Proof that closed forms are exact using Stokes' theorem. Terence Tao says that Stokes' theorem could be taken as a definition of the exterior derivative, and in this spirit I am looking for a proof that closed forms are exact using Stokes' theorem. The special case of 1-forms is fairly straightforward, and I'm wondering if there's a similar proof for higher differential forms.
Theorem: A closed differential form $\phi$ on a simply connected domain is exact.
Proof sketch when $\phi$ is a 1-form:
Choose some base point $x_0$ and define $$f(x) = \int_\alpha\phi$$
where $\alpha$ is any path from $x_0$ to $x$. The choice of $\alpha$ does not matter because say we had another path $\alpha'$ from $x_0$ to $x$. Then since our domain is simply connected we can take a surface $S$ who's boundary is $\alpha - \alpha'$. Then using Stokes' theorem:
$$\int_\alpha\phi - \int_{\alpha'}\phi = \int_{\alpha-\alpha'}\phi = \int_Sd\phi = 0$$
since $\phi$ is closed, so the integral is the same over both paths.
Now I claim $\phi = df$. Say $\beta$ is a path from $x$ to $y$, $\alpha$ a path from $x_0$ to $x$, and $\alpha'$ a path from $x_0$ to $y$. Then:
$$\int_\beta\phi = \int_{\alpha + \beta - \alpha'}\phi + \int_{\alpha'}\phi - \int_\alpha\phi = 0 + f(y) - f(x) = \int_{\partial\beta}f$$
since $\alpha + \beta - \alpha'$ is a closed loop and $\phi$ is closed, so by the same logic as in the previous paragraph that integral is 0. Thus, since we are using Stokes' theorem as our definition of the exterior derivative, $\phi = df$ as desired.
Note that I am totally brushing over the issue of whether the exterior derivative is well defined under this definition.
So, is there a way to adapt this proof to higher order forms? One idea I've been playing with is given an $n$-form $\phi$, define the $(n-1)$-form $\varphi$ by:
$$\varphi_x(v_1, \ldots, v_{n-1}) = \int_0^1 \phi_{\alpha(t)}(v_1, \ldots, v_{n-1}, \frac{d\alpha}{dt})dt$$
where $\alpha$ is again a path from $x_0$ to $x$, in this case parameterized by the interval $[0,1]$. The subscript on the forms is the basepoint, and the arguments are tangent vectors at that point. The $v_i$ are tangent vectors at the basepoint $x$, which means for that integral to make sense we need an isomorphism from the tangent space at $x$ to the tangent space at each point along $\alpha$. We can do this by choosing a basis for each tangent space in some continuous manner (we may have to upgrade our "simply connected" qualifier for this to work).
So, can this idea of integrating an $n$-form over a path to get an $(n-1)$-form work as a proof of this theorem? Is there a different way to prove this theorem using Stokes' theorem? I find the above proof for 1-forms very intuitive and I'd like to see a similarly natural proof of this theorem in general.
 A: Okay, I have some ideas towards a proof for 2-forms. Say $\phi$ is a closed 2-form, we will try to construct a 1-form $\varphi$ with $d\varphi = \phi$. We know that integral of $\varphi$ over any closed loop using Stokes' theorem, since the integral of $\phi$ over any surface with that loop for it's boundary is the same. So, say for any pair of points $x$ and $y$ we could choose a main path between them $\alpha_{xy}$, and say we chose the value of $\int_{\alpha_{xy}}\varphi$ for every $x$ and $y$. Then we could determine the value of $\int\varphi$ over any other path $\beta$ with endpoints $x$ and $y$ because we know the integral over $\alpha_{xy}$ and we know the difference between the integral over $\alpha_{xy}$ and the integral over $\beta$. Then by taking limits of paths as their endpoints approach each other we could find the value of $\varphi$ at individual points, so this would be enough information to construct $\varphi$.
But, we have to be careful how we choose our $\alpha$'s and how we choose the values of the integrals over the $\alpha$'s, because not every way of doing this is consistent. I don't know the theory of continuous spaces of paths, and maybe there's a nice way to do this that I don't know about. But here's my idea of how to achieve this: it would be nice if for any points $x$, $y$, $z$, we had $\alpha_{xy} + \alpha_{yz} = \alpha_{xz}$, because then we could choose some scalar field $f$ and let $\int_{\alpha_{xy}}\varphi = f(y) - f(x)$ for all $x$ and $y$. (and $df$ is like the $+C$ because we're integrating.) We can choose such $\alpha$'s in a not so nice way: take some fractal tree that gets arbitrarily close to every point in the space, and define $\alpha_{xy}$ to be the unique path between those two points traveling along the tree. Insert some analysis here to formalize this idea. I think integrating over these paths shouldn't be a problem because they only get weird at their ends, and those ends have arbitrarily small measure so they end up not mattering in the integral. This solution isn't very nice though, I think we'd have to choose $f$ in some discontinuous way if we want $\varphi$ to be continuous.
Now the question is how to extend this to 3-forms. Some kind of tree made of surfaces? I'm guessing there's a nicer solution using the theory of continuous spaces of curves/surfaces.
