Understanding the definition of interior product of differential forms as $(\iota_X\beta)(Y)=\beta(X,Y)$

Question

The interior product of a 2-form $\beta$ and vector field $X$ is defined by

$(i_X\beta)(Y)=\beta(X,Y)$

where $Y$ is a vector field.

This is the definition of a 2-form (and it's similar for a p-form, just extended) but I don't understand what this definition means by $Y$? Just any old vector field that we can choose? Surely $Y$ has to be subject to certain constraints and are these just assumed implicitly? e.g $Y$ has to the same dimension of $X$? Or is it slightly more subtle than this? My confusion gets even worse With a p-form, as then we have $p-1$ of these new vector fields!

I'll show you an example I was trying to understand before reverting back to this ambiguous (hopefully not for long) definition for clarity, and failing:

$\textit{Compute the interior product of X and} \space d\omega \space \textit{where}$

$X=y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}$,

$\omega=2zdx+3xdy-7zx^2dz$.

So I know how to find

$d\omega=3dx \wedge dy -(14zx+2) dx \wedge dz$

and as $d\omega$ is a 2-form, we use the formula right at the top. But what is $Y$ in this case??

Besides this even, what does it mean by, for example, $\omega(X)$? i.e the interior product of the 1-form $\omega$ with the vector field $X$.

Answer 1

It's useful to think of differential forms as antisymmetric multilinear mappings, i.e a $p$-form eats $p$ arbitrary vector fields and gives you an ordinary function of $n$ variables (where $n$ is the dimension of your manifold).

The interior product is a mapping from a $p$ form $\omega$ to a $(p-1)$ form, since you've fixed one argument of the $p$ form to be a particular vector field, say $X$ and as a result, $\iota_X \omega$ can only act on $(p-1)$ vector fields now.

To get the explicit formula in local coordinates for the interior product of a form with a vector field, you can use the formula for two forms $$(\iota_X \eta)_i = X^j\eta_{ji}$$(summation convention) to get $\iota_X d\omega$:

$(i_X d\omega) = X^j (d\omega)_{ji} dx^{i}.$

If this is still confusing, consider working your example out directly by the definition of everything:

$d\omega = 3dx\wedge dy - 14zxdx\wedge dz$, which written in terms of tensors is given by

$d\omega = 3(dx\otimes dy - dy \otimes dx) - 14zx (dx\otimes dz - dz\otimes dx)$

So then contracting in the first argument with the vector field X gives you:

$\iota_X d\omega = 3(dx(X)\otimes dy - dy(X) \otimes dx) - 14zx (dx(X)\otimes dz - dz(X)\otimes dx)$

$\iota_X d\omega = 3[dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dy - dy(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) \otimes dx] - 14zx [dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dz - dz(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dx]$

Now since $dx$ is a linear functional (field) that are dual to $\partial_x$ (and so on for $y$ and $z$), we get that

$dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) = y$

$dy(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) = 2z$

$dz(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) = 3xy$.

Using this, gives you

$\iota_X d\omega = [-3(2z)+ 14zx(3xy)]dx + (3y)dy - (14zxy)dz$

as your one-form.

Note that this is a laborious way and only serves to illustrate the definitions to you. In practice, it would be faster to get the answer by the formula given above.

Now you can take any arbitrary vector field $Y$, do the same contraction with this now one-form to get a function of $3$ variables.

Answer 2

$Y$ is arbitrary (in the sense that it could be any vector field of the correct dimension; if you have a 2-form built on $\mathbb R^3$, then the vector field should be $\mathbb R^3 \mapsto \mathbb R^3$).

So what you're discovering is that the interior product of a vector field and a 2-form field generates a 1-form field. Let $\sigma(Y) = i_X (d\omega)(Y)$. Then we have,

$$\sigma(Y) = d\omega(X, Y)$$

It should be clear looking at the left-hand side that this is indeed a 1-form.

What is the interior product of a one-form and a vector? Well, one-forms take vectors as arguments and return scalars. That's part and parcel to the idea of one-forms as linear functionals.

Answer 3

I will provide an alternate method to calculate the interor product, which I find is much more "hands-on". First, recall that given two vector fields $P = p_1 \partial_x + p_2 \partial_y + p_3\partial z$, and $Q = q_1 \partial_x + q_2 \partial_y + q_3 \partial_z$, and a two-form $\Omega = o_{12} dx \wedge dy + o_{23} dy \wedge dz + o_{13} dx \wedge dz$, the evaluation $\Omega(P, Q)$ is $\Omega(P, Q) = o_{12} (p_1 q_2 - p_2 q_1) + o_{23} (p_2 q_3 - q_2 p_3)+ o_{13} (p_1 q_3 - p_3 q_1)$. This can be derived as follows.

Now, we wish to find $\omega = i_P \Omega $. Using the definition, this means that $\omega(Q) = \Omega(P, Q)$. We Let $\omega = \alpha_1 dx + \alpha_2 dy + \alpha_3 dz$.

Recall that we evaluate evaluate a one-form $\omega(Q)$ as $\omega(Q) = \alpha_1 q_1 + \alpha_2 q_2 + \alpha_3 q_3$, since the $dx$ is the dual of $\partial_x$, and similarly for $dy$ and $dz$.

Now, we evaluate $\omega(Q) = \Omega(P, Q)$ and solve for the unknowns $\{\alpha_{1, 2, 3} \}.$

$$ \begin{align*} &\omega(Q) = \Omega(P, Q) \\ % &\alpha_1 q_1 + \alpha_2 q_2 + \alpha_3 q_3 = o_{12} (p_1 q_2 - p_2 q_1) + o_{23} (p_2 q_3 - q_2 p_3) + o_{13} (p_1 q_3 - p_3 q_1) \\ % &\alpha_1 q_1 + \alpha_2 q_2 + \alpha_3 q_3 = q_1(-o_{13} p_3 - o_{12} p_2 ) + q_2 (o_{12} p_1 - o_{23} p_3) + q_3 (o_{13} p_1 + o_{23} p_2) \end{align*} $$

Comparing coefficeients of $q_1, q_2, q_3$ yields \begin{align*} \alpha_1 = -o_{13} p_3 - o_{12} p_2 \qquad \alpha_2 = o_{12} p_1 - o_{23} p_3 \qquad \alpha_3 = o_{13} p_1 + o_{23} p_2 \end{align*}

Substituting terms from our given question, we have \begin{align*} &o_{12} = 3 \qquad o_{23} = 0 \qquad o_{13} = -(14zx + 2) \\ &p_1 = y \qquad p_2 = 2z \qquad p_3 = 3xy \end{align*}

\begin{align*} &\alpha_1 = -o_{13} p_3 - o_{12} - (- (14xz + 2)(3xy)) -3(2z) = (14xz + 2)(3xy) -3(2z) \\ &\alpha_2 = o_{12} p_1 - o_{23} p_3 = 3 y - 0 (3xy) = 3y \\ &\alpha_3 = o_{13} p_1 + o_{23} p_2 = -(14xz + 2) (y) + 0 (2z) = (-14xz + 2)y \end{align*}

Our final one-form is $\omega = \left[(14xz + 2)(3xy) -3(2z)\right] dx + (3y) dy + \left[(-14xz + 2)y \right] dz$

which agrees with calculations by @Ahsan (except for a $+2$ factor everywhere in the term $-(14xz + 2)$, because I believe they started with the incorrect value for $d\omega$).

Answer 4

It is purely a linear algebra notion used in differential calculus. Let $V$ be an $n$-dimensional vector space over some field $F$, and $\Lambda^k(V^*)$ the $k$-th exterior power of its dual, $k=0, 1, \ldots, n$. So for $k=1$ this is just the dual $V^*$. For $v\in V$, $i_v:V^*\to F$ is defined by $i_v(\alpha)=\alpha(v)$. This extends uniquely to $i_v:\Lambda^k(V^*)\to \Lambda^{k-1}(V^*),$ $k=2, \ldots,n$, by the anti-derivation property: $i_v( \alpha\wedge\beta)=(i_v\alpha)\wedge \beta+ (-1)^{deg(\alpha)}\alpha\wedge (i_v\beta).$ It is easy to write down explicit formulas using this definition.

Now take $V$ to be the tangent space of a manifold at some of its points.