# Question on the proof of Stokes' Theorem in Spivak

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

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.