Areas of polygons in the the language of Riemannian geometry

areadifferential-geometryriemannian-geometryvolume

I have a pedantic and somewhat vague question that has been bothering me a bit and that I was hoping for some clarification on. I'm sure it is a standard issue that there is a standard way of dealing with and I was hoping for some pointers. I've been thinking a bit about differential forms and volume of Riemannian manifolds which is what prompts the following question.

Suppose $X^2$ is for simplicity a compact Riemannian manifold of dimension 2. I know how to define the area of codimension 0 submanifold $M$ of $X$, I take the volume form on $X$ and I integrate it over $M$. What about when $M$ is not a submanifold but is still "reasonably nice"? For me, reasonably nice means, maybe $M$ is a compact subset whose boundary is a union of a bunch of smooth arcs. Or maybe the boundary is a union of geodesics – so we just have a polygon.

Well, then I imagine I can define the area of $M$ as the supremum of the areas of all subsets of $M$ that are submanifolds. Or maybe I should just develop all the necessary material for manifolds with corners so that it covers this case…

I was wondering what the proper way to formalize all of this might be and possibly if there was a reference that does this? I also don't know exactly what "reasonably nice" should mean exactly – I guess in dimension 2 I'd like it to cover poygons and in dimension 3 for it to cover polytopes….

Best Answer

On any Riemannian (or even pseudo-Riemannian) manifold $(M,g)$ there is a standard measure $\lambda_g$ defined on the Lebesgue $\sigma$-algebra of $M$. Therefore, if you have any Lebesgue-measurable set $A\subset M$, then you can compute its measure $\lambda_g(A)$ (this includes things like polygons, or even curved polygons, for example on the surface of the sphere). With respect to this measure, any submanifold $S$ of dimension strictly smaller than that of $M$ will have measure zero (just as the measure of hyperplanes in $\Bbb{R}^n$ have $n$-dimensional Lebesgue measure zero). Once you have this measure, of course all the machinery from measure theory can be applied here.

In the comments, you also mention Stokes theorem. One of the 'basic' versions of Stokes is:

Let $M$ be an oriented $n$-dimensional ($C^2$) manifold, and $\Omega\subset M$ an open set, $\partial\Omega\subset\text{bd}(\Omega)$ be the regular part of the boundary (so $\partial\Omega$ is a $C^2$, $(n-1)$-dimensional embedded submanifold of $M$, provided with the 'induced Stokes orientation'). If $\omega$ is a $C^1$, $(n-1)$-form on $M$ such that $\text{supp}(\omega)\cap \overline{\Omega}$ is compact and doesn't intersect the 'irregular boundary' $\text{bd}(\Omega)\setminus \partial\Omega$, then we have \begin{align} \int_{\Omega}d\omega&=\int_{\partial\Omega}\iota^*\omega, \end{align} where $\iota:\partial \Omega\to M$ is the inclusion.

This is just the basic version of Stokes, with a slightly careful tracking of how many times things are being differentiated. Also, the reason I wrote it this way is because generally, I think it's simpler to think of having one big manifold, and integrating over subsets of it, because trying to characterize the nature of singularities up to diffeomorphism and making a formal intrinsic definition can be slightly tricky. For instance, if you develop the machinery of smooth manifolds with corners, you'll still be able to prove a version of Stokes. However, a cone (with its pointy vertex, and the base circle) will not be a manifold-with-corners, so even such a 'simple' case is not covered by the machinery.

With a little extra effort, one can allow for the topological boundary $\text{bd}(\Omega)$ to have slightly wilder behavior. Essentially, one defines a class of 'negligible sets' (see Lang's Real and Functional Analysis, or Dieudonne's Treatise on Analysis Vol. IX for details). With this, one has a slightly improved version of Stokes which accomodates some singularities:

Let $M$ be an oriented $C^2$, $n$-dimensional manifold, $\Omega\subset M$ an open set, $\text{bd}(\Omega)$ its topological boundary and $\partial\Omega$ the regular part of the boundary (with its induced orientation). Let $\omega$ be a $C^1$, $(n-1)$-form on $M$. If

  • $\iota^*\omega$ is integrable on $\partial\Omega$ (where $\iota:\partial \Omega\to M$ denotes the inclusion),
  • the intersection $\text{supp}(\omega)\cap\overline{\Omega}$ is compact and the intersection of $\text{supp}(\omega)$ with the non-regular part of $\text{bd}(\Omega)$ is 'negligible',

then Stokes' formula holds: \begin{align} \int_{\Omega}d\omega&=\int_{\partial\Omega}\iota^*\omega. \end{align}

Proving this version of Stokes is pretty easy using the definition of 'negligible' provided in the texts above (it amounts to a pretty routine dominated convergence argument). The point with this version is that "negligible" includes many of the standard things we might encounter. For example, a sufficient condition for negligibility is to have a compact subset $C\subset M$ which is contained inside a submanifold of dimension $n-2$. Therefore, this version of Stokes applies to many of the practical situations we might be interested in:

  • For example think of a bounded open cone in $\Bbb{R}^3$. It's topological boundary consists of the pointy vertex, the lateral surface, and the base circle. Then, the regular part $\partial\Omega$ is the lateral surface. The non-regular part is the pointy vertex (0-dimensional) and the base circle (1-dimensional), which together are compact and contained in an $n-2=3-2=1$-dimensional submanifold of $\Bbb{R}^3$.
  • This applies to polyhedra in general.
  • It applies to the flat Pacman because the non-regular part of the boundary are the three vertices (0-dimensional) where the straight lines/circle meet.

Pacman

It also applied to a 'curved pacman' (eg if you somehow map pacman onto a sphere, it will still work).

This is just one way of generalizing Stokes' theorem, and I'm sure there are a ton of different approaches which I do not know. One thing you can be comforted by is that roughly speaking, if you can actually visualize and draw the set you have in mind, then that region is 'nice enough' for Stokes' theorem to be valid. Also, one may wish for weaker conditions on the differential forms, and so Sobolev-type forms might come in handy (but I'm getting out of my depth here, so I won't comment further :).

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