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.