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avoided confusing "G_st. G"
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YCor
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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 $G_{\mathrm{st}}$ in at mostonly 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?

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" $G_{\mathrm{st}}$. $G$ will differ from $G_{\mathrm{st}}$ in at most 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?

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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ncfr
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Reductive groups over number rings

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" $G_{\mathrm{st}}$. $G$ will differ from $G_{\mathrm{st}}$ in at most 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?