Restriction of scalars of simple algebraic groups - MathOverflow

Restriction of scalars of simple algebraic groups

2010-12-02T21:18:36Z

I'm trying to understand the following basic property of the restriction of scalars:

Given an absolutely simple algebraic groups $G$ defined over a number field $k$, are there at most finitely (up-to $k$-isomorphism) many absolutely simple algebraic groups $H$ defined over $k$ with $$Res_{k/\mathbb Q}G\cong_{\mathbb Q} Res_{k/\mathbb Q}H?$$ (i.e. their restrictions are isomorphic over $\mathbb Q$)

I understand that it follows that $G$ and $H$ are $k$-forms of each other and good understanding of the possible forms over the completions of $k$ should lead to an answer. This is because there is a theorem by Borel and Serre that establish the finiteness of forms that agree at all but finitely many completions.

Therefore my vague intuition is that the answer is yes, but I can't find a precise proof. Thinking about it also directed me to an even more basic question:

Given $G$ and $H$, two forms of each other (i.e., $G$ and $H$ both defined over $k$, and are isomorphic over the algebraic closure $\bar k$.) Is $G$ and $H$ isomorphic over $k_v$ for all but finitely many valuations $v$ of $k$?

I think that a positive answer to the latter should imply a positive answer to the former, but I think the latter might have a negative answer...

I'll be a happy if someone can shed some light about it and\or guide me to the relevant results in the literature. Thanks!

Answer by Lucio Guerberoff for Restriction of scalars of simple algebraic groups

2010-12-02T22:24:23Z

For the question in the comments, Kevin is right: every connected reductive group over k is an inner form of a unique quasi-split group, up to isomorphism. This follows from an argument using Galois cohomology and the splitting

$$ 1\to\operatorname{Int}(G)\to\operatorname{Aut}(G)\to\operatorname{Aut}(\Psi_0(G))\to 1,$$ where $\Psi_0(G)$ is the based root datum of G. More details can be found on the first article of Corvallis, and surely in some other place (maybe Borel-Tits?).

Answer by Aakumadula for Restriction of scalars of simple algebraic groups

2012-11-18T13:17:14Z

If I may be allowed, I will use a "high powered" theorem to deduce this is the case of number fields. It is indeed true that given an absolutely simple $k$-algebraic group $G$, the number of absolutely simple $k$-algebraic groups $H$ such that the restriction of scalars $R_{k/{\mathbb Q}}G$ and $R_{k/{\mathbb Q}}H$ are ${\mathbb Q}$-isomorphic, is finite. The hypothesis implies that as abstract groups $G(k)$ and $H(k)$ are isomorphic, since they are both ${\mathbb Q}$-rational points of the restriction of scalars group. Now by the Margulis superrigidity theorem (it was initially proved for ($S$)-arithmetic groups, but there is a version in his book for $k$-rational points which may be thought of as irreducible lattices in the adelic groups) such an abstract isomorphism $\theta $ arises from an isomorphism $\sigma:k \rightarrow k$ of the number field $k$, and a $k$-isomorphism $\phi: ^{\sigma }G \rightarrow H$ of $k$ algebraic groups. This means that $\theta (g)= (\phi (\sigma(g))$ for all $g\in G(k)$. $^{\sigma }G$ is the same group as $G$, twisted by the automorphism $\sigma$ on scalars.

In particular, $H$ is isomorphic to $^{\sigma }G$ for some $\sigma$. Since the number of the $\sigma$'s is finite, it follows that the number of $H$ is finite.