Suppose $G$ is a Lie group. Let $\omega$ be a left-invariant $n$-form on $G$. For any $a \in G$, show that $R_a ^∗ \omega$ is also left-invariant, where $R_a : G \to G$ is the right multiplication by $a$.
I have to show that for any $g \in G$ we have $L^∗ _g (R_a ^∗ \omega) = R_a ^∗ \omega$. How do I go about this?
On the one hand, $L_g ^* (R_a ^* \omega) = (L_g ^* R_a ^*) \omega = (R_a \circ L_g) ^* \omega = \ldots$.
On the other hand, $(R_a \circ L_g) (x) = gax = (L_g \circ R_a) (x)$, so the previous formula continues as $\ldots = (L_g \circ R_a) ^* \omega = (R_a ^* L_g ^*) \omega = R_a ^* (L_g ^* \omega) = R_a ^* \omega$ (the last equality because $\omega$ is left invariant). You have just proved that $L_g ^* (R_a ^* \omega) = R_a ^* \omega$, which shows that $R_a ^* \omega$ is left-invariant, too.