I have a question regarding the conceptual understanding of the exterior derivative. I've read that one can view a $k$-form on an $n$-dimensional manifold as a collection of infinitesimal (oriented) hyperplanes of dimension $n-k$. When given $k$ tangent vectors, this interpretation involves measuring the number of hyperplanes intersected by the infinitesimal parallelotope spanned by these vectors. For the derivative, this process extends to creating a $k+1$-dimensional parallelotope and observing the intersections of hyperplanes with its boundaries.
My confusion arises in the context of 'zoom levels.' At the highest zoom level, where everything appears flat, the distribution of hyperplanes seems uniform everywhere, leading to a derivative that always equals zero. However, when I zoom out slightly, to where linear effects start manifesting, the distribution of hyperplanes might change, but the curvature of the manifold also becomes apparent. This curvature complicates the concept of parallel transport, making it challenging to construct the parallelotope in a coordinate-independent way. Is there a resolution to this issue, or is the hyperplane analogy primarily an intuitive guide that shouldn't be taken too literally?
EDIT
User @Trebor pointed out that not every $k$-form can be visualized in terms of infinitesimal hyperplanes. This raises two new questions:
- Can I visualize those $k$-forms as some kind of weave of orthogonal, non-intersecting hyperplanes?
- Even if I abandon this concept, can I still use the idea that for the exterior derivative, I sum over the output of the $k$-form for all boundary blades of my $k+1$-dimensional parallelotope (with the question about zoom levels remaining relevant)?
ADDED:
My answer below did not address the concept of an exterior derivative. I like the definition of the exterior derivative as being exactly what is needed to generalize the Fundamental Theorem of Calculus to higher dimensions. The downside of this approach is that it uses coordinates and therefore coordinate independence of the definition has to be proved. This then leads naturally to Stokes' Theorem on a manifold with boundary.
ORIGINAL ANSWER:
Instead of using the word infinitesimal explicitly, let’s just do everything on the tangent space $T_p$ at a point $p$. And just call it $T$.
Then it all becomes just linear and multi linear algebra.
$k=1$: A $k$-form is a covector $\ell \in T^*$. The level sets of $\ell$ are parallel hyperplanes. Given a vector $v\in T$, $\ell(v)$ tell you which level set the top of the vector touches. It is a linear measurement, like a coordinate function. If you rescale $\ell$, the level sets remain unchanged but the values of $\ell$ get rescaled. That’s like changing the units of measurement (like inches to meters).
$\ell$ also defines units of measurement along a 1-dimensional linear subspace $L$. The vector $u\in L$ for which $\ell(u)=1$ is treated as this unit vector. When you integrate a line integral of a $1$-form field along a parameterized curve, you’re integrating the “length” with respect to the $1$-form of the velocity vector.
$k=2$: use idea from last paragraph. Analogy of a vector in a line is the parallelogram spanned by two vectors $v_1, v_2$ in a $2$-dimensional plane. A $2$-form $\alpha$ defines an area function on the set of all parallelograms, where the (oriented) area of the parallelogram spanned by $v_1, v_2$ is $$\alpha(v_1,v_2).$$ You can, if you want, pick one with area $1$ and call it a unit square. Again, I view a $2$-form to be a measuring instrument that defines units of area on the space of parallelograms in each $2$-plane.
The surface integral of a 2-form on a parametrized surface is the integral of over all points the oriented area with respect to the $2$-form of the parallelogram spanned by the two coordinate tangent vectors at each point.
$k\ge 1$: analogous to $k=2$.