# Integral conjugacy vs. Rational conjugacy

Let $G$ be an algebraic group over a field $F$. Let $g\in G(F)$, and write $C(g)$ for the centralizer of $g$ in $G$. Conjugacy over $F$ is of course not necessarily the same thing as conjugacy over an algebraic closure $\overline{F}$. The discrepancy between the two notions is parametrized by the following set: $$\ker\left(H^1(F,C(g))\to H^1(F, G)\right).$$ If $G$ is reductive and connected with simply-connected derived subgroup, the above kernel is in bijection with the set of conjugacy classes within the stable conjugacy class of $g$ (a very important notion in the Langlands program and everything related to the Arthur-Selberg trace formula).

My question is the following: if $F$ is a local non-archimedean field with ring of integers $\mathcal{O}$, $G$ is reductive, connected, and unramified, and $g\in G(\mathcal{O})$, is there a similar way to parametrize the $G(\mathcal{O})$-conjugacy classes within the $G(F)$-conjugacy class of $g$? By similar, I mean using cohomology groups.

• What group would be acting here? Whether or not two $G(F)$-conjugate elements are $G(\mathcal{O})$-conjugate is nothing to do with the Galois action, so it doesn't seem reasonable to expect that you can measure it with a Galois cohomology group. – David Loeffler Apr 19 '13 at 10:02
• @David Of course, Galois cohomology doesn't seem reasonable. But I'm still thinking these integral conjugacy classes may be torsors over some group... I was just wondering if this is a standard/well known result. – M Turgeon Apr 20 '13 at 21:11