The Unapologetic Mathematician

Mathematics for the interested outsider

Simply-Connected Spaces and Cohomology

We’ve seen that if a manifold M 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 M is simply-connected, so that any closed curve c can be written as the boundary c=\partial S of some surface S. Then we take any closed 1-form \omega with d\omega=0. Stokes’ theorem tells us that

\displaystyle\int\limits_c\omega=\int\limits_{\partial S}\omega=\int\limits_Sd\omega=\int\limits_S0=0

for any closed curve c. But this means that every closed 1-form \omega is path-independent, and path-independent 1-forms are exact. And so we conclude that H^1(M)=0, as asserted.

It will (eventually) turn out that the fact that both H_1(M) and H^1(M) 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.

December 17, 2011 - Posted by | Differential Topology, Topology


  1. 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 | Reply

    • 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 | Reply

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