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 $X$.
Best Answer
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.