# Extending surface independence of Stokes' theorem in elementary calculus to the Stokes' theorem in differential geometry

Okay so I discovered, here When is a surface integral equal to double integral over projection? A verification of Stokes' Theorem. Intuition and relation to Green's Theorem. (and also here Applying Stokes' theorem - what surface?), a view of Stokes' Theorem, at least in elementary calculus, as not only

Given a surface $$\Sigma$$, let $$C$$ be its boundary curve. If (insert assumptions), then $$\int_C = \int \int_{\Sigma}$$.

but also

Given a curve $$C$$, we have for any surface $$\Sigma$$ with $$C$$ as its boundary curve that if (insert assumptions), then $$\int_C = \int \int_{\Sigma}$$.

Thus,

Given a surface $$\Sigma$$, if (insert assumptions), then $$\int \int_{\Sigma} = \int \int_{\Sigma'}$$ for any $$\Sigma'$$ that has the same boundary curve as $$\Sigma$$ as long as (insert assumptions).

Question:

How can we extend this view to Stokes' theorem in differential geometry?

If I have 2 distinct smooth oriented $$n$$-manifolds with boundary $$M$$ and $$N$$ that actually turn out have the same manifold boundary $$\partial M=\partial N$$, then, under assumptions (insert here), I want to use Stokes' Theorem to say something like $$\int_M d \omega=\int_{\partial M} \omega = \int_{\partial N} \omega=\int_N d \omega \tag{A}$$ Not sure what $$\omega$$ would be though. It's has to be a smooth differential $$(n−1)$$-form with compact support on both $$N$$ and $$M$$. Also, if we're going to be strict about it, then $$(A)$$ should look more like

$$\int_M d \omega=\int_{\partial M} \iota_M^{*}\omega = \int_{\partial N} \iota_N^{*}\omega=\int_N d \omega \tag{B}$$

where $$(\cdot)^{*}$$ denotes pullback and $$\iota_M: \partial M \to M$$ and $$\iota_N: \partial N \to N$$ are inclusion maps in which case I'm not sure we can have the same '$$\omega$$' since we can't exactly have $$\omega: M \to \wedge(T^{*}M)$$ to be also $$\omega: N \to \wedge(T^{*}N)$$.

Some examples that might make sense out of $$\omega$$ are when $$M$$ is a submanifold with boundary of $$N$$ (in which case I guess $$M$$ is open in $$N$$ extending from the case for submanifolds without boundary of codimension zero) or when they have a submanifold with boundary in common or something. In the former example (assuming it works), we could have the '$$\omega$$' on $$N$$ to be just $$\omega$$ and then the '$$\omega$$' on $$M$$ is $$\omega|_{M}$$, in which case I hope that $$\omega(M) = \omega|_{M}(M)$$, a subset of $$\wedge(T^{*}N)$$, is a vector subbundle that is bundle isomorphic to the cotangent bundle $$\wedge(T^{*}M)$$. I think the latter example can work similarly.

It makes sense if you take both $$M$$ and $$N$$ submanifolds of an other one $$P$$, with (same) boundaries. Then, for $$\omega$$ a $$(k-1)$$-form on $$P$$ (where $$k$$ is the dimension of $$M$$ and $$N$$) it makes sense to write

$$\int_Md\omega=\int_{\partial M}\omega=\int_{\partial N}\omega=\int_Nd\omega$$

with the induced identifications given by $$i_M,i_N$$ and $$i_{\partial M}=i_{\partial N}$$, all with target $$P$$.

If you take $$P=\mathbb{R}^n$$ it should give something similar to your first statements.