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:


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


  1. […] Stokes’ Theorem on Manifolds […]

    Pingback by The Fundamental Theorem of Line Integrals « The Unapologetic Mathematician | October 24, 2011 | Reply

  2. […] Stokes’ theorem tells us […]

    Pingback by The Divergence Theorem « The Unapologetic Mathematician | November 22, 2011 | Reply

  3. […] last we come to the version of Stokes’ theorem that people learn with that name in calculus courses. Ironically, unlike the fundamental theorem […]

    Pingback by The Classical Stokes Theorem « The Unapologetic Mathematician | November 23, 2011 | Reply

  4. […] in the de Rham cohomology. But we know that it cannot also be exact, for if for some -form then Stokes’ theorem would tell us […]

    Pingback by Compact Oriented Manifolds without Boundary have Nontrivial Homology « The Unapologetic Mathematician | November 24, 2011 | Reply

  5. […] . The support of both and is contained in some large -dimensional parallelepiped , so we can use Stokes’ theorem to […]

    Pingback by Compactly Supported De Rham Cohomology « The Unapologetic Mathematician | December 6, 2011 | Reply

  6. […] curve can be written as the boundary of some surface . Then we take any closed -form with . Stokes’ theorem tells us […]

    Pingback by Simply-Connected Spaces and Cohomology « The Unapologetic Mathematician | December 17, 2011 | Reply

  7. […] for exactness: if for some -form , then Stokes’ theorem tells us […]

    Pingback by A Family of Nontrivial Homology Classes (part 3) « The Unapologetic Mathematician | December 27, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: