Does satisfying Stokes' Theorem imply that a form is linear?

Let $M$ be an $n$-manifold. A differential $k$-form $\omega \in \Omega^k M$ assigns to each point $x \in M$ a function $\omega_x : \Lambda^k T_x M \to \mathbb{R}$ which is linear.

My suspicion is that this pointwise statement of linearity is essentially equivalent to the global statement given by Stokes' theorem. To make the question precise requires a notion of "nonlinear differential form," which is provided by the notion of $k$-density, in the sense of Gelfand. For completeness I include the definition given in the link:

Definition: A $k$-density ("nonlinear* $k$-*form") on an $n$-manifold $M$ is a function assigning to each point $x \in M$ a function $\omega_x : S^k T_x M \to \mathbb{R}$ which is homogeneous (of degree 1). Here $S^k T_x M$ is the cone of simple (a.k.a. decomposable) elements of $\Lambda^k T_x M$.

A $k$-density can be pulled back to a $k$-submanifold $N \subset M$ and integrated once $N$ is given an orientation, in exactly the same way that a linear $k$-form is pulled back and integrated. The homogeneity condition ensures that this integration is independent of how $N$ is parameterized, and that the integral changes sign if the orientation of $N$ is reversed. I'm happy to discuss the fact that $\omega$ need only be defined on simple forms if it seems puzzling.

With these preliminaries, I can precisely state my question.

Question: Let $\omega$ be a $k$-density on an $n$-manifold $M$. Suppose there exists a $k+1$-density $\eta$ such that for every compact oriented $k+1$-submanifold with boundary $N \subseteq M$, we have $\int_{\partial N} \omega = \int_N \eta$ (where $\partial N$ is given the induced orientation). Does it follow that $\omega$ is a differential $k$-form, i.e. that $\omega_x$ is linear at each point $x$? Does it follow that $\eta$ is linear?

I'm pretty sure that if $\omega$ is linear, then $\eta$ can only be $\mathrm{d} \omega$, and so is also linear.

Throughout here, I haven't mentioned smoothness assumptions: I assume everything is smooth. But if the answer depends on the degree of regularity, I would find that very interesting. Also, the notion of homogeneity used could be weakened to positive homogeneity. If this affects the answer, again I would find this very interesting.

Motivation: (edit 11/06): I'm realizing I never really gave the motivation for this question. It more-or-less comes down to something that Ilya Grigoriev said a few years ago in this MO response. Basically, when asked for intuition about differential forms, he said "They're things you can integrate, plus they're linear." He admitted that the linearity seemed like a weird sacrifice of geometric intuition in favor of algebraic simplicity, since it rules out arc length / surface area and similar measures.

My question here grew out of the sense that differential forms ought to seem natural from some purely geometric perpspective -- otherwise, for one thing, it's hard to explain why they are so perfect for describing electromagnetism. In a comment to Ilya's answer, Haralde Hanche-Olsen pointed to Stokes' theorem as motivation for the linearity of differential forms: my question asked to make this precise. For me, Stokes' theorem is satisfyingly "geometric", even though it's starkly divorced from concepts like arc length and surface area. So with Anton's proof below, I for one would be happy to claim that differential forms are geometrically natural after all: Stokes' theorem carves them out as a separate "regime" of geometry, independent from arc length / surface area and related densities.

  • $\begingroup$ Although Anton doesn't mention it explicitly, his proof below continues to work if you only assume that $\omega, \eta$ are positively homogeneous. Full homogeneity will follow from the case $n=2$ in his argument (note that, unlike in the question statement, this "$n$" is not the dimension of $M$). $\endgroup$
    – Tim Campion
    Nov 6, 2013 at 1:40

2 Answers 2


Yes $\omega$ and $\eta$ have to be linear.

We will need the following generalization of Minkowski's theorem on the existence of polyhedra with prescribed surface normals; see Theorem 1, p. 475 in "Gaussian images of surfaces and ellipticity of surface area functionals" by D. Burago and S. Ivanov. Note that the Grassman space $G(k,n)$ of $k$-planes in $\mathbb R^n$ can be thought as a set of unit unit simple $k$-vectors of $\mathbb R^n$.

Let $\mu$ be a measure on $G(n,k)$ and $\varepsilon>0$. Assume $$\int\limits_{G(k,n)}x\cdot\mu=0.$$ Then there is a closed $k$-dimensional submanifold $S$ in $\mathbb R^n$ such that the pushforward measure of $\mathrm{\rm vol}_k$ along the Gauss map in $G(k,n)$ is $\varepsilon$-close to $\mu$.

Now let us start. First we prove that $\eta$ is linear. It will be sufficient for us that $\int_N\eta=0$ for any closed $(k+1)$-submanifold $N$.

It is straightforward to reduce the question to the Euclidean space with parallel $k$-density. Once it is done, note that linearity of $\eta$ is equivalent to the fact that if $s_1,s_2,\dots,s_n$ be a collection of simple $(k+1)$-vectors such that $$s_1+s_2+\dots+s_n=0$$ then $$\eta(s_1)+\eta(s_2)+\dots+\eta(s_n)=0.$$

To show that the later property holds for $\omega$, we use Burago--Ivanov theorem for the measure $\mu$ on the Grassman with support at $\{\tfrac{s_i}{|s_i|}\}$ and such that $\mu(\tfrac{s_i}{|s_i|})=|s_i|$ for each $i$.

Now we will use linarity of $\eta$ to show that $\omega$ is also linear.

Note that $d\eta=0$. By Poincaré lemma, there is a linear $k$-form $\omega'$ such that $d\omega'=\eta$. It follows that $\int_{K}(\omega-\omega')=0$ for any closed $k$-dimensional submanifold $K$. Repeating the same argument, we get that $\omega-\omega'$ is linear; whence the result follows.

  • $\begingroup$ Thanks for this answer. I'm having trouble digesting it, though. Is there a more elementary way to put this that doesn't involve the Gauss map on the Grassman space? $\endgroup$
    – Tim Campion
    Nov 1, 2013 at 23:59
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    $\begingroup$ Gauss map is a honeybun; it sends a point on the oriented $k$-manifold to its tangent space (which is a point in the Grassman space). $\endgroup$ Nov 2, 2013 at 5:09
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    $\begingroup$ Here are few more details. By Stokes' theorem $\int_M\omega=0$. Note that $$\left|\int_M\omega-\sum \omega(s_i)\right|\le \varepsilon\cdot\mathrm{Const}.$$ Since $\varepsilon$ is arbitrary, it implies that the sum of $\sum \omega(s_i)=0$. $\endgroup$ Nov 2, 2013 at 5:09
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    $\begingroup$ @TimCampion, yes the submanifold has to be almost polyhedral. $\endgroup$ Nov 2, 2013 at 18:10
  • 1
    $\begingroup$ Now the reference to the statement is included. $\endgroup$ Nov 2, 2013 at 18:11

Let me just record a few notes on Anton's very nice answer since my comments are already overflowing. This is mostly for my own benefit, and I'm making it community wiki.

Last things first, the argument proceeds by first showing that the "exact density" $\eta$ is linear, and then bootstraps to show that $\omega$ is linear too. In showing that $\eta$ is linear, the only fact used (besides the fact that $\eta$ is a density) is that $\eta$ integrates to 0 over any closed submanifold by the usual Stokes' Theorem-type argument: $\int_S \eta = \int_{\partial S} \omega = \int_{\emptyset} \omega = 0$. This implication is used twice in the bootstrap argument, applied first to $\eta$ and then to $\omega - \omega'$: very elegant.

To show that $\eta$ is linear, Anton formulates a very nice criterion of linearity, that $\sum_{i=1}^n s_i = 0$ should imply that $\sum_{i=1}^n \eta(s_i) = 0$. I believe that we can take $n=3$ here (in particular, let's forget that I used $n$ to denote the dimension of the ambient space way back in the problem statement).

The heart of the proof is to apply the Burago-Ivanov theorem. The point is that the integral in the theorem statement becomes just the sum $s_1 + \dots + s_n = 0$. So the theorem gives us a closed submanifold $S$ whose tangents are concentrated along the $s_i$, with appropriate weighting, except for a set of Euclidean measure $< \epsilon$. Following Anton's comment (with updated notation),

$\int_S \eta = \int_{S_1} \eta + \int_{S_2} \eta = \sum \eta(s_i) + O_{\eta}(\epsilon)$

where $S_1 \subset S$ is the set of points where $S$ is tangent to some $s_i$ and $S_2 \subset S$ is the remainder. Now, as advertized, we use the fact that $\eta$ integrates to 0 over the closed submanifold $S$: $\int_S \eta = 0$, so $\sum \eta(s_i) = O_{\eta}(\epsilon)$. Since $\epsilon>0$ was arbitrary, we have $ \sum \eta(s_i) = 0$.

The Burago-Ivanov theorem was a little intimidating for me. There is a measure on the Grassmanian and a weird $\Lambda^k(\mathbb{R}^n)$ - valued integral over the Grassmanian (note that "$x$" in the integral denotes the identity function when the Grassmanian is regarded as a subspace of $\Lambda^k(\mathbb{R}^n)$, or alternatively it denotes the inclusion map $G(k,n) \to \Lambda^k(\mathbb{R}^n)$). But it's not so bad. The point of the theorem is that it's actually easy to construct measures on the Grassmanian. The "geometric" way is to take some surface $S$ and pushforward its Euclidean surface area along the Gauss map: the resulting measure tells you how much of the surface is parallel to a given plane represented by a point in the Grassmanian. Burago and Ivanov call this the "weighted Gaussian image" of the surface $S$. The "algebraic" way is to simply pick a finite number of unit $k$-vectors $u_1, \dots, u_N$, and assign weights $\lambda_i$ to them to form a discrete measure $\mu$ with $\mu(u_i) = \lambda_i$. For example, Anton constructs a measure this way in his argument.

We can ask whether an algebraically-constructed measure also arises from the geometric construction, as a weighted Gaussian image. Any finitely-supported measure arises geometrically by taking a bunch of disconnected shapes of appropriate areas which lie in the appropriate planes. But according to Burago and Ivanov, we can do better than this (if we're willing to allow an $\epsilon$ of room): we can can construct such a surface whose boundary is controlled by the integral of the measure (which, again, for a discrete measure is a weighted sum of its supporting $k$-vectors). In particular, if the integral / sum comes out to zero, we can take the surface to be closed: this correspondence between an algebraic condition (related to linearity) and a geometric condition amenable to the Stokes' criterion is exactly what we need.

The $\epsilon$ of room corresponds to parts of the surface whose tangent planes weren't supported by our measure. This is needed because for a general set of planes we can't quite construct a closed polytope with sides parallel to exactly those planes. For example, the planes might be linearly independent, so there's no lower-dimensional subspace for them to intersect along. Given that there are such geometric obstructions, it's pretty remarkable that the Burago-Ivanov theorem actually holds! What does such surfaces look like? Burago and Ivanov provide one example: consider a triangular prism: by stretching the prism on one axis and shrinking on the other, you can shrink the end caps to be arbitrarily small while maintaining constant area of the other sides. This is the sort of thing that has to happen.


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