Maurer Cartan 1-form under the pull-back of the right action of a Lie group

Question

Let $G$ be a Lie group and $\mathfrak{g}$ be the associated Lie algebra. Let $\omega \in \Omega^1 (G, \mathfrak{g})$ be the $\textbf{Maurer Cartan}$ 1-form defined as follows \begin{equation*} \omega (v) = dl_{g^{-1}} (v), \, \, \forall v \in T_g G, \end{equation*} where $dl_{g^{-1}}$ denotes the push forward of the left action of the group $l_g : G \longrightarrow G, h \mapsto gh$. Prove that for all $v \in T_gG$, $r_h^* \omega (v) = Ad(h) \omega (v)$, where $r_{h}: G \longrightarrow G, g \mapsto gh^{-1}$ denotes the right action of the group while $Ad: G \rightarrow GL(\mathfrak{g}), h \mapsto Ad(h)$, $Ad(h)\omega(v) = h \omega(v) h^{-1}$ is the adjoint representation of the group.

By definition of pull-back $r_h^* \omega (v) = \omega (dr_h(v))$ and inserting the definition of $\omega$ one gets $r_h^* \omega (v) = dl_{g^{-1}} (dr_h (v))$. At this point, I get stuck since I don't know how to recover the adjoint representation of the group. My doubt concerns how to explicit the action of the push forward of $r_h$ on $v \in T_g G$.

If anyone could give me a suggestion on how to proceed, it would be greatly appreciated.

Answer 1

Actually, I realized it was quite straightforward to prove. As I wrote, $r_h^* \omega (v) = \omega (dr_h v)$. Besides, since $dr_h : T_g G \rightarrow T_{gh^{-1}} G$, $dr_h v$ belongs to $T_{gh^{-1}} G$ and thus $r_h^* \omega (v) = \omega (dr_h v) = dl_{hg^{-1}} (dr_h v) = hg^{-1} v h^{-1} = Ad(h) \omega(v)$ (with a slight abuse of notation).

Answer 2

If you use the abuses of notation $gv$ and $vg$ for ${\rm d}(L_g)(v)$ and ${\rm d}(R_g)(v)$, where the differentials are evaluated at the base point of $v$, it becomes trivial: $$ R_g^*\omega(v) = \omega(vg) = g^{-1}vg = {\rm Ad}(g^{-1})v,$$since the chain rule says that the "associativity" $(gv)h = g(vh)$ holds, and ${\rm Ad}(g)$ becomes conjugation by $g$.