Interior product between differential forms and vector fields

Question

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?

Answer 1

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.

Answer 2

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?

Answer 3

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

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