I will phrase my question on $\mathbb R^3$ for concreteness.
Given a one-form $\omega = f_x(x, y, z) dx + f_y (x, y, z) dy + f_z(x, y, z) dz$, we can evaluate this on a vector field $X = g_x(x, y, z) \frac{d}{dx} + g_y (x, y, z) \frac{d}{dy} + g_z(x, y, z) \frac{d}{dz}$ as $\omega X = f_x g_x + f_y g_y + f_z g_z$ due to $\frac{d}{dx_i}$ and $dx_i$ being dual to each other.
Now, given a two-form $\Omega = f_{xy} dx \wedge dy + f_{yz} dy \wedge dz + f_{xz} dx \wedge dz$, how do I evaluate this on two vector fields, $X = g_x \frac{d}{dx} + g_y \frac{d}{dy} + g_z \frac{d}{dz}$ and $Y = h_x \frac{d}{dx} + h_y \frac{d}{dy} + h_z \frac{d}{dz}$?
Intuitively, I would have expected the answer to be $\Omega(X, Y) = f_{xy} g_x h_y + f_{yx} g_y h_z + f_{xz} g_x h_z$, since each component of the wedge in $\Omega$ will pick up the corresponding component from the vector field.
However, this seems to contradict the answer found in this question, so I clearly don't understand how to evaluate two forms.
Could someone show me the explicit computation in this case of $\Omega, X, Y$, the general formula, and if possible, the geometric meaning of evaluating a form $\Omega$ on the vector fields $X, Y$?
EDIT: I believe I now know how to perform the evaluation, I would like someone to verify my computation. We first get $9$ terms by the multi-linearity of $\Omega$, and then simplify using anti-symmetry:
\begin{align*}
&\Omega(X, Y) = \Omega(g_x \partial_x + g_y \partial_y + g_z \partial_z, h_x \partial_x + h_y \partial_y + h_z \partial_z) =\\
&g_x h_x \Omega(\partial_x, \partial_x) +
g_x h_y\Omega (\partial_x, \partial_y) +
g_x h_z\Omega (\partial_x, \partial_z) + \\
%
&g_y h_x \Omega(\partial_y, \partial_x) +
g_y h_y\Omega (\partial_y, \partial_y) +
g_y h_z\Omega (\partial_y, \partial_z) + \\
%
&g_z h_x \Omega(\partial_z, \partial_x) +
g_z h_y\Omega (\partial_z, \partial_y) +
g_z h_z\Omega (\partial_z \partial_z)
\end{align*}
We know that $\Omega(\partial_x, \partial_x) = \Omega(\partial_y, \partial_y) = \Omega(\partial_z, \partial_z) = 0$ by anti-symmetry.
Also, $\Omega(\partial_y, \partial_x) = – \Omega(\partial_x, \partial_y)$, and similar rules hold for $\partial_y, \partial_z$.
Using these, we simplify the above expression of $\Omega(X, Y)$ as:
\begin{align*}
\Omega(X, Y) &=
(g_x h_y – g_y h_x) \Omega(\partial_x, \partial_y) +
(g_x h_z – g_z h_x) \Omega(\partial_x, \partial_z) +
(g_y h_z – g_z h_y) \Omega(\partial_y, \partial_z) \\
&= (g_x h_y – g_y h_x) f_{xy} + (g_y h_z – g_z h_y)f_{yz} + (g_x h_z – g_z h_x) f_{xz}
\end{align*}
However, I don't understand the geometric content of this — how do I visualize what values $\Omega$ will give me for vector fields $X$ and $Y$?
Best Answer
Let's ask a simpler question: What is the (geometric) meaning of $Q = dx \wedge dy (v, w)$, where $v$ and $w$ are arbitrary vectors?
There's a version of this that makes sense in 3-space, but not more generally, and I'll come back to that.
Let's look at the more general case. First, if $v$ and $w$ are collinear, then one is a multiple of the other, and the antisymmetry makes $Q$ be zero. So let's look at the case where $v$ and $w$ are not collinear, so they span some plane $P$, which we can orient by saying that going from $v$ towards $w$ is the counterclockwise direction. Roughly speaking, we want to think about how this plane "looks like" the $xy$-plane (because we're computing $dx \wedge dy$). The vectors $v$ and $w$ span a little parallelogram, $S$ in the plane $P$. (If $v$ and $w$ were standard basis vectors, this parallelogram would be a square.) So I'm just going to think of the pair $(v, w)$ as representing that little oriented parallelogram in the plane $P$.
If you project $S$ onto the $xy$-plane orthogonally, you get a little (oriented) parallelogram in the $xy$-plane, and you can compute its (oriented) area (where a negative area indicates reverse-orientation from the standard orientation). That oriented area is $dx \wedge dy (v, w)$ [up to a factor of $\frac12$, depending on how your text defines the wedge product.]
Let me digress for a moment about Plucker coordinates. Suppose you have a vector in 3-space, and you compute the lengths of its projections onto the three lines spanned by the standard unit vectors (i.e., the oriented lengths of the projections onto the x, y, z axes). From those numbers, you can reconstruct the vector itself. (Indeed, if your vector is $[a,b,c]$ those numbers will be $a$, $b$, and $c$, so it's particularly easy.)
But let me say it differently: suppose you have an oriented LINE in 3-space. Pick a vector $v$ in that line, compute the three projections, and from them, you can recover the oriented line. Let's say the vector $v$ you picked has length $s$. Then the numbers $a/s, b/s, c/s$ are independent of the choice of $v$, and are called the Plucker coordinates for the line.
Now let's do the same thing for an oriented PLANE in 3-space. Start with a tiny oriented parallelogram in the plane with one vertex at the origin, and area $A$. Compute its (oriented) projected areas on the $xy$, $yz$, and $zx$ planes, and divide each by $A$. Those three numbers are enough to let you reconstruct the plane in which that parallelogram lies, and are indepedent of your choice of parallelogram, as long as the area $A$ is nonzero. (If the numbers are $p,q,r$, then the vector $(p,q,r)$ is the unit normal vector to the plane in 3-space.)
All this requires proof of course, and I'm not going to give proofs here. But Plucker more or less observed that this idea works for $k$-planes in $n$-space: if you take a tiny paralleliped in the plane, and compute its oriented projected areas (or volumes, etc.) on each of the coordinate $k$-planes, and divide each by the area or volume of the original parallelipiped, you get $n \choose k$ numbers called the Plucker coordinates of the $k$-plane, from which you can recover the $k$-plane itself. Each of the plucker coordinates tells you "how much does your $k$-plane look like this particular coordinate $k$-plane?"
So back to $dx \wedge dy$: when applied to the pair $(v, w)$, it produces the "xy"-plane Plucker coordinate for the plane spanned by $(v, w)$, but multiplied by the area of the the oriented parallelogram spanned by $v$ and $w$.
There's a closely related question which is "why would you want this in the first place?" Well, if you go back to area integrals, where you write stuff like $$ \iint_C f dA $$ you often compute those by parameterizing the domain of integration, with some parameterization $S: [0, 1] \times [0, 1] \to C$, and then instead compute the oridinary double-integral $$ \int_0^1 \int_0^1 f(S(u, v)) \left|\left| \frac{\partial S}{\partial u}(u,v) \times \frac{\partial S}{\partial v}(u,v)\right|\right| du~ dv $$ where the stuff in the $\|\cdot\|$ signs is a "change of area" term, right?
Well, here's a different way to think of it: that's the size of the parallelogram spanned by the two partial-derivative vectors, i.e., it's almost exactly a plucker coordinate.
===
OK, I've said enough. I hope this is of some value, and that you can tie together the last few things I said to make some sense.