# The Unapologetic Mathematician

## Stokes’ Theorem on Manifolds

Now we come back to Stokes’ theorem, but in the context of manifolds with boundary.

If $M$ is such a manifold of dimension $n$, and if $\omega$ is a compactly-supported $n$-form, then as usual we can use a partition of unity to break up the form into pieces, each of which is supported within the image of an orientation-preserving singular $n$-cube. For each singular cube $c$, either the image $c([0,1]^n)$ is contained totally within the interior of $M$, or it runs up against the boundary. In the latter case, without loss of generality, we can assume that $c([0,1]^n)\cap M$ is exactly the face $c_{n,0}([0,1]^{n-1})$ of $c$ where the $n$th coordinate is zero.

In the first case, our work is easy:

$\displaystyle\int\limits_Md\omega=\int\limits_cd\omega=\int\limits_{\partial d}\omega=\int\limits_{\partial M}\omega$

since $\omega$ is zero everywhere along the image of $\partial c$, and along $\partial M$.

In the other case, the vector fields $\frac{\partial}{\partial u^i}$ — in order — give positively-oriented basss of the tangent spaces of the standard $n$-cube. As $c$ is orientation, preserving, the ordered collection $\left(c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^n}\right)$ gives positively-oriented bases of the tangent spaces of the image of $c$. The basis $\left(c_*\left(-\frac{\partial}{\partial u^n}\right),c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^{n-1}}\right)$ is positively-oriented if and only if $n$ is even, since we have to pull the $n$th vector past $n-1$ others, picking up a negative sign for each one. But for a point $(a,0)$ with $a\in[0,1]^{-1}$, we see that

$\displaystyle c_{*(a,0)}\left(\frac{\partial}{\partial u^i}\right)=(c_{n,0})_{*a}\left(\frac{\partial}{\partial u^i}\right)$

for all $1\leq i\leq n-1$. That is, these image vectors are all within the tangent space of the boundary, and in this order. And since $c_*\left(-\frac{\partial}{\partial u^n}\right)$ is outward-pointing, this means that $c_{n,0}:[0,1]^{n-1}\to\partial M$ is orientation-preserving if and only if $n$ is even.

Now we can calculate

\displaystyle\begin{aligned}\int\limits_Md\omega&=\int\limits_cd\omega\\&=\int\limits_{\partial c}\omega\\&=\int\limits_{(-1)^nc_{n,0}}\omega\\&=(-1)^n\int\limits_{c_{n,0}}\omega\\&=(-1)^n(-1)^n\int\limits_{\partial M}\omega\\&=\int\limits_{\partial M}\omega\end{aligned}

where we use the fact that integrals over orientation-reversing singular cubes pick up negative signs, along with the sign that comes attached to the $(n,0)$ face of a singular $n$-cube to cancel each other off.

So in general we find

\displaystyle\begin{aligned}\int\limits_{\partial M}\omega&=\sum\limits_{\phi\in\Phi}\int\limits_{\partial M}\phi\omega\\&=\sum\limits_{\phi\in\Phi}\int\limits_Md(\phi\omega)\\&=\sum\limits_{\phi\in\Phi}\int\limits_M\left(d\phi\wedge\omega+\phi d\omega\right)\\&=\sum\limits_{\phi\in\Phi}\int\limits_Md\phi\wedge\omega+\int\limits_Md\omega\end{aligned}

The last sum is finite, since on of the support of $\omega$ all but finitely many of the $\phi$ are constantly zero, meaning that their differentials are zero as well. Since the sum is (locally) finite, we have no problem pulling it all the way inside:

$\displaystyle\sum\limits_{\phi\in\Phi}\int\limits_Md\phi\wedge\omega=\int\limits_Md\left(\sum\limits_{\phi\in\Phi}\phi\right)\wedge\omega=\int\limits_Md\left(1\right)\wedge\omega=0$

so the sum cancels off, leaving just the integral, as we’d expect. That is, under these circumstances,

$\displaystyle\int\limits_Md\omega=\int\limits_{\partial M}\omega$

which is Stokes’ theorem on manifolds.

September 16, 2011 - Posted by | Differential Topology, Topology

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