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Let $G$ be a connected, reductive group over a field $k$. Let $A_G$ be the split component of $G$. If necessary, assume $k$ is perfect. Let $g \in G(k)$ be a semisimple element. Then the connected centralizer $Z_G(g)^0$ is defined over $k$. It is reductive (Borel, LAG, 13.19). Its Lie algebra is

$$\{ X \in \mathfrak g : \operatorname{Ad}(g)X = X\}$$

I'm trying to understand why the following statement is true:

$A_G$ is the maximal $k$-split torus of $Z_G(g)^0$ if and only if $g$ is not contained in any proper parabolic $k$-subgroups of $G$.

This is claimed in James Arthur's book, An Introduction to the Trace Formula, $\S 10$.

I might have a better idea of how to go about this if I had a handle on how to describe the parabolic $k$-subgroups of $Z_G(g)^0$. For example, the claim that $A_G$ is a maximal $k$-split torus of $Z_G(g)^0$ implies that $Z_G(g)^0$ cannot have any proper parabolic $k$-subgroups.

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What you want is a specific part of something that fits into a wider framework due to Borel--Tits. Specifically, you're looking for Theorem 4.15 and Corollary 4.16 of "Groupes réductifs", Inst. Hautes Études Sci. Publ. Math., 1965. Part of this result says that if $P \leqslant G$ is a $k$-parabolic subgroup then the $k$-Levi complements of $P$ are of the form $C_G(T)$ where $T \leqslant R(P)$, the radical of $P$, is a $k$-split maximal torus of $R(P)$. Moreover, you have $C_G(T) = G$ if and only if $T \leqslant Z(G)$. $\space$

I'll say $L \leqslant G$ is a Levi subgroup of $G$ if it is the Levi complement of a parabolic subgroup $P \leqslant G$. Now let $S \leqslant C_G^{\circ}(g)$ be a maximal torus, which is necessarily a maximal torus of $G$. Being connected reductive if $C_G^{\circ}(g)$ is contained in a parabolic subgroup $P$ then it's contained in a Levi complement of $P$. Hence, it suffices to show that $C_G^{\circ}(g)$ is not contained in any proper Levi subgroup of $G$.

As the intersection of two Levi subgroups containing a common maximal torus is again a Levi subgroup there is a unique minimal Levi subgroup containing $C_G^{\circ}(g)$, namely the intersection of all such Levi subgroups. More explicitly, this Levi subgroup is given by $C_G(Z^{\circ}(C_G^{\circ}(g)))$. Hence, $C_G^{\circ}(g)$ is contained in no proper parabolic subgroup of $G$ if and only if $Z^{\circ}(C_G^{\circ}(g)) = Z^{\circ}(G)$.

Borel--Tits' result allows you to do the same over $k$. Namely, there's a unique minimal $k$-Levi subgroup containing $C_G^{\circ}(g)$. It's realised as $C_G(S)$ where $S \leqslant Z^{\circ}(C_G^{\circ}(g))$ is a $k$-split maximal torus of $Z^{\circ}(C_G^{\circ}(g))$. Hence, we have $C_G(S) = G$ if and only $S$ is a $k$-split maximal torus of $Z^{\circ}(G)$.

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  • $\begingroup$ Thanks for your answer. Your last paragraph seems to show just one direction: $A_G$ is a maximal split torus of $C_G^{\circ}(g)$ $\Rightarrow$ $A_G$ is the split component of $C_G^{\circ}(g)$ $\Rightarrow$ $C_G(A_G) =G$ is the unique minimal Levi containing $C_G^{\circ}(g)$. It follows that $C_G^{\circ}(g)$, hence $g$, is not a member of any proper $k$-parabolic subgroup. $\endgroup$ – D_S Jan 9 at 17:06
  • $\begingroup$ But so far I don't see how we can get the converse implication. $\endgroup$ – D_S Jan 9 at 17:08
  • $\begingroup$ @D_S, if $A$ is a non-central split torus in $C_G(g)^\circ$, then $C_G(A)$ is the Levi component of a proper ($k$-)parabolic subgroup of $G$ containing $g$. $\endgroup$ – LSpice Feb 6 at 15:37

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