# A Interior products, exterior derivatives and one forms

1. Oct 29, 2016

### 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?

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

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

2. Nov 4, 2016

### Greg Bernhardt

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Nov 6, 2016

### 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

Last edited: Nov 8, 2016
4. Nov 8, 2016

### zinq

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

5. Nov 8, 2016

### dextercioby

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