Simply-Connected Spaces and Cohomology
We’ve seen that if a manifold is simply-connected then the first degree of cubic singular homology is trivial. I say that the same is true of the first degree of de Rham cohomology.
Indeed, say that is simply-connected, so that any closed curve can be written as the boundary of some surface . Then we take any closed -form with . Stokes’ theorem tells us that
for any closed curve . But this means that every closed -form is path-independent, and path-independent -forms are exact. And so we conclude that , as asserted.
It will (eventually) turn out that the fact that both and vanish together is not a coincidence, but is in fact an example of a much deeper correspondence between homology and cohomology — between topology and analysis.
This is wrong, Stokes theorem doesn´t say that, the theorem is just for (n-1)-forms in n-dimensional manifolds, so what is proven here works only for 2-dimensional manifolds.
Comment by Kalejandro | November 9, 2017 |
Or… you could actually click through that link (this one, just to help you out a little more) and learn about the generalization of Stokes’ theorem to all other dimensions, which subsumes the divergence theorem, the fundamental theorem of calculus, and more!
Comment by John Armstrong | November 9, 2017 |