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Let me take an example for special orthogonal subgroup $SO(3)$.

Let $F$ be a number field and $\alpha,\beta \in F^{\times}$ such that $x^2+\alpha y^2+ \beta z^2$ does not represent $0$.

Let $J$ be a $3 \times 3$ diagonal matrix who diagonal entries are $1,\alpha,\beta$.

Let $V$ be the quadratic space over $F$ with symmetric matrix $J$ which determines a quadratic form of $SO(V)$.

Then $SO(V)$ is non-split. I am wondering whether $SO(V)$ can be split at some place $v$ of $F$. If it is, what is the condition of $\alpha, \beta$ for $SO(V)_v$ would be split?

Thanks in advance!

Andrew
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  • What does "$x^2+\alpha y^2+\beta z^2$ does not represent $0$" mean? – Arthur Dec 31 '21 at 13:53
  • @Arthur, oh, it means there are no $(x,y,z) \ne (0,0,0)$ such that $x^2+\alpha y^2 + \beta z^2=0$. – Andrew Dec 31 '21 at 14:20
  • It represents $0$ at least in all finite place not above $2$ where $v(\alpha)=v(\beta)=0$. What is the meaning of split again? – reuns Dec 31 '21 at 15:42
  • @reuns, Thanks for the comment! When $V$ arises as a trace zero space from a central simple algebra $B$ of dimension 4, $V$ is called split or not when $B$ is split or not. And $SO(V)$ is split if and only if $V$ is split. In the meanwhile, if $SO(V)$ is non-split, then I am wondering whether $SO(V)_v$ is split at most only for finitely many places $v$? – Andrew Dec 31 '21 at 18:05
  • The norm form on the quaternion algebra $B_v$ represents $0$ for all but finitely places so $B_v \cong M_2(F_v)$ for all but finitely many places – reuns Dec 31 '21 at 18:51
  • @reuns, Oh, though a global $SO(V)$’s is non-split, it is split almost everywhere places. Thank you! – Andrew Dec 31 '21 at 19:59

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