The Unapologetic Mathematician

Mathematics for the interested outsider

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 nth 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 nth 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.

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

7 Comments »

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