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In this question I was trying to see why $k$-forms are selected as the way to generalize vector calculus rather than $k$-vectors and a comment providing links to other questions made me end up with another doubt on the geometric interpretation of $k$-forms and $k$-vectors.

Considering for a while $\Lambda^k(V)$ and $\Lambda^k(V^\ast)$ without regard to manifolds, I've convinced myself very well that an element of $\Lambda^k(V)$ represents simply pieces of $k$-dimensional planes. So for instance, if $v,w\in V$ then $v\wedge w\in \Lambda^2(V)$ would be simply the oriented paralelogram generated by $v$ and $w$ and it would carry it's area as information.

My understanding of an element of $\Lambda^k(V^\ast)$ was that of an object which can do measures with objects from $\Lambda^k(V)$. So that if $\omega \in \Lambda^1(V)=V^\ast$, then putting $\hat{\omega}=\omega^{-1}(1)$ we see that $\hat{\omega}$ can represent $\omega$ in the sense that $\omega$ defines what means for a vector to cross $1$ unit along a direction without regard to metrics.

Now, my doubt is that in that answers, people consider elements from $\Lambda^k(V^\ast)$ as pieces of $k$-planes also. So that $dx\wedge dy$ would be consider as the paralelogram $e_1\wedge e_2$. This confuses me a lot, and I think the problem is that I'm really failing to get the interpretation of $k$-forms.

Is this right to represent $dx\wedge dy$ by that paralelogram? If so, why is that? How to really get these ideas?

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I think the reason people identify these $k$-forms with geometric objects is because of integration. People see $ dx^1, dx^2, \ldots$ in integrations and think that they represent the manifold being integrated over, that basis forms are what you need in order to write integrals.

But this is not true at all! An integral of a $k$-form-field involves integrating the scalar function that arises from the $k$-form field eating a tangent $k$-vector at every point on the integration manifold. The differentials we write in integrals, $dx^1, dx^2, \ldots$ are more closely associated with basis vectors, not basis forms. That's why integral notation can be very misleading, because sometimes we mean $dx^1$ to be the exterior derivative of the coordinate function (thus giving a basis form) when in integrals, it's very misleading to take it to mean this.

The customary notation for integrals implies a relationship between basis forms and the integral differential that isn't really there.

Muphrid
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  • "integral differential" $dx$ is a basis 1-form. – rych Mar 08 '14 at 04:08
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    $dx$ meaning just "integrate over $x$" is empty and without meaning. $dx$ when used with wedge products is a form, sure. But those basis forms do not intrinsically denote what variables are being integrated over (which is why I think the $dx$ notation for basis forms is misleading), and that masks the fact that these basis forms are eating tangent vectors from the manifold of integration. – Muphrid Mar 08 '14 at 04:21
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Using the forms we have separated the notion, the function, from its arguments. For example, the volume form calculates the signed volume, a number, of a given n-vector. It is useful in curved spaces, where the forms take "infinitesimal" vectors (at a given point), hence the name differential form (field).

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