 at 11:12
@Geoff: Do we ever have $SL_n(\mathbb{F}_p)/Z$ non-simple when $n \geq 3$? – S. Carnahan Jul 14 '11 at 11:31
@Neil: You seem to assume that for every homomorphism from $SL_n(\mathbb{Z})$ to $GL_{n-1}(\mathbb{Z})$ must take the $p$-congruence subgroup into the $p$-congruence subgroup. Why is that true? – Mark Sapir Jul 14 '11 at 11:53
@Mark: the OP asks about Z/p in the second half of the question. I do not know an argument for the Z case. – Neil Strickland Jul 14 '11 at 12:07
Mark: I think his answer only works for finite fileds, as he use the facts of simple groups. – yeshengkui Jul 14 '11 at 12:09

By the Margulis superrigidity theorem, any homomorphism from $SL(n,\mathbb{Z})$ to $SL(n-1,\mathbb{R})$ with infinite image has to extend to a homomorphism from $SL(n,\mathbb{R})$ to $SL(n-1,\mathbb{R})$. But any non-trivial representation of $SL(n,R)$ has to have dimension at least $n$, so this is impossible.

The best online source for the Margulis superrigidity theorem is Dave Witte Morris's unfinished book

The above argument uses Theorem 12.1 and Theorem 12.3.

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@Alex Eskin: Don't you need to say "virtually extends", since the image need not be Zariski dense? The argument still works verbatim though. – Tom Church Jul 14 '11 at 18:59
@Tom Church: I have added the words "with infinite image". Is this what you meant? Also since SL(n,Z) is a nonuniform lattice, Margulis says that the Zariski closure of the image is semisimple with no compact factors. – Alex Eskin Jul 14 '11 at 20:06

S. Weinberger proved in Lemma 3 of [ SL(n,Z) cannot act on small tori. Geometric topology (Athens, GA, (1993), 406--408, AMS/IP Stud. Adv. Math., 2.1, Amer. Math. Soc., Providence, RI,1997.] that

If $2 < n > m$ then every homomorphism $SL(n,\mathbb{Z}) \rightarrow SL(m, \mathbb{Z})$ is trivial.

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