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.
Best Answer
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 $\mathbb R^n$.
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.