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.
Best Answer
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 $g$.