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 $m>2$?
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 $m1$. 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 nontrivial 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 ebook on arithmetic groups (not very sure). 


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whose coordinates add up to $0$, and which have order exactly $d$. – Keerthi Madapusi Pera Feb 28 '12 at 17:51