# Maurer-Cartan form left invariance

Let $G$ be a lie group and $\mathfrak{g}$ its lie algebra, that is, the $\mathbb R$-space of left invariant vector fields on $G$. Recall the isomorphism $\mathfrak{g}\simeq T_eG$.

The Maurer-cartan form on $G$ is the $1$-form $\omega\in \Omega^1(G; \mathfrak{g})$ defined through the composition $$\omega_g:T_gG\stackrel{(dL_{g^{-1}})_g}{\longrightarrow} T_e G\stackrel{\simeq}{\longrightarrow} \mathfrak{g}.$$ Notice $$\omega_g(v)=u\Leftrightarrow u_e=(dL_{g^{-1}})_g(v).$$

I was meant to show $\omega$ is left invariant, that is, $L_g^*(\omega)=\omega$ for every $g\in G$ where $L_g^*(\omega)$ is the pullback form. The computation goes as follows:

We show initially: \begin{align*} \displaystyle L^*_g(\omega)_{g^{-1} h}=\omega_{g^{-1} h}, \end{align*} for every $g, h\in G$. In fact: \begin{align*} \displaystyle L_g^*(\omega)_{g^{-1} h}(v)&=\omega_{h}((dL_g)_{g^{-1} h}(v)). \end{align*} But: \begin{align*} \displaystyle \omega_h((dL_g)_{g^{-1} h}(v))=u\Leftrightarrow u_e&=d(L_{h^{-1}})_h((dL_{g})_{g^{-1} h}(v))\\ &=d(L_{h^{-1}}\circ L_g)_{g^{-1} h}(v)\\ &=(dL_{(g^{-1} h)^{-1}})_{g^{-1} h}(v)\\ &\Leftrightarrow \omega_{g^{-1} h}(v)=u. \end{align*} Hence: \begin{align*} \displaystyle L_g^*(\omega)_{g^{-1} h}(v)=\omega_{g^{-1} h}(v), \end{align*} that is, $L^*_g(\omega)_{g^{-1} h}=\omega_{g^{-1} h}$. Now, given $g, h\in G$ we might write $h=g^{-1} k$ (it suffices taking $k=gh$). Therefore: \begin{align*} \displaystyle L_g^*(\omega)_h&=L^*_g(\omega)_{g^{-1} k}=\omega_{g^{-1} k}=\omega_{h}. \end{align*} Is my proof ok?

Obs. I know this is a classical result but in every place I see the proof the notation is different from what I'm using and I end up confused.