Question

I have two problems: Let it $\Omega^{*}_{G}:=(\mathbb{C}[\mathfrak{g}]\otimes\Omega^{*}(M))^{G}$ be the complex of equivariant differential forms on a differential manifold $M$ (in which acts a Lie group $G$ and we suppose that $Lie(G)=\mathfrak{g}$). We can define the equivariant differential $$ d_\mathfrak{g}: (\mathbb{C}[\mathfrak{g}]\otimes\Omega^\ast(M))^{G} \rightarrow (\mathbb{C}[\mathfrak{g}]\otimes\Omega^{\ast}(M))^{G} $$ as $ (d_{\mathfrak{g}}\alpha)(X)=d(\alpha(X))-\iota_{X}\alpha(X)$ where $\iota$ denotes the inner product and $X \in \mathfrak{g}$. So I have to proove this assertion: Suppose $\alpha \in \Omega^\ast_{G}(M)$ is $d_{\mathfrak{g}}$-closed. Let $X \in \mathfrak{g}$ be such that the zero set $M_{0}(X)$ is finite. Then $\alpha(X)_{[n]}$ is $d$-exact on $M\setminus M_{0}(X)$. In order to prove the proposition I define the operator $d_{X}=d-\iota_{X}$. Then define $\theta \in \Omega^{1}(M)$ to be the dual of $X$ in the $G$-invariant metrics $\theta(\xi)=(X,\xi)$ for all $\xi \in \Gamma(M,TM)$. What means this notation? (how can I use the identification induced from the action of $\mathfrak{g}$ on $C^{\infty}(M)$ given by $(X.\phi)(x)=\frac{d}{dt}\phi(e^{-tX}.x)\arrowvert_{t=0}$?). Now denoting as $L(x)$ as Lie derivative we compute $$(L(X)\theta)(\xi)=\frac{d}{dt}\phi(e^{tX}.\theta)(\xi)\arrowvert_{t=0}=\frac{d}{dt}\theta(e^{tX}.\xi)\arrowvert_{t=0}=\frac{d}{dt}(e^{-tX}.X,\xi)\arrowvert_{t=0}=0.$$ (I didn't understood the last two steps) Hence $\theta$ is $X$-invariant. Now how I compute $d_{X}\theta$? And how can I invert this form?