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?


1 Answer 1


I 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!

  • $\begingroup$ This is a good point. I won't accept the answer yet, since I know that many people will shout at me $\endgroup$
    – Joe Tait
    May 8, 2014 at 7:27
  • $\begingroup$ Dear Joe, I'm certainly not pleading for you to accept my answer, but, just out of curiosity, why would people shout at you if you did ?? $\endgroup$ May 8, 2014 at 7:33
  • $\begingroup$ Sorry, that comment wasn't completed - I intended to use an example (if I could find one) in an internal seminar, for motivation. There are a lot of topologists in the dept, and I think they will complain that it is silly to work from the defn, there are lots of nice theorems, etc etc. I was going to claim (in a hyperbolic fashion) that people would shout/complain in the seminar $\endgroup$
    – Joe Tait
    May 8, 2014 at 11:39
  • $\begingroup$ Dear Joe, in my opinion one should always try to work from the definition: nice theorems are there only for the case that one cannot do that. Anyway, thanks for your explanation and good luck for your internal seminar ! $\endgroup$ May 8, 2014 at 11:54

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