Motivation/Background:
I wish to understand the more practical/computable aspects of integration of differential forms on manifolds. In this question I shall usually implicitly include "manifolds with boundary" and "manifolds with corners" into the term "manifold".
Most textbooks on differential geometry introduce two related but a priori seemingly distinct theories of integration, integration on manifolds (with boundary/corners), and integration on chains. The exact relationship between the two is unclear to me.
Usually integration on manifolds is defined by assuming that the differential form has compact support (to avoid convergence issues), but on manifolds with boundary/corners this does mean that the support can intersect the boundary.
The integral is defined in terms of coordinates, and is globalized by using a locally finite cover by coordinate neighborhoods and a partition of unity subordinate to this cover. In practice, this does not provide a method amenable to calculations because constructing partitions of unity explicitly can be difficult/impossible.
By contrast, the integration on chains goes as follows. I will use cubic chains in this question as opposed to simplicical chains, as I find it is easier to do calculations with cubes. As far as I am aware, the two theories are equivalent.
An elementary $r$-cube is defined to be $I^r\subset\mathbb R^r$, where $I=[0,1]$. As a subspace of $\mathbb R^r$, the elementary $r$-cube is a manifold with corners. An $r$-cube in an ($n$ dimensional) manifold $M$ is defined to be a smooth map $\sigma:I^r\rightarrow M$. This map does not need to be an immersion or injective or anything, it might be quite degenerate.
An $r$-chain $C$ on $M$ is defined to be a formal finite linear combination of $r$-cubes: $$ C=\sum_i \alpha_i\sigma_i.$$
If $\omega\in \Omega^r(M)$ is a (smooth) $r$-form, the integral of $\omega$ over an $r$-cube $\sigma$ is defined as $$ \int_\sigma\omega=\int_{I^r}\sigma^\ast\omega, $$ where in the last integral the standard orientation of $\mathbb R^r$ is used. In other words, we may write $\sigma^\ast\omega=\rho(u^1,...,u^r)\mathrm du^1\wedge...\wedge\mathrm du^r$, where $u^1,...,u^r$ are the standard coordinates on $\mathbb R^r$, and the integral is $$ \int_\sigma\omega=\int_0^1...\int_0^1\rho(u^1,...,u^r)\mathrm du^1...\mathrm du^r. $$
For a general $r$-chain $C=\sum_i\alpha_i\sigma_i$, we have $$ \int_C\omega=\sum_i\alpha_i\int_{\sigma_i}\omega. $$
For the elementary $r$-cube $I^r$, the $i$th positive boundary is defined as $$ \partial_i^+I^r=\{u\in I^r:\ u^i=1 \}, $$ while the $i$th negative boundary is defined as $$ \partial_i^- I^r=\{ u\in I^r:\ u^i=0\}, $$ and the boundary of $I^r$ is $$ \partial I^r=\sum_{i=1}^r\partial_i^+ I^r-\sum_{i=1}^r\partial_i^- I^r. $$ If $\sigma:I^r\rightarrow M$ is a general $r$-cube, then $$ \partial_i^+\sigma=\sigma|_{\partial_i^+ I^r},\quad\partial_i^-\sigma=\sigma|_{\partial^-_i I^r} $$ and $$ \partial\sigma=\sum_{i=1}^r\partial_i^+\sigma -\sum_{i=1}^r\partial^-_i\sigma. $$ The boundary of a general $r$-chain is defined by linearity.
When one integrates over curves, since every 1 dimensional manifold is diffeomorphic to either $\mathbb R$, $S^1$, or a closed interval, one may always effectively parametrize the curve using a single coordinate system, which might be degenerate at a single point (in case of $S^1$), which is of measure zero, so this degeneracy does not affect the integral.
It seems to me that chains realize this ideal for higher dimensional manifolds, since integration over a chain is always a linear combination of integrals over a cube, which is calculatable in principle (I mean this in the sense that the parametrization itself is not a part of the problem - the integrals of course might be impossible to explicitly compute), and when we integrate over surfaces in physics and vector calculus, we usually actually integrate on chains, rather than manifolds.
To give an explicit example, a sphere of radius $r_0$ in $\mathbb R^3$ is a submanifold, which cannot be described using a single chart. However if we consider the map $$ \sigma:I^2\rightarrow \mathbb R^3,\ \sigma(u^1,u^2)=\left(\begin{matrix} r_0\sin(\pi u^1)\cos(2\pi u^2) \\ r_0\sin(\pi u^1)\sin(2\pi u^2) \\ r_0\cos(\pi u^1)\end{matrix}\right), $$ this map fails to be an immersion at $u^1=0$ and $u^1=1$, but it does cover the entire sphere as point sets. The set of points where this map fails to be an immersion is a set of zero measure, thus the entire sphere can be described for the purposes of integration using a single $2$-cube.
Moreover, this also describes the boundary, since the $u^1=1$ and $u^1=0$ boundaries are single points, so they are sets of measure zero as $1$-manifolds, and the $u^2=1$ and $u^2=0$ boundaries are the same line segments but with the opposite orientation, so $1$-integrals over these $1$-cubes cancel. Hence, the boundary of the sphere is zero even from the chain point of view.
Questions: For the practical parametrization of submanifolds and the explicit computability of integrals, it would be very beneficial if the situation outlined with the sphere embedded in $\mathbb R^3$ would be general. In other words, a theorem of the following form should exist:
Hopefully existing theorem: Given an $r$ dimensional "submanifold with boundary/corners" $S$ in $M$, there is an $r$-chain $C$ in $M$ that is "equivalent" to $S\ \blacksquare$.
I have put "submanifold" into quotes because there are multiple inequivalent possible definitions (injective immersion, embedding etc.) and I do not know which one should be used, and I have put "equivalent" into quotes because I do not know what definition should such an equivalence satisfy. I would assume that as point sets the image of $S$ and $C$ should agree and it should be true that for any $\omega\in\Omega^r(M)$ $r$-form we should have $$ \int_S\omega=\int_C\omega, $$ where the first expression is integration of $\omega$ in the sense of manifolds and the second expression is integration of $\omega$ in the sense of chains. The manifolds in question are orientable and oriented whenever necessary.
So...
- Does such a result exist? In other words, given a submanifold, is it always possible to replace the submanifold with a chain for the purposes of integration?
- I would be terribly appreciative if references (papers and/or textbooks) would be given to me where such results are treated/proven. If the answer to 1) is affirmative, then the proofs/treatment of that result, and if the answer to 1) is negative, then for example sources that treat/explain when exactly can integration over manifolds be replaced with integration over chains.