# I Interior product with differential forms

1. Apr 2, 2016

### dyn

Hi. I'm trying to self-study differential geometry and have come across interior products of vectors and differential forms. I will use brackets to show the interior product and I would just like to check I am understanding something correctly. Do I need to manipulate the differential form to get the differential to be differentiated at the front of the form ? ie
( ∂/∂x , dx∧dy ) = dy and ( ∂/∂x , dy∧dx ) = -dy

( ∂/∂z , dx∧dy∧dz ) = dx∧dy and ( ∂/∂z , dx∧dz∧dy ) = -dx∧dy

Have I got this right ? Thanks

2. Apr 2, 2016

### Orodruin

Staff Emeritus
It is really difficult to tell unless you first specify what notation you are using and how you have defined your interior product. As it appears, you have defined the interior product in such a way that $(X,\omega)(X_1,X_2,\ldots) = \omega(X,X_1,X_2,\ldots)$, where $\omega$ is a tensor of type $(0,n)$, i.e., essentially using $X$ as the first argument of $\omega$ (seen as a linear map from $(T_pM)^n$ to the real numbers).

You can then use the fact that $dx \wedge dy = dx \otimes dy - dy \otimes dx$ to verify your identity. When you have $\partial/\partial x$ you would get
$$\left(\frac{\partial}{\partial x}, dy \wedge dx\right) = \underbrace{dy(\partial_x)}_{=0} dx - \underbrace{dx(\partial_x)}_{=1} dy = -dy.$$
Of course, when you have the differential form expressed in the coordinate differentials, you can generally just anti-commute the relevant coordinate differential to the first position. All terms where it is not in the first position will vanish identically and the interior product will just correspond to anti-commuting the relevant coordinate differential to the front and then removing it from the exterior product.

3. Apr 2, 2016

### dyn

Thanks for your reply. I don't understand everything in your post but it sounds like I am right in the calculations .