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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 $m>2$?

flag	asked Feb 28 2012 at 16:42 ronggang 331●5

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The only such forms are $SL_{m+1}(D)$, where $D$ is an indefinite division algebra over $\mathbb{Q}$ of degree $n/(m+1)$. So the question becomes: are there finitely many isomorphism classes of indefinite division algebras of fixed degree $d$? This is far from being true. Translating into Brauer groups, you're essentially looking for elements of \bigoplus_{p}\mathbb{Z}/d\mathbb{Z}) whose co-ordinates add up to $0$, and which have order exactly $d$. – Keerthi Madapusi Pera Feb 28 2012 at 17:51

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Also, if $\mathbb Q$ is replaced by a field $K$ with underfield $k$ with $[K:k]=2$, there are rational forms that are "unitary groups". And more complicated versions made from algebras with involutions of second kind. These constructions also manufacture infinitely-many, over global fields. Nevertheless, as in the other comment, there is a fairly systematic classification scheme. – paul garrett Feb 29 2012 at 2:10

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Those unitary groups, aren't they groups over $k$ itself? It seems to me that if $G$ is a unitary group for some hermitian form over some real quadratic extension of $\mathbb{Q}$, then it's a group over $\mathbb{Q}$ which is isomorphic to $\text{SL}_n$ over $\mathbb{R}$. – Yves Cornulier Feb 29 2012 at 4:39

Yves--You are correct. These would also be $\mathbb{Q}$-forms of $SL_{n,\mathbb{R}}$. – Keerthi Madapusi Pera Feb 29 2012 at 6:09

Aakumadula · Answer 1 · 2013-01-12 11:17:30Z UTC

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 $D\otimes {\mathbb R}=M_2({\mathbb R})$. 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 central division algebra over $E$ with an involution of the second kind such that the involution restricted to $E/Q$ is the non-trivial element of the Galois group of $E/{\mathbb Q}$, and $h:D^m\times D^m \rightarrow D$ is Hermitian with respect to this involution. Furthermore, $D\otimes E_v$ is the matrix algebra $M_2(E_v)$ for both the archimedean embeddings $E_v$ of $E$.

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

how many Q-forms of SL_n(R) are there for a given Q-rank

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