# Generators for a certain congruence subgroup of SL(n,Z)

I'm looking for a reference (or quick proof) of the following fact. Fix some $n \geq 3$ and some $\ell \geq 2$. Set $\Gamma_n(\ell) = \text{ker}(\text{SL}_n(\mathbb{Z}) \rightarrow \text{SL}_n(\mathbb{Z}/\ell \mathbb{Z}))$. Next, set

$$\Gamma_n'(\ell) = \{\text{A \in \Gamma_n(\ell) | for all diagonal entries a_{ii} of A, we have \ell^2 | (a_{ii}-1)}\}.$$

It is an easy exercise to show that $\Gamma_n'(\ell)$ is a subgroup of $\Gamma_n(\ell)$. Finally, for all $1 \leq i,j \leq n$ with $i \neq j$, let $e_{ij}$ be the elementary matrix obtained by taking the identity matrix and placing a $1$ at position $(i,j)$.

Observe that $e_{ij}^{\ell} \in \Gamma_n'(\ell)$. The fact I'm interested in is the fact that $\Gamma_n'(\ell)$ is generated by the set of all the $e_{ij}^{\ell}$. I've seen this asserted without proof in print before (though I don't recall exactly where), but I've been unable to work out a proof myself.

• One can probably break up the proof into two steps: show it for $n=3$, and then show that $\Gamma_{n+1}'(\ell)$ is generated by the $n+1$ copies of $\Gamma_n'(\ell)$ sitting inside it in the natural way. I'm not sure how much easier the case $n=3$ is than the general case though (and the statement is false for $n=2$). – Ian Agol Apr 8 '13 at 5:20

There is a Theorem of Tits which says that the group generated by $e_{ij}^l$ has finite index in $SL_n({\mathbb Z})$ if $n\geq 3$. The paper is a Comptes rendus announcement (generating systems of congruence subgroups, CR. ACad. Sci 283 (1976), no. 9 A693-A695 (see the following review http://www.ams.org/mathscinet/search/publdoc.html?amp=&loc=revcit&revcit=244257&vfpref=html&r=16&mx-pid=424966). By a result of Mennicke, finite index subgroups of $SL_n({\mathbb Z})$ are congruence subgroups ($n\geq 3$). So, you need only check that the congruence closure of the group generated by $e_{ij}^l$ in $SL_n({\mathbb Z})$ is of the desired type. This is a purely local checking (prime by prime, for primes dividing $l$) and can be done easily.

ADDED: what you are asking for is (seemingly) stronger than Tits' theorem: not only do the $e_{ij}^l$ generate a finite index subgroup, they generate a specific congruence subgroup. But, as the previous paragraph shows, this is $equivalent$ to the result of Tits.

I guess the folllowing is a short explicit proof of Tits' result for this very special case, assuming CSP for $\rm{SL}_n(\bf{Z})$ (so this should be seen as a lenghty comment on the above answer). The key step is to show that the group $\Lambda=\langle e_{ij}^\ell\rangle$ contains $\Gamma_n(\ell^2)$: the commutator $[e_{ij}^\ell,e_{km}^\ell]$ is equal to $1+\ell^2( [n_{ij},n_{km}] - n_{ij}^2 - n_{km}^2)+ O(\ell^3)$ where $n_{ab}=e_{ab}-1$ is a nilpotent matrix. On the other hand you have $(e_{ij}^\ell)^\ell=1+\ell^2 n_{ij}$, and as $n_{ij}^2=n_{i,j+1}$ it follows that $\Lambda$ contains in fact also the matrices $1+\ell^2 [n_{ij},n_{km}]$ modulo $\ell^3$. So you finally get that $\Lambda\Gamma_n(\ell^3)$ contains all matrices $1+\ell^2a$ where $a$ is a sum of $n_{ij}$s and their Lie brackets, and these generate the $\bf{Z}/\ell$-module $\mathfrak{sl}_n(\bf{Z}/\ell)$. As remarked by Aakumadula below (and apparently the fact that $\Gamma_n'(\ell)$ contains $\Gamma_n(\ell^2)$ also follows from a theorem of Tits, see her/his answer to this question), it remains to show that for a prime $p$ not dividing $\ell$ the $e_{ij}^\ell$ generate $\Gamma_n$ modulo $\Gamma_n(p)$: in this case we have $\Lambda\Gamma_n(p)=\langle e_{ij}\rangle\Gamma_n(p)$ and it is well-known that the images of the $e_{ij}$ generate $\rm{SL}_n(\bf{F}_p)$. So we have that the congruence closure of $\Lambda$ contains $\Gamma_n(\ell^2)$ and by CSP it follows that $\Lambda\supset\Gamma_n(\ell^2)$.

To conclude: now you have only to show that the $e_{ij}^l$ generate $\Gamma_n'(\ell)$ modulo $\ell^2$. But this is obvious as $\Gamma_n'(\ell)/\Gamma_n(\ell^2)$ is just the subspace of the abelian group $\Gamma_n(\ell)/\Gamma_n(\ell^2)\cong \mathfrak{sl}_n(\bf{Z}/\ell)$ of matrices with zeroes on the diagonal.

• I agree that the key is to show that the group contains $\Gamma_n(\ell^2)$. But I don't understand your argument for this. Can you expand it a bit? Thanks! – Edward Cooper Apr 6 '13 at 2:43
• I do not understand Cooper's remark. The theorem of Tits quoted above shows that $e_{ij}^l$ generate a congruence subgroup, which can only be $\Gamma _n(l^2)$. – Venkataramana Apr 6 '13 at 4:08
• @Cooper: I edited to give more details. – Jean Raimbault Apr 6 '13 at 12:09
• @jean: your argument only proves the local statement that in $SL_n$ of the $l$-adic integers (suppose $l$ is a prime), the $e_{ij}^l$ generate the appropriate congruence closure. It does not prove that at the global level (i.e. the integral points of $SL_n$, ), the $e_{ij}^l$ generate the appropriate congruence subgroup. You do need the theorem of Tits (and that is a non-trivial result). – Venkataramana Apr 6 '13 at 13:44
• @Aakumadula: you're right, there is also need to check that at primes $p$ not dividing $\ell$ the $e_{ij}^\ell$ generate all of $\rm{SL}_n(\bf{Z}_p)$. However this can be done in an elementary way since the $e_{ij}$ generate $\rm{SL}_n(\bf{F}_p)$. Edited to add that. – Jean Raimbault Apr 6 '13 at 14:20

FIrst I'd clarify that your notation $e_{ij}^\ell$ actually refers to the matrix with diagonal entries 1, the off-diagonal $(i,j)$ entry equal to $\ell$, and other entries 0.

I don't know what you've read, but since these matrix calculations are quite old and also deeply embedded in the study of the Congruence Subgroup Problem, it's a good idea to look into some of the relevant older literature. On a concrete level, the emphasis is on the group $\Gamma: = \mathrm{SL}_n(\mathbb{Z})$ and its subgroups of finite index, when $n \geq 3$ (the case $n=2$ being much more complicated). Here the key players are the (normal) principal congruence subgroups you've denoted by $\Gamma_n(\ell)$ and the interleaved elementary subgroups: inverse images of finite groups generated by elementary/unipotent matrices.

Key references available online include the 1964 announcement by Bass-Lazard-Serre here and the detailed follow-up by Bass-Milnor-Serre here.

Some of the concrete calculations you are looking for are also written down in section 17 of my (typewritten) 1980 Springer Lecture Notes Arithmetic Groups. The older lecture notes Algebraic K-Theory by Bass (1968) contain a vast amount of detail, and of course there are newer treatises including those by Hahn and O'Meara along with a new book by Weibel.

ADDED: The short paper by Tits cited by Aakumadula is definitely helpful for your question, though it's dependent on the earlier work and is not readily available online (nor is the ancient review I wrote). The literature on congruence subgroups is extensive and often far more general than what you need, but I don't see a direct computational proof of the result you read somewhere. (Advice: Keep track of those sources.) Also, notation varies in the subject, but your choice of $e_{ij}$ is unfortunate since that symbol usually means the matrix with a single nonzero entry $1$. A more usual convention is to write something like $x_{ij}(\ell)$ for your unipotent matrix.

• You my also need to use Lemma 3.1 of arXiv:1301.6082 to deal with the distinction between the subgroup generated by the $e^\ell_{ij}$ and the normal subgroup generated by them, meaning the smallest normal subgroup of the elementary group that contains them. In the congruence subgroup problem one tends to use normal subgroups. – Wilberd van der Kallen Apr 3 '13 at 13:26
• As van der Kallen alluded to, Bass-Milnor-Serre (and its antecedents) only proved that the $e_{ij}^{\ell}$ (which, by the way, seems pretty unambiguous to me; it means the $\ell^{\text{th}}$ power of $e_{ij}$) normally generate the whole group $\Gamma_n(\ell)$. This isn't quite what I want, which is a somewhat finer result. – Edward Cooper Apr 3 '13 at 14:39
• @Wilberd van der Kallen : Are you sure you got the arXiv identifier right? When I went there, I got a paper entitled "Reaction-diffusion model Monte Carlo simulations on the GPU", which doesn't sound right at all. – Edward Cooper Apr 3 '13 at 14:40
• @Edward : Bass-Milnor-Serre and its "antecedents" is not quite right. There s this paper of Tits (which uses Bass-Milnor-Serre) where he proves that the group generated by $e_{ij}^l$ is a finite index subgroup for $n\geq 3$. – Venkataramana Apr 4 '13 at 1:16
• @Edward Cooper. Sorry. Try arXiv:1303.60882 by Stepanov. – Wilberd van der Kallen Apr 10 '13 at 6:57