The following is a quick outline of the proof of Stokes' Theorem as found in a Comprehensive Introduction to Differential Geometry Vol. 1 by Spivak.

Theorem (Local Stokes' Theorem). Let $M$ be a smooth manifold, $c$ a singular $k$-chain and $\omega$ a $k - 1$-form on $M$. Then $$\int_c d\omega = \int_{\partial c} \omega.$$

Theorem (Stokes' Theorem). Let $M^n$ be an oriented smooth manifold with boundary and $\omega \in \Omega^{n - 1}_c(M)$. Then $$\int_M d\omega = \int_{\partial M}\omega$$ where $\partial M$ is given the induced orientation.

Proof. Suppose that the support of $\omega$ is contained in the interior of some positively oriented singular cube $c$ with $\operatorname{im} c \cap \partial M = \varnothing$. Then we can apply the local Stokes' theorem to conclude. Indeed, we have that $$\int_M d\omega = \int_c d\omega = \int_{\partial c}\omega = 0.$$ Shouldn't it really be $\operatorname{int} M$ in the first integral instead of just $M$? Because the local Stokes' theorem only applies for manifolds with boundaries. However, the next step is to consider a singular cube $c$ such that $\partial M \cap \operatorname{im}c = \operatorname{im}F_1c$, where $F_1c$ is the first front face. Spivak then proceeds again by using the local version of Stokes' theorem: $$\int_M d\omega = \int_c d\omega = \int_{\partial c}\omega = ...$$ Why can we use the local version here? Again, I mean, the local version applies for manifolds without boundary.


1 Answer 1


The local version of Stokes Theorem is one of those results which can be generalised to hold for the case of a manifold with boundary with pretty low effort.


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