Restriction of scalars of simple algebraic groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:56:25Z http://mathoverflow.net/feeds/question/48098 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48098/restriction-of-scalars-of-simple-algebraic-groups Restriction of scalars of simple algebraic groups Menny 2010-12-02T21:18:36Z 2012-11-18T15:04:39Z <p>I'm trying to understand the following basic property of the restriction of scalars:</p> <p>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 $\mathbb Q$)</p> <p>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.</p> <p>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:</p> <p>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 $k$?</p> <p>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...</p> <p>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!</p> http://mathoverflow.net/questions/48098/restriction-of-scalars-of-simple-algebraic-groups/48109#48109 Answer by Lucio Guerberoff for Restriction of scalars of simple algebraic groups Lucio Guerberoff 2010-12-02T22:24:23Z 2010-12-02T22:24:23Z <p>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</p> <p>$$1\to\operatorname{Int}(G)\to\operatorname{Aut}(G)\to\operatorname{Aut}(\Psi_0(G))\to 1,$$ 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?).</p> http://mathoverflow.net/questions/48098/restriction-of-scalars-of-simple-algebraic-groups/112759#112759 Answer by Aakumadula for Restriction of scalars of simple algebraic groups Aakumadula 2012-11-18T13:17:14Z 2012-11-18T15:04:39Z <p>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. </p> <p>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. </p>