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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!

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For the latter question, take G=SL_3 and H=SU_3. They're isomorphic over the algebraic closure, but are only isomorphic at half of the completions, hence answer is no. –  Peter McNamara Dec 2 '10 at 21:26
    
Indeed. But these are outer forms? (or maybe not? I'm quite confused). Is it also not true for inner forms? –  Menny Dec 2 '10 at 21:44
    
[They are indeed outer forms] –  Kevin Buzzard Dec 2 '10 at 21:45
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Arithmetic & Galois actions on root data are irrelevant. First I give easy proof of something very general: for any field $K$, finite etale $K$-algebras $K'$ and $K''$, and ss gps $G'$ and $G''$ over $K'$ and $K''$ resp. with all fibers connd and abs. simple, every $K$-gp isom. R$_{K'/K}(G')\simeq$ R$_{K''/K}(G'')$ comes from a unique pair $(f,\alpha)$ consisting of a $K$-algebra isom $\alpha:K''\simeq K'$ and a gp isom $f:G'\simeq G''$ over Spec($\alpha$). Pf: By uniqueness and Galois descent, enough to check after base change to $K_s$. Then $K'$ and $K''$ are powers of $K$, so it's obvious! –  BCnrd Dec 3 '10 at 6:38
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OK, that being said, we now dispose of the original finiteness question using nothing subtle or arithmetic. Let $K'/K$ be a finite separable extension of fields and let $G'$ and $H'$ be two connected semisimple $K'$-groups that are abs. simple. Suppose the $K$-groups R$_{K'/K}(G')$ and R$_{K'/K}(H')$ are $K$-isomorphic. Previous comment shows $H'$ is an Aut($K'/K$)-twist of $G'$ (which is clearly "best possible"), and Aut($K'/K$) is a finite group. Hence, only finitely many possibilities for $H'$, and we've nailed down exactly which ones actually occur (e.g., if Aut($K'/K) = 1$ then $H'=G'$!) –  BCnrd Dec 3 '10 at 6:43
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2 Answers 2

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?).

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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.

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Don't the comments to the original question provide a short elementary proof of the finiteness over any field using nothing beyond Galois descent (in particular, no input from number theory)? –  user28172 Nov 18 '12 at 16:11
    
Indeed, but I find this answer very interesting also. Thanks, Askumadula! –  Menny Nov 19 '12 at 20:57
    
@Menny: OK, but doesn't Margulis' theorem (say in its adelic form) require an isotropicity hypothesis (or more)? It would be good to clarify this issue. –  user28172 Nov 20 '12 at 4:14
    
No, Margulis' theorem does not require isotropy hypothesis; if $G$ is isotropic over $K$, then this result is a theorem of Borel and Tits (abstract homomorphisms paper). The Margulis theorem, on the other hand, does not use cocompact or non-cocompact (the proof in his book). –  Venkataramana Nov 20 '12 at 9:00
    
And thanks Menny for your remarks. –  Venkataramana Nov 20 '12 at 9:01
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