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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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7
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Se http://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 $n-2$. |
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11
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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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8
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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 http://people.uleth.ca/~dave.morris/books/IntroArithGroups.html The above argument uses Theorem 12.1 and Theorem 12.3. |
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3
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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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