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Let $q$ be a quadratic form over a nonarchimedean local field $F$, and let $\operatorname{O}(q)$ be the corresponding orthogonal group. Let $g\in\operatorname{O}(q)$ be semisimple.

Can we know anything about the eigenvalues (in $\overline{F}$) of $g$?

To be more specific, I would like to know if there is any analogue of the well-knonw theorem for the real case that the eigenvalues of an orthogonal matrix have modulus 1.

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    $\begingroup$ In general eigenvalues $\neq 1,-1$ come by inverse pairs ($t$ has same multiplicity as $t^{-1}$ for each $t$). This is the only constraint since in the split case, using a split torus one can prescribe arbitrarily the eigenvalues in the given field, with this only constraint. $\endgroup$
    – YCor
    Commented Sep 27, 2022 at 9:42
  • $\begingroup$ One can prove that the eigenvalues of the elements of $O(q)$ are always $\pm 1$ iff the quadratic form $q$ is anisotropic : $v=0$ is the unique solution of $q(v)=0$. There exist anisotropic quadratic forms in dimensions $1$ to $4$. Any quadratic form in dimension $>4$ is isotropic. $\endgroup$
    – Paul Broussous
    Commented Nov 27, 2022 at 15:59

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