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Let $F$ be a number field.

If $G$ is a reductive group over $\mathcal{O}_F$ then we can look where $G\otimes \mathbb{C}$ fits in the classification of complex reductive groups and get a "standard model" $G_{\mathrm{st}}$. The algebraic group $G$ will differ from $G_{\mathrm{st}}$ at only finitely many places.

Is it true that if we fix $G\otimes \mathbb{C}$ and fix the places of difference then there is only finitely many reductive groups over $\mathcal{O}_F$ we could have started with?

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    $\begingroup$ Welcome new contributor. You will find the answer to your question, and presumably even more, in arxiv.org/abs/1501.04526 $\endgroup$
    – Ariyan Javanpeykar
    Commented Sep 10, 2021 at 9:01
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    $\begingroup$ @AriyanJavanpeykar You could post that comment as an answer. $\endgroup$
    – David Loeffler
    Commented Sep 10, 2021 at 9:07

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Yes.

See Javanpeykar and Loughran - Good reduction of algebraic groups and flag varieties.

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