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While reading the answers to Geometric understanding of differential forms., I stumbled upon this one, which, among other things, makes the case that $df\wedge dg$ can be interpreted as representing a "density of curves" of constant $f$ and $g$. Here $f,g\in C^\infty(M)$ for some manifold $M$.

More generally, they state that $\alpha\wedge\beta$, for generic 1-forms $\alpha,\beta$, represents the density of curves formed by the intersection of the surfaces represented by $\alpha$ and $\beta$.

I don't quite understand how this picture works. I can understand a 1-form $\alpha$ as characterising surfaces, in that at every point $p\in M$, $\alpha_p$ is a linear functional, and thus represents a plane via (the dual of) its normal vector. Similarly, $df_p$ represents a plane orthogonal to the direction of maximum variation of $f$, and thus I understand $df$ as representing surfaces of constant $f$.

But how does this picture work when we take exterior products?

For example, say $M=\mathbb R^3$ and $\alpha=xdx+ydy$ and $\beta=xdx$. Then at every point I can picture $\alpha$ as the plane orthogonal to the vector $(x,y,0)$, and $\beta$ as the plane orthogonal to the vector $(x,0,0)$. Now, $$\alpha\wedge\beta = yx \,dy\wedge dx.$$ This is now a 2-form, which takes two tangent vectors as input. How do I visualise it geometrically as a "density of curves"?

Returning to the specific case with $df\wedge dg$, a toy example could be $$f(x,y,z) = x+y^2, \qquad g(x,y,z) = yz, \\ df = dx + 2y \, dy, \qquad dg= z\, dy + y \, dz, \\ df\wedge dg = z \, dx\wedge dy + y \, dx\wedge dz + 2y^2 \, dy\wedge dz.$$

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  • $\begingroup$ This is sloppy indeed. A general $1$-form $\alpha$ in $\Bbb R^3$ does not correspond to a surface. You can get a plane at each point $p$, but very rarely will those planes integrate to give you a surface, even locally. $\endgroup$ – Ted Shifrin May 4 at 16:56
  • $\begingroup$ @TedShifrin what do you mean exactly with "integrate to give a surface" here? $\endgroup$ – glS May 5 at 11:19
  • $\begingroup$ For example, if $\alpha = dz - x\,dy$, there is no surface, even locally, whose tangent planes are given by $\alpha=0$. $\endgroup$ – Ted Shifrin May 5 at 16:49

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