# The centralizer of a semisimple element which is not contained in any proper parabolic subgroups

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.

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

• 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.
– D_S
Jan 9 '19 at 17:06
• But so far I don't see how we can get the converse implication.
– D_S
Jan 9 '19 at 17:08
• @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 $g$. Feb 6 '19 at 15:37