Show that $i_Yi_Xd\omega=d\omega(X,Y)$ for $\omega$ a $1$-form

Question

If $\omega$ is a $1$-form, how does $i_Yi_Xd\omega=d\omega(X,Y)$?

I get that $d\omega$ is a 2-form. So $i_X(d\omega)=d\omega(X,v_{2})$. So how do we proceed?

I dont see how the step $i_Y(\alpha(X,.))=\alpha(X,Y)$ could be justified. Surely iterating the contraction again would "refix" the first entry in the k-form. Then we would get something like $i_Y(\alpha(X,.))=\alpha(Y,\cdot)$.

Possibly a lack of understanding of what $d\omega(X,Y)$ is notating.

Answer 1

If $\alpha$ is a $1$-form then $i_X\alpha=\alpha(X).$ If $\alpha$ is a $2$-form then $i_X\alpha$ is the $1$-form $\alpha(X,\cdot).$ Thus, $$i_Yi_X\alpha=i_Y(i_X\alpha)=i_Y(\alpha(X,.))=\alpha(X,Y).$$

In the particulat case of $\alpha=d\omega$ you get the desired result.

Note that $i_X(d\omega)$ is a $1$-form and thus $(i_X(d\omega))(\cdot)=d\omega(X,\cdot).$