# Has a conjugation of SL2(Z) finite index in SL2(Z)? (Modular group)

Dear all,

I have a probably rather simple question: Suppose we have a Matrix $M\in SL_2(\mathbb{Q})$. Does the group $M^{-1} SL_2(\mathbb{Z}) M \cap SL_2(\mathbb{Z})$ then always have finite index in $SL_2(\mathbb{Z})$? Why? Why not?

I really was not able to solve this problem!

All the best Karl

-
Is this homework? I have set this question for homework in the past :-) I agree that it can be a little tricky though :-) – Kevin Buzzard Oct 22 '10 at 17:13
Hint: look at the effect of $M$-conjugation on matrices sufficiently congruent to 1. (Think in terms of ${\rm{SL}}_n$ rather than ${\rm{SL}}_2$, or even any flat affine group scheme of finite type over $\mathbf{Z}$, to force clean thinking rather than explicit matrix manipulations.) – BCnrd Oct 22 '10 at 17:36
Hint: this group is the joint stabiliser of 2 lattices in $\mathbb{Q}^2$. – Tony Scholl Oct 22 '10 at 17:38
@Brian: oh, you beat me to it – Tony Scholl Oct 22 '10 at 17:39
Seriously, BCnrd? :) – David Hansen Oct 22 '10 at 18:35

I, for one, am less than thrilled with snobbish kibitzing in the comments. Just answer the question already instead of dropping hints and passing judgment.

The answer is yes for $\text{SL}(n,\mathbb{Z})$. Let $d$ be the product of the denominators in the matrices $M$ and $M^{-1}$. Let $\Gamma_d \subseteq \text{SL}(n,\mathbb{Z})$ be the subgroup of matrices of the form $I+dA$. This subgroup has finite index because it is the kernel of the congruence homomorphism $$\text{SL}(n,\mathbb{Z}) \longrightarrow \text{SL}(n,\mathbb{Z}/d),$$ whose target is a finite group. On the other hand, $M\Gamma_dM^{-1} \subseteq \text{SL}(n,\mathbb{Z})$ because $MIM^{-1} = I$ and $dMAM^{-1}$ is an integer matrix. Thus the intersection in question has finite index because it contains $\Gamma_d$ as a subgroup.

The argument is quite general: You can replace $\text{SL}$ by other algebraic groups defined over $\mathbb{Z}$, and you can replace $\mathbb{Z}$ by any number field ring and $\mathbb{Q}$ by the corresponding number field.

-
So this proves that the commensurator of $SL_2(\mathbb{Z})$ inside $GL_2(\mathbb{Q})$ is equal to $GL_2(\mathbb{Q})$, does it? – Karl Oct 23 '10 at 9:26
Yeah, I think so. – Greg Kuperberg Oct 23 '10 at 11:59
For affine $O_K$-groups $G$ of finite type (say with $K$ a number field), we can express Greg's congruential argument in a topological form that bypasses "explicit" matrix arguments and instead puts the congruential aspects entirely into the structure of the locally compact $K$-algebra $A_f$ of finite adeles for $K$. To be precise, the compact open subring $\widehat{O} = \prod_{v \nmid \infty} O_{K_v}$ satisfies $O_K = K \cap \widehat{O}$ inside of $A_f$, and it has a base of opens around 0 given by $I \widehat{O}$ for nonzero ideals $I$ of $O_K$. Hence, entirely for functorial reasons,... – BCnrd Oct 23 '10 at 21:54
...we have $G(O_K) = G(K) \cap G(\widehat{O})$ inside $G(A_f)$, where $G(\widehat{O})$ is compact open subgp of loc. compact topological gp $G(A_f)$ and has base of opens around 1 given by the open subgps $G_I =\ker(G(\widehat{O}) \rightarrow G(\widehat{O}/I\widehat{O})$ (of finite index). Really $A_f$ carries all the work, no matrices. For $g \in G(K)$, the open subgp $g G_I g^{-1}$ is contained in the compact open $G(\widehat{O})$ for suitable $I$ since $G(A_f)$ is topological gp, and so must have finite index. Intersect with $G(K)$ to see $g G(O_K) g^{-1}$ and $G(O_K)$ are commensurable.QED – BCnrd Oct 23 '10 at 21:59
Clarification: in the preceding comment, "(of finite index)" refers to the $G_I$ as they sit inside of $G(\widehat{O})$, not $G(A_f)$ of course. – BCnrd Oct 23 '10 at 22:02

There is a second proof which Tony Scholl hints at in the comments. This is probably secretly equivalent to the argument Greg writes up, but I find it easier to think about.

$SL_2(\mathbb{Z})$ is the subgroup of $SL_2(\mathbb{Q})$ preserving the lattice $L_1:=\mathbb{Z}^2$ inside $\mathbb{Q}^2$. Similarly, $M SL_2(\mathbb{Z}) M^{-1}$ is the gorup of matrices preserving $L_2 := M \cdot L_1$. So the group we are interested in is the group of matrices sending $L_1$ and $L_2$ to themselves.

Choose an integer $N$ such that $L_1 \cap L_2 \supseteq N L_1$ and $L_1 + L_2 \subseteq N^{-1} L_1$. Let $\Gamma$ be the subgroup of $SL_2(\mathbb{Z})$ which acts trivially on $L_1/ N^2 L_1$. The subgroup $\Gamma$ has finite index as it is the kernel of $SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}/N^2)$.

Now, $\Gamma$ stabilizes $N L_1$ and $N^{-1} L_1$, and acts trivially on $(N^{-1} L_1)/(N L_1)$. In particular, any lattice $K$ with $N^{-1} L_1 \supseteq K \supseteq N L_1$ will be taken to itself by $\Gamma$. We chose $N$ so that $L_2$ lies between $N^{-1} L_1$ and $N L_1$. So $\Gamma$ takes $L_2$ to itself, and we deduce that $\Gamma$ is contained in the group we are interested in. So the group we are interested in has index $\leq [SL_2(\mathbb{Z}) : \Gamma]$ in $SL_2(\mathbb{Z})$.

-