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I read in the paper " From Laplace to Langlands via representations of orthogonal groups" by Benedict Gross and Mark Reeder that there are, up to isomorphism, two orthogonal groups of the (non-degenerate) quadratic forms on the $3$-dimensional $p$-adic vector space ${\mathbb Q}_p^3$, one compact and one non-compact. The reference provided is Serrs's "a course in arithmetic", Chapter IV, in which I was not able to find a proof. As far as I know, up to equivalence and scaling, there are three anisotropic and three isotropic such quadratic forms. Is it obvious that the corresponding orthogonal groups in each class are isomorphic?

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    $\begingroup$ No, since the dimension is odd, up to scaling you can suppose the determinant equal to $-1$, so the only remaining invariant is $\epsilon\in\{\pm 1\}$, and then for $\epsilon=1$ the form is isotropic (hence the group non-compact) and for $\epsilon=-1$ its anisotropic (and hence the group compact). $\endgroup$
    – YCor
    Aug 5 '14 at 13:44
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    $\begingroup$ Forms of ${\rm{SO}}_3$ are the same as forms of ${\rm{PGL}}_2$, and those in turn correspond to isomorphism classes of quaternion algebras. By our knowledge of the Brauer group of local fields, there are exactly two such algebras over any local field. The corresponding groups are split (for the matrix algebra) and anisotropic (for the division algebra). $\endgroup$
    – user27920
    Aug 5 '14 at 15:10
  • $\begingroup$ user52824: Thank you for the enlightening comment. I soon have to deal with several other cases (including different ${\mathbf Q}_p$-forms of ${\mathrm{SO}}_4$.) It would be great if you could please provide a reference that explains how to systematically find all these forms. $\endgroup$ Aug 6 '14 at 14:23
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If $q$ is a quadratic form over a field $k$, the orthogonal group $\mathrm{O}(q)$ doesn't change if you replace $q$ by $\lambda q$ for $\lambda\in k^*$. In rank 3, since $\lambda ^3\equiv \lambda $ (mod. $k^{*2}$), this implies that you can consider only forms of discriminant 1. Over $\mathbb{Q}_p$ $(p>2)$ this leaves only the invariant $\varepsilon $ (defined in terms of the Hilbert symbol), equal to $\pm 1$.

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