I'm struggling with the definition of an arithmetic subgroup of an algebraic group defined over $\mathbb{Q}$. In Wikipedia you can read:

If $\mathrm G$ is an algebraic subgroup of $\mathrm{GL}_n(\mathbb{Q})$ for some $n$ then we can define an arithmetic subgroup of $\mathrm G(\mathbb{Q})$ as the group of integer points $\Gamma = \mathrm{GL}_n(\mathbb{Z}) \cap \mathrm G(\mathbb{Q}).$ [...] The subgroup defined above can change when we take different embeddings $\mathrm G \to \mathrm{GL}_n(\mathbb{Q}).$
Thus a better notion is to take for definition of an arithmetic subgroup of $\mathrm G(\mathbb{Q})$ any group $\Lambda$ which is commensurable [...] to a group $\Gamma$ defined as above (with respect to any embedding into $ \mathrm{GL}_n$). With this definition, to the algebraic group $\mathrm G$ is associated a collection of "discrete" subgroups all commensurable to each other.

This definition imply that for two embeddings $\mathrm G \to \mathrm{GL}_n(\mathbb{Q})$ the groups $\mathrm{GL}_n(\mathbb{Z}) \cap \mathrm G(\mathbb{Q})$ are commensurable. Why is it true?


1 Answer 1


Write the coordinates $a_{ij}$ of one embedding into $GL_n$ as polynomial functions, defined over $\mathbb Q$, in the coordinates $b_{ij}$ of a different embedding into $GL_n$. We can do this because, by definition, embeddings give an isomorphism of algebraic varieties. (I guess we should allow the inverse of the determinant as one of our coordinates. I will still write them as $a_{ij}, b_{ij}$ for notational simplicity.)

Let $N$ be the greatest common denominator of all rational numbers appearing as coefficients of these polynomials.

Now if $b_{ij} = \delta_{ij}$ for all $i,j$ then the group element goes to the identity under the second embedding, so it is the identity, so $a_{ij} =\delta_{ij}$ for all $i,j$. It follows, by elementary algebra, that if $b_{ij}$ is congruent to $\delta_{ij}$ modulo $N$ for all $i,j$, then $a_{ij}$ is an integer for all $i,j$.

Since matrices congruent to the identity mod $N$ are a finite index subgroup, this shows the matrices with $a_{ij}$ integral contain a finite index subgroup of the matrices with $b_{ij}$-integral. Handling the other direction symmetrically, they are commensurable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.