Verifying that a differential is surjective

Question

I've been reading "Weakly commensurable arithmetic groups and locally symmetric spaces" (Prasad and Rapinchuk, 2009). I'm having some trouble showing the following fact:

Let $K_v$ be a local field, $G$ a connected absolutely almost simple group defined over $K_v$ and let $T \subset G$ be a maximal $K_v$-torus. Let $\phi$ be the map. $$ \phi: G(K_v) \times T(K_v) \longrightarrow G(K_v), \ \ (g,t) \mapsto gtg^{-1} $$

Show that the differential $d_{( g,t)} \phi$ is surjective for any $( g, t) ∈ G(K_v ) × T_\mathrm{reg} (K_v)$ (where $T_\mathrm{reg}$ is the Zariski-open subvariety of regular elements in $T$).

I understand why this should be true, but I'm having trouble proving it, or even getting started with the proof. Can anyone point me in the right direction?

Answer 1

You can identify $d\phi$ with the map $(X,Y)\mapsto \operatorname{Ad}(g^{-1})(\operatorname{Ad}(t)X-X+Y)$ from $\operatorname{Lie}(G)\times\operatorname{Lie}(T)$ to $\operatorname{Lie}(G)$. Since $t$ is semisimple, $\operatorname{Ad}(t)$ acts diagonalisably on $\operatorname{Lie}(G)$ (passing to an algebraic closure if necessary, which is harmeless). Since $t$ is regular, the 1-eigenspace of this action is precisely $\operatorname{Lie}(T)$. Surjectivity is now clear.