Let $G$ be a connected, reductive group over a field $k$. Let $S$ be a maximal $k$-split torus of $G$, and $T$ a maximal torus of $G$ which contains $S$ and which is defined over $k$. Let $A$ be the split component of $G$, the largest split torus inside the center of $G$.

I recall a result like this:

$$A = (\bigcap\limits_{\alpha \in \Phi} \operatorname{Ker} \alpha)^{\circ}$$

where $\Phi = \Phi(S,G)$, the roots of $S$ in $G$. However, I can't seem to remember why this is true. I would appreciate a proof or reference.

When $G$ is split, a stronger result is true:

$$Z_G = \bigcap\limits_{\alpha \in \Phi}\operatorname{Ker} \alpha$$

which implies the result I want.


1 Answer 1


$(S \cap Z_G)^{\circ}$ is a split torus contained in the center of $G$, so $(S \cap Z_G)^0 \subseteq A$. Let $\widetilde{\Delta}$ be a set of simple roots of $T$ in $G$, chosen so that the set $\Delta = \{ \alpha|_S : \alpha \in \widetilde{\Delta}\} \backslash \{0\}$ is a set of simple roots of $S$ in $G$. Since

$$Z_G = \bigcap\limits_{\alpha \in \Phi(T,G)}\operatorname{Ker}\alpha= \bigcap\limits_{\alpha \in \widetilde{\Delta}}\operatorname{Ker}\alpha$$

we have

$$S \cap Z_G = \bigcap\limits_{\alpha \in \Delta} \operatorname{Ker}\alpha = \bigcap\limits_{\alpha \in \Phi(S,G)} \operatorname{Ker}\alpha$$

which implies that

$$(\bigcap\limits_{\alpha \in \Phi(S,G)} \operatorname{Ker}\alpha)^{\circ} \subseteq A $$

On the other hand, the right hand side is clearly contained in the left hand side, since $t \in A$ implies $\operatorname{Ad}t$ is trivial.


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