Is there an intuitve motivation for the wedge product in differential geometry?

I've recently started studying differential forms and have been looking at differential forms. I'm struggling to understand the motivation for introducing the notion of the wedge product. Does it simply arise when generalising the notion of a "signed area/volume" in higher dimensional spaces, or is there a deeper reasoning behind it?

If it is just a generalisation of a "signed area/volume" in higher dimensional spaces then my understanding is that the "area" spanned by two tangent vectors $X,Y$ is given by the wedge product between their associated differential forms. Thus, in one-dimension, if we have a one-form $\omega$ expressed in a local coordinate basis as $\omega =f_{i}(x)dx^{i}$, then $$\omega\wedge\omega = f_{i}(x)f_{j}(x)dx^{i}\wedge dx^{j}$$ and so from this, if X=Y, then the "area" spanned by them should be zero and so, $$\omega\wedge\omega (X,X)=0=f_{i}(x)f_{j}(x)dx^{i}(X)\wedge dx^{j}(X)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ \quad\;\;=\frac{1}{2}\left[f_{i}(x)f_{j}(x)dx^{i}(X)\wedge dx^{j}(X)+f_{j}(x)f_{i}(x)dx^{j}(X)\wedge dx^{i}(X)\right]$$ and this implies that $$dx^{i}(X)\wedge dx^{j}(X)=-dx^{j}(X)\wedge dx^{i}(X)$$ I'm unsure whether my understanding here is correct or not?

• Just to start, if $\omega$ is a one-form, then $\omega \wedge \omega = 0$. – Muphrid Mar 17 '15 at 21:11
• I was wondering what the intuition behind this is though? – Perpetual learner Mar 17 '15 at 22:41

Your guess is indeed correct, as is your conclusion that the wedge product is anti-symmetric. A concrete example that shows that this is the right motivation is the case of a wedge product between two one-forms in a three dimensional Riemannian manifold. In that case, if $\omega_1$ is the one-form corresponding to a vector $v_1$, and $\omega_2$ the one corresponding to $v_2$, then the Hodge dual $\star \omega_1\wedge \omega_2$ corresponds to the cross product $v_1\times v_2$. It is a basic result in linear algebra that this latter product is a signed area element.
If the previous seems cryptic (which certainly does, if you're just starting,) the basic point is that in a three dimensional Riemannian manifold there is a natural isomorphism between the two-forms and one forms (this is the $\star$ map above.)