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.


1 Answer 1


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 $Z^{\circ}(G)$.

  • $\begingroup$ 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
    Commented Jan 9, 2019 at 17:06
  • $\begingroup$ But so far I don't see how we can get the converse implication. $\endgroup$
    – D_S
    Commented Jan 9, 2019 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
    Commented Feb 6, 2019 at 15:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.