The other day, I was writing exam problems for my students in my multivariable calculus class, and I was really trying to hammer home this idea that Green’s theorems allow you to compute area using a line integral. In particular, you can have your pick of 2-dimensional vector fields \(\mathbf{F}\) for which
\[ \int_{\partial R} \mathbf{F} \cdot \mathbf{T} \, ds = \textrm{Area}(R) \]
So one naturally wonders if you can adapt this idea for the surface area of an (oriented) surface with (oriented) boundary. The calculus student would probably point to the classical Stokes’ theorem to look for a solution, and eventually find that we’d need a vector field \(\mathbf{F}\) for which \(\textrm{curl}\, \mathbf{F}=\mathbf{n}\), where \(\mathbf{n}\) is the unit normal vector field which orients the surface \(\Sigma\). We immediately run into our first (and only) obstruction to this idea, which is that such a vector field exists if and only if \(\text{div}\, \mathbf{n} = 0\). Note that here we are assuming that we extend \(\mathbf{n}\) into a tubular neighborhood \(\Sigma\times [0,1] \subseteq \mathbb{R}^3\).
By an abuse of notation, let’s say use \(\mathbf{n}\) to mean an extension on such a neighborhood. The calculus student might then argue that the normal vector field defines a smooth 2-plane field given by the orthogonal complement of \(\mathbf{n}\) and that our construction guarantees that there exists an embedded surface \(\Sigma\) which is tangent to this 2-plane field.
This property of plane fields is called being locally integrable. At every point where \(\mathbf{n}\) is defined, there is an embedded surface (namely, parallel copies of \(\Sigma\)) tangent to the plane field it defines.
A creative extension of this idea may be to ask for a plane field for which at any point, there are no embedded surfaces tangent to it. One will soon arrive at the definition of a contact structure in \(\mathbb{R}^3\). Perhaps this could make for a good DRP in the future… If that does happen, I wonder how much of differential forms we can sidestep before it becomes to cumbersome to do calculations.
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