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 $\int_C = \int \int_{\Sigma}$.


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).


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.


1 Answer 1


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.


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