Coordinate-free definition of integration of differential forms? Let $\omega$ be an $n$-form on an oriented $n$-manifold $M$. To integrate $\omega$, we choose an atlas $(O_\alpha, (x^1_\alpha,\dots, x^n_\alpha))_\alpha$ for $M$ and a partition of unity $\phi_\alpha$ subordinate to the atlas. Then we write $\omega|_{O_\alpha} = f_\alpha \mathrm{d}x^1 \wedge \dots \wedge \mathrm{d}x^n$ and define $\int_M \omega = \sum_\alpha \int_{O_\alpha} \phi_\alpha f_\alpha dx^1\cdots dx^n$, where now the "d"'s represent the Lebesgue measure rather than the exterior derivative of differential forms. Then we show that the result doesn't depend on the choice of atlas or partition of unity.
Is there an alternate definition that avoids the coordinates? It seems to me that one should be able to define integration of a differential form in a coordinate-independent way and then derive the above formula as a consequence.
It's not actually the partition of unity that bugs me the most. What really puzzles me is the way we use coordinates to "magically" transform our differential form into a measure. This transformation doesn't depend on a choice of coordinates, so why should we have to use coordinates to describe it?
 A: I've got a partial answer that in particular addresses Zhen Lin's objections. It requires relating integration in different dimensions in two different ways.
We rely on two principles:


*

*External product: $\int_{X\times Y} \omega \boxtimes \eta = (\int_X \omega) \cdot (\int_Y \eta)$, whenever $\omega$ is a top-dimensional form on $X$ and $\eta$ is a top-dimensional form on $Y$. By $\omega \boxtimes \eta$ I mean the external product, $\omega\boxtimes \eta = \pi_X^*(\omega)\wedge \pi_Y^*(\eta)$. 

*Stokes' theorem: $\int_{\partial \Omega} \omega = \int_{\Omega} \mathrm{d}\omega$


Suppose we've gotten as far as agreeing that for 1-dimensional integrals, $\int_{\Omega} f \mathrm{d}x = c\int_{\Omega} f dx$ for some scalar $c$. Then consider the area of the unit square $\int_{I \times I} \mathrm{d}x \wedge \mathrm{d}y$.


*

*Using (1), $\int_{I \times I} \mathrm{d}x \wedge \mathrm{d}y = (\int_I \mathrm{d}x)(\int_I \mathrm{d}y) = (c\int_I dx)(c\int_I dy) = c^2$

*Using (2), $\int_{I \times I} \mathrm{d}x \wedge \mathrm{d}y = \int_{I \times I} \mathrm{d}x \wedge \mathrm{d}y = \int_{\partial (I\times I)} x\mathrm{d}y = c \int_{\partial (I\times I)} x dy = c$ (using that $\mathrm{d}(x\mathrm{d}y) = \mathrm{d}x \wedge \mathrm{d}y$)


So we have $c^2 = c$, and so $c = 0$ or $c=1$. Of course $c = 0$ can be eliminated as a degenerate case. From this we can conclude that $\int f \mathrm{d}x^1 \wedge \cdots \wedge \mathrm{d}x^n = \int f dx^1\cdots dx^n$ by approximating $f$ by polynomials, say (since the integral of a monomial can be integrated by the external product rule, and so the integral of a polynomial can be calculated by linearity from there. Some continuity principle is needed.), and using pullback and linearity principles we can derive the value of the integral in general, say by the partition of unity argument.
Edit
Here's a way to derive the condition $\int f\mathrm{d}x = c\int f dx$ from a weaker assumption. Assume that $\int_{\Omega} f\mathrm{d}x = \int_{\Omega} fg dx$ for some function $g$, where $\Omega \subseteq \mathrm{R}^n$ is the closure of a bounded open subdomain of $\mathbb{R}^n$, although we only need subdomains of $I=[0,1]$, the unit interval. (Plausibly this can be concluded from some general continuity and naturality conditions on the integration operator.) Consider the integral $\int_{[0,t]} \mathrm{d}x$:


*

*By the assumption, $\int_{[0,t]} \mathrm{d}x = \int_{[0,t]} g(x) dx = G(t)$ where $G$ is the antiderivative of $g$ (with $G(0) = 0$), using the ordinary fundamental theorem of calculus.

*Let $\phi: I \to [0,t]$ be the multiply-by-$t$ map. By pullback, $\int_{[0,t]} \mathrm{d}x = \int_I \phi^*(\mathrm{d}x) = \int_I t\mathrm{d}x = t G(1)$.
So $G(t) = tc$ where $c = G(1)$. By differentiating, $g(t) = c$, and we have $\int f \mathrm{d}x = c \int f dx$ as desired.
A: There is a coordinate free approach to integration in the book Global Calculus by Ramanan, Chapter 3. In particular, the change of variable formula is deduced from abstract nonsense in Corollary 2.9. Unfortunately, I don't really understand the abstract nonsense carried out earlier. I cannot even precisely pinpoint where the "magic" is happening. I will try to summarize my understanding of what he does.
Roughly, Ramanan constructs the sheaf of densities as a subsheaf of the sheaf of Borel measures in Def. 2.6. To me, this definition is a bit unclear, because it uses a flat homomorphism from top dimensional forms to Borel measures even though it is remarked in Rem. 2.3 that this does not in general exist.
He claims that it is obvious that one has a canonical isomorphism of the sheaf of densities with the tensor product of top dimensional differential forms and the orientation sheaf (i.e. with the usual definition of densities found for example on Wikipedia. I cannot confirm that this is obvious, but if it is obvious, this suggests that the magic is happening somewhere else.
Borel measures can be integrated by definition (by pairing with the constant unit function), so viewing densities as a subsheaf of Borel measures, one can integrate densities.
On the other hand, densities on $\mathbb{R}^n$ have the form $f dx_1 \dots dx_n$.
I don't know where he proves that the integral of this density agrees with the Lebesgue integral of $f$.
A: It's probably worth trying to write down a precise statement. I think many variations should be possible.
Notation:


*

*Let $Man$ be the category of compact oriented smooth manifolds $(M,o_M)$ with boundary, with orientation-preserving smooth embeddings as morphisms. Here $o_M$ is an orientation form, well-defined up to multiplication by an everywhere-positive function. The category $Man$ is symmetric monoidal under cartesian product $\times$. The unit is the positively-oriented one point manifold $\bullet$. If $M_1,M_2 \to M$ are maps which are jointly surjective and $M_1 \cap M_2$ is lower-dimensional, then say that $M = M_1 \cup M_2$ is a partition.

*Let $\iota Man \subset Man$ be the symmetric monoidal subcategory with the same objects whose morphisms are the orientation-preserving diffeomorphisms.

*Let $Vect$ be category of real vector spaces equipped with a closed cone of "nonnegative" vectors. A morphism is a linear map preserving nonnegativity. The category $Vect$ is symmetric monoidal under tensor product $\otimes$. The unit is $\mathbb R$ with the usual positive cone.

*Let $\Omega^{topdim}: Man^{op} \to Vect$ be defined by $(M^n,o_M) \mapsto \Omega^n(M)$. A form is in the positive cone if it is of the form $fo_M$ where $f$ is an everywhere-positive smooth function. The functor $\Omega^{topdim}$ is lax symmetric-monoidal via the maps $\Omega^{topdim}(\bullet) \cong \mathbb R$ and $\boxtimes: \Omega^{topdim}(M) \otimes \Omega^{topdim}(N) \to \Omega^{topdim}(M \times N)$. 

*Let $\mathbb R$ also denote the functor $Man^{op} \to Vect$ which is constant at $\mathbb R$. This functor is lax symmetric monoidal in a natural way.
Definition:


*

*An integral operator is a monoidal natural transformation $\int: \Omega^{topdim}|_{\iota Man} \to \mathbb R$ such that $\int_M \omega = \int_{M_1} i_1^\ast \omega + \int_{M_2} i_2^\ast \omega$ for every partition $M = M_1 \cup M_2$ (where $i_j: M_j \to M$ is the inclusion). 

*An integral operator $\int$ is said to satisfy Stokes' theorem if $\int_M \mathrm{d} \omega = \int_{\partial M} \omega$ for every $M \in Man$ and $\omega \in \Omega^{topdim}(\partial M)$.
Claims:


*

*The integral operators are in bijection with positive real numbers $c$. The bijection sends $c$ the map $c \int: (M^n,o_M) \mapsto (\omega \mapsto c^n \int_M \omega)$ where $\int_M \omega$ is the usual integration of differential forms.

*The usual integration operator (i.e. the case $c=1$ from Claim 1) is the unique integration operator satisfying Stokes' theorem.
Remark:
Note that Stoke's Theorem is used only in a very weak sense as a normalization condition. Perhaps some other principle could be substituted here.
Proof Sketch:


*

*Existence: use the standard theory of integration of differential forms and verify that these properties hold in that case.

*The partition property implies a locality property: if a form vanishes in a region, then the integral over that region is zero.

*Therefore, the value of an integral operator may be computed via a partition of unity, reducing to the case where $M = D^n$ is a disk.

*It also follows from locality that a integrating a form is given by writing it in coordinates and integrating against some absolutely continuous measure.

*Deduce as above that this absolutely continuous measure is a scalar multiple of Lebesgue measure.

*Conclude from monoidalness that the scalar for dimension $n$ is given by $c^n$ where $c$ is the scalar for dimension 1.
