Interior products, exterior derivatives and one forms

[spaghetti3451]

If $\bf{v}$ is a vector and $\alpha$ is a $p$-form, their interior product $(p-1)$-form $i_{\bf{v}}\alpha$ is defined by$i_{\bf{v}}\alpha^{0}=\bf{0}$$i_{\bf{v}}\alpha^{1}=\alpha({\bf{v}})$$i_{\bf{v}}\alpha^{p}({\bf{w}}_{2},\dots,{\bf{w}}_{p})=\alpha^{p}({\bf{v}},{\bf{w}}_{2},\dots,{\bf{w}}_{p})$Now consider the following expressions, where $\bf{d}$ is the exterior derivative and $\nu$ is a $1$=form.How do you prove that the two expressions are equal to each other?$i_{\bf{v}}\textbf{d}(\textbf{d}\nu)+\textbf{d}i_{\bf{v}}(\textbf{d}\nu)$

$\textbf{d}(i_{\bf{v}}\textbf{d}\nu)+\textbf{d}(\textbf{d}i_{\bf{v}}\nu)$

 

[dextercioby]

The exterior differentiation is nilpotent to a degree of 2. It won't commute with the interior differentiation*, therefore you need to consider the right order of operators, that is acting to the right (pun intended!).

*See below

 

[zinq]

dextercioby, what do you mean by "interior differentiation" ?

 

[dextercioby]

Sorry, interior product. I've amended my post.