Let $G$ be a linear algebraic group defined over $\mathbb Q$. Suppose that $G$ is isomorphic to $SL_n$ over $\mathbb R$. Suppose the $\mathbb Q$-rank of $G$ is fixed, say $m$. How many types are there for $G$ up to $\mathbb Q$-isomorphism? Are they finite especially for $m>2$?

## 1 Answer

I would refer to the very nice article by Jacques Tits in "Algebraic Groups and Discontinuous Subgroups" AMS Symposiain Pure Math Vol 9 (Boulder Conference) 1966. The title of the article by Tits is "Classification of Algebraic Groups", and gives the $k$-forms of simple algebraic groups over the separable algebraic closure of $k$ for any field $k$.

From the tables in this article it is not difficult to deduce that the only $\mathbb Q$-forms of $SL_n$ which over $\mathbb R$ become isomorphic to $SL_n({\mathbb R})$ are the following

(1) The groups $SL_m(D)$ where $D$ is a central division algebra over $\mathbb Q$ of degree $d$ such that $dm=n$ and $D\otimes {\mathbb R}=M_d({\mathbb R})$. The $\mathbb Q$-rank of $G$ is $m-1$. As Keerti Madapusi has observed, there are infinitely many such.

(2) The groups $SU_m(D,h)$ where $K/{\mathbb Q}$ is a real quadratic extension, $D$ is a degree $d$ central division algebra over $K$ with an involution of the second kind such that the involution restricted to $K/Q$ is the non-trivial element of the Galois group of $K/{\mathbb Q}$, and $h:D^m\times D^m \rightarrow D$ is Hermitian with respect to this involution. Furthermore, $D\otimes K_v$ is the matrix algebra $M_d(K_v)$ for both the archimedean embeddings $K_v$ of $K$. The $\mathbb Q$ rank of $G$ is the maximal number of (see Tits' article for a discussion) hyperbolic planes in $h$; it can again be proved that there are infinitely many with a fixed $\mathbb Q$-rank, by varying either the field $K$ or the hermitian form $h$.

I think these may even be worked out in Dave Witte's e-book on arithmetic groups (not very sure).

`\bigoplus_{p}\mathbb{Z}/d\mathbb{Z})`

whose co-ordinates add up to $0$, and which have order exactly $d$. $\endgroup$