# Absolute roots restricting to a given root form a Galois orbit

Let $G$ be a quasisplit, connected reductive group over a field $k$. Let $S$ be a maximal split torus of $G$, and $T = Z_G(S)$, which is a maximal torus of $G$ which is defined over $k$. Let $B$ be a Borel subgroup of $G$ containing $T$ which is defined over $k$. Let $\Delta, \tilde{\Delta}$ be the simple roots for $\Phi = \Phi(S,G), \tilde{\Phi} = \Phi(T,G)$ corresponding to $B$.

Let $\Gamma = \textrm{Gal}(k_s/k)$. Then $\Gamma$ permutes $\Phi^+$ and $\Delta$, we have $\tilde{\Delta}|S = \Delta$, and the fibers under the restriction map $\tilde{\Delta} \rightarrow \Delta$ are exactly the orbits of $\tilde{\Delta}$ under the action of $\Gamma$. This is proved in Section 12 of Brian Conrad's notes on reductive groups.

What can we say about general (nonsimple) roots? If $\alpha \in \Phi^+ = \Phi(S,B)$, do all the roots in $\tilde{\Phi}^+$ restricting to $\alpha$ form a Galois orbit?