# Commensurator of a subgroup of matrices

Let $$k$$ be a totally real number field and let $$\mathcal{O}_k$$ denote its ring of integers. If $$H$$ is a subgroup of $$\text{GL}(n, \mathbb{R})$$ let denote with $$H(k)$$ and $$H(\mathcal{O}_k)$$ the intersections $$H\cap \text{GL}(n, k)$$ and $$H\cap \text{GL}(n, \mathcal{O}_k)$$.

Let $$H$$ be a semisimple algebraic group defined over $$k$$. I need to show that $$H(k)$$ commensurates $$H(\mathcal{O}_k)$$, i.e. that for every $$\gamma\in H(k)$$ the subgroup $$\gamma H(\mathcal{O}_k)\gamma^{-1}$$ is commensurable to $$H(\mathcal{O}_k)$$.

I do not have any idea of how to proceed. The only observation I have done is that since the commensurator of a subgroup is group, and the constant matrices commensurate $$H(\mathcal{O}_k)$$, I can suppose that $$\gamma$$ has entries in $$\mathcal{O}_k$$, but in this way I probably lose the information of $$\gamma$$ being an element of $$H$$.

• Also posted in [math.stackexchange]( math.stackexchange.com/questions/3365855/…) – Arturo Magidin Sep 22 '19 at 18:41
• Please do not post the same question to both sites at the same time. Pick one site (the one you think it most appropriate), polst, and wait for a response. Wait longer than 40 minutes on a weekend. If you fail to secure responses in one site and you think you might get them in the other, then indicate in your post that it is a cross-post, to avoid possible duplication of effort. – Arturo Magidin Sep 22 '19 at 18:42
• I deleted the one in stack exchange, sorry :) – Diego95 Sep 22 '19 at 19:05
• For the question to make sense, you need to fix an embedding of $k$ into $\mathbf{R}$. Also, are you assuming that $H$ is a $k$-subgroup? Otherwise I'm surprised by the generality. – YCor Sep 22 '19 at 20:18
• No, I mean that $H$ is defined by algebraic equations with coefficients in $k$. In your generality, $H$ could be a dense subgroup, far from a Zariski-closed one. "$H$ is semisimple": this doesn't make sense for an arbitrary subgroup, but for an algebraic subgroup, or a closed one. – YCor Sep 23 '19 at 9:26

This is well-known, and you don't need semisimplicity for this to hold. You can prove it by considering congruence subgroups of $$H(\mathcal O_k)$$ as follows. Let $$\mathfrak n$$ be an ideal of $$\mathcal O_k$$, assume that $$H$$ is embedded in $$\mathrm{GL}_d$$, so we have a well-defined reduction modulo $$\mathfrak n$$ from $$H(\mathcal O_k)$$ to $$\mathrm{GL}_d(\mathcal O_k/\mathfrak n)$$ and define : $$\Gamma(\mathfrak n) = \{ g \in H(\mathcal O_k), g = \mathrm{Id} \pmod {\mathfrak n}\}.$$ This is a finite-index subgroup of $$H(\mathcal O_k)$$. Then if $$\gamma = (a_{ij}) \in H(k)$$, and we write $$\gamma^{-1} = a^{ij}$$ there exists $$\mathfrak n$$ such that $$a_{ij}\mathfrak n, a^{ij}\mathfrak n \subset \mathcal O_k$$ for all $$i, j$$ and it follows that $$\gamma \Gamma(\mathfrak n^2)\gamma^{-1} \subset H(\mathcal O_k)$$. In particular $$H(\mathcal O_k) \cap \gamma H(\mathcal O_k) \gamma^{-1}$$ contains $$\Gamma(\mathfrak n^2)$$ and so has finite index in $$H(\mathcal O_k)$$.