De Rham's theorem states that for any smooth manifold $M$ the singular cohomology and de Rham cohomology of $M$ are isomorphic.
Are there any examples of manifolds for which it is easier to compute the de Rham cohomology?
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Sign up to join this communityI would say that de Rham cohomology is always easier to calculate than singular cohomology!
Singular cohomology is essentially impossible to calculate if you just use the definition: you need general theorems.
You don't believe me? I challenge you to calculate $H^2_{\operatorname {sing}}((0,1), \mathbb R)$ from the definition : good luck with your non-denumerable set of singular $2$-simplices in $(0,1)$ and the $\mathbb R$-vector space they generate!
By contrast $H^2_{\operatorname {de Rham}}((0,1), \mathbb R)=0$ is obvious because the only differential form of degree $2$ on $(0,1)$ is zero.
So you can make this "calculation" one picosecond after you have learned the definition of de Rham cohomology!