# The Unapologetic Mathematician

## 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