The following is the concrete lemma
Suppose $X$ is a vector field with no zeros and it's generated by a group action with no fixed points. Then there exists a 1-form $\omega$ such that $$ \iota_{X} \omega=1$$ and $$\iota_{X} d\omega=0$$
We can choose any $\omega$ satisfy the first condition since $X$ has no zeros. However, to see the second one, the text said the following
average the form over group action then we can obtain an invariant such form
I found this explanation really unclear and confusing. What does it mean to average the form over group action without changing its first condition? how does this relate to the second condition?
Thanks for any comment.
In the comments it is specified that the acting group $G$ is the circle group.
The operation of "averaging over the group action" is just the map $$A : \omega \mapsto \int_G L_g^*\omega \,dg,$$ where for any $g \in G$ the map $L_g : M \to M$ is the action of $g$, namely, $p \mapsto g \cdot p$ and $dg$ is the unique $G$-invariant $1$-form on $G$ that satisfies $\int_G dg = 1$. Unwinding definitions then shows that if $\omega$ satisfies the condition $\omega(X) = 1$ then $(A(\omega))(X) = 1$, too, that is, the averaging operation preserves the first property.
Cartan's Magic Formula asserts that $$\mathcal L_X \omega = \iota_X d\omega + d(\iota_X \omega) .$$ Thus, to prove that an already-averaged $1$-form $\omega$ satisfies $\iota_X d\omega = 0$, it's enough to show that $\mathcal L_X \omega = 0$, and this again amounts to unwinding definitions.