I am considering the following form of Stokes's theorem:

Let $\omega$ be an $n-1$ differential form with compact support on an oriented manifold of dimension $n$. Let us consider the boundary $\partial M$ of $M$ with the induced orientation. Then $$\int_{M}d\omega = \int_{\partial M}\omega.$$

In Green's theorem we have a positively oriented, piecewise smooth, simple closed curve $C$ in a plane, and $D$ be the region bounded by $C$. So $C$ plays the role of $\partial M$ and $D$ of $M$.

On the other hand in Stokes's theorem the boundary $\partial M$ is a smooth manifold of dimension $n-1$.

My question is the following: is there a way to deduce Green's theorem from Stokes's theorem?

  • $\begingroup$ In what form would you like Green's theorem? $\endgroup$ – Chappers Apr 15 '15 at 17:14
  • $\begingroup$ For a simple closed curve $C$ which is piecewise smooth but not necessarily smooth. $\endgroup$ – JdiNirv Apr 15 '15 at 17:15
  • $\begingroup$ So $\int_D \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) \, dx \, dy = \int_C ( L \, dx + M \, dy )$? $\endgroup$ – Chappers Apr 15 '15 at 17:19
  • $\begingroup$ Yes but it seems to me that in Stokes theorem (at least the version I wrote in the question) the boundary $\partial M$ which in this case is $C$ should be smooth $\endgroup$ – JdiNirv Apr 15 '15 at 17:22
  • $\begingroup$ Ah, that you can deal with by taking a smooth approximation to it (basically, round off the corners), then take the limit. (And since we're doing integrals, everything'll work, much like it works for the Gauss-Bonnet theorem.) $\endgroup$ – Chappers Apr 15 '15 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.