# Interior product between differential forms and vector fields

I don't understand what is meant when someone writes that forms (or form fields) "eat" vectors (or vector fields). For example when I have a one form field ω=3dx+5dy+3xdz and a vector field X=3x∂x+5y∂y+3∂z, then their interior product $\iota_{X}\omega=\omega(X)= (3dx+5dy+3xdz)(3x\partial_{x}+5y\partial_{y}+3\partial_{z})=6x(\partial_{x}dx)+25y(\partial_{y}dy)+6x(\partial_{z}dz)=9x+25y+6x=15x+25y$

I think this is right. So in this case ω(X) doesn’t mean that ω is function of X, it just means multiplication. Now how does this work with two forms? For example X=y∂x+2z∂y+3xy∂z and the two form ω=3dx∧dy−(14zx+2)dx∧dz

$\iota_{X}\omega=\omega(X,V)=(3dx\wedge dy-(14zx+2)dx\wedge dz)(y\partial_{x}+2z\partial_{y}+3xy\partial_{z}, V)$

Now how do I proceed from here?

• Well, $3\cdot 3=9$.. Commented Aug 6, 2014 at 21:09
• The interior product sends a $p$-form to a $(p-1)$-form by putting the vector field in the first argument of the $p$-form. When people say forms eat vector fields, they mean that forms are defined as linear functions on fields: it is just a dual construction, in the same way a dual vector eats a vector. Commented Aug 6, 2014 at 21:14
• In your first example, you have a $1$-form being mapped to a $0$-form (aka a function) by just evaluating the form on the given field. Saying it "just means multiplication" isn't right. You have to know how, e.g., $dx$ acts on $\partial_x$ and $\partial_y$ as a function. Commented Aug 6, 2014 at 21:16
• Well, yes a 1-form $\omega$ is a linear function of $X$, satisfying $dx(\partial_x)=1$ and $dx(\partial_y)=dx(\partial_z)=0$. Similarly, for a 2-form, all you have to know is $$(dx\land dy)\big(\partial_x,\partial_y\big)=1$$ and that $dx\land dy$ is antisymmetric. Commented Aug 6, 2014 at 21:18

Interior product is defined like this: if $\omega$ is a $k$-form and $X$ is a vector field, then $\iota_X \omega$ is a $(k-1)$-form defined by (remember, $\iota_X \omega$ "eats" $k-1$ vectors and returns a function): $$[\iota_X \omega](V_1, V_2, \dots, V_{k-1}) := \omega(X,V_1, V_2, \dots, V_{k-1}) ,$$ i.e., we just put $X$ in the first argument of $\omega$, leaving $k-1$ remaining slots.

Here's an example: let's say $\omega$ is the two-form $\omega = dx \wedge dy$. Let's first just work out what $\omega$ is on some pairs of vector fields (you should essentially take all of the following to be part of the definition of $dx \wedge dy$):

1. $\omega(\partial_x, \partial_y) = 1$.

2. $\omega(\partial_y, \partial_x) = - \omega(\partial_x, \partial_y) = -1$.

3. $\omega(\partial_x, \partial_z) = 0$. Similarly, if we plug in $\partial_z$ into either slot in $\omega$ (no matter what is in the other slot), we get zero, since $dz$ does not appear in $\omega$.

4. $\omega(\partial_x, \partial_x) = - \omega(\partial_x, \partial_x)$ by the antisymmetry property of forms, so $\omega(\partial_x, \partial_x) = 0$. (By the same reasoning, $\omega(X, X) = 0$ for any $X$ and any two-form $\omega$.)

5. An exercise in linearity: $\omega(2 \partial_x + \partial_z, 3 \partial_x - 2\partial_y) = -4$ (do you see why?).

Now let's work out $\iota_{\partial_x} \omega$, the contraction of $\omega = dx \wedge dy$ with $\partial_x$. To do so, we need to find what $\iota_{\partial_x} \omega$ is on each of the three basis vectors $\partial_x$, $\partial_y$, and $\partial_z$:

1. $[\iota_{\partial_x} \omega] (\partial_x) = \omega(\partial_x, \partial_x) = 0$.

2. $[\iota_{\partial_x} \omega] (\partial_y) = \omega(\partial_x, \partial_y) = 1$.

3. $[\iota_{\partial_x} \omega] (\partial_z) = \omega(\partial_x, \partial_z) = 0$.

Put succinctly, the above information tells us simply that $$\iota_{\partial_x} \omega = dy.$$

In hindsight, we didn't have to write out all of the above. Instead, we could have seen that $\iota_{\partial_x} \omega = dy$ just by "plucking off" the $dx$ (which "eats" the vector $\partial_x$ we're contracting with) from the front of $dx \wedge dy$ to leave $dy$.

You can apply this somewhat formal and mindless "plucking off rule" to compute other contractions, but you have to be careful with signs. For example, $$\iota_{\partial_y} \omega = \iota_{\partial_y} (dx \wedge dy) = \iota_{\partial_y} (- dy \wedge dx) = -dx.$$ The $\partial_y$ plucks off the $dy$, leaving the $dx$, but we had to move the $dy$ to the first position in order to pluck it off, introducing a minus sign. I would encourage you to check that this is correct using the first (longer) technique above of evaluating $\iota_{\partial_y} \omega$ on each of $\partial_x$, $\partial_y$, and $\partial_z$.

I hope that if you understand this example, you will be able to do your problem simply by using linearity.

Thank you for your answers and Phillip Andreae has the best answer so far. But I want to answer the question myself because I found an easier and more intuitive way to look at this problem. I just think of differential forms as alternating multilinear functions, for example the determinant ! Let's take a simpler example:

$\omega = 3x dx\wedge dy$

$X = 2y\partial_{x}+4xy\partial_{y}$

Now the I can write the two form as a determinant with a factor $\omega = 3x dx\wedge dy(X,V) = 3x\begin{vmatrix} dx(X) & dx(V) \\ dy(X) & dy(V) \end{vmatrix}$
This equals $\omega = 3x[dx(X)dy(V)-dx(V)dy(X)]$

$dx(X)= 2y$ and $dy(X)=4xy$.

So $\iota_{X}\omega= 6xydy-12x^{2}ydx$

The same goes with three forms and so on. Any objections?

• So in this case $dx(V)$ and $dy(V)$ become simply $dx$ and $dy$? And if yes why? Commented Jun 1, 2020 at 17:37

A 2-form eats 2 vectors, not 1. So if $\omega$ is a 2-form and $X$ is a vector field, $i_X\omega$ is not a function, but a 1-form. Alternatively, if $X,Y$ are vector fields, then $\omega(X,Y)$ is a function.

More generally, a $k$-form eats $k$ vector fields and returns a function. The most important thing about differential forms is that they are linear and skew symmetric, i.e. if $\omega$ is a $k$-form, $X_1,\ldots,X_k$ are vector fields and $\sigma$ is a permutation, then

$\omega(X_{\sigma(1)},\ldots,X_{\sigma(k)})=(-1)^{sgn(\sigma)}\omega(X_1,\ldots,X_k)$.