3

I think of differential $k$-forms $\omega$ roughly as things that "eat oriented $k$-dimensional parallelopiped segments of a manifold $M$ and poop out a number" (in a linear fashion, yada yada). Maybe this is a "mesoscale" interpretation of differential forms; not micro as in the tangent space conception, but still these guys should only eat small oriented segments. I also think of the integral of such forms as the result of the following recipe:

$$\int_M \omega := \begin{cases} \text{cut $M$ into small $k$-dim. (oriented) segments} \\ \text{apply $\omega$ to those segments, getting numbers out} \\ \text{sum those numbers} \\ \text{take the limit as the size of the segments $\to 0$} \end{cases}$$

(I learned these interpretations from this wonderful MSE answer). However, I'm having hard time proving rigorously this limit exists, i.e. that this definition is well-defined.

I think a major benefit to this conception of the integral is its coordinate independence: the standard definition of integration of differential forms involves pulling back to $\mathbb R^n$, and using the standard Riemann/Lebesgue integral. (Many people find this distasteful and worth complaining about on MSE.) I think this standard definition "philosophically misses the point"; one does not need to go back to $\mathbb R^n$ and use Riemann/Lebesgue to "measure", because the $k$-form $\omega$ is already the thing that can "measure".

It is also very tangible/concrete: if I give someone a funky surface in $\mathbb R^3$ (like just some stiff cloth in some arrangement), and a formula for a vector field $F : \mathbb R^3 \to \mathbb R^3$, they can just analyze a bunch of tiny chunks of the cloth, and calculate using the above recipe, without needing to know a 2D-parameterization of the cloth surface.

Here are some of my ideas for formalization. If there are existing sources, please tell me. Otherwise, I would appreciate any guidance in where complications may arise and/or how to complete this argument in a rigorous but clean manner.

  • Every smooth manifold $M$ has arbitrarily good PL approximations $M'$.
  • A continuous differential $k$-form $\omega$ on $M$ (eating infinitessimal $k$-dimensional oriented parallelopipeds or similar polyhedra --- infinitessimal since the standard definition of differential forms is stuck on the tangent space) I think gives some sort of $k$-"mesoform" $\omega'$ on $M'$, which now eat small (but not infinitesimal) $k$-dimensional oriented parallelopipeds or similar polyhedra.
  • I then propose defining the integral $\int_M \omega$ to be the result of the following recipe: take a PL approximation to $M$, apply $\omega'$ to these polyhedral pieces, sum over all the pieces in the PL approximation, and take the limit as the PL approximation gets better and better.
D.R.
  • 10,556
  • 3
    You're going to have to be exceedingly careful in treating how "the PL approximation gets better and better." – Ted Shifrin May 31 '25 at 22:25
  • As a for-instance of Ted's comment: if your parallelipipeds get thin but not small (i.e., they thin out along one direction but not the other) then their area goes to zero, but your discrete sum will not, in general, converge to the integral. – John Hughes May 31 '25 at 23:33
  • @TedShifrin ah right, the Schwarz lantern pathologies. Still, maybe there's some result restricting to considering "good" PL approximations, like those satisfying some condition analogous to say the normal vectors to the parallelogram segments being aligned with the normals to the surface. – D.R. May 31 '25 at 23:45
  • 2
    You're probably going to need convergence in the appropriate norm on integral currents. I think the technicalities will make it clear why we don't take this approach "in real life." – Ted Shifrin May 31 '25 at 23:47
  • 1
    If memory serves me right, there is the notion of cohomology-homology pairing to define integration. Obviously it has no limits, but maybe you can show that taking finer and finer PL approximations (using a fixed riemannian metric by defining finer as the diameter going to 0 and then showing the choice of metric does not matter) and summing over the pieces is the same as that. Very nice question by the way! – HackR Jun 01 '25 at 02:11
  • Isn't taking a PL approximation essentially equivalent to "going back" to $\mathbf{R}^n$? – Alexey Jun 01 '25 at 07:10
  • Diff. $k$-form $\omega$ calculates volume of tangent $k$-vectors of arbitrary sizes, and if we take them small and project them back onto the manifold using $exp$ map we get next evaluation points for our limit summation. In that limit process the $k$-polytopes start approximating $M$ better and tighter. – rych Jun 01 '25 at 12:53

0 Answers0