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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

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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
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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
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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
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Seriously, BCnrd? :) –  David Hansen Oct 22 '10 at 18:35
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2 Answers

up vote 21 down vote accepted

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.

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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
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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
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...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
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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 $SL_2(\mathbb{Z})$.

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